3.207 \(\int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{(a+b x^4)^4} \, dx\)

Optimal. Leaf size=462 \[ -\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}+\frac {(a h+5 b d) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{32 a^{7/2} b^{3/2}}+\frac {x \left (7 (a g+11 b c)+12 x (a h+5 b d)+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (2 x (a h+5 b d)+a g+11 b c+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {x \left (x (b d-a h)-a g+b c+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3} \]

[Out]

1/12*x*(b*c-a*g+(-a*h+b*d)*x+b*e*x^2+b*f*x^3)/a/b/(b*x^4+a)^3+1/384*x*(7*a*g+77*b*c+12*(a*h+5*b*d)*x+45*b*e*x^
2)/a^3/b/(b*x^4+a)+1/96*(-8*a*f+x*(11*b*c+a*g+2*(a*h+5*b*d)*x+9*b*e*x^2))/a^2/b/(b*x^4+a)^2+1/32*(a*h+5*b*d)*a
rctan(x^2*b^(1/2)/a^(1/2))/a^(7/2)/b^(3/2)-1/1024*ln(-a^(1/4)*b^(1/4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2))*(77*b*c+7
*a*g-15*e*a^(1/2)*b^(1/2))/a^(15/4)/b^(5/4)*2^(1/2)+1/1024*ln(a^(1/4)*b^(1/4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2))*(
77*b*c+7*a*g-15*e*a^(1/2)*b^(1/2))/a^(15/4)/b^(5/4)*2^(1/2)+1/512*arctan(-1+b^(1/4)*x*2^(1/2)/a^(1/4))*(77*b*c
+7*a*g+15*e*a^(1/2)*b^(1/2))/a^(15/4)/b^(5/4)*2^(1/2)+1/512*arctan(1+b^(1/4)*x*2^(1/2)/a^(1/4))*(77*b*c+7*a*g+
15*e*a^(1/2)*b^(1/2))/a^(15/4)/b^(5/4)*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.62, antiderivative size = 462, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {1858, 1854, 1855, 1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}+\frac {(a h+5 b d) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{32 a^{7/2} b^{3/2}}-\frac {8 a f-x \left (2 x (a h+5 b d)+a g+11 b c+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {x \left (7 (a g+11 b c)+12 x (a h+5 b d)+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}+\frac {x \left (x (b d-a h)-a g+b c+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(a + b*x^4)^4,x]

[Out]

(x*(b*c - a*g + (b*d - a*h)*x + b*e*x^2 + b*f*x^3))/(12*a*b*(a + b*x^4)^3) + (x*(7*(11*b*c + a*g) + 12*(5*b*d
+ a*h)*x + 45*b*e*x^2))/(384*a^3*b*(a + b*x^4)) - (8*a*f - x*(11*b*c + a*g + 2*(5*b*d + a*h)*x + 9*b*e*x^2))/(
96*a^2*b*(a + b*x^4)^2) + ((5*b*d + a*h)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(32*a^(7/2)*b^(3/2)) - ((77*b*c + 15*S
qrt[a]*Sqrt[b]*e + 7*a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(5/4)) + ((77*b*c +
 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(5/4)) - ((77*
b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15
/4)*b^(5/4)) + ((77*b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2]
)/(512*Sqrt[2]*a^(15/4)*b^(5/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
 + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
 x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
 Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1854

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[((a*Coeff[Pq, x, q] -
 b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q, x])*(a + b*x^n)^(p + 1))/(a*b*n*(p + 1)), x] + Dist[1/(a*n*(p + 1))
, Int[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^(p + 1), x], x] /; q == n - 1] /
; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
 x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rule 1858

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]] /; G
eQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1876

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(x^ii*(Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2)))/(a + b*x^n), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

Rubi steps

\begin {align*} \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^4\right )^4} \, dx &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}-\frac {\int \frac {-b (11 b c+a g)-2 b (5 b d+a h) x-9 b^2 e x^2-8 b^2 f x^3}{\left (a+b x^4\right )^3} \, dx}{12 a b^2}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {\int \frac {7 b (11 b c+a g)+12 b (5 b d+a h) x+45 b^2 e x^2}{\left (a+b x^4\right )^2} \, dx}{96 a^2 b^2}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}-\frac {\int \frac {-21 b (11 b c+a g)-24 b (5 b d+a h) x-45 b^2 e x^2}{a+b x^4} \, dx}{384 a^3 b^2}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}-\frac {\int \left (-\frac {24 b (5 b d+a h) x}{a+b x^4}+\frac {-21 b (11 b c+a g)-45 b^2 e x^2}{a+b x^4}\right ) \, dx}{384 a^3 b^2}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}-\frac {\int \frac {-21 b (11 b c+a g)-45 b^2 e x^2}{a+b x^4} \, dx}{384 a^3 b^2}+\frac {(5 b d+a h) \int \frac {x}{a+b x^4} \, dx}{16 a^3 b}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {\left (77 b c+15 \sqrt {a} \sqrt {b} e+7 a g\right ) \int \frac {\sqrt {a} \sqrt {b}+b x^2}{a+b x^4} \, dx}{256 a^{7/2} b^{3/2}}-\frac {\left (15 \sqrt {b} e-\frac {7 (11 b c+a g)}{\sqrt {a}}\right ) \int \frac {\sqrt {a} \sqrt {b}-b x^2}{a+b x^4} \, dx}{256 a^3 b^{3/2}}+\frac {(5 b d+a h) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^2\right )}{32 a^3 b}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {(5 b d+a h) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{32 a^{7/2} b^{3/2}}-\frac {\left (77 b c-15 \sqrt {a} \sqrt {b} e+7 a g\right ) \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx}{512 \sqrt {2} a^{15/4} b^{5/4}}-\frac {\left (77 b c-15 \sqrt {a} \sqrt {b} e+7 a g\right ) \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx}{512 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\left (77 b c+15 \sqrt {a} \sqrt {b} e+7 a g\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{512 a^{7/2} b^{3/2}}+\frac {\left (77 b c+15 \sqrt {a} \sqrt {b} e+7 a g\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{512 a^{7/2} b^{3/2}}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {(5 b d+a h) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{32 a^{7/2} b^{3/2}}-\frac {\left (77 b c-15 \sqrt {a} \sqrt {b} e+7 a g\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\left (77 b c-15 \sqrt {a} \sqrt {b} e+7 a g\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\left (77 b c+15 \sqrt {a} \sqrt {b} e+7 a g\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}-\frac {\left (77 b c+15 \sqrt {a} \sqrt {b} e+7 a g\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{12 a b \left (a+b x^4\right )^3}+\frac {x \left (7 (11 b c+a g)+12 (5 b d+a h) x+45 b e x^2\right )}{384 a^3 b \left (a+b x^4\right )}-\frac {8 a f-x \left (11 b c+a g+2 (5 b d+a h) x+9 b e x^2\right )}{96 a^2 b \left (a+b x^4\right )^2}+\frac {(5 b d+a h) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{32 a^{7/2} b^{3/2}}-\frac {\left (77 b c+15 \sqrt {a} \sqrt {b} e+7 a g\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\left (77 b c+15 \sqrt {a} \sqrt {b} e+7 a g\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt {2} a^{15/4} b^{5/4}}-\frac {\left (77 b c-15 \sqrt {a} \sqrt {b} e+7 a g\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}+\frac {\left (77 b c-15 \sqrt {a} \sqrt {b} e+7 a g\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{512 \sqrt {2} a^{15/4} b^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 461, normalized size = 1.00 \[ \frac {-6 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (16 a^{5/4} h+15 \sqrt {2} \sqrt {a} b^{3/4} e+80 \sqrt [4]{a} b d+7 \sqrt {2} a \sqrt [4]{b} g+77 \sqrt {2} b^{5/4} c\right )+6 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (-16 a^{5/4} h+15 \sqrt {2} \sqrt {a} b^{3/4} e-80 \sqrt [4]{a} b d+7 \sqrt {2} a \sqrt [4]{b} g+77 \sqrt {2} b^{5/4} c\right )-\frac {256 a^{11/4} \sqrt {b} (a (f+x (g+h x))-b x (c+x (d+e x)))}{\left (a+b x^4\right )^3}+\frac {32 a^{7/4} \sqrt {b} x (a (g+2 h x)+11 b c+b x (10 d+9 e x))}{\left (a+b x^4\right )^2}+\frac {8 a^{3/4} \sqrt {b} x \left (7 a g+12 a h x+77 b c+60 b d x+45 b e x^2\right )}{a+b x^4}-3 \sqrt {2} \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )+3 \sqrt {2} \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-15 \sqrt {a} \sqrt {b} e+7 a g+77 b c\right )}{3072 a^{15/4} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(a + b*x^4)^4,x]

[Out]

((8*a^(3/4)*Sqrt[b]*x*(77*b*c + 7*a*g + 60*b*d*x + 12*a*h*x + 45*b*e*x^2))/(a + b*x^4) + (32*a^(7/4)*Sqrt[b]*x
*(11*b*c + b*x*(10*d + 9*e*x) + a*(g + 2*h*x)))/(a + b*x^4)^2 - (256*a^(11/4)*Sqrt[b]*(-(b*x*(c + x*(d + e*x))
) + a*(f + x*(g + h*x))))/(a + b*x^4)^3 - 6*(77*Sqrt[2]*b^(5/4)*c + 80*a^(1/4)*b*d + 15*Sqrt[2]*Sqrt[a]*b^(3/4
)*e + 7*Sqrt[2]*a*b^(1/4)*g + 16*a^(5/4)*h)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 6*(77*Sqrt[2]*b^(5/4)*c
- 80*a^(1/4)*b*d + 15*Sqrt[2]*Sqrt[a]*b^(3/4)*e + 7*Sqrt[2]*a*b^(1/4)*g - 16*a^(5/4)*h)*ArcTan[1 + (Sqrt[2]*b^
(1/4)*x)/a^(1/4)] - 3*Sqrt[2]*b^(1/4)*(77*b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^
(1/4)*x + Sqrt[b]*x^2] + 3*Sqrt[2]*b^(1/4)*(77*b*c - 15*Sqrt[a]*Sqrt[b]*e + 7*a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/
4)*b^(1/4)*x + Sqrt[b]*x^2])/(3072*a^(15/4)*b^(3/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.20, size = 521, normalized size = 1.13 \[ \frac {\sqrt {2} {\left (40 \, \sqrt {2} \sqrt {a b} b^{2} d + 8 \, \sqrt {2} \sqrt {a b} a b h + 77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g + 15 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{512 \, a^{4} b^{3}} + \frac {\sqrt {2} {\left (40 \, \sqrt {2} \sqrt {a b} b^{2} d + 8 \, \sqrt {2} \sqrt {a b} a b h + 77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g + 15 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{512 \, a^{4} b^{3}} + \frac {\sqrt {2} {\left (77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g - 15 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{1024 \, a^{4} b^{3}} - \frac {\sqrt {2} {\left (77 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g - 15 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{1024 \, a^{4} b^{3}} + \frac {45 \, b^{3} x^{11} e + 60 \, b^{3} d x^{10} + 12 \, a b^{2} h x^{10} + 77 \, b^{3} c x^{9} + 7 \, a b^{2} g x^{9} + 126 \, a b^{2} x^{7} e + 160 \, a b^{2} d x^{6} + 32 \, a^{2} b h x^{6} + 198 \, a b^{2} c x^{5} + 18 \, a^{2} b g x^{5} + 113 \, a^{2} b x^{3} e + 132 \, a^{2} b d x^{2} - 12 \, a^{3} h x^{2} + 153 \, a^{2} b c x - 21 \, a^{3} g x - 32 \, a^{3} f}{384 \, {\left (b x^{4} + a\right )}^{3} a^{3} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x, algorithm="giac")

[Out]

1/512*sqrt(2)*(40*sqrt(2)*sqrt(a*b)*b^2*d + 8*sqrt(2)*sqrt(a*b)*a*b*h + 77*(a*b^3)^(1/4)*b^2*c + 7*(a*b^3)^(1/
4)*a*b*g + 15*(a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^3) + 1/512*s
qrt(2)*(40*sqrt(2)*sqrt(a*b)*b^2*d + 8*sqrt(2)*sqrt(a*b)*a*b*h + 77*(a*b^3)^(1/4)*b^2*c + 7*(a*b^3)^(1/4)*a*b*
g + 15*(a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^3) + 1/1024*sqrt(2)
*(77*(a*b^3)^(1/4)*b^2*c + 7*(a*b^3)^(1/4)*a*b*g - 15*(a*b^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(
a/b))/(a^4*b^3) - 1/1024*sqrt(2)*(77*(a*b^3)^(1/4)*b^2*c + 7*(a*b^3)^(1/4)*a*b*g - 15*(a*b^3)^(3/4)*e)*log(x^2
 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4*b^3) + 1/384*(45*b^3*x^11*e + 60*b^3*d*x^10 + 12*a*b^2*h*x^10 + 77*
b^3*c*x^9 + 7*a*b^2*g*x^9 + 126*a*b^2*x^7*e + 160*a*b^2*d*x^6 + 32*a^2*b*h*x^6 + 198*a*b^2*c*x^5 + 18*a^2*b*g*
x^5 + 113*a^2*b*x^3*e + 132*a^2*b*d*x^2 - 12*a^3*h*x^2 + 153*a^2*b*c*x - 21*a^3*g*x - 32*a^3*f)/((b*x^4 + a)^3
*a^3*b)

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maple [A]  time = 0.07, size = 607, normalized size = 1.31 \[ \frac {h \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{32 \sqrt {a b}\, a^{2} b}+\frac {5 d \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{32 \sqrt {a b}\, a^{3}}+\frac {15 \sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{512 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3} b}+\frac {15 \sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{512 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3} b}+\frac {15 \sqrt {2}\, e \ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{1024 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3} b}+\frac {7 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, g \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{512 a^{3} b}+\frac {7 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, g \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{512 a^{3} b}+\frac {7 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, g \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{1024 a^{3} b}+\frac {77 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{512 a^{4}}+\frac {77 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{512 a^{4}}+\frac {77 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{1024 a^{4}}+\frac {\frac {15 b^{2} e \,x^{11}}{128 a^{3}}+\frac {\left (a h +5 b d \right ) b \,x^{10}}{32 a^{3}}+\frac {21 b e \,x^{7}}{64 a^{2}}+\frac {7 \left (a g +11 b c \right ) b \,x^{9}}{384 a^{3}}+\frac {\left (a h +5 b d \right ) x^{6}}{12 a^{2}}+\frac {113 e \,x^{3}}{384 a}+\frac {3 \left (a g +11 b c \right ) x^{5}}{64 a^{2}}-\frac {\left (a h -11 b d \right ) x^{2}}{32 a b}-\frac {f}{12 b}-\frac {\left (7 a g -51 b c \right ) x}{128 a b}}{\left (b \,x^{4}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x)

[Out]

(15/128/a^3*b^2*e*x^11+1/32*(a*h+5*b*d)/a^3*b*x^10+7/384*(a*g+11*b*c)/a^3*b*x^9+21/64/a^2*b*e*x^7+1/12/a^2*(a*
h+5*b*d)*x^6+3/64/a^2*(a*g+11*b*c)*x^5+113/384/a*e*x^3-1/32*(a*h-11*b*d)/a/b*x^2-1/128*(7*a*g-51*b*c)/a/b*x-1/
12/b*f)/(b*x^4+a)^3+7/512/a^3/b*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)*g+77/512/a^4*c*(a/b)^(1/4)
*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)+7/1024*(a/b)^(1/4)*2^(1/2)/a^3/b*g*ln((x^2+(a/b)^(1/4)*2^(1/2)*x+(a/b
)^(1/2))/(x^2-(a/b)^(1/4)*2^(1/2)*x+(a/b)^(1/2)))+77/1024/a^4*c*(a/b)^(1/4)*2^(1/2)*ln((x^2+(a/b)^(1/4)*2^(1/2
)*x+(a/b)^(1/2))/(x^2-(a/b)^(1/4)*2^(1/2)*x+(a/b)^(1/2)))+7/512*(a/b)^(1/4)*2^(1/2)/a^3/b*g*arctan(2^(1/2)/(a/
b)^(1/4)*x+1)+77/512*(a/b)^(1/4)*2^(1/2)/a^4*c*arctan(2^(1/2)/(a/b)^(1/4)*x+1)+1/32/a^2/b/(a*b)^(1/2)*arctan((
1/a*b)^(1/2)*x^2)*h+5/32/(a*b)^(1/2)/a^3*d*arctan((1/a*b)^(1/2)*x^2)+15/1024/a^3*e/b/(a/b)^(1/4)*2^(1/2)*ln((x
^2-(a/b)^(1/4)*2^(1/2)*x+(a/b)^(1/2))/(x^2+(a/b)^(1/4)*2^(1/2)*x+(a/b)^(1/2)))+15/512/a^3*e/b/(a/b)^(1/4)*2^(1
/2)*arctan(2^(1/2)/(a/b)^(1/4)*x+1)+15/512/a^3*e/b/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)

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maxima [A]  time = 3.13, size = 517, normalized size = 1.12 \[ \frac {45 \, b^{3} e x^{11} + 126 \, a b^{2} e x^{7} + 12 \, {\left (5 \, b^{3} d + a b^{2} h\right )} x^{10} + 7 \, {\left (11 \, b^{3} c + a b^{2} g\right )} x^{9} + 113 \, a^{2} b e x^{3} + 32 \, {\left (5 \, a b^{2} d + a^{2} b h\right )} x^{6} + 18 \, {\left (11 \, a b^{2} c + a^{2} b g\right )} x^{5} - 32 \, a^{3} f + 12 \, {\left (11 \, a^{2} b d - a^{3} h\right )} x^{2} + 3 \, {\left (51 \, a^{2} b c - 7 \, a^{3} g\right )} x}{384 \, {\left (a^{3} b^{4} x^{12} + 3 \, a^{4} b^{3} x^{8} + 3 \, a^{5} b^{2} x^{4} + a^{6} b\right )}} + \frac {\frac {\sqrt {2} {\left (77 \, b^{\frac {3}{2}} c - 15 \, \sqrt {a} b e + 7 \, a \sqrt {b} g\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (77 \, b^{\frac {3}{2}} c - 15 \, \sqrt {a} b e + 7 \, a \sqrt {b} g\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} + \frac {2 \, {\left (77 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + 15 \, \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + 7 \, \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g - 80 \, \sqrt {a} b^{\frac {3}{2}} d - 16 \, a^{\frac {3}{2}} \sqrt {b} h\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}} + \frac {2 \, {\left (77 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + 15 \, \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + 7 \, \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g + 80 \, \sqrt {a} b^{\frac {3}{2}} d + 16 \, a^{\frac {3}{2}} \sqrt {b} h\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}}}{1024 \, a^{3} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x, algorithm="maxima")

[Out]

1/384*(45*b^3*e*x^11 + 126*a*b^2*e*x^7 + 12*(5*b^3*d + a*b^2*h)*x^10 + 7*(11*b^3*c + a*b^2*g)*x^9 + 113*a^2*b*
e*x^3 + 32*(5*a*b^2*d + a^2*b*h)*x^6 + 18*(11*a*b^2*c + a^2*b*g)*x^5 - 32*a^3*f + 12*(11*a^2*b*d - a^3*h)*x^2
+ 3*(51*a^2*b*c - 7*a^3*g)*x)/(a^3*b^4*x^12 + 3*a^4*b^3*x^8 + 3*a^5*b^2*x^4 + a^6*b) + 1/1024*(sqrt(2)*(77*b^(
3/2)*c - 15*sqrt(a)*b*e + 7*a*sqrt(b)*g)*log(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(3/
4)) - sqrt(2)*(77*b^(3/2)*c - 15*sqrt(a)*b*e + 7*a*sqrt(b)*g)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sq
rt(a))/(a^(3/4)*b^(3/4)) + 2*(77*sqrt(2)*a^(1/4)*b^(7/4)*c + 15*sqrt(2)*a^(3/4)*b^(5/4)*e + 7*sqrt(2)*a^(5/4)*
b^(3/4)*g - 80*sqrt(a)*b^(3/2)*d - 16*a^(3/2)*sqrt(b)*h)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(
1/4))/sqrt(sqrt(a)*sqrt(b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(3/4)) + 2*(77*sqrt(2)*a^(1/4)*b^(7/4)*c + 15*sq
rt(2)*a^(3/4)*b^(5/4)*e + 7*sqrt(2)*a^(5/4)*b^(3/4)*g + 80*sqrt(a)*b^(3/2)*d + 16*a^(3/2)*sqrt(b)*h)*arctan(1/
2*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(3/4
)))/(a^3*b)

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mupad [B]  time = 6.08, size = 1743, normalized size = 3.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(a + b*x^4)^4,x)

[Out]

symsum(log((123200*b^3*c*d^2 - 3375*a*b^2*e^3 - 88935*b^3*c^2*e + 448*a^3*g*h^2 + 11200*a*b^2*d^2*g + 4928*a^2
*b*c*h^2 - 735*a^2*b*e*g^2 + 49280*a*b^2*c*d*h - 16170*a*b^2*c*e*g + 4480*a^2*b*d*g*h)/(2097152*a^9*b) - root(
68719476736*a^15*b^6*z^4 + 1211105280*a^8*b^5*c*e*z^2 + 335544320*a^9*b^4*d*h*z^2 + 110100480*a^9*b^4*e*g*z^2
+ 838860800*a^8*b^5*d^2*z^2 + 33554432*a^10*b^3*h^2*z^2 - 88309760*a^5*b^4*c*d*g*z - 17661952*a^6*b^3*c*g*h*z
- 485703680*a^4*b^5*c^2*d*z - 97140736*a^5*b^4*c^2*h*z - 802816*a^7*b^2*g^2*h*z + 3686400*a^6*b^3*e^2*h*z - 40
14080*a^6*b^3*d*g^2*z + 18432000*a^5*b^4*d*e^2*z - 268800*a^3*b^2*d*e*g*h - 2956800*a^2*b^3*c*d*e*h - 672000*a
^2*b^3*d^2*e*g - 295680*a^3*b^2*c*e*h^2 + 485100*a^2*b^3*c*e^2*g - 26880*a^4*b*e*g*h^2 - 7392000*a*b^4*c*d^2*e
 + 81920*a^4*b*d*h^3 + 12782924*a*b^4*c^3*g + 614400*a^3*b^2*d^2*h^2 + 22050*a^3*b^2*e^2*g^2 + 1743126*a^2*b^3
*c^2*g^2 + 2048000*a^2*b^3*d^3*h + 105644*a^3*b^2*c*g^3 + 2668050*a*b^4*c^2*e^2 + 50625*a^2*b^3*e^4 + 2401*a^4
*b*g^4 + 2560000*a*b^4*d^4 + 4096*a^5*h^4 + 35153041*b^5*c^4, z, k)*(root(68719476736*a^15*b^6*z^4 + 121110528
0*a^8*b^5*c*e*z^2 + 335544320*a^9*b^4*d*h*z^2 + 110100480*a^9*b^4*e*g*z^2 + 838860800*a^8*b^5*d^2*z^2 + 335544
32*a^10*b^3*h^2*z^2 - 88309760*a^5*b^4*c*d*g*z - 17661952*a^6*b^3*c*g*h*z - 485703680*a^4*b^5*c^2*d*z - 971407
36*a^5*b^4*c^2*h*z - 802816*a^7*b^2*g^2*h*z + 3686400*a^6*b^3*e^2*h*z - 4014080*a^6*b^3*d*g^2*z + 18432000*a^5
*b^4*d*e^2*z - 268800*a^3*b^2*d*e*g*h - 2956800*a^2*b^3*c*d*e*h - 672000*a^2*b^3*d^2*e*g - 295680*a^3*b^2*c*e*
h^2 + 485100*a^2*b^3*c*e^2*g - 26880*a^4*b*e*g*h^2 - 7392000*a*b^4*c*d^2*e + 81920*a^4*b*d*h^3 + 12782924*a*b^
4*c^3*g + 614400*a^3*b^2*d^2*h^2 + 22050*a^3*b^2*e^2*g^2 + 1743126*a^2*b^3*c^2*g^2 + 2048000*a^2*b^3*d^3*h + 1
05644*a^3*b^2*c*g^3 + 2668050*a*b^4*c^2*e^2 + 50625*a^2*b^3*e^4 + 2401*a^4*b*g^4 + 2560000*a*b^4*d^4 + 4096*a^
5*h^4 + 35153041*b^5*c^4, z, k)*((20185088*a^7*b^4*c + 1835008*a^8*b^3*g)/(2097152*a^9*b) - (x*(655360*a^7*b^4
*d + 131072*a^8*b^3*h))/(131072*a^9*b)) + (614400*a^4*b^3*d*e + 122880*a^5*b^2*e*h)/(2097152*a^9*b) + (x*(1897
28*a^3*b^4*c^2 - 7200*a^4*b^3*e^2 + 1568*a^5*b^2*g^2 + 34496*a^4*b^3*c*g))/(131072*a^9*b)) + (x*(4000*b^3*d^3
+ 32*a^3*h^3 - 5775*b^3*c*d*e + 2400*a*b^2*d^2*h + 480*a^2*b*d*h^2 - 1155*a*b^2*c*e*h - 525*a*b^2*d*e*g - 105*
a^2*b*e*g*h))/(131072*a^9*b))*root(68719476736*a^15*b^6*z^4 + 1211105280*a^8*b^5*c*e*z^2 + 335544320*a^9*b^4*d
*h*z^2 + 110100480*a^9*b^4*e*g*z^2 + 838860800*a^8*b^5*d^2*z^2 + 33554432*a^10*b^3*h^2*z^2 - 88309760*a^5*b^4*
c*d*g*z - 17661952*a^6*b^3*c*g*h*z - 485703680*a^4*b^5*c^2*d*z - 97140736*a^5*b^4*c^2*h*z - 802816*a^7*b^2*g^2
*h*z + 3686400*a^6*b^3*e^2*h*z - 4014080*a^6*b^3*d*g^2*z + 18432000*a^5*b^4*d*e^2*z - 268800*a^3*b^2*d*e*g*h -
 2956800*a^2*b^3*c*d*e*h - 672000*a^2*b^3*d^2*e*g - 295680*a^3*b^2*c*e*h^2 + 485100*a^2*b^3*c*e^2*g - 26880*a^
4*b*e*g*h^2 - 7392000*a*b^4*c*d^2*e + 81920*a^4*b*d*h^3 + 12782924*a*b^4*c^3*g + 614400*a^3*b^2*d^2*h^2 + 2205
0*a^3*b^2*e^2*g^2 + 1743126*a^2*b^3*c^2*g^2 + 2048000*a^2*b^3*d^3*h + 105644*a^3*b^2*c*g^3 + 2668050*a*b^4*c^2
*e^2 + 50625*a^2*b^3*e^4 + 2401*a^4*b*g^4 + 2560000*a*b^4*d^4 + 4096*a^5*h^4 + 35153041*b^5*c^4, z, k), k, 1,
4) + ((113*e*x^3)/(384*a) - f/(12*b) + (3*x^5*(11*b*c + a*g))/(64*a^2) + (x^6*(5*b*d + a*h))/(12*a^2) + (7*b*x
^9*(11*b*c + a*g))/(384*a^3) + (x*(51*b*c - 7*a*g))/(128*a*b) + (b*x^10*(5*b*d + a*h))/(32*a^3) + (15*b^2*e*x^
11)/(128*a^3) + (x^2*(11*b*d - a*h))/(32*a*b) + (21*b*e*x^7)/(64*a^2))/(a^3 + b^3*x^12 + 3*a^2*b*x^4 + 3*a*b^2
*x^8)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/(b*x**4+a)**4,x)

[Out]

Timed out

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