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

Optimal. Leaf size=413 \[ -\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{64 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{64 \sqrt {2} a^{11/4} b^{5/4}}+\frac {(a h+3 b d) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}}-\frac {4 a f-x \left (2 x (a h+3 b d)+a g+7 b c+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {x \left (x (b d-a h)-a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2} \]

[Out]

1/8*x*(b*c-a*g+(-a*h+b*d)*x+b*e*x^2+b*f*x^3)/a/b/(b*x^4+a)^2+1/32*(-4*a*f+x*(7*b*c+a*g+2*(a*h+3*b*d)*x+5*b*e*x
^2))/a^2/b/(b*x^4+a)+1/16*(a*h+3*b*d)*arctan(x^2*b^(1/2)/a^(1/2))/a^(5/2)/b^(3/2)-1/256*ln(-a^(1/4)*b^(1/4)*x*
2^(1/2)+a^(1/2)+x^2*b^(1/2))*(21*b*c+3*a*g-5*e*a^(1/2)*b^(1/2))/a^(11/4)/b^(5/4)*2^(1/2)+1/256*ln(a^(1/4)*b^(1
/4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2))*(21*b*c+3*a*g-5*e*a^(1/2)*b^(1/2))/a^(11/4)/b^(5/4)*2^(1/2)+1/128*arctan(-1
+b^(1/4)*x*2^(1/2)/a^(1/4))*(21*b*c+3*a*g+5*e*a^(1/2)*b^(1/2))/a^(11/4)/b^(5/4)*2^(1/2)+1/128*arctan(1+b^(1/4)
*x*2^(1/2)/a^(1/4))*(21*b*c+3*a*g+5*e*a^(1/2)*b^(1/2))/a^(11/4)/b^(5/4)*2^(1/2)

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Rubi [A]  time = 0.49, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {1858, 1854, 1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{64 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{64 \sqrt {2} a^{11/4} b^{5/4}}+\frac {(a h+3 b d) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}}-\frac {4 a f-x \left (2 x (a h+3 b d)+a g+7 b c+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {x \left (x (b d-a h)-a g+b c+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2} \]

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)^3,x]

[Out]

(x*(b*c - a*g + (b*d - a*h)*x + b*e*x^2 + b*f*x^3))/(8*a*b*(a + b*x^4)^2) - (4*a*f - x*(7*b*c + a*g + 2*(3*b*d
+ a*h)*x + 5*b*e*x^2))/(32*a^2*b*(a + b*x^4)) + ((3*b*d + a*h)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(16*a^(5/2)*b^(
3/2)) - ((21*b*c + 5*Sqrt[a]*Sqrt[b]*e + 3*a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(64*Sqrt[2]*a^(11/4)*
b^(5/4)) + ((21*b*c + 5*Sqrt[a]*Sqrt[b]*e + 3*a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(64*Sqrt[2]*a^(11/
4)*b^(5/4)) - ((21*b*c - 5*Sqrt[a]*Sqrt[b]*e + 3*a*g)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/
(128*Sqrt[2]*a^(11/4)*b^(5/4)) + ((21*b*c - 5*Sqrt[a]*Sqrt[b]*e + 3*a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)
*x + Sqrt[b]*x^2])/(128*Sqrt[2]*a^(11/4)*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 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 )^3} \, dx &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2}-\frac {\int \frac {-b (7 b c+a g)-2 b (3 b d+a h) x-5 b^2 e x^2-4 b^2 f x^3}{\left (a+b x^4\right )^2} \, dx}{8 a b^2}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2}-\frac {4 a f-x \left (7 b c+a g+2 (3 b d+a h) x+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {\int \frac {3 b (7 b c+a g)+4 b (3 b d+a h) x+5 b^2 e x^2}{a+b x^4} \, dx}{32 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 )}{8 a b \left (a+b x^4\right )^2}-\frac {4 a f-x \left (7 b c+a g+2 (3 b d+a h) x+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {\int \left (\frac {4 b (3 b d+a h) x}{a+b x^4}+\frac {3 b (7 b c+a g)+5 b^2 e x^2}{a+b x^4}\right ) \, dx}{32 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 )}{8 a b \left (a+b x^4\right )^2}-\frac {4 a f-x \left (7 b c+a g+2 (3 b d+a h) x+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {\int \frac {3 b (7 b c+a g)+5 b^2 e x^2}{a+b x^4} \, dx}{32 a^2 b^2}+\frac {(3 b d+a h) \int \frac {x}{a+b x^4} \, dx}{8 a^2 b}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2}-\frac {4 a f-x \left (7 b c+a g+2 (3 b d+a h) x+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e+3 a g\right ) \int \frac {\sqrt {a} \sqrt {b}-b x^2}{a+b x^4} \, dx}{64 a^{5/2} b^{3/2}}+\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e+3 a g\right ) \int \frac {\sqrt {a} \sqrt {b}+b x^2}{a+b x^4} \, dx}{64 a^{5/2} b^{3/2}}+\frac {(3 b d+a h) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^2\right )}{16 a^2 b}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2}-\frac {4 a f-x \left (7 b c+a g+2 (3 b d+a h) x+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {(3 b d+a h) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}}-\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e+3 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}{128 \sqrt {2} a^{11/4} b^{5/4}}-\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e+3 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}{128 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e+3 a g\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{128 a^{5/2} b^{3/2}}+\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e+3 a g\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{128 a^{5/2} b^{3/2}}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2}-\frac {4 a f-x \left (7 b c+a g+2 (3 b d+a h) x+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {(3 b d+a h) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}}-\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e+3 a g\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e+3 a g\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e+3 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 )}{64 \sqrt {2} a^{11/4} b^{5/4}}-\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e+3 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 )}{64 \sqrt {2} a^{11/4} b^{5/4}}\\ &=\frac {x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{8 a b \left (a+b x^4\right )^2}-\frac {4 a f-x \left (7 b c+a g+2 (3 b d+a h) x+5 b e x^2\right )}{32 a^2 b \left (a+b x^4\right )}+\frac {(3 b d+a h) \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}}-\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e+3 a g\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\left (21 b c+5 \sqrt {a} \sqrt {b} e+3 a g\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} b^{5/4}}-\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e+3 a g\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}+\frac {\left (21 b c-5 \sqrt {a} \sqrt {b} e+3 a g\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} b^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 411, normalized size = 1.00 \[ \frac {-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (8 a^{5/4} h+5 \sqrt {2} \sqrt {a} b^{3/4} e+24 \sqrt [4]{a} b d+3 \sqrt {2} a \sqrt [4]{b} g+21 \sqrt {2} b^{5/4} c\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (-8 a^{5/4} h+5 \sqrt {2} \sqrt {a} b^{3/4} e-24 \sqrt [4]{a} b d+3 \sqrt {2} a \sqrt [4]{b} g+21 \sqrt {2} b^{5/4} c\right )-\frac {32 a^{7/4} \sqrt {b} (a (f+x (g+h x))-b x (c+x (d+e x)))}{\left (a+b x^4\right )^2}+\frac {8 a^{3/4} \sqrt {b} x (a (g+2 h x)+7 b c+b x (6 d+5 e x))}{a+b x^4}+\sqrt {2} \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (5 \sqrt {a} \sqrt {b} e-3 a g-21 b c\right )+\sqrt {2} \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-5 \sqrt {a} \sqrt {b} e+3 a g+21 b c\right )}{256 a^{11/4} b^{3/2}} \]

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)^3,x]

[Out]

((8*a^(3/4)*Sqrt[b]*x*(7*b*c + b*x*(6*d + 5*e*x) + a*(g + 2*h*x)))/(a + b*x^4) - (32*a^(7/4)*Sqrt[b]*(-(b*x*(c
+ x*(d + e*x))) + a*(f + x*(g + h*x))))/(a + b*x^4)^2 - 2*(21*Sqrt[2]*b^(5/4)*c + 24*a^(1/4)*b*d + 5*Sqrt[2]*
Sqrt[a]*b^(3/4)*e + 3*Sqrt[2]*a*b^(1/4)*g + 8*a^(5/4)*h)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 2*(21*Sqrt[
2]*b^(5/4)*c - 24*a^(1/4)*b*d + 5*Sqrt[2]*Sqrt[a]*b^(3/4)*e + 3*Sqrt[2]*a*b^(1/4)*g - 8*a^(5/4)*h)*ArcTan[1 +
(Sqrt[2]*b^(1/4)*x)/a^(1/4)] + Sqrt[2]*b^(1/4)*(-21*b*c + 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*Log[Sqrt[a] - Sqrt[2]*a
^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + Sqrt[2]*b^(1/4)*(21*b*c - 5*Sqrt[a]*Sqrt[b]*e + 3*a*g)*Log[Sqrt[a] + Sqrt[2]
*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(256*a^(11/4)*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)^3,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.20, size = 459, normalized size = 1.11 \[ \frac {\sqrt {2} {\left (12 \, \sqrt {2} \sqrt {a b} b^{2} d + 4 \, \sqrt {2} \sqrt {a b} a b h + 21 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{128 \, a^{3} b^{3}} + \frac {\sqrt {2} {\left (12 \, \sqrt {2} \sqrt {a b} b^{2} d + 4 \, \sqrt {2} \sqrt {a b} a b h + 21 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{128 \, a^{3} b^{3}} + \frac {\sqrt {2} {\left (21 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{256 \, a^{3} b^{3}} - \frac {\sqrt {2} {\left (21 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a b g - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{256 \, a^{3} b^{3}} + \frac {5 \, b^{2} x^{7} e + 6 \, b^{2} d x^{6} + 2 \, a b h x^{6} + 7 \, b^{2} c x^{5} + a b g x^{5} + 9 \, a b x^{3} e + 10 \, a b d x^{2} - 2 \, a^{2} h x^{2} + 11 \, a b c x - 3 \, a^{2} g x - 4 \, a^{2} f}{32 \, {\left (b x^{4} + a\right )}^{2} a^{2} b} \]

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)^3,x, algorithm="giac")

[Out]

1/128*sqrt(2)*(12*sqrt(2)*sqrt(a*b)*b^2*d + 4*sqrt(2)*sqrt(a*b)*a*b*h + 21*(a*b^3)^(1/4)*b^2*c + 3*(a*b^3)^(1/
4)*a*b*g + 5*(a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^3*b^3) + 1/128*sq
rt(2)*(12*sqrt(2)*sqrt(a*b)*b^2*d + 4*sqrt(2)*sqrt(a*b)*a*b*h + 21*(a*b^3)^(1/4)*b^2*c + 3*(a*b^3)^(1/4)*a*b*g
+ 5*(a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^3*b^3) + 1/256*sqrt(2)*(2
1*(a*b^3)^(1/4)*b^2*c + 3*(a*b^3)^(1/4)*a*b*g - 5*(a*b^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b)
)/(a^3*b^3) - 1/256*sqrt(2)*(21*(a*b^3)^(1/4)*b^2*c + 3*(a*b^3)^(1/4)*a*b*g - 5*(a*b^3)^(3/4)*e)*log(x^2 - sqr
t(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^3*b^3) + 1/32*(5*b^2*x^7*e + 6*b^2*d*x^6 + 2*a*b*h*x^6 + 7*b^2*c*x^5 + a*b*
g*x^5 + 9*a*b*x^3*e + 10*a*b*d*x^2 - 2*a^2*h*x^2 + 11*a*b*c*x - 3*a^2*g*x - 4*a^2*f)/((b*x^4 + a)^2*a^2*b)

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maple [A]  time = 0.06, size = 561, normalized size = 1.36 \[ \frac {h \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{16 \sqrt {a b}\, a b}+\frac {3 d \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{16 \sqrt {a b}\, a^{2}}+\frac {5 \sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{128 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2} b}+\frac {5 \sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{128 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2} b}+\frac {5 \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 )}{256 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2} b}+\frac {3 \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 )}{128 a^{2} b}+\frac {3 \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 )}{128 a^{2} b}+\frac {3 \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 )}{256 a^{2} b}+\frac {21 \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 )}{128 a^{3}}+\frac {21 \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 )}{128 a^{3}}+\frac {21 \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 )}{256 a^{3}}+\frac {\frac {5 b e \,x^{7}}{32 a^{2}}+\frac {\left (a h +3 b d \right ) x^{6}}{16 a^{2}}+\frac {9 e \,x^{3}}{32 a}+\frac {\left (a g +7 b c \right ) x^{5}}{32 a^{2}}-\frac {\left (a h -5 b d \right ) x^{2}}{16 a b}-\frac {f}{8 b}-\frac {\left (3 a g -11 b c \right ) x}{32 a b}}{\left (b \,x^{4}+a \right )^{2}} \]

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)^3,x)

[Out]

(5/32/a^2*b*e*x^7+1/16*(a*h+3*b*d)/a^2*x^6+1/32*(a*g+7*b*c)/a^2*x^5+9/32/a*e*x^3-1/16*(a*h-5*b*d)/a/b*x^2-1/32
*(3*a*g-11*b*c)/a/b*x-1/8/b*f)/(b*x^4+a)^2+3/128*(a/b)^(1/4)*2^(1/2)/a^2/b*g*arctan(2^(1/2)/(a/b)^(1/4)*x-1)+2
1/128*(a/b)^(1/4)*2^(1/2)/a^3*c*arctan(2^(1/2)/(a/b)^(1/4)*x-1)+3/256*(a/b)^(1/4)*2^(1/2)/a^2/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)))+21/256*(a/b)^(1/4)*2^(1/2)/a^3*c*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)))+3/128*(a/b)^(1/4)*2^(1/2)/a^2/b
*g*arctan(2^(1/2)/(a/b)^(1/4)*x+1)+21/128*(a/b)^(1/4)*2^(1/2)/a^3*c*arctan(2^(1/2)/(a/b)^(1/4)*x+1)+1/16/a/b/(
a*b)^(1/2)*arctan((1/a*b)^(1/2)*x^2)*h+3/16/(a*b)^(1/2)/a^2*d*arctan((1/a*b)^(1/2)*x^2)+5/256/(a/b)^(1/4)*2^(1
/2)/a^2/b*e*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)))+5/128/(a/b)^(1
/4)*2^(1/2)/a^2/b*e*arctan(2^(1/2)/(a/b)^(1/4)*x+1)+5/128/(a/b)^(1/4)*2^(1/2)/a^2/b*e*arctan(2^(1/2)/(a/b)^(1/
4)*x-1)

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maxima [A]  time = 3.07, size = 446, normalized size = 1.08 \[ \frac {5 \, b^{2} e x^{7} + 2 \, {\left (3 \, b^{2} d + a b h\right )} x^{6} + 9 \, a b e x^{3} + {\left (7 \, b^{2} c + a b g\right )} x^{5} - 4 \, a^{2} f + 2 \, {\left (5 \, a b d - a^{2} h\right )} x^{2} + {\left (11 \, a b c - 3 \, a^{2} g\right )} x}{32 \, {\left (a^{2} b^{3} x^{8} + 2 \, a^{3} b^{2} x^{4} + a^{4} b\right )}} + \frac {\frac {\sqrt {2} {\left (21 \, b^{\frac {3}{2}} c - 5 \, \sqrt {a} b e + 3 \, a \sqrt {b} g\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (21 \, b^{\frac {3}{2}} c - 5 \, \sqrt {a} b e + 3 \, a \sqrt {b} g\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} + \frac {2 \, {\left (21 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + 5 \, \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + 3 \, \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g - 24 \, \sqrt {a} b^{\frac {3}{2}} d - 8 \, a^{\frac {3}{2}} \sqrt {b} h\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}} + \frac {2 \, {\left (21 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + 5 \, \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + 3 \, \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g + 24 \, \sqrt {a} b^{\frac {3}{2}} d + 8 \, a^{\frac {3}{2}} \sqrt {b} h\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}}}{256 \, a^{2} b} \]

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)^3,x, algorithm="maxima")

[Out]

1/32*(5*b^2*e*x^7 + 2*(3*b^2*d + a*b*h)*x^6 + 9*a*b*e*x^3 + (7*b^2*c + a*b*g)*x^5 - 4*a^2*f + 2*(5*a*b*d - a^2
*h)*x^2 + (11*a*b*c - 3*a^2*g)*x)/(a^2*b^3*x^8 + 2*a^3*b^2*x^4 + a^4*b) + 1/256*(sqrt(2)*(21*b^(3/2)*c - 5*sqr
t(a)*b*e + 3*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)*(
21*b^(3/2)*c - 5*sqrt(a)*b*e + 3*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)) + 2*(21*sqrt(2)*a^(1/4)*b^(7/4)*c + 5*sqrt(2)*a^(3/4)*b^(5/4)*e + 3*sqrt(2)*a^(5/4)*b^(3/4)*g - 24*sq
rt(a)*b^(3/2)*d - 8*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*(21*sqrt(2)*a^(1/4)*b^(7/4)*c + 5*sqrt(2)*a^(3/4)*b^(5/
4)*e + 3*sqrt(2)*a^(5/4)*b^(3/4)*g + 24*sqrt(a)*b^(3/2)*d + 8*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^2*b)

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mupad [B]  time = 5.69, size = 1686, normalized size = 4.08 \[ \text {result too large to display} \]

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)^3,x)

[Out]

((9*e*x^3)/(32*a) - f/(8*b) + (x^5*(7*b*c + a*g))/(32*a^2) + (x^6*(3*b*d + a*h))/(16*a^2) + (x*(11*b*c - 3*a*g
))/(32*a*b) + (x^2*(5*b*d - a*h))/(16*a*b) + (5*b*e*x^7)/(32*a^2))/(a^2 + b^2*x^8 + 2*a*b*x^4) + symsum(log((3
024*b^3*c*d^2 - 125*a*b^2*e^3 - 2205*b^3*c^2*e + 48*a^3*g*h^2 + 432*a*b^2*d^2*g + 336*a^2*b*c*h^2 - 45*a^2*b*e
*g^2 + 2016*a*b^2*c*d*h - 630*a*b^2*c*e*g + 288*a^2*b*d*g*h)/(32768*a^6*b) - root(268435456*a^11*b^6*z^4 + 314
5728*a^7*b^4*d*h*z^2 + 983040*a^7*b^4*e*g*z^2 + 6881280*a^6*b^5*c*e*z^2 + 524288*a^8*b^3*h^2*z^2 + 4718592*a^6
*b^5*d^2*z^2 - 258048*a^5*b^3*c*g*h*z - 774144*a^4*b^4*c*d*g*z - 18432*a^6*b^2*g^2*h*z + 51200*a^5*b^3*e^2*h*z
- 903168*a^4*b^4*c^2*h*z - 55296*a^5*b^3*d*g^2*z + 153600*a^4*b^4*d*e^2*z - 2709504*a^3*b^5*c^2*d*z - 5760*a^
3*b^2*d*e*g*h - 40320*a^2*b^3*c*d*e*h - 8640*a^2*b^3*d^2*e*g - 6720*a^3*b^2*c*e*h^2 + 6300*a^2*b^3*c*e^2*g - 9
60*a^4*b*e*g*h^2 - 60480*a*b^4*c*d^2*e + 3072*a^4*b*d*h^3 + 111132*a*b^4*c^3*g + 13824*a^3*b^2*d^2*h^2 + 450*a
^3*b^2*e^2*g^2 + 23814*a^2*b^3*c^2*g^2 + 27648*a^2*b^3*d^3*h + 2268*a^3*b^2*c*g^3 + 22050*a*b^4*c^2*e^2 + 625*
a^2*b^3*e^4 + 81*a^4*b*g^4 + 20736*a*b^4*d^4 + 256*a^5*h^4 + 194481*b^5*c^4, z, k)*(root(268435456*a^11*b^6*z^
4 + 3145728*a^7*b^4*d*h*z^2 + 983040*a^7*b^4*e*g*z^2 + 6881280*a^6*b^5*c*e*z^2 + 524288*a^8*b^3*h^2*z^2 + 4718
592*a^6*b^5*d^2*z^2 - 258048*a^5*b^3*c*g*h*z - 774144*a^4*b^4*c*d*g*z - 18432*a^6*b^2*g^2*h*z + 51200*a^5*b^3*
e^2*h*z - 903168*a^4*b^4*c^2*h*z - 55296*a^5*b^3*d*g^2*z + 153600*a^4*b^4*d*e^2*z - 2709504*a^3*b^5*c^2*d*z -
5760*a^3*b^2*d*e*g*h - 40320*a^2*b^3*c*d*e*h - 8640*a^2*b^3*d^2*e*g - 6720*a^3*b^2*c*e*h^2 + 6300*a^2*b^3*c*e^
2*g - 960*a^4*b*e*g*h^2 - 60480*a*b^4*c*d^2*e + 3072*a^4*b*d*h^3 + 111132*a*b^4*c^3*g + 13824*a^3*b^2*d^2*h^2
+ 450*a^3*b^2*e^2*g^2 + 23814*a^2*b^3*c^2*g^2 + 27648*a^2*b^3*d^3*h + 2268*a^3*b^2*c*g^3 + 22050*a*b^4*c^2*e^2
+ 625*a^2*b^3*e^4 + 81*a^4*b*g^4 + 20736*a*b^4*d^4 + 256*a^5*h^4 + 194481*b^5*c^4, z, k)*((344064*a^5*b^4*c +
49152*a^6*b^3*g)/(32768*a^6*b) - (x*(24576*a^5*b^4*d + 8192*a^6*b^3*h))/(4096*a^6*b)) + (15360*a^3*b^3*d*e +
5120*a^4*b^2*e*h)/(32768*a^6*b) + (x*(7056*a^2*b^4*c^2 - 400*a^3*b^3*e^2 + 144*a^4*b^2*g^2 + 2016*a^3*b^3*c*g)
)/(4096*a^6*b)) + (x*(216*b^3*d^3 + 8*a^3*h^3 - 315*b^3*c*d*e + 216*a*b^2*d^2*h + 72*a^2*b*d*h^2 - 105*a*b^2*c
*e*h - 45*a*b^2*d*e*g - 15*a^2*b*e*g*h))/(4096*a^6*b))*root(268435456*a^11*b^6*z^4 + 3145728*a^7*b^4*d*h*z^2 +
983040*a^7*b^4*e*g*z^2 + 6881280*a^6*b^5*c*e*z^2 + 524288*a^8*b^3*h^2*z^2 + 4718592*a^6*b^5*d^2*z^2 - 258048*
a^5*b^3*c*g*h*z - 774144*a^4*b^4*c*d*g*z - 18432*a^6*b^2*g^2*h*z + 51200*a^5*b^3*e^2*h*z - 903168*a^4*b^4*c^2*
h*z - 55296*a^5*b^3*d*g^2*z + 153600*a^4*b^4*d*e^2*z - 2709504*a^3*b^5*c^2*d*z - 5760*a^3*b^2*d*e*g*h - 40320*
a^2*b^3*c*d*e*h - 8640*a^2*b^3*d^2*e*g - 6720*a^3*b^2*c*e*h^2 + 6300*a^2*b^3*c*e^2*g - 960*a^4*b*e*g*h^2 - 604
80*a*b^4*c*d^2*e + 3072*a^4*b*d*h^3 + 111132*a*b^4*c^3*g + 13824*a^3*b^2*d^2*h^2 + 450*a^3*b^2*e^2*g^2 + 23814
*a^2*b^3*c^2*g^2 + 27648*a^2*b^3*d^3*h + 2268*a^3*b^2*c*g^3 + 22050*a*b^4*c^2*e^2 + 625*a^2*b^3*e^4 + 81*a^4*b
*g^4 + 20736*a*b^4*d^4 + 256*a^5*h^4 + 194481*b^5*c^4, z, k), k, 1, 4)

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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)**3,x)

[Out]

Timed out

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