3.120 \(\int \frac {c+d x}{(a+b x^4)^3} \, dx\)

Optimal. Leaf size=266 \[ -\frac {21 c \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 c \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}-\frac {21 c \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 c \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b}}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {1855, 1876, 211, 1165, 628, 1162, 617, 204, 275, 205} \[ \frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}-\frac {21 c \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 c \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}-\frac {21 c \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 c \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b}}+\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*x^4)^3,x]

[Out]

(x*(c + d*x))/(8*a*(a + b*x^4)^2) + (x*(7*c + 6*d*x))/(32*a^2*(a + b*x^4)) + (3*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]
])/(16*a^(5/2)*Sqrt[b]) - (21*c*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(64*Sqrt[2]*a^(11/4)*b^(1/4)) + (21*c
*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(64*Sqrt[2]*a^(11/4)*b^(1/4)) - (21*c*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*
b^(1/4)*x + Sqrt[b]*x^2])/(128*Sqrt[2]*a^(11/4)*b^(1/4)) + (21*c*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqr
t[b]*x^2])/(128*Sqrt[2]*a^(11/4)*b^(1/4))

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, 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 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 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}{\left (a+b x^4\right )^3} \, dx &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}-\frac {\int \frac {-7 c-6 d x}{\left (a+b x^4\right )^2} \, dx}{8 a}\\ &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {\int \frac {21 c+12 d x}{a+b x^4} \, dx}{32 a^2}\\ &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {\int \left (\frac {21 c}{a+b x^4}+\frac {12 d x}{a+b x^4}\right ) \, dx}{32 a^2}\\ &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {(21 c) \int \frac {1}{a+b x^4} \, dx}{32 a^2}+\frac {(3 d) \int \frac {x}{a+b x^4} \, dx}{8 a^2}\\ &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {(21 c) \int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx}{64 a^{5/2}}+\frac {(21 c) \int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx}{64 a^{5/2}}+\frac {(3 d) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^2\right )}{16 a^2}\\ &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b}}+\frac {(21 c) \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} \sqrt {b}}+\frac {(21 c) \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} \sqrt {b}}-\frac {(21 c) \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} \sqrt [4]{b}}-\frac {(21 c) \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} \sqrt [4]{b}}\\ &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b}}-\frac {21 c \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 c \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {(21 c) \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} \sqrt [4]{b}}-\frac {(21 c) \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} \sqrt [4]{b}}\\ &=\frac {x (c+d x)}{8 a \left (a+b x^4\right )^2}+\frac {x (7 c+6 d x)}{32 a^2 \left (a+b x^4\right )}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {b}}-\frac {21 c \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 c \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} \sqrt [4]{b}}-\frac {21 c \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {21 c \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{128 \sqrt {2} a^{11/4} \sqrt [4]{b}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 249, normalized size = 0.94 \[ \frac {\frac {32 a^{7/4} x (c+d x)}{\left (a+b x^4\right )^2}+\frac {8 a^{3/4} x (7 c+6 d x)}{a+b x^4}-\frac {6 \left (8 \sqrt [4]{a} d+7 \sqrt {2} \sqrt [4]{b} c\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt {b}}+\frac {6 \left (7 \sqrt {2} \sqrt [4]{b} c-8 \sqrt [4]{a} d\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt {b}}-\frac {21 \sqrt {2} c \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{\sqrt [4]{b}}+\frac {21 \sqrt {2} c \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{\sqrt [4]{b}}}{256 a^{11/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + b*x^4)^3,x]

[Out]

((32*a^(7/4)*x*(c + d*x))/(a + b*x^4)^2 + (8*a^(3/4)*x*(7*c + 6*d*x))/(a + b*x^4) - (6*(7*Sqrt[2]*b^(1/4)*c +
8*a^(1/4)*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/Sqrt[b] + (6*(7*Sqrt[2]*b^(1/4)*c - 8*a^(1/4)*d)*ArcTan[
1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/Sqrt[b] - (21*Sqrt[2]*c*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^
2])/b^(1/4) + (21*Sqrt[2]*c*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/b^(1/4))/(256*a^(11/4))

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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((d*x+c)/(b*x^4+a)^3,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.18, size = 256, normalized size = 0.96 \[ \frac {21 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} c \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{256 \, a^{3} b} - \frac {21 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} c \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{256 \, a^{3} b} + \frac {3 \, \sqrt {2} {\left (4 \, \sqrt {2} \sqrt {a b} b d + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{128 \, a^{3} b^{2}} + \frac {3 \, \sqrt {2} {\left (4 \, \sqrt {2} \sqrt {a b} b d + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{128 \, a^{3} b^{2}} + \frac {6 \, b d x^{6} + 7 \, b c x^{5} + 10 \, a d x^{2} + 11 \, a c x}{32 \, {\left (b x^{4} + a\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*x^4+a)^3,x, algorithm="giac")

[Out]

21/256*sqrt(2)*(a*b^3)^(1/4)*c*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^3*b) - 21/256*sqrt(2)*(a*b^3)^(
1/4)*c*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^3*b) + 3/128*sqrt(2)*(4*sqrt(2)*sqrt(a*b)*b*d + 7*(a*b^
3)^(1/4)*b*c)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^3*b^2) + 3/128*sqrt(2)*(4*sqrt(2)
*sqrt(a*b)*b*d + 7*(a*b^3)^(1/4)*b*c)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^3*b^2) +
1/32*(6*b*d*x^6 + 7*b*c*x^5 + 10*a*d*x^2 + 11*a*c*x)/((b*x^4 + a)^2*a^2)

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maple [A]  time = 0.05, size = 222, normalized size = 0.83 \[ \frac {d \,x^{2}}{8 \left (b \,x^{4}+a \right )^{2} a}+\frac {c x}{8 \left (b \,x^{4}+a \right )^{2} a}+\frac {3 d \,x^{2}}{16 \left (b \,x^{4}+a \right ) a^{2}}+\frac {7 c x}{32 \left (b \,x^{4}+a \right ) a^{2}}+\frac {3 d \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{16 \sqrt {a b}\, a^{2}}+\frac {21 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{128 a^{3}}+\frac {21 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{128 a^{3}}+\frac {21 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{256 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(b*x^4+a)^3,x)

[Out]

1/8*c*x/a/(b*x^4+a)^2+7/32*c/a^2*x/(b*x^4+a)+21/256*c/a^3*(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)))+21/128*c/a^3*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4
)*x+1)+21/128*c/a^3*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)+1/8*d*x^2/a/(b*x^4+a)^2+3/16*d/a^2*x^2
/(b*x^4+a)+3/16*d/a^2/(a*b)^(1/2)*arctan((1/a*b)^(1/2)*x^2)

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maxima [A]  time = 3.06, size = 269, normalized size = 1.01 \[ \frac {6 \, b d x^{6} + 7 \, b c x^{5} + 10 \, a d x^{2} + 11 \, a c x}{32 \, {\left (a^{2} b^{2} x^{8} + 2 \, a^{3} b x^{4} + a^{4}\right )}} + \frac {3 \, {\left (\frac {7 \, \sqrt {2} c \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {7 \, \sqrt {2} c \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} + \frac {2 \, {\left (7 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} c - 8 \, \sqrt {a} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {1}{4}}} + \frac {2 \, {\left (7 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} c + 8 \, \sqrt {a} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {1}{4}}}\right )}}{256 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*x^4+a)^3,x, algorithm="maxima")

[Out]

1/32*(6*b*d*x^6 + 7*b*c*x^5 + 10*a*d*x^2 + 11*a*c*x)/(a^2*b^2*x^8 + 2*a^3*b*x^4 + a^4) + 3/256*(7*sqrt(2)*c*lo
g(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)) - 7*sqrt(2)*c*log(sqrt(b)*x^2 - sqrt(2)
*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)) + 2*(7*sqrt(2)*a^(1/4)*b^(1/4)*c - 8*sqrt(a)*d)*arctan(1/2*sqr
t(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^(1/4)) +
2*(7*sqrt(2)*a^(1/4)*b^(1/4)*c + 8*sqrt(a)*d)*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^(1/4)))/a^2

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mupad [B]  time = 4.99, size = 315, normalized size = 1.18 \[ \frac {\frac {5\,d\,x^2}{16\,a}+\frac {11\,c\,x}{32\,a}+\frac {7\,b\,c\,x^5}{32\,a^2}+\frac {3\,b\,d\,x^6}{16\,a^2}}{a^2+2\,a\,b\,x^4+b^2\,x^8}+\left (\sum _{k=1}^4\ln \left (\frac {b^2\,\left (63\,c\,d^2+36\,d^3\,x-{\mathrm {root}\left (268435456\,a^{11}\,b^2\,z^4+4718592\,a^6\,b\,d^2\,z^2-2709504\,a^3\,b\,c^2\,d\,z+194481\,b\,c^4+20736\,a\,d^4,z,k\right )}^2\,a^5\,b\,c\,7168-\mathrm {root}\left (268435456\,a^{11}\,b^2\,z^4+4718592\,a^6\,b\,d^2\,z^2-2709504\,a^3\,b\,c^2\,d\,z+194481\,b\,c^4+20736\,a\,d^4,z,k\right )\,a^2\,b\,c^2\,x\,1176+{\mathrm {root}\left (268435456\,a^{11}\,b^2\,z^4+4718592\,a^6\,b\,d^2\,z^2-2709504\,a^3\,b\,c^2\,d\,z+194481\,b\,c^4+20736\,a\,d^4,z,k\right )}^2\,a^5\,b\,d\,x\,4096\right )\,3}{a^6\,2048}\right )\,\mathrm {root}\left (268435456\,a^{11}\,b^2\,z^4+4718592\,a^6\,b\,d^2\,z^2-2709504\,a^3\,b\,c^2\,d\,z+194481\,b\,c^4+20736\,a\,d^4,z,k\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*x^4)^3,x)

[Out]

((5*d*x^2)/(16*a) + (11*c*x)/(32*a) + (7*b*c*x^5)/(32*a^2) + (3*b*d*x^6)/(16*a^2))/(a^2 + b^2*x^8 + 2*a*b*x^4)
+ symsum(log((3*b^2*(63*c*d^2 + 36*d^3*x - 7168*root(268435456*a^11*b^2*z^4 + 4718592*a^6*b*d^2*z^2 - 2709504
*a^3*b*c^2*d*z + 194481*b*c^4 + 20736*a*d^4, z, k)^2*a^5*b*c - 1176*root(268435456*a^11*b^2*z^4 + 4718592*a^6*
b*d^2*z^2 - 2709504*a^3*b*c^2*d*z + 194481*b*c^4 + 20736*a*d^4, z, k)*a^2*b*c^2*x + 4096*root(268435456*a^11*b
^2*z^4 + 4718592*a^6*b*d^2*z^2 - 2709504*a^3*b*c^2*d*z + 194481*b*c^4 + 20736*a*d^4, z, k)^2*a^5*b*d*x))/(2048
*a^6))*root(268435456*a^11*b^2*z^4 + 4718592*a^6*b*d^2*z^2 - 2709504*a^3*b*c^2*d*z + 194481*b*c^4 + 20736*a*d^
4, z, k), k, 1, 4)

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sympy [A]  time = 1.99, size = 192, normalized size = 0.72 \[ \operatorname {RootSum} {\left (268435456 t^{4} a^{11} b^{2} + 4718592 t^{2} a^{6} b d^{2} - 2709504 t a^{3} b c^{2} d + 20736 a d^{4} + 194481 b c^{4}, \left (t \mapsto t \log {\left (x + \frac {- 67108864 t^{3} a^{9} b d^{2} - 9633792 t^{2} a^{6} b c^{2} d - 589824 t a^{4} d^{4} - 2765952 t a^{3} b c^{4} + 423360 a c^{2} d^{3}}{193536 a c d^{4} - 453789 b c^{5}} \right )} \right )\right )} + \frac {11 a c x + 10 a d x^{2} + 7 b c x^{5} + 6 b d x^{6}}{32 a^{4} + 64 a^{3} b x^{4} + 32 a^{2} b^{2} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*x**4+a)**3,x)

[Out]

RootSum(268435456*_t**4*a**11*b**2 + 4718592*_t**2*a**6*b*d**2 - 2709504*_t*a**3*b*c**2*d + 20736*a*d**4 + 194
481*b*c**4, Lambda(_t, _t*log(x + (-67108864*_t**3*a**9*b*d**2 - 9633792*_t**2*a**6*b*c**2*d - 589824*_t*a**4*
d**4 - 2765952*_t*a**3*b*c**4 + 423360*a*c**2*d**3)/(193536*a*c*d**4 - 453789*b*c**5)))) + (11*a*c*x + 10*a*d*
x**2 + 7*b*c*x**5 + 6*b*d*x**6)/(32*a**4 + 64*a**3*b*x**4 + 32*a**2*b**2*x**8)

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