3.126 \(\int \frac {(a+b \tan ^{-1}(c x^3))^3}{x^7} \, dx\)

Optimal. Leaf size=146 \[ b^2 c^2 \log \left (2-\frac {2}{1-i c x^3}\right ) \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {1}{2} i b c^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )^2-\frac {1}{6} c^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )^3-\frac {b c \left (a+b \tan ^{-1}\left (c x^3\right )\right )^2}{2 x^3}-\frac {\left (a+b \tan ^{-1}\left (c x^3\right )\right )^3}{6 x^6}-\frac {1}{2} i b^3 c^2 \text {Li}_2\left (\frac {2}{1-i c x^3}-1\right ) \]

[Out]

-1/2*I*b*c^2*(a+b*arctan(c*x^3))^2-1/2*b*c*(a+b*arctan(c*x^3))^2/x^3-1/6*c^2*(a+b*arctan(c*x^3))^3-1/6*(a+b*ar
ctan(c*x^3))^3/x^6+b^2*c^2*(a+b*arctan(c*x^3))*ln(2-2/(1-I*c*x^3))-1/2*I*b^3*c^2*polylog(2,-1+2/(1-I*c*x^3))

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Rubi [F]  time = 1.67, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (a+b \tan ^{-1}\left (c x^3\right )\right )^3}{x^7} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*ArcTan[c*x^3])^3/x^7,x]

[Out]

(3*a*b^2*c^2*Log[x])/4 - (b*c*(1 - I*c*x^3)*(2*a + I*b*Log[1 - I*c*x^3])^2)/(16*x^3) + (I/16)*b*c^2*Log[I*c*x^
3]*(2*a + I*b*Log[1 - I*c*x^3])^2 - (c^2*(2*a + I*b*Log[1 - I*c*x^3])^3)/48 - (2*a + I*b*Log[1 - I*c*x^3])^3/(
48*x^6) + (b^3*c*(1 + I*c*x^3)*Log[1 + I*c*x^3]^2)/(16*x^3) + (I/16)*b^3*c^2*Log[(-I)*c*x^3]*Log[1 + I*c*x^3]^
2 - (I/48)*b^3*c^2*Log[1 + I*c*x^3]^3 - ((I/48)*b^3*Log[1 + I*c*x^3]^3)/x^6 + (I/8)*b^3*c^2*PolyLog[2, (-I)*c*
x^3] - (I/8)*b^3*c^2*PolyLog[2, I*c*x^3] - (b^2*c^2*(2*a + I*b*Log[1 - I*c*x^3])*PolyLog[2, 1 - I*c*x^3])/8 +
(I/8)*b^3*c^2*Log[1 + I*c*x^3]*PolyLog[2, 1 + I*c*x^3] + (I/8)*b^3*c^2*PolyLog[3, 1 - I*c*x^3] - (I/8)*b^3*c^2
*PolyLog[3, 1 + I*c*x^3] + (I/8)*b*Defer[Subst][Defer[Int][(((-2*I)*a + b*Log[1 - I*c*x])^2*Log[1 + I*c*x])/x^
3, x], x, x^3] - (I/8)*b^2*Defer[Subst][Defer[Int][(((-2*I)*a + b*Log[1 - I*c*x])*Log[1 + I*c*x]^2)/x^3, x], x
, x^3]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}\left (c x^3\right )\right )^3}{x^7} \, dx &=\int \left (\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{8 x^7}+\frac {3 i b \left (-2 i a+b \log \left (1-i c x^3\right )\right )^2 \log \left (1+i c x^3\right )}{8 x^7}-\frac {3 i b^2 \left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log ^2\left (1+i c x^3\right )}{8 x^7}+\frac {i b^3 \log ^3\left (1+i c x^3\right )}{8 x^7}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{x^7} \, dx+\frac {1}{8} (3 i b) \int \frac {\left (-2 i a+b \log \left (1-i c x^3\right )\right )^2 \log \left (1+i c x^3\right )}{x^7} \, dx-\frac {1}{8} \left (3 i b^2\right ) \int \frac {\left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log ^2\left (1+i c x^3\right )}{x^7} \, dx+\frac {1}{8} \left (i b^3\right ) \int \frac {\log ^3\left (1+i c x^3\right )}{x^7} \, dx\\ &=\frac {1}{24} \operatorname {Subst}\left (\int \frac {(2 a+i b \log (1-i c x))^3}{x^3} \, dx,x,x^3\right )+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{24} \left (i b^3\right ) \operatorname {Subst}\left (\int \frac {\log ^3(1+i c x)}{x^3} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} (b c) \operatorname {Subst}\left (\int \frac {(2 a+i b \log (1-i c x))^2}{x^2 (1-i c x)} \, dx,x,x^3\right )-\frac {1}{16} \left (b^3 c\right ) \operatorname {Subst}\left (\int \frac {\log ^2(1+i c x)}{x^2 (1+i c x)} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{16} (i b) \operatorname {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{x \left (-\frac {i}{c}+\frac {i x}{c}\right )^2} \, dx,x,1-i c x^3\right )+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} \left (i b^3\right ) \operatorname {Subst}\left (\int \frac {\log ^2(x)}{x \left (\frac {i}{c}-\frac {i x}{c}\right )^2} \, dx,x,1+i c x^3\right )\\ &=-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{16} (i b) \operatorname {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{\left (-\frac {i}{c}+\frac {i x}{c}\right )^2} \, dx,x,1-i c x^3\right )+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} \left (i b^3\right ) \operatorname {Subst}\left (\int \frac {\log ^2(x)}{\left (\frac {i}{c}-\frac {i x}{c}\right )^2} \, dx,x,1+i c x^3\right )-\frac {1}{16} (b c) \operatorname {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{x \left (-\frac {i}{c}+\frac {i x}{c}\right )} \, dx,x,1-i c x^3\right )+\frac {1}{16} \left (b^3 c\right ) \operatorname {Subst}\left (\int \frac {\log ^2(x)}{x \left (\frac {i}{c}-\frac {i x}{c}\right )} \, dx,x,1+i c x^3\right )\\ &=-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{16} (b c) \operatorname {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{-\frac {i}{c}+\frac {i x}{c}} \, dx,x,1-i c x^3\right )+\frac {1}{8} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {2 a+i b \log (x)}{-\frac {i}{c}+\frac {i x}{c}} \, dx,x,1-i c x^3\right )+\frac {1}{16} \left (b^3 c\right ) \operatorname {Subst}\left (\int \frac {\log ^2(x)}{\frac {i}{c}-\frac {i x}{c}} \, dx,x,1+i c x^3\right )-\frac {1}{8} \left (b^3 c\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{\frac {i}{c}-\frac {i x}{c}} \, dx,x,1+i c x^3\right )-\frac {1}{16} \left (i b c^2\right ) \operatorname {Subst}\left (\int \frac {(2 a+i b \log (x))^2}{x} \, dx,x,1-i c x^3\right )-\frac {1}{16} \left (i b^3 c^2\right ) \operatorname {Subst}\left (\int \frac {\log ^2(x)}{x} \, dx,x,1+i c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} i b c^2 \log \left (i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}+\frac {1}{16} i b^3 c^2 \log \left (-i c x^3\right ) \log ^2\left (1+i c x^3\right )-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} i b^3 c^2 \text {Li}_2\left (-i c x^3\right )+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (b^3 c\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{-\frac {i}{c}+\frac {i x}{c}} \, dx,x,1-i c x^3\right )-\frac {1}{16} c^2 \operatorname {Subst}\left (\int x^2 \, dx,x,2 a+i b \log \left (1-i c x^3\right )\right )+\frac {1}{8} \left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x) (2 a+i b \log (x))}{x} \, dx,x,1-i c x^3\right )-\frac {1}{16} \left (i b^3 c^2\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\log \left (1+i c x^3\right )\right )-\frac {1}{8} \left (i b^3 c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x) \log (x)}{x} \, dx,x,1+i c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} i b c^2 \log \left (i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {1}{48} c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right )^3-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}+\frac {1}{16} i b^3 c^2 \log \left (-i c x^3\right ) \log ^2\left (1+i c x^3\right )-\frac {1}{48} i b^3 c^2 \log ^3\left (1+i c x^3\right )-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} i b^3 c^2 \text {Li}_2\left (-i c x^3\right )-\frac {1}{8} i b^3 c^2 \text {Li}_2\left (i c x^3\right )-\frac {1}{8} b^2 c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right ) \text {Li}_2\left (1-i c x^3\right )+\frac {1}{8} i b^3 c^2 \log \left (1+i c x^3\right ) \text {Li}_2\left (1+i c x^3\right )+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{8} \left (i b^3 c^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-i c x^3\right )-\frac {1}{8} \left (i b^3 c^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+i c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} i b c^2 \log \left (i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {1}{48} c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right )^3-\frac {\left (2 a+i b \log \left (1-i c x^3\right )\right )^3}{48 x^6}+\frac {b^3 c \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{16 x^3}+\frac {1}{16} i b^3 c^2 \log \left (-i c x^3\right ) \log ^2\left (1+i c x^3\right )-\frac {1}{48} i b^3 c^2 \log ^3\left (1+i c x^3\right )-\frac {i b^3 \log ^3\left (1+i c x^3\right )}{48 x^6}+\frac {1}{8} i b^3 c^2 \text {Li}_2\left (-i c x^3\right )-\frac {1}{8} i b^3 c^2 \text {Li}_2\left (i c x^3\right )-\frac {1}{8} b^2 c^2 \left (2 a+i b \log \left (1-i c x^3\right )\right ) \text {Li}_2\left (1-i c x^3\right )+\frac {1}{8} i b^3 c^2 \log \left (1+i c x^3\right ) \text {Li}_2\left (1+i c x^3\right )+\frac {1}{8} i b^3 c^2 \text {Li}_3\left (1-i c x^3\right )-\frac {1}{8} i b^3 c^2 \text {Li}_3\left (1+i c x^3\right )+\frac {1}{8} (i b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^3} \, dx,x,x^3\right )\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 196, normalized size = 1.34 \[ -\frac {a \left (a \left (a+3 b c x^3\right )-6 b^2 c^2 x^6 \log \left (\frac {c x^3}{\sqrt {c^2 x^6+1}}\right )\right )+3 b^2 \tan ^{-1}\left (c x^3\right )^2 \left (a c^2 x^6+a+b c x^3 \left (1+i c x^3\right )\right )+3 b \tan ^{-1}\left (c x^3\right ) \left (a \left (a c^2 x^6+a+2 b c x^3\right )-2 b^2 c^2 x^6 \log \left (1-e^{2 i \tan ^{-1}\left (c x^3\right )}\right )\right )+3 i b^3 c^2 x^6 \text {Li}_2\left (e^{2 i \tan ^{-1}\left (c x^3\right )}\right )+b^3 \left (c^2 x^6+1\right ) \tan ^{-1}\left (c x^3\right )^3}{6 x^6} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c*x^3])^3/x^7,x]

[Out]

-1/6*(3*b^2*(a + a*c^2*x^6 + b*c*x^3*(1 + I*c*x^3))*ArcTan[c*x^3]^2 + b^3*(1 + c^2*x^6)*ArcTan[c*x^3]^3 + 3*b*
ArcTan[c*x^3]*(a*(a + 2*b*c*x^3 + a*c^2*x^6) - 2*b^2*c^2*x^6*Log[1 - E^((2*I)*ArcTan[c*x^3])]) + a*(a*(a + 3*b
*c*x^3) - 6*b^2*c^2*x^6*Log[(c*x^3)/Sqrt[1 + c^2*x^6]]) + (3*I)*b^3*c^2*x^6*PolyLog[2, E^((2*I)*ArcTan[c*x^3])
])/x^6

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (c x^{3}\right )^{3} + 3 \, a b^{2} \arctan \left (c x^{3}\right )^{2} + 3 \, a^{2} b \arctan \left (c x^{3}\right ) + a^{3}}{x^{7}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))^3/x^7,x, algorithm="fricas")

[Out]

integral((b^3*arctan(c*x^3)^3 + 3*a*b^2*arctan(c*x^3)^2 + 3*a^2*b*arctan(c*x^3) + a^3)/x^7, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{3}}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))^3/x^7,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^3) + a)^3/x^7, x)

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maple [F]  time = 0.88, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctan \left (c \,x^{3}\right )\right )^{3}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^3))^3/x^7,x)

[Out]

int((a+b*arctan(c*x^3))^3/x^7,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left ({\left (c \arctan \left (c x^{3}\right ) + \frac {1}{x^{3}}\right )} c + \frac {\arctan \left (c x^{3}\right )}{x^{6}}\right )} a^{2} b + \frac {1}{2} \, {\left ({\left (\arctan \left (c x^{3}\right )^{2} - \log \left (c^{2} x^{6} + 1\right ) + 6 \, \log \relax (x)\right )} c^{2} - 2 \, {\left (c \arctan \left (c x^{3}\right ) + \frac {1}{x^{3}}\right )} c \arctan \left (c x^{3}\right )\right )} a b^{2} - \frac {a b^{2} \arctan \left (c x^{3}\right )^{2}}{2 \, x^{6}} + \frac {\frac {3}{8} \, {\left (8 \, x^{6} \int -\frac {84 \, c^{2} x^{6} \arctan \left (c x^{3}\right ) \log \left (c^{2} x^{6} + 1\right ) - 180 \, c x^{3} \arctan \left (c x^{3}\right )^{2} - 392 \, {\left (c^{2} x^{6} + 1\right )} \arctan \left (c x^{3}\right )^{3} + 21 \, {\left (c x^{3} - 2 \, {\left (c^{2} x^{6} + 1\right )} \arctan \left (c x^{3}\right )\right )} \log \left (c^{2} x^{6} + 1\right )^{2}}{8 \, {\left (c^{2} x^{13} + x^{7}\right )}}\,{d x} - 20 \, \arctan \left (c x^{3}\right )^{3} + 7 \, \arctan \left (c x^{3}\right ) \log \left (c^{2} x^{6} + 1\right )^{2}\right )} b^{3}}{192 \, x^{6}} - \frac {a^{3}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))^3/x^7,x, algorithm="maxima")

[Out]

-1/2*((c*arctan(c*x^3) + 1/x^3)*c + arctan(c*x^3)/x^6)*a^2*b + 1/2*((arctan(c*x^3)^2 - log(c^2*x^6 + 1) + 6*lo
g(x))*c^2 - 2*(c*arctan(c*x^3) + 1/x^3)*c*arctan(c*x^3))*a*b^2 - 1/2*a*b^2*arctan(c*x^3)^2/x^6 + 1/192*(192*x^
6*integrate(-1/64*(12*c^2*x^6*arctan(c*x^3)*log(c^2*x^6 + 1) - 12*c*x^3*arctan(c*x^3)^2 - 56*(c^2*x^6 + 1)*arc
tan(c*x^3)^3 + 3*(c*x^3 - 2*(c^2*x^6 + 1)*arctan(c*x^3))*log(c^2*x^6 + 1)^2)/(c^2*x^13 + x^7), x) - 4*arctan(c
*x^3)^3 + 3*arctan(c*x^3)*log(c^2*x^6 + 1)^2)*b^3/x^6 - 1/6*a^3/x^6

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^3}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))^3/x^7,x)

[Out]

int((a + b*atan(c*x^3))^3/x^7, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{3}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))**3/x**7,x)

[Out]

Integral((a + b*atan(c*x**3))**3/x**7, x)

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