Optimal. Leaf size=206 \[ a^2 x-\frac {3 a b \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 i a b \sqrt [3]{x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+2 i a b x-\frac {3 i b^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{2/3}}{d}-b^2 x \]
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Rubi [A] time = 0.35, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {3739, 3722, 3719, 2190, 2531, 2282, 6589, 3720, 2279, 2391, 30} \[ a^2 x+\frac {6 i a b \sqrt [3]{x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 a b \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+2 i a b x-\frac {3 i b^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{2/3}}{d}-b^2 x \]
Antiderivative was successfully verified.
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Rule 30
Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3719
Rule 3720
Rule 3722
Rule 3739
Rule 6589
Rubi steps
\begin {align*} \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx &=3 \operatorname {Subst}\left (\int x^2 (a+b \tan (c+d x))^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (a^2 x^2+2 a b x^2 \tan (c+d x)+b^2 x^2 \tan ^2(c+d x)\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=a^2 x+(6 a b) \operatorname {Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )+\left (3 b^2\right ) \operatorname {Subst}\left (\int x^2 \tan ^2(c+d x) \, dx,x,\sqrt [3]{x}\right )\\ &=a^2 x+2 i a b x+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-(12 i a b) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^2}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )-\left (3 b^2\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\sqrt [3]{x}\right )-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int x \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(12 a b) \operatorname {Subst}\left (\int x \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d}+\frac {\left (12 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {6 i a b \sqrt [3]{x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(6 i a b) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}\\ &=-\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {6 i a b \sqrt [3]{x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(3 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}\\ &=-\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {3 i b^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 i a b \sqrt [3]{x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 a b \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}\\ \end {align*}
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Mathematica [A] time = 2.13, size = 185, normalized size = 0.90 \[ x \left (a^2+2 a b \tan (c)-b^2\right )+\frac {b \left (\frac {6 i b d^2 x^{2/3}-4 i a d^3 x}{1+e^{2 i c}}+3 i \left (b-2 a d \sqrt [3]{x}\right ) \text {Li}_2\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+6 d \sqrt [3]{x} \left (b-a d \sqrt [3]{x}\right ) \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-3 a \text {Li}_3\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )}{d^3}+\frac {3 b^2 x^{2/3} \sec (c) \sin \left (d \sqrt [3]{x}\right ) \sec \left (c+d \sqrt [3]{x}\right )}{d} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.56, size = 318, normalized size = 1.54 \[ \frac {6 \, b^{2} d^{2} x^{\frac {2}{3}} \tan \left (d x^{\frac {1}{3}} + c\right ) + 2 \, {\left (a^{2} - b^{2}\right )} d^{3} x - 3 \, a b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 3 \, a b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} - 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) + {\left (-6 i \, a b d x^{\frac {1}{3}} + 3 i \, b^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) + {\left (6 i \, a b d x^{\frac {1}{3}} - 3 i \, b^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) - 6 \, {\left (a b d^{2} x^{\frac {2}{3}} - b^{2} d x^{\frac {1}{3}}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 6 \, {\left (a b d^{2} x^{\frac {2}{3}} - b^{2} d x^{\frac {1}{3}}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right )}{2 \, d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.29, size = 0, normalized size = 0.00 \[ \int \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} x + \frac {6 \, b^{2} x^{\frac {2}{3}} \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) - {\left (b^{2} d \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + b^{2} d \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + 2 \, b^{2} d \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + b^{2} d\right )} x - \frac {{\left (-8 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} a b + 4 \, {\left (6 i \, a b c + 3 i \, b^{2}\right )} {\left (d x^{\frac {1}{3}} + c\right )}^{2} + 12 \, a b {\rm Li}_{3}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) + 2 \, {\left (-12 i \, a b c^{2} - 12 i \, b^{2} c\right )} {\left (d x^{\frac {1}{3}} + c\right )} + {\left (24 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} a b + 24 i \, a b c^{2} + 24 i \, b^{2} c + 2 \, {\left (-24 i \, a b c - 12 i \, b^{2}\right )} {\left (d x^{\frac {1}{3}} + c\right )}\right )} \arctan \left (\sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ), \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + 1\right ) + {\left (-24 i \, {\left (d x^{\frac {1}{3}} + c\right )} a b + 24 i \, a b c + 12 i \, b^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}\right ) + 12 \, {\left ({\left (d x^{\frac {1}{3}} + c\right )}^{2} a b + a b c^{2} + b^{2} c - {\left (2 \, a b c + b^{2}\right )} {\left (d x^{\frac {1}{3}} + c\right )}\right )} \log \left (\cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + 1\right )\right )} {\left (d \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + d\right )}}{4 \, d^{3}}}{d \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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