Optimal. Leaf size=206 \[ -\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {3 i b^2 \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 i a b \sqrt [3]{x} \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 a b \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d} \]
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Rubi [A]
time = 0.25, 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 = {3824, 3803,
3800, 2221, 2611, 2320, 6724, 3801, 2317, 2438, 30} \begin {gather*} 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 \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2221
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 3800
Rule 3801
Rule 3803
Rule 3824
Rule 6724
Rubi steps
\begin {align*} \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx &=3 \text {Subst}\left (\int x^2 (a+b \tan (c+d x))^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \text {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) \text {Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )+\left (3 b^2\right ) \text {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) \text {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 ) \text {Subst}\left (\int x^2 \, dx,x,\sqrt [3]{x}\right )-\frac {\left (6 b^2\right ) \text {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) \text {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 ) \text {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) \text {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 ) \text {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) \text {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 ) \text {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 = 2.27, size = 193, normalized size = 0.94 \begin {gather*} \frac {b \left (\frac {2 i d^2 e^{2 i c} \left (-3 b+2 a d \sqrt [3]{x}\right ) x^{2/3}}{1+e^{2 i c}}+6 d \left (b-a d \sqrt [3]{x}\right ) \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )-3 i \left (b-2 a d \sqrt [3]{x}\right ) \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )-3 a \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )}{d^3}+\frac {3 b^2 x^{2/3} \sec (c) \sec \left (c+d \sqrt [3]{x}\right ) \sin \left (d \sqrt [3]{x}\right )}{d}+x \left (a^2-b^2+2 a b \tan (c)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.82, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 320, normalized size = 1.55 \begin {gather*} \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 ) - 3 \, {\left (2 i \, a b d x^{\frac {1}{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 ) - 3 \, {\left (-2 i \, a b d x^{\frac {1}{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}} \end {gather*}
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 \begin {gather*} \int \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )^{2}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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