Optimal. Leaf size=408 \[ \frac {a^2 x^2}{2}+\frac {45 i a b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {45 a b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i a b x^2-\frac {45 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{5/3}}{d}-\frac {1}{2} b^2 x^2 \]
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Rubi [A] time = 0.57, antiderivative size = 408, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3747, 3722, 3719, 2190, 2531, 6609, 2282, 6589, 3720, 30} \[ \frac {a^2 x^2}{2}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {45 i a b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 a b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {45 i a b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i a b x^2+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{5/3}}{d}-\frac {1}{2} b^2 x^2 \]
Antiderivative was successfully verified.
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Rule 30
Rule 2190
Rule 2282
Rule 2531
Rule 3719
Rule 3720
Rule 3722
Rule 3747
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx &=3 \operatorname {Subst}\left (\int x^5 (a+b \tan (c+d x))^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (a^2 x^5+2 a b x^5 \tan (c+d x)+b^2 x^5 \tan ^2(c+d x)\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {a^2 x^2}{2}+(6 a b) \operatorname {Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )+\left (3 b^2\right ) \operatorname {Subst}\left (\int x^5 \tan ^2(c+d x) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {a^2 x^2}{2}+i a b x^2+\frac {3 b^2 x^{5/3} \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^5}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )-\left (3 b^2\right ) \operatorname {Subst}\left (\int x^5 \, dx,x,\sqrt [3]{x}\right )-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int x^4 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(30 a b) \operatorname {Subst}\left (\int x^4 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d}+\frac {\left (30 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^4}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(60 i a b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}-\frac {\left (60 b^2\right ) \operatorname {Subst}\left (\int x^3 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}\\ &=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {30 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {30 a b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(90 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3}+\frac {\left (90 i b^2\right ) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3}\\ &=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {30 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i a b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(90 i a b) \operatorname {Subst}\left (\int x \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4}-\frac {\left (90 b^2\right ) \operatorname {Subst}\left (\int x \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4}\\ &=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {30 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {45 a b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(45 a b) \operatorname {Subst}\left (\int \text {Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^5}-\frac {\left (45 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^5}\\ &=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {30 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {45 a b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(45 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}-\frac {\left (45 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}\\ &=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {30 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {45 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {45 a b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {45 i a b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}\\ \end {align*}
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Mathematica [A] time = 3.47, size = 383, normalized size = 0.94 \[ \frac {1}{2} \left (x^2 \left (a^2+2 a b \tan (c)-b^2\right )+b \left (-\frac {30 \left (2 a d x-3 b x^{2/3}\right ) \text {Li}_3\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 i \left (a d x^{4/3}-2 b x\right ) \text {Li}_2\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i a \text {Li}_6\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {90 a \sqrt [3]{x} \text {Li}_5\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {90 i a x^{2/3} \text {Li}_4\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {12 a x^{5/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {4 i a x^2}{1+e^{2 i c}}-\frac {45 b \text {Li}_5\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {90 i b \sqrt [3]{x} \text {Li}_4\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {30 b x^{4/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {12 i b x^{5/3}}{d+e^{2 i c} d}\right )+\frac {6 b^2 x^{5/3} \sec (c) \sin \left (d \sqrt [3]{x}\right ) \sec \left (c+d \sqrt [3]{x}\right )}{d}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 2 \, a b x \tan \left (d x^{\frac {1}{3}} + c\right ) + a^{2} x, x\right ) \]
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 \[ \int {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}^{2} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.28, size = 0, normalized size = 0.00 \[ \int x \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.44, size = 2402, normalized size = 5.89 \[ \text {result too large to display} \]
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 \[ \int x\,{\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right )}^2 \,d x \]
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 \[ \int x \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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