Optimal. Leaf size=408 \[ \frac{15 i a b x^{4/3} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{45 i a b x^{2/3} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac{30 a b x \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{45 a b \sqrt [3]{x} \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac{45 i a b \text{PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac{45 b^2 x^{2/3} \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac{30 i b^2 x \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{45 i b^2 \sqrt [3]{x} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac{45 b^2 \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac{a^2 x^2}{2}-\frac{6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i a b x^2+\frac{15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac{3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac{3 i b^2 x^{5/3}}{d}-\frac{1}{2} b^2 x^2 \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.574565, antiderivative size = 408, normalized size of antiderivative = 1., 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.
[In]
[Out]
Rule 3747
Rule 3722
Rule 3719
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 3720
Rule 30
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*}
Mathematica [A] time = 3.17856, size = 383, normalized size = 0.94 \[ \frac{1}{2} \left (b \left (-\frac{30 i \left (a d x^{4/3}-2 b x\right ) \text{PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{30 \left (2 a d x-3 b x^{2/3}\right ) \text{PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{90 i a x^{2/3} \text{PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{90 a \sqrt [3]{x} \text{PolyLog}\left (5,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac{45 i a \text{PolyLog}\left (6,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{90 i b \sqrt [3]{x} \text{PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac{45 b \text{PolyLog}\left (5,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{12 a x^{5/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac{4 i a x^2}{1+e^{2 i c}}+\frac{30 b x^{4/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac{12 i b x^{5/3}}{d+e^{2 i c} d}\right )+x^2 \left (a^2+2 a b \tan (c)-b^2\right )+\frac{6 b^2 x^{5/3} \sec (c) \sin \left (d \sqrt [3]{x}\right ) \sec \left (c+d \sqrt [3]{x}\right )}{d}\right ) \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.118, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\tan \left ( c+d\sqrt [3]{x} \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [B] time = 3.00737, size = 3244, normalized size = 7.95 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \tan{\left (c + d \sqrt [3]{x} \right )}\right )^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x^{\frac{1}{3}} + c\right ) + a\right )}^{2} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]