Integrand size = 18, antiderivative size = 408 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=-\frac {3 i b^2 x^{5/3}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {30 i b^2 x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {45 b^2 \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {45 a b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {45 i a b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d} \]
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Time = 0.87 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3832, 3803, 3800, 2221, 2611, 6744, 2320, 6724, 3801, 30} \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\frac {a^2 x^2}{2}+\frac {45 i a b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {45 a b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i a b x^2-\frac {45 b^2 \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {45 i b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {45 b^2 x^{2/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 i b^2 x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 b^2 x^{4/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{5/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{5/3}}{d}-\frac {1}{2} b^2 x^2 \]
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Rule 30
Rule 2221
Rule 2320
Rule 2611
Rule 3800
Rule 3801
Rule 3803
Rule 3832
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = 3 \text {Subst}\left (\int x^5 (a+b \tan (c+d x))^2 \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {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) \text {Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )+\left (3 b^2\right ) \text {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) \text {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 ) \text {Subst}\left (\int x^5 \, dx,x,\sqrt [3]{x}\right )-\frac {\left (15 b^2\right ) \text {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) \text {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 ) \text {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} \operatorname {PolyLog}\left (2,-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) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}-\frac {\left (60 b^2\right ) \text {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 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {30 a b x \operatorname {PolyLog}\left (3,-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) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3}+\frac {\left (90 i b^2\right ) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (2,-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 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i a b x^{2/3} \operatorname {PolyLog}\left (4,-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) \text {Subst}\left (\int x \operatorname {PolyLog}\left (4,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4}-\frac {\left (90 b^2\right ) \text {Subst}\left (\int x \operatorname {PolyLog}\left (3,-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 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {45 a b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-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) \text {Subst}\left (\int \operatorname {PolyLog}\left (5,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^5}-\frac {\left (45 i b^2\right ) \text {Subst}\left (\int \operatorname {PolyLog}\left (4,-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 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {45 a b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-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) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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 ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i a b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {45 b^2 x^{2/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {30 a b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 i b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {45 i a b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {45 b^2 \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^6}+\frac {45 a b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {45 i a b \operatorname {PolyLog}\left (6,-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*}
Time = 3.85 (sec) , antiderivative size = 571, normalized size of antiderivative = 1.40 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\frac {1}{2} \left (-\frac {i b e^{2 i c} \left (-12 b d^5 e^{-2 i c} x^{5/3}+4 a d^6 e^{-2 i c} x^2+30 i b d^4 e^{-2 i c} \left (1+e^{2 i c}\right ) x^{4/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-12 i a d^5 e^{-2 i c} \left (1+e^{2 i c}\right ) x^{5/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-60 b d^3 \left (1+e^{-2 i c}\right ) x \operatorname {PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+30 a d^4 \left (1+e^{-2 i c}\right ) x^{4/3} \operatorname {PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+90 i b d^2 e^{-2 i c} \left (1+e^{2 i c}\right ) x^{2/3} \operatorname {PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-60 i a d^3 e^{-2 i c} \left (1+e^{2 i c}\right ) x \operatorname {PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+90 b d \left (1+e^{-2 i c}\right ) \sqrt [3]{x} \operatorname {PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-90 a d^2 \left (1+e^{-2 i c}\right ) x^{2/3} \operatorname {PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-45 i b e^{-2 i c} \left (1+e^{2 i c}\right ) \operatorname {PolyLog}\left (5,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+90 i a d e^{-2 i c} \left (1+e^{2 i c}\right ) \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+45 a \left (1+e^{-2 i c}\right ) \operatorname {PolyLog}\left (6,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )}{d^6 \left (1+e^{2 i c}\right )}+\frac {6 b^2 x^{5/3} \sec (c) \sec \left (c+d \sqrt [3]{x}\right ) \sin \left (d \sqrt [3]{x}\right )}{d}+x^2 \left (a^2-b^2+2 a b \tan (c)\right )\right ) \]
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\[\int x {\left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )}^{2}d x\]
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\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}^{2} x \,d x } \]
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\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int x \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )^{2}\, dx \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2421 vs. \(2 (320) = 640\).
Time = 0.55 (sec) , antiderivative size = 2421, normalized size of antiderivative = 5.93 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\text {Too large to display} \]
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\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}^{2} x \,d x } \]
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Timed out. \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int x\,{\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right )}^2 \,d x \]
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