Integrand size = 16, antiderivative size = 203 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \]
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Time = 0.36 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {14, 3832, 3800, 2221, 2611, 6744, 2320, 6724} \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{2} i b x^2 \]
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Rule 14
Rule 2221
Rule 2320
Rule 2611
Rule 3800
Rule 3832
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \int \left (a x+b x \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx \\ & = \frac {a x^2}{2}+b \int x \tan \left (c+d \sqrt [3]{x}\right ) \, dx \\ & = \frac {a x^2}{2}+(3 b) \text {Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-(6 i 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 ) \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {(15 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 {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {(30 i 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 {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {(45 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 {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {(45 i 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 {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac {(45 b) \text {Subst}\left (\int \operatorname {PolyLog}\left (5,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{2 d^5} \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {(45 i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(5,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \\ & = \frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \]
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\[\int x \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )d x\]
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\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x \,d x } \]
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\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )\, dx \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 618 vs. \(2 (150) = 300\).
Time = 0.40 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.04 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {5 \, {\left (d x^{\frac {1}{3}} + c\right )}^{6} a + 5 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{6} b - 30 \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} a c - 30 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b c + 75 \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} a c^{2} + 75 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c^{2} - 100 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} a c^{3} - 100 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{3} + 75 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} a c^{4} + 75 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{4} - 30 \, {\left (d x^{\frac {1}{3}} + c\right )} a c^{5} - 30 \, b c^{5} \log \left (\sec \left (d x^{\frac {1}{3}} + c\right )\right ) + 2 \, {\left (-48 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b + 150 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c - 200 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{2} + 150 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{3} - 75 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{4}\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 ) + 15 \, {\left (16 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b - 40 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c + 40 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{2} - 20 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{3} + 5 i \, b c^{4}\right )} {\rm Li}_2\left (-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}\right ) - {\left (48 \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b - 150 \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c + 200 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{2} - 150 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{3} + 75 \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{4}\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 ) + 360 i \, b {\rm Li}_{6}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) + 90 \, {\left (8 \, {\left (d x^{\frac {1}{3}} + c\right )} b - 5 \, b c\right )} {\rm Li}_{5}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) + 60 \, {\left (-12 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b + 15 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c - 5 i \, b c^{2}\right )} {\rm Li}_{4}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) - 30 \, {\left (16 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b - 30 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c + 20 \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{2} - 5 \, b c^{3}\right )} {\rm Li}_{3}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )})}{10 \, d^{6}} \]
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\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x \,d x } \]
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Timed out. \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x\,\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right ) \,d x \]
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