Optimal. Leaf size=195 \[ \frac {a x^3}{3}+\frac {15 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {15 b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 i b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{3} i b x^3 \]
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Rubi [A] time = 0.28, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {14, 3747, 3719, 2190, 2531, 6609, 2282, 6589} \[ \frac {a x^3}{3}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {15 i b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {15 b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {15 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{3} i b x^3 \]
Antiderivative was successfully verified.
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Rule 14
Rule 2190
Rule 2282
Rule 2531
Rule 3719
Rule 3747
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^2 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^2+b x^2 \tan \left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^3}{3}+b \int x^2 \tan \left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^3}{3}+(2 b) \operatorname {Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-(4 i b) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^5}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {(10 b) \operatorname {Subst}\left (\int x^4 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(20 i b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(30 b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {15 i b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(30 i b) \operatorname {Subst}\left (\int x \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {15 i b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {15 b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(15 b) \operatorname {Subst}\left (\int \text {Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {15 i b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {15 b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {(15 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}\\ &=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {15 i b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {15 b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {15 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 195, normalized size = 1.00 \[ \frac {a x^3}{3}+\frac {15 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {15 b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 i b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {10 b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {5 i b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{3} i b x^3 \]
Antiderivative was successfully verified.
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{2} \tan \left (d \sqrt {x} + c\right ) + a x^{2}, x\right ) \]
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 \[ \int {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.83, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a +b \tan \left (c +d \sqrt {x}\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.01, size = 618, normalized size = 3.17 \[ \frac {5 \, {\left (d \sqrt {x} + c\right )}^{6} a + 5 i \, {\left (d \sqrt {x} + c\right )}^{6} b - 30 \, {\left (d \sqrt {x} + c\right )}^{5} a c - 30 i \, {\left (d \sqrt {x} + c\right )}^{5} b c + 75 \, {\left (d \sqrt {x} + c\right )}^{4} a c^{2} + 75 i \, {\left (d \sqrt {x} + c\right )}^{4} b c^{2} - 100 \, {\left (d \sqrt {x} + c\right )}^{3} a c^{3} - 100 i \, {\left (d \sqrt {x} + c\right )}^{3} b c^{3} + 75 \, {\left (d \sqrt {x} + c\right )}^{2} a c^{4} + 75 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{4} - 30 \, {\left (d \sqrt {x} + c\right )} a c^{5} - 30 \, b c^{5} \log \left (\sec \left (d \sqrt {x} + c\right )\right ) - {\left (96 i \, {\left (d \sqrt {x} + c\right )}^{5} b - 300 i \, {\left (d \sqrt {x} + c\right )}^{4} b c + 400 i \, {\left (d \sqrt {x} + c\right )}^{3} b c^{2} - 300 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{3} + 150 i \, {\left (d \sqrt {x} + c\right )} b c^{4}\right )} \arctan \left (\sin \left (2 \, d \sqrt {x} + 2 \, c\right ), \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) + 1\right ) - {\left (-240 i \, {\left (d \sqrt {x} + c\right )}^{4} b + 600 i \, {\left (d \sqrt {x} + c\right )}^{3} b c - 600 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} + 300 i \, {\left (d \sqrt {x} + c\right )} b c^{3} - 75 i \, b c^{4}\right )} {\rm Li}_2\left (-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}\right ) - {\left (48 \, {\left (d \sqrt {x} + c\right )}^{5} b - 150 \, {\left (d \sqrt {x} + c\right )}^{4} b c + 200 \, {\left (d \sqrt {x} + c\right )}^{3} b c^{2} - 150 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{3} + 75 \, {\left (d \sqrt {x} + c\right )} b c^{4}\right )} \log \left (\cos \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + \sin \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) + 1\right ) + 360 i \, b {\rm Li}_{6}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) + 90 \, {\left (8 \, {\left (d \sqrt {x} + c\right )} b - 5 \, b c\right )} {\rm Li}_{5}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) - {\left (720 i \, {\left (d \sqrt {x} + c\right )}^{2} b - 900 i \, {\left (d \sqrt {x} + c\right )} b c + 300 i \, b c^{2}\right )} {\rm Li}_{4}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) - 30 \, {\left (16 \, {\left (d \sqrt {x} + c\right )}^{3} b - 30 \, {\left (d \sqrt {x} + c\right )}^{2} b c + 20 \, {\left (d \sqrt {x} + c\right )} b c^{2} - 5 \, b c^{3}\right )} {\rm Li}_{3}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )})}{15 \, d^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )\right ) \,d x \]
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 \[ \int x^{2} \left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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