Optimal. Leaf size=402 \[ -\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {20 i b^2 x^{3/2} \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 i a b x^2 \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {30 b^2 x \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 a b x^{3/2} \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 i b^2 \sqrt {x} \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {30 i a b x \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {15 b^2 \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 a b \sqrt {x} \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {15 i a b \text {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d} \]
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Rubi [A]
time = 0.43, antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps
used = 20, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3832, 3803,
3800, 2221, 2611, 6744, 2320, 6724, 3801, 30} \begin {gather*} \frac {a^2 x^3}{3}+\frac {15 i a b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 a b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {30 i a b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 a b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 i a b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2}{3} i a b x^3-\frac {15 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{5/2}}{d}-\frac {b^2 x^3}{3} \end {gather*}
Antiderivative was successfully verified.
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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*} \int x^2 \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \text {Subst}\left (\int x^5 (a+b \tan (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \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 {x}\right )\\ &=\frac {a^2 x^3}{3}+(4 a b) \text {Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^5 \tan ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-(8 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 {x}\right )-\left (2 b^2\right ) \text {Subst}\left (\int x^5 \, dx,x,\sqrt {x}\right )-\frac {\left (10 b^2\right ) \text {Subst}\left (\int x^4 \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(20 a b) \text {Subst}\left (\int x^4 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (20 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 {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 i a b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(40 i a b) \text {Subst}\left (\int x^3 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (40 b^2\right ) \text {Subst}\left (\int x^3 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 i a b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(60 a b) \text {Subst}\left (\int x^2 \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {\left (60 i b^2\right ) \text {Subst}\left (\int x^2 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 i a b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 a b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {30 i a b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(60 i a b) \text {Subst}\left (\int x \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {\left (60 b^2\right ) \text {Subst}\left (\int x \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 i a b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 a b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {30 i a b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 a b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(30 a b) \text {Subst}\left (\int \text {Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}-\frac {\left (30 i b^2\right ) \text {Subst}\left (\int \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 i a b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 a b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {30 i a b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 a b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(15 i a b) \text {Subst}\left (\int \frac {\text {Li}_5(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {\left (15 b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_4(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 i a b x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {30 b^2 x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 a b x^{3/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {30 i a b x \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {15 b^2 \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 a b \sqrt {x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {15 i a b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {2 b^2 x^{5/2} \tan \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}
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Mathematica [A]
time = 2.13, size = 395, normalized size = 0.98 \begin {gather*} -\frac {2 i b e^{2 i c} \left (2 b x^{5/2}-\frac {2}{3} a d x^3+\frac {e^{-2 i c} \left (1+e^{2 i c}\right ) \left (10 i b d^4 x^2 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )-4 i a d^5 x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )-10 d^3 \left (-2 b+a d \sqrt {x}\right ) x^{3/2} \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-10 i d^2 \left (-3 b+2 a d \sqrt {x}\right ) x \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-30 b d \sqrt {x} \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )+30 a d^2 x \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-15 i b \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )+30 i a d \sqrt {x} \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-15 a \text {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )\right )}{2 d^5}\right )}{d \left (1+e^{2 i c}\right )}+\frac {2 b^2 x^{5/2} \sec (c) \sec \left (c+d \sqrt {x}\right ) \sin \left (d \sqrt {x}\right )}{d}+\frac {1}{3} x^3 \left (a^2-b^2+2 a b \tan (c)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.90, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \tan \left (c +d \sqrt {x}\right )\right )^{2}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] 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 = 0.75, size = 2421, normalized size = 6.02 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^{2} \left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^2\,{\left (a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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