Optimal. Leaf size=421 \[ -\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {PolyLog}\left (2,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {PolyLog}\left (3,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {6 i b^2 \text {PolyLog}\left (4,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 a b \text {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5} \]
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Rubi [A]
time = 0.38, antiderivative size = 421, normalized size of antiderivative = 1.00, number of steps
used = 21, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {4290, 4275,
4268, 2611, 6744, 2320, 6724, 4269, 3798, 2221} \begin {gather*} \frac {2}{5} a^2 x^{5/2}+\frac {96 a b \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {48 a b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 i b^2 \text {Li}_4\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {12 i b^2 x \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^2}{d} \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 3798
Rule 4268
Rule 4269
Rule 4275
Rule 4290
Rule 6724
Rule 6744
Rubi steps
\begin {align*} \int x^{3/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \text {Subst}\left (\int x^4 (a+b \csc (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \left (a^2 x^4+2 a b x^4 \csc (c+d x)+b^2 x^4 \csc ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}+(4 a b) \text {Subst}\left (\int x^4 \csc (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^4 \csc ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}-\frac {(16 a b) \text {Subst}\left (\int x^3 \log \left (1-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(16 a b) \text {Subst}\left (\int x^3 \log \left (1+e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (8 b^2\right ) \text {Subst}\left (\int x^3 \cot (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(48 i a b) \text {Subst}\left (\int x^2 \text {Li}_2\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(48 i a b) \text {Subst}\left (\int x^2 \text {Li}_2\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (16 i b^2\right ) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^3}{1-e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {48 a b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(96 a b) \text {Subst}\left (\int x \text {Li}_3\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(96 a b) \text {Subst}\left (\int x \text {Li}_3\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (24 b^2\right ) \text {Subst}\left (\int x^2 \log \left (1-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(96 i a b) \text {Subst}\left (\int \text {Li}_4\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(96 i a b) \text {Subst}\left (\int \text {Li}_4\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}+\frac {\left (24 i b^2\right ) \text {Subst}\left (\int x \text {Li}_2\left (e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(96 a b) \text {Subst}\left (\int \frac {\text {Li}_4(-x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(96 a b) \text {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {\left (12 b^2\right ) \text {Subst}\left (\int \text {Li}_3\left (e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 a b \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {\left (6 i b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^2 \cot \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {6 i b^2 \text {Li}_4\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 a b \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}\\ \end {align*}
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Mathematica [A]
time = 11.82, size = 686, normalized size = 1.63 \begin {gather*} \frac {2 a^2 x^{5/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \sin ^2\left (c+d \sqrt {x}\right )}{5 \left (b+a \sin \left (c+d \sqrt {x}\right )\right )^2}+\frac {2 b \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \left (-\frac {2 i b d^4 e^{2 i c} x^2}{-1+e^{2 i c}}+2 a d^4 x^2 \log \left (1-e^{i \left (c+d \sqrt {x}\right )}\right )-2 a d^4 x^2 \log \left (1+e^{i \left (c+d \sqrt {x}\right )}\right )+4 b d^3 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )+8 i a d^3 x^{3/2} \text {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )-8 i a d^3 x^{3/2} \text {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )-6 i b d^2 x \text {PolyLog}\left (2,e^{2 i \left (c+d \sqrt {x}\right )}\right )-24 a d^2 x \text {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )+24 a d^2 x \text {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )+6 b d \sqrt {x} \text {PolyLog}\left (3,e^{2 i \left (c+d \sqrt {x}\right )}\right )-48 i a d \sqrt {x} \text {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )+48 i a d \sqrt {x} \text {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )+3 i b \text {PolyLog}\left (4,e^{2 i \left (c+d \sqrt {x}\right )}\right )+48 a \text {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )-48 a \text {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )\right ) \sin ^2\left (c+d \sqrt {x}\right )}{d^5 \left (b+a \sin \left (c+d \sqrt {x}\right )\right )^2}+\frac {b^2 x^2 \csc \left (\frac {c}{2}\right ) \csc \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \sin ^2\left (c+d \sqrt {x}\right ) \sin \left (\frac {d \sqrt {x}}{2}\right )}{d \left (b+a \sin \left (c+d \sqrt {x}\right )\right )^2}+\frac {b^2 x^2 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right ) \sin ^2\left (c+d \sqrt {x}\right ) \sin \left (\frac {d \sqrt {x}}{2}\right )}{d \left (b+a \sin \left (c+d \sqrt {x}\right )\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int x^{\frac {3}{2}} \left (a +b \csc \left (c +d \sqrt {x}\right )\right )^{2}\, dx\]
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. 2836 vs. \(2 (334) = 668\).
time = 0.42, size = 2836, normalized size = 6.74 \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^{\frac {3}{2}} \left (a + b \csc {\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^{3/2}\,{\left (a+\frac {b}{\sin \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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