Optimal. Leaf size=258 \[ \frac {2}{5} a x^{5/2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 b \text {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \text {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5} \]
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Rubi [A]
time = 0.16, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {14, 4290,
4268, 2611, 6744, 2320, 6724} \begin {gather*} \frac {2}{5} a x^{5/2}+\frac {48 b \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 i b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2320
Rule 2611
Rule 4268
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 ) \, dx &=\int \left (a x^{3/2}+b x^{3/2} \csc \left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {2}{5} a x^{5/2}+b \int x^{3/2} \csc \left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {2}{5} a x^{5/2}+(2 b) \text {Subst}\left (\int x^4 \csc (c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a x^{5/2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {(8 b) \text {Subst}\left (\int x^3 \log \left (1-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(8 b) \text {Subst}\left (\int x^3 \log \left (1+e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(24 i 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 {(24 i 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 {2}{5} a x^{5/2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(48 b) \text {Subst}\left (\int x \text {Li}_3\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(48 b) \text {Subst}\left (\int x \text {Li}_3\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(48 i b) \text {Subst}\left (\int \text {Li}_4\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(48 i b) \text {Subst}\left (\int \text {Li}_4\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {2}{5} a x^{5/2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(48 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 {(48 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 {2}{5} a x^{5/2}-\frac {4 b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 b \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 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 = 0.61, size = 286, normalized size = 1.11 \begin {gather*} \frac {2 \left (a d^5 x^{5/2}+5 b d^4 x^2 \log \left (1-e^{i \left (c+d \sqrt {x}\right )}\right )-5 b d^4 x^2 \log \left (1+e^{i \left (c+d \sqrt {x}\right )}\right )+20 i b d^3 x^{3/2} \text {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )-20 i b d^3 x^{3/2} \text {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )-60 b d^2 x \text {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )+60 b d^2 x \text {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )-120 i b d \sqrt {x} \text {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )+120 i b d \sqrt {x} \text {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )+120 b \text {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )-120 b \text {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )\right )}{5 d^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int x^{\frac {3}{2}} \left (a +b \csc \left (c +d \sqrt {x}\right )\right )\, dx\]
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. 730 vs. \(2 (200) = 400\).
time = 0.34, size = 730, normalized size = 2.83 \begin {gather*} \frac {2 \, {\left (d \sqrt {x} + c\right )}^{5} a - 10 \, {\left (d \sqrt {x} + c\right )}^{4} a c + 20 \, {\left (d \sqrt {x} + c\right )}^{3} a c^{2} - 20 \, {\left (d \sqrt {x} + c\right )}^{2} a c^{3} + 10 \, {\left (d \sqrt {x} + c\right )} a c^{4} - 10 \, b c^{4} \log \left (\cot \left (d \sqrt {x} + c\right ) + \csc \left (d \sqrt {x} + c\right )\right ) + 10 \, {\left (-i \, {\left (d \sqrt {x} + c\right )}^{4} b + 4 i \, {\left (d \sqrt {x} + c\right )}^{3} b c - 6 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} + 4 i \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \arctan \left (\sin \left (d \sqrt {x} + c\right ), \cos \left (d \sqrt {x} + c\right ) + 1\right ) + 10 \, {\left (-i \, {\left (d \sqrt {x} + c\right )}^{4} b + 4 i \, {\left (d \sqrt {x} + c\right )}^{3} b c - 6 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} + 4 i \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \arctan \left (\sin \left (d \sqrt {x} + c\right ), -\cos \left (d \sqrt {x} + c\right ) + 1\right ) + 40 \, {\left (i \, {\left (d \sqrt {x} + c\right )}^{3} b - 3 i \, {\left (d \sqrt {x} + c\right )}^{2} b c + 3 i \, {\left (d \sqrt {x} + c\right )} b c^{2} - i \, b c^{3}\right )} {\rm Li}_2\left (-e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 40 \, {\left (-i \, {\left (d \sqrt {x} + c\right )}^{3} b + 3 i \, {\left (d \sqrt {x} + c\right )}^{2} b c - 3 i \, {\left (d \sqrt {x} + c\right )} b c^{2} + i \, b c^{3}\right )} {\rm Li}_2\left (e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) - 5 \, {\left ({\left (d \sqrt {x} + c\right )}^{4} b - 4 \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} + 2 \, \cos \left (d \sqrt {x} + c\right ) + 1\right ) + 5 \, {\left ({\left (d \sqrt {x} + c\right )}^{4} b - 4 \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} - 2 \, \cos \left (d \sqrt {x} + c\right ) + 1\right ) + 240 \, b {\rm Li}_{5}(-e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 240 \, b {\rm Li}_{5}(e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 240 \, {\left (-i \, {\left (d \sqrt {x} + c\right )} b + i \, b c\right )} {\rm Li}_{4}(-e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 240 \, {\left (i \, {\left (d \sqrt {x} + c\right )} b - i \, b c\right )} {\rm Li}_{4}(e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 120 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c + b c^{2}\right )} {\rm Li}_{3}(-e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 120 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c + b c^{2}\right )} {\rm Li}_{3}(e^{\left (i \, d \sqrt {x} + i \, c\right )})}{5 \, d^{5}} \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 )\, 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 ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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