Optimal. Leaf size=312 \[ -\frac {128 a^2 \sqrt {a+i a \sinh (c+d x)}}{15 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}+\frac {64 a^2 x \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh ^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {32 a^2 x \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d} \]
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Rubi [A] time = 0.21, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3319, 3310, 3296, 2638} \[ -\frac {128 a^2 \sqrt {a+i a \sinh (c+d x)}}{15 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}+\frac {64 a^2 x \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh ^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {32 a^2 x \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d} \]
Antiderivative was successfully verified.
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Rule 2638
Rule 3296
Rule 3310
Rule 3319
Rubi steps
\begin {align*} \int x (a+i a \sinh (c+d x))^{5/2} \, dx &=\left (4 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int x \sinh ^5\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx\\ &=-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}-\frac {1}{5} \left (16 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int x \sinh ^3\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx\\ &=-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {32 a^2 x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {1}{15} \left (32 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int x \sinh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx\\ &=-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {32 a^2 x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {64 a^2 x \sqrt {a+i a \sinh (c+d x)} \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{15 d}-\frac {\left (64 a^2 \text {csch}\left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}\right ) \int \cosh \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{15 d}\\ &=-\frac {128 a^2 \sqrt {a+i a \sinh (c+d x)}}{15 d^2}-\frac {64 a^2 \cosh ^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{45 d^2}-\frac {16 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{25 d^2}+\frac {32 a^2 x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {8 a^2 x \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{5 d}+\frac {64 a^2 x \sqrt {a+i a \sinh (c+d x)} \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{15 d}\\ \end {align*}
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Mathematica [A] time = 1.27, size = 218, normalized size = 0.70 \[ \frac {a^2 (\sinh (c+d x)-i)^2 \sqrt {a+i a \sinh (c+d x)} \left (-2250 d x \sinh \left (\frac {1}{2} (c+d x)\right )+4500 i \sinh \left (\frac {1}{2} (c+d x)\right )+375 d x \sinh \left (\frac {3}{2} (c+d x)\right )+250 i \sinh \left (\frac {3}{2} (c+d x)\right )+45 d x \sinh \left (\frac {5}{2} (c+d x)\right )-18 i \sinh \left (\frac {5}{2} (c+d x)\right )+2250 (2-i d x) \cosh \left (\frac {1}{2} (c+d x)\right )+(-250-375 i d x) \cosh \left (\frac {3}{2} (c+d x)\right )+45 i d x \cosh \left (\frac {5}{2} (c+d x)\right )-18 \cosh \left (\frac {5}{2} (c+d x)\right )\right )}{450 d^2 \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \]
Antiderivative was successfully verified.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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 \[ \int {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.04, size = 0, normalized size = 0.00 \[ \int x \left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} x\,{d x} \]
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 \[ \int x\,{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]
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 \[ \int x \left (i a \left (\sinh {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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