Optimal. Leaf size=301 \[ \frac {2 i \sqrt {2} a \left (4 a^2+b^2\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{15 d \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {i \left (92 a^2-9 b^2\right ) \sqrt {2 a+b \sinh (2 c+2 d x)} E\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{60 \sqrt {2} d \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {2 \sqrt {2} a b \cosh (2 c+2 d x) \sqrt {2 a+b \sinh (2 c+2 d x)}}{15 d} \]
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Rubi [A] time = 0.39, antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2666, 2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 i \sqrt {2} a \left (4 a^2+b^2\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{15 d \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {i \left (92 a^2-9 b^2\right ) \sqrt {2 a+b \sinh (2 c+2 d x)} E\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{60 \sqrt {2} d \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {2 \sqrt {2} a b \cosh (2 c+2 d x) \sqrt {2 a+b \sinh (2 c+2 d x)}}{15 d} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2656
Rule 2661
Rule 2663
Rule 2666
Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int (a+b \cosh (c+d x) \sinh (c+d x))^{5/2} \, dx &=\int \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^{5/2} \, dx\\ &=\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {2}{5} \int \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)} \left (\frac {1}{8} \left (20 a^2-3 b^2\right )+2 a b \sinh (2 c+2 d x)\right ) \, dx\\ &=\frac {2 \sqrt {2} a b \cosh (2 c+2 d x) \sqrt {2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {4}{15} \int \frac {\frac {1}{16} a \left (60 a^2-17 b^2\right )+\frac {1}{32} b \left (92 a^2-9 b^2\right ) \sinh (2 c+2 d x)}{\sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}} \, dx\\ &=\frac {2 \sqrt {2} a b \cosh (2 c+2 d x) \sqrt {2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {1}{60} \left (92 a^2-9 b^2\right ) \int \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)} \, dx-\frac {1}{15} \left (2 a \left (4 a^2+b^2\right )\right ) \int \frac {1}{\sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}} \, dx\\ &=\frac {2 \sqrt {2} a b \cosh (2 c+2 d x) \sqrt {2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {\left (\left (92 a^2-9 b^2\right ) \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}\right ) \int \sqrt {\frac {a}{a-\frac {i b}{2}}+\frac {b \sinh (2 c+2 d x)}{2 \left (a-\frac {i b}{2}\right )}} \, dx}{60 \sqrt {\frac {a+\frac {1}{2} b \sinh (2 c+2 d x)}{a-\frac {i b}{2}}}}-\frac {\left (2 a \left (4 a^2+b^2\right ) \sqrt {\frac {a+\frac {1}{2} b \sinh (2 c+2 d x)}{a-\frac {i b}{2}}}\right ) \int \frac {1}{\sqrt {\frac {a}{a-\frac {i b}{2}}+\frac {b \sinh (2 c+2 d x)}{2 \left (a-\frac {i b}{2}\right )}}} \, dx}{15 \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}}\\ &=\frac {2 \sqrt {2} a b \cosh (2 c+2 d x) \sqrt {2 a+b \sinh (2 c+2 d x)}}{15 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}-\frac {i \left (92 a^2-9 b^2\right ) E\left (\frac {1}{2} \left (2 i c-\frac {\pi }{2}+2 i d x\right )|\frac {2 b}{2 i a+b}\right ) \sqrt {2 a+b \sinh (2 c+2 d x)}}{60 \sqrt {2} d \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac {2 i \sqrt {2} a \left (4 a^2+b^2\right ) F\left (\frac {1}{2} \left (2 i c-\frac {\pi }{2}+2 i d x\right )|\frac {2 b}{2 i a+b}\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}{15 d \sqrt {2 a+b \sinh (2 c+2 d x)}}\\ \end {align*}
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Mathematica [A] time = 1.41, size = 239, normalized size = 0.79 \[ \frac {-32 i a \left (4 a^2+b^2\right ) \sqrt {\frac {2 a+b \sinh (2 (c+d x))}{2 a-i b}} F\left (\frac {1}{4} (-4 i c-4 i d x+\pi )|-\frac {2 i b}{2 a-i b}\right )+b \left (88 a^2 \cosh (2 (c+d x))+b \sinh (4 (c+d x)) (28 a+3 b \sinh (2 (c+d x)))\right )+2 \left (184 i a^3+92 a^2 b-18 i a b^2-9 b^3\right ) \sqrt {\frac {2 a+b \sinh (2 (c+d x))}{2 a-i b}} E\left (\frac {1}{4} (-4 i c-4 i d x+\pi )|-\frac {2 i b}{2 a-i b}\right )}{120 d \sqrt {4 a+2 b \sinh (2 (c+d x))}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2}\right )} \sqrt {b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 1.11, size = 1260, normalized size = 4.19 \[ \text {result too large to display} \]
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 \[ \int {\left (b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]
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 \[ \int {\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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