Optimal. Leaf size=259 \[ \frac {2}{105} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}-\frac {2 i \left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{105 b \sqrt {a+b \sinh (x)}}+\frac {2 i \left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{105 b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}+\frac {2}{35} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2} \]
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Rubi [A] time = 0.45, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2753, 2752, 2663, 2661, 2655, 2653} \[ \frac {2}{105} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}-\frac {2 i \left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{105 b \sqrt {a+b \sinh (x)}}+\frac {2 i \left (161 a^2 A b+15 a^3 B-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{105 b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}+\frac {2}{35} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx &=\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {2}{7} \int (a+b \sinh (x))^{3/2} \left (\frac {1}{2} (7 a A-5 b B)+\frac {1}{2} (7 A b+5 a B) \sinh (x)\right ) \, dx\\ &=\frac {2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {4}{35} \int \sqrt {a+b \sinh (x)} \left (\frac {1}{4} \left (35 a^2 A-21 A b^2-40 a b B\right )+\frac {1}{4} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \sinh (x)\right ) \, dx\\ &=\frac {2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {8}{105} \int \frac {\frac {1}{8} \left (105 a^3 A-119 a A b^2-135 a^2 b B+25 b^3 B\right )+\frac {1}{8} \left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) \sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx\\ &=\frac {2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}-\frac {\left (\left (a^2+b^2\right ) \left (56 a A b+15 a^2 B-25 b^2 B\right )\right ) \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx}{105 b}+\frac {\left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) \int \sqrt {a+b \sinh (x)} \, dx}{105 b}\\ &=\frac {2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {\left (\left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) \sqrt {a+b \sinh (x)}\right ) \int \sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}} \, dx}{105 b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (\left (a^2+b^2\right ) \left (56 a A b+15 a^2 B-25 b^2 B\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}} \, dx}{105 b \sqrt {a+b \sinh (x)}}\\ &=\frac {2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {2 i \left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {a+b \sinh (x)}}{105 b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i \left (a^2+b^2\right ) \left (56 a A b+15 a^2 B-25 b^2 B\right ) F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{105 b \sqrt {a+b \sinh (x)}}\\ \end {align*}
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Mathematica [A] time = 0.80, size = 241, normalized size = 0.93 \[ \frac {\cosh (x) (a+b \sinh (x)) \left (90 a^2 B+6 b \sinh (x) (15 a B+7 A b)+154 a A b+15 b^2 B \cosh (2 x)-65 b^2 B\right )+\frac {2 i \sqrt {\frac {a+b \sinh (x)}{a-i b}} \left (b \left (105 a^3 A-135 a^2 b B-119 a A b^2+25 b^3 B\right ) F\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )+\left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \left ((a-i b) E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )-a F\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )\right )\right )}{b}}{105 \sqrt {a+b \sinh (x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{2} \sinh \relax (x)^{3} + A a^{2} + {\left (2 \, B a b + A b^{2}\right )} \sinh \relax (x)^{2} + {\left (B a^{2} + 2 \, A a b\right )} \sinh \relax (x)\right )} \sqrt {b \sinh \relax (x) + a}, x\right ) \]
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 \[ \int {\left (B \sinh \relax (x) + A\right )} {\left (b \sinh \relax (x) + a\right )}^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.16, size = 1893, normalized size = 7.31 \[ \text {result too large to display} \]
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 \[ \int {\left (B \sinh \relax (x) + A\right )} {\left (b \sinh \relax (x) + a\right )}^{\frac {5}{2}}\,{d x} \]
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 \[ \int \left (A+B\,\mathrm {sinh}\relax (x)\right )\,{\left (a+b\,\mathrm {sinh}\relax (x)\right )}^{5/2} \,d x \]
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 \[ \int \left (A + B \sinh {\relax (x )}\right ) \left (a + b \sinh {\relax (x )}\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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