Optimal. Leaf size=232 \[ \frac{4 i a (a-b) (2 a-b) \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1} \text{EllipticF}\left (i c+i d x,\frac{b}{a}\right )}{15 d \sqrt{a+b \sinh ^2(c+d x)}}-\frac{i \left (23 a^2-23 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(c+d x)} E\left (i c+i d x\left |\frac{b}{a}\right .\right )}{15 d \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1}}+\frac{b \sinh (c+d x) \cosh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{4 b (2 a-b) \sinh (c+d x) \cosh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d} \]
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Rubi [A] time = 0.303784, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3180, 3170, 3172, 3178, 3177, 3183, 3182} \[ -\frac{i \left (23 a^2-23 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(c+d x)} E\left (i c+i d x\left |\frac{b}{a}\right .\right )}{15 d \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1}}+\frac{b \sinh (c+d x) \cosh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{4 b (2 a-b) \sinh (c+d x) \cosh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{4 i a (a-b) (2 a-b) \sqrt{\frac{b \sinh ^2(c+d x)}{a}+1} F\left (i c+i d x\left |\frac{b}{a}\right .\right )}{15 d \sqrt{a+b \sinh ^2(c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3180
Rule 3170
Rule 3172
Rule 3178
Rule 3177
Rule 3183
Rule 3182
Rubi steps
\begin{align*} \int \left (a+b \sinh ^2(c+d x)\right )^{5/2} \, dx &=\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{1}{5} \int \sqrt{a+b \sinh ^2(c+d x)} \left (a (5 a-b)+4 (2 a-b) b \sinh ^2(c+d x)\right ) \, dx\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{1}{15} \int \frac{a \left (15 a^2-11 a b+4 b^2\right )+b \left (23 a^2-23 a b+8 b^2\right ) \sinh ^2(c+d x)}{\sqrt{a+b \sinh ^2(c+d x)}} \, dx\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}-\frac{1}{15} (4 a (a-b) (2 a-b)) \int \frac{1}{\sqrt{a+b \sinh ^2(c+d x)}} \, dx+\frac{1}{15} \left (23 a^2-23 a b+8 b^2\right ) \int \sqrt{a+b \sinh ^2(c+d x)} \, dx\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}+\frac{\left (\left (23 a^2-23 a b+8 b^2\right ) \sqrt{a+b \sinh ^2(c+d x)}\right ) \int \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}} \, dx}{15 \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}}-\frac{\left (4 a (a-b) (2 a-b) \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}\right ) \int \frac{1}{\sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}} \, dx}{15 \sqrt{a+b \sinh ^2(c+d x)}}\\ &=\frac{4 (2 a-b) b \cosh (c+d x) \sinh (c+d x) \sqrt{a+b \sinh ^2(c+d x)}}{15 d}+\frac{b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^{3/2}}{5 d}-\frac{i \left (23 a^2-23 a b+8 b^2\right ) E\left (i c+i d x\left |\frac{b}{a}\right .\right ) \sqrt{a+b \sinh ^2(c+d x)}}{15 d \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}}+\frac{4 i a (a-b) (2 a-b) F\left (i c+i d x\left |\frac{b}{a}\right .\right ) \sqrt{1+\frac{b \sinh ^2(c+d x)}{a}}}{15 d \sqrt{a+b \sinh ^2(c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.39443, size = 208, normalized size = 0.9 \[ \frac{64 i a \left (2 a^2-3 a b+b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (c+d x))-b}{a}} \text{EllipticF}\left (i (c+d x),\frac{b}{a}\right )+\sqrt{2} b \sinh (2 (c+d x)) \left (88 a^2+28 b (2 a-b) \cosh (2 (c+d x))-88 a b+3 b^2 \cosh (4 (c+d x))+25 b^2\right )-16 i a \left (23 a^2-23 a b+8 b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (c+d x))-b}{a}} E\left (i (c+d x)\left |\frac{b}{a}\right .\right )}{240 d \sqrt{2 a+b \cosh (2 (c+d x))-b}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.085, size = 609, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \sinh \left (d x + c\right )^{4} + 2 \, a b \sinh \left (d x + c\right )^{2} + a^{2}\right )} \sqrt{b \sinh \left (d x + c\right )^{2} + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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