Optimal. Leaf size=158 \[ -\frac {2 \sqrt {2} b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {2 i \sqrt {2} \sqrt {2 a+b \sinh (2 c+2 d x)} E\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{d \left (4 a^2+b^2\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}} \]
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Rubi [A] time = 0.11, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2666, 2664, 21, 2655, 2653} \[ -\frac {2 \sqrt {2} b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {2 i \sqrt {2} \sqrt {2 a+b \sinh (2 c+2 d x)} E\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{d \left (4 a^2+b^2\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}} \]
Antiderivative was successfully verified.
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Rule 21
Rule 2653
Rule 2655
Rule 2664
Rule 2666
Rubi steps
\begin {align*} \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^{3/2}} \, dx &=\int \frac {1}{\left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^{3/2}} \, dx\\ &=-\frac {2 \sqrt {2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {8 \int \frac {-\frac {a}{2}-\frac {1}{4} b \sinh (2 c+2 d x)}{\sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}} \, dx}{4 a^2+b^2}\\ &=-\frac {2 \sqrt {2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt {2 a+b \sinh (2 c+2 d x)}}+\frac {4 \int \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)} \, dx}{4 a^2+b^2}\\ &=-\frac {2 \sqrt {2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt {2 a+b \sinh (2 c+2 d x)}}+\frac {\left (4 \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}{\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}}}}\\ &=-\frac {2 \sqrt {2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {2 i \sqrt {2} 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)}}{\left (4 a^2+b^2\right ) d \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}\\ \end {align*}
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Mathematica [A] time = 0.44, size = 119, normalized size = 0.75 \[ \frac {2 \left (-b \cosh (2 (c+d x))+(b+2 i a) \sqrt {\frac {2 a+b \sinh (2 (c+d x))}{2 a-i b}} E\left (\frac {1}{4} (-4 i c-4 i d x+\pi )|-\frac {2 i b}{2 a-i b}\right )\right )}{d \left (4 a^2+b^2\right ) \sqrt {a+\frac {1}{2} b \sinh (2 (c+d x))}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}}{b^{2} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2}}, x\right ) \]
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 = 1.09, size = 630, normalized size = 3.99 \[ \frac {16 \sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}\, \sqrt {\frac {\left (-\sinh \left (2 d x +2 c \right )+i\right ) b}{i b +2 a}}\, \sqrt {\frac {\left (\sinh \left (2 d x +2 c \right )+i\right ) b}{i b -2 a}}\, \EllipticF \left (\sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}, \sqrt {-\frac {i b -2 a}{i b +2 a}}\right ) a^{2}+4 \sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}\, \sqrt {\frac {\left (-\sinh \left (2 d x +2 c \right )+i\right ) b}{i b +2 a}}\, \sqrt {\frac {\left (\sinh \left (2 d x +2 c \right )+i\right ) b}{i b -2 a}}\, \EllipticF \left (\sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}, \sqrt {-\frac {i b -2 a}{i b +2 a}}\right ) b^{2}-16 \sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}\, \sqrt {\frac {\left (-\sinh \left (2 d x +2 c \right )+i\right ) b}{i b +2 a}}\, \sqrt {\frac {\left (\sinh \left (2 d x +2 c \right )+i\right ) b}{i b -2 a}}\, \EllipticE \left (\sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}, \sqrt {-\frac {i b -2 a}{i b +2 a}}\right ) a^{2}-4 \sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}\, \sqrt {\frac {\left (-\sinh \left (2 d x +2 c \right )+i\right ) b}{i b +2 a}}\, \sqrt {\frac {\left (\sinh \left (2 d x +2 c \right )+i\right ) b}{i b -2 a}}\, \EllipticE \left (\sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}, \sqrt {-\frac {i b -2 a}{i b +2 a}}\right ) b^{2}-4 b^{2} \left (\sinh ^{2}\left (2 d x +2 c \right )\right )-4 b^{2}}{\left (4 a^{2}+b^{2}\right ) b \cosh \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sinh \left (2 d x +2 c \right )}\, d} \]
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 \[ \int \frac {1}{{\left (b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]
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 \[ \int \frac {1}{\left (a + b \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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