Optimal. Leaf size=330 \[ -\frac {2 (a-b) (2 a+b) \sinh (e+f x) \cosh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {(4 a-b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 a^2 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\left (8 a^2-3 a b-2 b^2\right ) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 a^2 b^3 f}-\frac {\left (8 a^2-3 a b-2 b^2\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 a^2 b^3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {(a-b) \sinh (e+f x) \cosh ^3(e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.33, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3192, 413, 526, 531, 418, 492, 411} \[ \frac {\left (8 a^2-3 a b-2 b^2\right ) \tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 a^2 b^3 f}-\frac {2 (a-b) (2 a+b) \sinh (e+f x) \cosh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {(4 a-b) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 a^2 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {\left (8 a^2-3 a b-2 b^2\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{3 a^2 b^3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {(a-b) \sinh (e+f x) \cosh ^3(e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 411
Rule 413
Rule 418
Rule 492
Rule 526
Rule 531
Rule 3192
Rubi steps
\begin {align*} \int \frac {\cosh ^6(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac {(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+x^2} \left (a+2 b+(4 a-b) x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 a b f}\\ &=-\frac {(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {-a (4 a-b)+\left (-8 a^2+3 a b+2 b^2\right ) x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 a^2 b^2 f}\\ &=-\frac {(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\left ((4 a-b) \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 a b^2 f}-\frac {\left (\left (-8 a^2+3 a b+2 b^2\right ) \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 a^2 b^2 f}\\ &=-\frac {(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {(4 a-b) F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 a^2 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\left (8 a^2-3 a b-2 b^2\right ) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 a^2 b^3 f}+\frac {\left (\left (-8 a^2+3 a b+2 b^2\right ) \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 a^2 b^3 f}\\ &=-\frac {(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {\left (8 a^2-3 a b-2 b^2\right ) E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 a^2 b^3 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {(4 a-b) F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{3 a^2 b^2 f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\left (8 a^2-3 a b-2 b^2\right ) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 a^2 b^3 f}\\ \end {align*}
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Mathematica [C] time = 2.24, size = 206, normalized size = 0.62 \[ \frac {\frac {1}{2} (a-b) \left (-2 \sqrt {2} b \sinh (2 (e+f x)) \left (8 a^2+b (5 a+2 b) \cosh (2 (e+f x))+a b-2 b^2\right )+4 i a^2 (8 a+b) \left (\frac {2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )\right )-2 i a^2 \left (8 a^2-3 a b-2 b^2\right ) \left (\frac {2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )}{6 a^2 b^3 f (2 a+b \cosh (2 (e+f x))-b)^{3/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sinh \left (f x + e\right )^{2} + a} \cosh \left (f x + e\right )^{6}}{b^{3} \sinh \left (f x + e\right )^{6} + 3 \, a b^{2} \sinh \left (f x + e\right )^{4} + 3 \, a^{2} b \sinh \left (f x + e\right )^{2} + a^{3}}, x\right ) \]
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 = 0.22, size = 812, normalized size = 2.46 \[ -\frac {\left (5 \sqrt {-\frac {b}{a}}\, a^{2} b -3 \sqrt {-\frac {b}{a}}\, a \,b^{2}-2 \sqrt {-\frac {b}{a}}\, b^{3}\right ) \sinh \left (f x +e \right ) \left (\cosh ^{4}\left (f x +e \right )\right )+\left (4 \sqrt {-\frac {b}{a}}\, a^{3}-6 \sqrt {-\frac {b}{a}}\, a^{2} b +2 \sqrt {-\frac {b}{a}}\, b^{3}\right ) \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )+\sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, b \left (4 \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a^{2}-2 \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a b -2 \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b^{2}-8 \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a^{2}+3 \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a b +2 \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b^{2}\right ) \left (\cosh ^{2}\left (f x +e \right )\right )+4 \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a^{3}-6 \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a^{2} b +2 \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b^{3}-8 \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a^{3}+11 \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a^{2} b -\sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) a \,b^{2}-2 \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right ) b^{3}}{3 a^{2} \left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {-\frac {b}{a}}\, b^{2} \cosh \left (f x +e \right ) f} \]
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 \[ \int \frac {\cosh \left (f x + e\right )^{6}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]
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 \[ \int \frac {{\mathrm {cosh}\left (e+f\,x\right )}^6}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]
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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