Optimal. Leaf size=165 \[ \frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f} \]
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Rubi [A] time = 0.15, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3194, 89, 78, 50, 63, 208} \[ \frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}-\frac {\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 78
Rule 89
Rule 208
Rule 3194
Rubi steps
\begin {align*} \int \cot ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(1-x)^2 \sqrt {a+b x}}{x^3} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {1}{2} (-8 a-b)+2 a x\right ) \sqrt {a+b x}}{x^2} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\left (8 a^2-8 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sin ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\left (8 a^2-8 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 a f}\\ &=\frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\left (8 a^2-8 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{8 a b f}\\ &=-\frac {\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}\\ \end {align*}
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Mathematica [A] time = 0.57, size = 103, normalized size = 0.62 \[ \frac {\left (-8 a^2+8 a b+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a} \sqrt {a+b \sin ^2(e+f x)} \left ((8 a-b) \csc ^2(e+f x)-2 a \csc ^4(e+f x)+8 a\right )}{8 a^{3/2} f} \]
Antiderivative was successfully verified.
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fricas [A] time = 2.06, size = 415, normalized size = 2.52 \[ \left [-\frac {{\left ({\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 8 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (8 \, a^{2} \cos \left (f x + e\right )^{4} - {\left (24 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{2} + 14 \, a^{2} - a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}, \frac {{\left ({\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 8 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) + {\left (8 \, a^{2} \cos \left (f x + e\right )^{4} - {\left (24 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{2} + 14 \, a^{2} - a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}\right ] \]
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 [A] time = 1.87, size = 230, normalized size = 1.39 \[ \frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{f}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right ) b}{f \sqrt {a}}-\frac {b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{8 f a \sin \left (f x +e \right )^{2}}+\frac {b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{8 f \,a^{\frac {3}{2}}}+\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{f \sin \left (f x +e \right )^{2}}-\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{4 f \sin \left (f x +e \right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.38, size = 215, normalized size = 1.30 \[ -\frac {8 \, \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - \frac {8 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - \frac {b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} - 8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a} + \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}}{a^{2}} - \frac {8 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{2}} - \frac {{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a^{2} \sin \left (f x + e\right )^{2}} + \frac {2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{4}}}{8 \, f} \]
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 \[ \int {\mathrm {cot}\left (e+f\,x\right )}^5\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]
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 \[ \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \cot ^{5}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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