Optimal. Leaf size=208 \[ \frac {(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.20, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3194, 89, 78, 51, 63, 208} \[ \frac {8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 78
Rule 89
Rule 208
Rule 3194
Rubi steps
\begin {align*} \int \frac {\cot ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(1-x)^2}{x^3 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (-8 a-7 b)+2 a x}{x^2 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=\frac {(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac {8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^3 f}\\ &=\frac {8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (8 a^2+40 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^4 f}\\ &=\frac {8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (8 a^2+40 a b+35 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^4 b f}\\ &=-\frac {\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}
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Mathematica [C] time = 0.84, size = 117, normalized size = 0.56 \[ \frac {\left (8 a^2+40 a b+35 b^2\right ) \csc ^2(e+f x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \sin ^2(e+f x)}{a}+1\right )+3 a \csc ^4(e+f x) \left (-2 a \csc ^2(e+f x)+8 a+7 b\right )}{24 a^3 f \sqrt {a+b \sin ^2(e+f x)} \left (a \csc ^2(e+f x)+b\right )} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.67, size = 984, normalized size = 4.73 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.99, size = 1411, normalized size = 6.78 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 5.60, size = 901, normalized size = 4.33 \[ \frac {7 \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 f \,a^{2} \sqrt {-a b}\, \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )}+\frac {13 b \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{6 f \,a^{3} \sqrt {-a b}\, \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )}+\frac {19 \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}\, b^{2}}{12 f \,a^{4} \sqrt {-a b}\, \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )}-\frac {\sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 f \,a^{2} b \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{6 f \,a^{3} \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {b \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 f \,a^{4} \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 f \,a^{2} b \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{6 f \,a^{3} \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {b \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 f \,a^{4} \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {7 \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 f \,a^{2} \sqrt {-a b}\, \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )}-\frac {13 b \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{6 f \,a^{3} \sqrt {-a b}\, \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )}-\frac {19 \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}\, b^{2}}{12 f \,a^{4} \sqrt {-a b}\, \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )}-\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 )}{f \,a^{\frac {5}{2}}}-\frac {5 \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 \,a^{\frac {7}{2}}}-\frac {35 \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^{2}}{8 f \,a^{\frac {9}{2}}}+\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{f \,a^{3} \sin \left (f x +e \right )^{2}}+\frac {11 b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{8 f \,a^{4} \sin \left (f x +e \right )^{2}}-\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{4 f \,a^{3} \sin \left (f x +e \right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 280, normalized size = 1.35 \[ -\frac {\frac {24 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} + \frac {120 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {7}{2}}} + \frac {105 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {9}{2}}} - \frac {24}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {8}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} - \frac {120 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3}} - \frac {40 \, b}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {105 \, b^{2}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{4}} - \frac {35 \, b^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {24}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \sin \left (f x + e\right )^{2}} - \frac {21 \, b}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \sin \left (f x + e\right )^{2}} + \frac {6}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \sin \left (f x + e\right )^{4}}}{24 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
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 \[ \int \frac {\cot ^{5}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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