Optimal. Leaf size=318 \[ \frac {4 a b^2 \tan (e+f x)}{f \left (a^2-b^2\right )^2 \sqrt {a+b \sec (e+f x)}}+\frac {b^2 \tan (e+f x)}{f \left (a^2-b^2\right ) (a+b \sec (e+f x))^{3/2}}-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}-\frac {(3 a-b) \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{f (a-b) (a+b)^{3/2}}+\frac {4 a \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{f (a-b) (a+b)^{3/2}} \]
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Rubi [A] time = 0.52, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3875, 3833, 4003, 4005, 3832, 4004} \[ \frac {4 a b^2 \tan (e+f x)}{f \left (a^2-b^2\right )^2 \sqrt {a+b \sec (e+f x)}}+\frac {b^2 \tan (e+f x)}{f \left (a^2-b^2\right ) (a+b \sec (e+f x))^{3/2}}-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}-\frac {(3 a-b) \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{f (a-b) (a+b)^{3/2}}+\frac {4 a \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{f (a-b) (a+b)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3832
Rule 3833
Rule 3875
Rule 4003
Rule 4004
Rule 4005
Rubi steps
\begin {align*} \int \frac {\csc ^2(e+f x)}{(a+b \sec (e+f x))^{3/2}} \, dx &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}-\frac {1}{2} (3 b) \int \frac {\sec (e+f x)}{(a+b \sec (e+f x))^{5/2}} \, dx\\ &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {b \int \frac {\sec (e+f x) \left (-\frac {3 a}{2}+\frac {1}{2} b \sec (e+f x)\right )}{(a+b \sec (e+f x))^{3/2}} \, dx}{a^2-b^2}\\ &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {4 a b^2 \tan (e+f x)}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}}-\frac {(2 b) \int \frac {\sec (e+f x) \left (\frac {1}{4} \left (3 a^2+b^2\right )+a b \sec (e+f x)\right )}{\sqrt {a+b \sec (e+f x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {4 a b^2 \tan (e+f x)}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}}-\frac {((3 a-b) b) \int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)}} \, dx}{2 (a-b) (a+b)^2}-\frac {\left (2 a b^2\right ) \int \frac {\sec (e+f x) (1+\sec (e+f x))}{\sqrt {a+b \sec (e+f x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {4 a \cot (e+f x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{(a-b) (a+b)^{3/2} f}-\frac {(3 a-b) \cot (e+f x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{(a-b) (a+b)^{3/2} f}-\frac {\cot (e+f x)}{f (a+b \sec (e+f x))^{3/2}}+\frac {b^2 \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {4 a b^2 \tan (e+f x)}{\left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 8.49, size = 259, normalized size = 0.81 \[ \frac {-2 b \left (3 a^2+4 a b+b^2\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {\frac {a \cos (e+f x)+b}{(a+b) (\cos (e+f x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )-(a-b) \csc (e+f x) (a (a-3 b) \cos (e+f x)+b (3 a-b))+8 a b (a+b) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {\frac {a \cos (e+f x)+b}{(a+b) (\cos (e+f x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )}{f \left (a^2-b^2\right )^2 \sqrt {a+b \sec (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right ) + a} \csc \left (f x + e\right )^{2}}{b^{2} \sec \left (f x + e\right )^{2} + 2 \, a b \sec \left (f x + e\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.32, size = 1065, normalized size = 3.35 \[ \text {result too large to display} \]
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 \[ \int \frac {\csc \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\sin \left (e+f\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]
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 \[ \int \frac {\csc ^{2}{\left (e + f x \right )}}{\left (a + b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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