Optimal. Leaf size=187 \[ -\frac {3 (a-5 b) (a-b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{8 a^{7/2} f}-\frac {b (13 a-15 b) \sec (e+f x)}{8 a^3 f \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a+b \sec ^2(e+f x)-b}} \]
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Rubi [A] time = 0.24, antiderivative size = 187, 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 = {3664, 470, 527, 12, 377, 207} \[ -\frac {b (13 a-15 b) \sec (e+f x)}{8 a^3 f \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {3 (a-5 b) (a-b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{8 a^{7/2} f}-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a+b \sec ^2(e+f x)-b}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 207
Rule 377
Rule 470
Rule 527
Rule 3664
Rubi steps
\begin {align*} \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^3 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {-a+b-4 (a-b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{4 a f}\\ &=-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {-(3 a-5 b) (a-b)+10 (a-b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{8 a^2 f}\\ &=-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int -\frac {3 (a-5 b) (a-b)^2}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 a^3 (a-b) f}\\ &=-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {(3 (a-5 b) (a-b)) \operatorname {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{8 a^3 f}\\ &=-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {(3 (a-5 b) (a-b)) \operatorname {Subst}\left (\int \frac {1}{-1+a x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 a^3 f}\\ &=-\frac {3 (a-5 b) (a-b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 a^{7/2} f}-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt {a-b+b \sec ^2(e+f x)}}\\ \end {align*}
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Mathematica [A] time = 5.09, size = 350, normalized size = 1.87 \[ \frac {\frac {\csc ^4(e+f x) \sec (e+f x) \left (\left (-8 a^2+52 a b-60 b^2\right ) \cos (2 (e+f x))+(a-b) (3 (a-5 b) \cos (4 (e+f x))-11 a-45 b)\right )}{4 \sqrt {2} a^3 \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}-\frac {3 (a-5 b) (a-b) \cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\tanh ^{-1}\left (\frac {a-(a-2 b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )\right )}{2 a^{7/2} \sqrt {\sec ^4\left (\frac {1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}}}{8 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.12, size = 705, normalized size = 3.77 \[ \left [\frac {3 \, {\left ({\left (a^{3} - 7 \, a^{2} b + 11 \, a b^{2} - 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (2 \, a^{3} - 15 \, a^{2} b + 28 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + a^{2} b - 6 \, a b^{2} + 5 \, b^{3} + {\left (a^{3} - 9 \, a^{2} b + 23 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (3 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} - 31 \, a^{2} b + 30 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (13 \, a^{2} b - 15 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{16 \, {\left ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac {3 \, {\left ({\left (a^{3} - 7 \, a^{2} b + 11 \, a b^{2} - 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (2 \, a^{3} - 15 \, a^{2} b + 28 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + a^{2} b - 6 \, a b^{2} + 5 \, b^{3} + {\left (a^{3} - 9 \, a^{2} b + 23 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + {\left (3 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} - 31 \, a^{2} b + 30 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (13 \, a^{2} b - 15 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, {\left ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\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 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.25, size = 10582, normalized size = 56.59 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]
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 \[ \int \frac {\csc ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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