Optimal. Leaf size=196 \[ \frac {b (3 a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}+\frac {\sqrt {b} (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f} \]
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Rubi [A] time = 0.17, antiderivative size = 196, 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 = {3663, 462, 453, 277, 195, 217, 206} \[ \frac {b (3 a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}+\frac {\sqrt {b} (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f} \]
Antiderivative was successfully verified.
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Rule 195
Rule 206
Rule 217
Rule 277
Rule 453
Rule 462
Rule 3663
Rubi steps
\begin {align*} \int \csc ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2 \left (a+b x^2\right )^{3/2}}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac {\operatorname {Subst}\left (\int \frac {\left (10 a+5 a x^2\right ) \left (a+b x^2\right )^{3/2}}{x^4} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac {(3 a+4 b) \operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x^2} \, dx,x,\tan (e+f x)\right )}{3 a f}\\ &=-\frac {(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac {(b (3 a+4 b)) \operatorname {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=\frac {b (3 a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}-\frac {(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac {(b (3 a+4 b)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {b (3 a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}-\frac {(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}+\frac {(b (3 a+4 b)) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}\\ &=\frac {\sqrt {b} (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b (3 a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}-\frac {(3 a+4 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 a f}\\ \end {align*}
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Mathematica [C] time = 2.17, size = 213, normalized size = 1.09 \[ \frac {\sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (-\frac {2 \left (8 a^2+34 a b+3 b^2\right ) \cot (e+f x)}{a}-4 (2 a+3 b) \cot (e+f x) \csc ^2(e+f x)+\frac {15 \sqrt {2} (3 a+4 b) \cot (e+f x) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{\sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}-6 a \cot (e+f x) \csc ^4(e+f x)+15 b \tan (e+f x)\right )}{30 \sqrt {2} f} \]
Antiderivative was successfully verified.
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fricas [A] time = 5.22, size = 655, normalized size = 3.34 \[ \left [\frac {15 \, {\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (16 \, a^{2} + 83 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{6} - {\left (40 \, a^{2} + 193 \, a b + 12 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (30 \, a^{2} + 125 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, {\left (a f \cos \left (f x + e\right )^{5} - 2 \, a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}, -\frac {15 \, {\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, {\left ({\left (16 \, a^{2} + 83 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{6} - {\left (40 \, a^{2} + 193 \, a b + 12 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (30 \, a^{2} + 125 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, a b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, {\left (a f \cos \left (f x + e\right )^{5} - 2 \, a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}\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 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{6}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.83, size = 6988, normalized size = 35.65 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.69, size = 202, normalized size = 1.03 \[ \frac {45 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) + 60 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) + 45 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right ) + \frac {60 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b^{2} \tan \left (f x + e\right )}{a} - \frac {30 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )} - \frac {40 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a \tan \left (f x + e\right )} - \frac {20 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}{a \tan \left (f x + e\right )^{3}} - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}{a \tan \left (f x + e\right )^{5}}}{30 \, f} \]
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 \[ \int \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^6} \,d x \]
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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