Optimal. Leaf size=183 \[ -\frac {2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 f (a+b)^4 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 f (a+b)^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\cot ^5(e+f x)}{5 f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{15 f (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}} \]
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Rubi [A] time = 0.18, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4132, 462, 453, 271, 191} \[ -\frac {2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 f (a+b)^4 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 f (a+b)^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\cot ^5(e+f x)}{5 f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{15 f (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}} \]
Antiderivative was successfully verified.
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Rule 191
Rule 271
Rule 453
Rule 462
Rule 4132
Rubi steps
\begin {align*} \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {2 (5 a+2 b)+5 (a+b) x^2}{x^4 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{15 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\left (15 a^2-10 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac {\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 (a+b)^3 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{15 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (2 b \left (15 a^2-10 a b-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^3 f}\\ &=-\frac {\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 (a+b)^3 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{15 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\cot ^5(e+f x)}{5 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 (a+b)^4 f \sqrt {a+b+b \tan ^2(e+f x)}}\\ \end {align*}
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Mathematica [A] time = 0.92, size = 126, normalized size = 0.69 \[ -\frac {\tan (e+f x) \sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (4 a \left (a^2-4 a b-5 b^2\right ) \csc ^2(e+f x)-8 a^2 (a-5 b)+3 (a+b)^3 \csc ^6(e+f x)+(a-5 b) (a+b)^2 \csc ^4(e+f x)\right )}{30 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 4.67, size = 314, normalized size = 1.72 \[ -\frac {{\left (8 \, {\left (a^{3} - 5 \, a^{2} b\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (5 \, a^{3} - 26 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} - 85 \, a^{2} b + 49 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b - 10 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} + 7 \, a^{4} b + 8 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 2 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} + 2 \, a^{4} b - 2 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 7 \, a b^{4} - 2 \, b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )} \]
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 \[ \int \frac {\csc \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 2.11, size = 204, normalized size = 1.11 \[ -\frac {\left (8 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3}-40 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b -20 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3}+104 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2} b -20 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{2}+15 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3}-85 a^{2} \left (\cos ^{2}\left (f x +e \right )\right ) b +49 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{2}+5 \left (\cos ^{2}\left (f x +e \right )\right ) b^{3}+30 a^{2} b -20 b^{2} a -2 b^{3}\right ) \left (\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}\right )^{\frac {3}{2}} \left (\cos ^{3}\left (f x +e \right )\right )}{15 f \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2} \sin \left (f x +e \right )^{5} \left (a +b \right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 282, normalized size = 1.54 \[ -\frac {\frac {30 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {48 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{4}} + \frac {15}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )} - \frac {40 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {24 \, b^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3} \tan \left (f x + e\right )} + \frac {10}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {6 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]
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 \[ \int \frac {\csc ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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