Optimal. Leaf size=218 \[ \frac {10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac {10 a \left (3 a^2+2 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}} \]
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Rubi [A] time = 0.17, antiderivative size = 218, 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 = {3512, 739, 778, 199, 231} \[ \frac {10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac {10 a \left (3 a^2+2 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 199
Rule 231
Rule 739
Rule 778
Rule 3512
Rubi steps
\begin {align*} \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx &=\frac {\sec ^2(e+f x)^{3/4} \operatorname {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{15/4}} \, dx,x,b \tan (e+f x)\right )}{b d^4 f (d \sec (e+f x))^{3/2}}\\ &=-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b \sec ^2(e+f x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {(a+x) \left (\frac {1}{2} \left (4+\frac {9 a^2}{b^2}\right )+\frac {5 a x}{2 b^2}\right )}{\left (1+\frac {x^2}{b^2}\right )^{11/4}} \, dx,x,b \tan (e+f x)\right )}{11 d^4 f (d \sec (e+f x))^{3/2}}\\ &=-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac {\left (15 a \left (2+\frac {3 a^2}{b^2}\right ) b \sec ^2(e+f x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{7/4}} \, dx,x,b \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}\\ &=\frac {10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac {\left (5 a \left (2+\frac {3 a^2}{b^2}\right ) b \sec ^2(e+f x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}\\ &=\frac {10 a \left (3 a^2+2 b^2\right ) F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac {10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}\\ \end {align*}
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Mathematica [A] time = 6.46, size = 296, normalized size = 1.36 \[ \frac {\sec ^3(e+f x) (a+b \tan (e+f x))^3 \left (\frac {a \left (347 a^2+103 b^2\right ) \sin (2 (e+f x))}{1232}+\frac {1}{308} a \left (16 a^2-15 b^2\right ) \sin (4 (e+f x))+\frac {1}{176} a \left (a^2-3 b^2\right ) \sin (6 (e+f x))-\frac {b \left (315 a^2+71 b^2\right ) \cos (2 (e+f x))}{1232}-\frac {1}{616} b \left (63 a^2+b^2\right ) \cos (4 (e+f x))-\frac {1}{176} b \left (3 a^2-b^2\right ) \cos (6 (e+f x))-\frac {1}{616} b \left (105 a^2+31 b^2\right )\right )}{f (d \sec (e+f x))^{11/2} (a \cos (e+f x)+b \sin (e+f x))^3}+\frac {10 a \left (3 a^2+2 b^2\right ) F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) (a+b \tan (e+f x))^3}{77 f \cos ^{\frac {5}{2}}(e+f x) (d \sec (e+f x))^{11/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{3} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \sec \left (f x + e\right )}}{d^{6} \sec \left (f x + e\right )^{6}}, x\right ) \]
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 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {11}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.10, size = 430, normalized size = 1.97 \[ -\frac {2 \left (21 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b -7 \left (\cos ^{6}\left (f x +e \right )\right ) b^{3}-7 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{3}+21 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a \,b^{2}-15 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a^{3}-10 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a \,b^{2}+11 \left (\cos ^{4}\left (f x +e \right )\right ) b^{3}-9 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{3}-6 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a \,b^{2}-15 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{3}-10 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a \,b^{2}-15 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}-10 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}\right )}{77 f \cos \left (f x +e \right )^{6} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {11}{2}}} \]
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] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{11/2}} \,d x \]
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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