Optimal. Leaf size=198 \[ \frac {2 a \left (7 a^2-6 b^2\right ) d^2 F\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right ) \sqrt {d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f} \]
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Rubi [A]
time = 0.11, antiderivative size = 198, 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 = {3593, 757, 794,
201, 237} \begin {gather*} \frac {2 a d^2 \left (7 a^2-6 b^2\right ) \sqrt {d \sec (e+f x)} F\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}+\frac {2 a d^2 \left (7 a^2-6 b^2\right ) \tan (e+f x) \sqrt {d \sec (e+f x)}}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 201
Rule 237
Rule 757
Rule 794
Rule 3593
Rubi steps
\begin {align*} \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx &=\frac {\left (d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int (a+x)^3 \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {\left (2 b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int (a+x) \left (\frac {1}{2} \left (-4+\frac {9 a^2}{b^2}\right )+\frac {13 a x}{2 b^2}\right ) \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{9 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac {\left (a \left (6-\frac {7 a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{7 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac {\left (a \left (6-\frac {7 a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 a \left (7 a^2-6 b^2\right ) d^2 F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt {d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}\\ \end {align*}
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Mathematica [A]
time = 1.87, size = 157, normalized size = 0.79 \begin {gather*} -\frac {2 d (d \sec (e+f x))^{3/2} \left (63 b \left (-3 a^2+b^2\right ) \cos ^2(e+f x)-15 a \left (7 a^2-6 b^2\right ) \cos ^{\frac {9}{2}}(e+f x) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )-15 a \left (7 a^2-6 b^2\right ) \cos ^3(e+f x) \sin (e+f x)-\frac {5}{2} b^2 (14 b+27 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{315 f (a \cos (e+f x)+b \sin (e+f x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains complex when optimal does not.
time = 0.62, size = 414, normalized size = 2.09
method | result | size |
default | \(\frac {2 \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2} \left (105 i \left (\cos ^{5}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-90 i \left (\cos ^{5}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+105 i \left (\cos ^{4}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-90 i \left (\cos ^{4}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+105 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{3}-90 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a \,b^{2}+189 a^{2} \left (\cos ^{2}\left (f x +e \right )\right ) b -63 b^{3} \left (\cos ^{2}\left (f x +e \right )\right )+135 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}+35 b^{3}\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{315 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4}}\) | \(414\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.13, size = 214, normalized size = 1.08 \begin {gather*} \frac {-15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (35 \, b^{3} d^{2} + 63 \, {\left (3 \, a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right )^{2} + 15 \, {\left (9 \, a b^{2} d^{2} \cos \left (f x + e\right ) + {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{2} \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{315 \, f \cos \left (f x + e\right )^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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