Optimal. Leaf size=129 \[ \frac {2 a \left (a^2-2 b^2\right ) F\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right ) \sqrt {d \sec (e+f x)}}{f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac {2 b \sqrt {d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f} \]
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Rubi [A]
time = 0.08, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3593, 757, 794,
237} \begin {gather*} \frac {2 a \left (a^2-2 b^2\right ) \sqrt {d \sec (e+f x)} F\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b \sqrt {d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f}+\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 237
Rule 757
Rule 794
Rule 3593
Rubi steps
\begin {align*} \int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx &=\frac {\sqrt {d \sec (e+f x)} \text {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac {\left (2 b \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {(a+x) \left (\frac {1}{2} \left (-4+\frac {5 a^2}{b^2}\right )+\frac {9 a x}{2 b^2}\right )}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{5 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac {2 b \sqrt {d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f}-\frac {\left (a \left (2-\frac {a^2}{b^2}\right ) b \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 )}{f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 a \left (a^2-2 b^2\right ) F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt {d \sec (e+f x)}}{f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac {2 b \sqrt {d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f}\\ \end {align*}
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Mathematica [A]
time = 2.21, size = 132, normalized size = 1.02 \begin {gather*} -\frac {2 \sqrt {d \sec (e+f x)} \left (5 b \left (-3 a^2+b^2\right ) \cos ^3(e+f x)-5 a \left (a^2-2 b^2\right ) \cos ^{\frac {7}{2}}(e+f x) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )-\frac {1}{2} b^2 \cos (e+f x) (2 b+5 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{5 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 = 1.12, size = 373, normalized size = 2.89
method | result | size |
default | \(\frac {2 \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2} \left (5 i \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-10 i \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+5 i \left (\cos ^{2}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-10 i \left (\cos ^{2}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+15 a^{2} \left (\cos ^{2}\left (f x +e \right )\right ) b -5 b^{3} \left (\cos ^{2}\left (f x +e \right )\right )+5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}+b^{3}\right ) \sqrt {\frac {d}{\cos \left (f x +e \right )}}}{5 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4}}\) | \(373\) |
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 = 175, normalized size = 1.36 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (5 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b^{3} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, f \cos \left (f x + e\right )^{2}} \end {gather*}
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 \begin {gather*} \int \sqrt {d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )^{3}\, dx \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 \sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,{\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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