Optimal. Leaf size=326 \[ -\frac {2 (b c-a d)^2 \tan (e+f x) \sqrt {\frac {a+b \sec (e+f x)}{a+b}} \Pi \left (\frac {2 d}{c+d};\sin ^{-1}\left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right )|\frac {2 b}{a+b}\right )}{c d f (c+d) \sqrt {-\tan ^2(e+f x)} \sqrt {a+b \sec (e+f x)}}-\frac {2 a \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{c f}+\frac {2 b \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d f} \]
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Rubi [A] time = 0.36, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3928, 3921, 3784, 3832, 3973} \[ -\frac {2 (b c-a d)^2 \tan (e+f x) \sqrt {\frac {a+b \sec (e+f x)}{a+b}} \Pi \left (\frac {2 d}{c+d};\sin ^{-1}\left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right )|\frac {2 b}{a+b}\right )}{c d f (c+d) \sqrt {-\tan ^2(e+f x)} \sqrt {a+b \sec (e+f x)}}-\frac {2 a \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{c f}+\frac {2 b \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d f} \]
Antiderivative was successfully verified.
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Rule 3784
Rule 3832
Rule 3921
Rule 3928
Rule 3973
Rubi steps
\begin {align*} \int \frac {(a+b \sec (e+f x))^{3/2}}{c+d \sec (e+f x)} \, dx &=\frac {\int \frac {a^2 d+b^2 c \sec (e+f x)}{\sqrt {a+b \sec (e+f x)}} \, dx}{c d}-\frac {(b c-a d)^2 \int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx}{c d}\\ &=-\frac {2 (b c-a d)^2 \Pi \left (\frac {2 d}{c+d};\sin ^{-1}\left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sec (e+f x)}{a+b}} \tan (e+f x)}{c d (c+d) f \sqrt {a+b \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}+\frac {a^2 \int \frac {1}{\sqrt {a+b \sec (e+f x)}} \, dx}{c}+\frac {b^2 \int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)}} \, dx}{d}\\ &=\frac {2 b \sqrt {a+b} \cot (e+f x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{d f}-\frac {2 a \sqrt {a+b} \cot (e+f x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{c f}-\frac {2 (b c-a d)^2 \Pi \left (\frac {2 d}{c+d};\sin ^{-1}\left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sec (e+f x)}{a+b}} \tan (e+f x)}{c d (c+d) f \sqrt {a+b \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}\\ \end {align*}
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Mathematica [A] time = 6.31, size = 230, normalized size = 0.71 \[ -\frac {4 \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {\cos (e+f x)}{\cos (e+f x)+1}} \sqrt {\frac {a \cos (e+f x)+b}{(a+b) (\cos (e+f x)+1)}} \sqrt {a+b \sec (e+f x)} \left (c (a-b)^2 (c+d) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )-2 \left (a^2 \left (c^2-d^2\right ) \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )+(b c-a d)^2 \Pi \left (\frac {c-d}{c+d};\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )\right )\right )}{c f (c-d) (c+d) (a \cos (e+f x)+b)} \]
Antiderivative was successfully verified.
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fricas [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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{d \sec \left (f x + e\right ) + c}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.70, size = 581, normalized size = 1.78 \[ -\frac {2 \sqrt {\frac {b +a \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \left (1+\cos \left (f x +e \right )\right )^{2} \left (\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} c^{2}+\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} c d -2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a b \,c^{2}-2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a b c d +\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2} c^{2}+\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2} c d -2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \frac {c -d}{c +d}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} d^{2}+4 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \frac {c -d}{c +d}, \sqrt {\frac {a -b}{a +b}}\right ) a b c d -2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \frac {c -d}{c +d}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2} c^{2}-2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, -1, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} c^{2}+2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, -1, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} d^{2}\right ) \left (-1+\cos \left (f x +e \right )\right )}{f \left (b +a \cos \left (f x +e \right )\right ) \sin \left (f x +e \right )^{2} c \left (c -d \right ) \left (c +d \right )} \]
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 \[ \int \frac {{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{d \sec \left (f x + e\right ) + c}\,{d x} \]
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 \[ \int \frac {{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{c+\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]
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 \[ \int \frac {\left (a + b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{c + d \sec {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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