Optimal. Leaf size=210 \[ -\frac {2 (A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 d}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b d}-\frac {2 a (A b-a B) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d}+\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b d} \]
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Rubi [A]
time = 0.46, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 9, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4118, 4187,
4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} \frac {2 (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b^2 d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}-\frac {2 a (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d (a+b)}+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 b d}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3856
Rule 3872
Rule 3934
Rule 4118
Rule 4187
Rule 4191
Rubi steps
\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx &=\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b d}+\frac {2 \int \frac {\sqrt {\sec (c+d x)} \left (\frac {a B}{2}+\frac {1}{2} b B \sec (c+d x)+\frac {3}{2} (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b}\\ &=\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b d}+\frac {4 \int \frac {-\frac {3}{4} a (A b-a B)-\frac {1}{4} b (3 A b-4 a B) \sec (c+d x)-\frac {1}{4} \left (3 a A b-3 a^2 B-b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 b^2}\\ &=\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b d}+\frac {4 \int \frac {-\frac {3}{4} a^2 (A b-a B)-\left (\frac {1}{4} a b (3 A b-4 a B)-\frac {3}{4} a b (A b-a B)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^2 b^2}-\frac {(a (A b-a B)) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{b^2}\\ &=\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b d}+\frac {B \int \sqrt {\sec (c+d x)} \, dx}{3 b}-\frac {(A b-a B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{b^2}-\frac {\left (a (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^2}\\ &=-\frac {2 a (A b-a B) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d}+\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b d}+\frac {\left (B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b}-\frac {\left ((A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{b^2}\\ &=-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 d}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b d}-\frac {2 a (A b-a B) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d}+\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b d}\\ \end {align*}
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Mathematica [A]
time = 33.70, size = 225, normalized size = 1.07 \begin {gather*} -\frac {\cot (c+d x) \left (-b^2 B \sec ^{\frac {5}{2}}(c+d x)+b^2 B \cos (2 (c+d x)) \sec ^{\frac {5}{2}}(c+d x)-6 b (A b-a B) E\left (\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}-2 \left (3 a^2 B+b^2 (-3 A+B)+3 a b (-A+B)\right ) F\left (\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}-6 a A b \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+6 a^2 B \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{3 b^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.74, size = 439, normalized size = 2.09
method | result | size |
default | \(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {2 B \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{6 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}\right )}{b}+\frac {2 \left (A b -B a \right ) a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{b^{2} \left (a^{2}-b a \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}+\frac {2 \left (A b -B a \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{b^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(439\) |
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 [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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 = -0.00 \begin {gather*} \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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