Optimal. Leaf size=315 \[ \frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {(A b-a B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^2 (a+b)^2 d}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
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Rubi [A]
time = 0.55, antiderivative size = 315, 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 = {4114, 4187,
4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} \frac {a (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b^2 d \left (a^2-b^2\right )}+\frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) \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 \left (a^2-b^2\right )}+\frac {\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \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) (a+b)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3856
Rule 3872
Rule 3934
Rule 4114
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))^2} \, dx &=\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{2} a (A b-a B)-b (A b-a B) \sec (c+d x)-\frac {1}{2} \left (a A b-3 a^2 B+2 b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {2 \int \frac {\frac {1}{4} a \left (a A b-3 a^2 B+2 b^2 B\right )+\frac {1}{2} b \left (a A b-2 a^2 B+b^2 B\right ) \sec (c+d x)+\frac {1}{4} \left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {2 \int \frac {\frac {1}{4} a^2 \left (a A b-3 a^2 B+2 b^2 B\right )-\left (-\frac {1}{2} a b \left (a A b-2 a^2 B+b^2 B\right )+\frac {1}{4} a b \left (a A b-3 a^2 B+2 b^2 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{a^2 b^2 \left (a^2-b^2\right )}+\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {(A b-a B) \int \sqrt {\sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}+\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac {\left (\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \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}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \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)}}{(a-b) b^2 (a+b)^2 d}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left ((A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 b \left (a^2-b^2\right )}+\frac {\left (\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {(A b-a B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \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)}}{(a-b) b^2 (a+b)^2 d}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(680\) vs. \(2(315)=630\).
time = 37.06, size = 680, normalized size = 2.16 \begin {gather*} -\frac {\frac {2 \left (-3 a^2 A b+4 A b^3+9 a^3 B-10 a b^2 B\right ) \cos ^2(c+d x) \left (F\left (\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-4 a A b^2+8 a^2 b B-4 b^3 B\right ) \cos ^2(c+d x) \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-a^2 A b+3 a^3 B-2 a b^2 B\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) F\left (\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{4 (a-b) b^2 (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{b^2 \left (-a^2+b^2\right )}+\frac {-a A b \sin (c+d x)+a^2 B \sin (c+d x)}{b \left (-a^2+b^2\right ) (b+a \cos (c+d x))}\right )}{d} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(849\) vs.
\(2(377)=754\).
time = 4.35, size = 850, normalized size = 2.70
method | result | size |
default | \(\text {Expression too large to display}\) | \(850\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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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}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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