Optimal. Leaf size=400 \[ -\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac {\left (a^3 A b-7 a A b^3+3 a^4 B-5 a^2 b^2 B+8 b^4 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^3 (a+b)^3 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))} \]
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Rubi [A]
time = 0.62, antiderivative size = 400, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 10, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3039, 4112,
4185, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} -\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}+\frac {\left (a^3 B+3 a^2 A b-7 a b^2 B+3 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 b d \left (a^2-b^2\right )^2 (a \sec (c+d x)+b)}-\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac {\left (3 a^4 B+a^3 A b-5 a^2 b^2 B-7 a A b^3+8 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {\left (3 a^5 B+a^4 A b-6 a^3 b^2 B-10 a^2 A b^3+15 a b^4 B-3 A b^5\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3039
Rule 3856
Rule 3872
Rule 3934
Rule 4112
Rule 4185
Rule 4191
Rubi steps
\begin {align*} \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx &=\int \frac {\sqrt {\sec (c+d x)} (B+A \sec (c+d x))}{(b+a \sec (c+d x))^3} \, dx\\ &=-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\int \frac {\frac {1}{2} (A b-a B)+2 (a A-b B) \sec (c+d x)-\frac {3}{2} (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (-a^2 A b-5 A b^3-3 a^3 B+9 a b^2 B\right )-b \left (3 a A b-a^2 B-2 b^2 B\right ) \sec (c+d x)+\frac {1}{4} \left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{4} b \left (-a^2 A b-5 A b^3-3 a^3 B+9 a b^2 B\right )-\left (b^2 \left (3 a A b-a^2 B-2 b^2 B\right )+\frac {1}{4} a \left (-a^2 A b-5 A b^3-3 a^3 B+9 a b^2 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )^2}-\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}+\frac {\left (a^3 A b-7 a A b^3+3 a^4 B-5 a^2 b^2 B+8 b^4 B\right ) \int \sqrt {\sec (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}-\frac {\left (\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^3 (a+b)^3 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}-\frac {\left (\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2}+\frac {\left (\left (a^3 A b-7 a A b^3+3 a^4 B-5 a^2 b^2 B+8 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a 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)}}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac {\left (a^3 A b-7 a A b^3+3 a^4 B-5 a^2 b^2 B+8 b^4 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^3 (a+b)^3 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 37.03, size = 786, normalized size = 1.96 \begin {gather*} -\frac {\frac {2 \left (-5 a^2 A b-A b^3+a^3 B+5 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 {a}{b};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (24 a A b^2-8 a^2 b B-16 b^3 B\right ) \cos ^2(c+d x) \Pi \left (-\frac {a}{b};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \cos (2 (c+d x)) (b+a \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 (2 a-b) 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)}-4 a^2 \Pi \left (-\frac {a}{b};\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 b^2 \Pi \left (-\frac {a}{b};\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 b^2 (a+b \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 )}}{16 (a-b)^2 b (a+b)^2 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2}-\frac {-a^2 A b \sin (c+d x)+a^3 B \sin (c+d x)}{2 b^2 \left (-a^2+b^2\right ) (a+b \cos (c+d x))^2}+\frac {a^3 A b \sin (c+d x)-7 a A b^3 \sin (c+d x)-5 a^4 B \sin (c+d x)+11 a^2 b^2 B \sin (c+d x)}{4 b^2 \left (-a^2+b^2\right )^2 (a+b \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. \(1936\) vs.
\(2(452)=904\).
time = 1.55, size = 1937, normalized size = 4.84
method | result | size |
default | \(\text {Expression too large to display}\) | \(1937\) |
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(-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 = -0.00 \begin {gather*} \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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