Optimal. Leaf size=427 \[ \frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]
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Rubi [A]
time = 0.67, antiderivative size = 427, 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 = {4115, 4185,
4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} \frac {b (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {\left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {\left (-8 a^5 B+24 a^4 A b+5 a^3 b^2 B-33 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac {b \left (-15 a^5 B+35 a^4 A b+6 a^3 b^2 B-38 a^2 A b^3-3 a b^4 B+15 A b^5\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 )}{4 a^4 d (a-b)^2 (a+b)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3856
Rule 3872
Rule 3934
Rule 4115
Rule 4185
Rule 4191
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx &=\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\int \frac {\frac {1}{2} \left (-4 a^2 A+5 A b^2-a b B\right )+2 a (A b-a B) \sec (c+d x)-\frac {3}{2} b (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right )-a \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right ) \sec (c+d x)+\frac {1}{4} b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {\frac {1}{4} a \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right )-\left (a^2 \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right )+\frac {1}{4} b \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{2 a^4 \left (a^2-b^2\right )^2}+\frac {\left (b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2}-\frac {\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \int \sqrt {\sec (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac {\left (b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 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}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac {b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 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)}}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}-\frac {\left (\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a 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 a^4 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 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 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a 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 a^4 \left (a^2-b^2\right )^2 d}+\frac {b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 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)}}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 37.35, size = 818, normalized size = 1.92 \begin {gather*} \frac {\frac {2 \left (8 a^4 A-7 a^2 A b^2+5 A b^4-5 a^3 b B-a b^3 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 (-32 a^3 A b+8 a A b^3+16 a^4 B+8 a^2 b^2 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 (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 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 )}}{16 a^2 (a-b)^2 (a+b)^2 d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {b \left (-13 a^2 A b+7 A b^3+9 a^3 B-3 a b^2 B\right ) \sin (c+d x)}{4 a^3 \left (-a^2+b^2\right )^2}-\frac {-A b^4 \sin (c+d x)+a b^3 B \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {-15 a^2 A b^3 \sin (c+d x)+9 A b^5 \sin (c+d x)+11 a^3 b^2 B \sin (c+d x)-5 a b^4 B \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 (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. \(1999\) vs.
\(2(479)=958\).
time = 8.64, size = 2000, normalized size = 4.68
method | result | size |
default | \(\text {Expression too large to display}\) | \(2000\) |
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(-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+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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