Optimal. Leaf size=253 \[ \frac {\left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b \left (a^2-b^2\right )^2 d}+\frac {3 \left (a^2+b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a \left (a^2-b^2\right )^2 d}+\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b (a+b)^3 d}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {a \left (a^2-7 b^2\right ) \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \]
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Rubi [A]
time = 0.51, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4349, 3930,
4185, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} \frac {3 \left (a^2+b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a d \left (a^2-b^2\right )^2}+\frac {\left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b d \left (a^2-b^2\right )^2}-\frac {a^2 \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {a \left (a^2-7 b^2\right ) \sin (c+d x)}{4 b d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b 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 3930
Rule 3934
Rule 4185
Rule 4191
Rule 4349
Rubi steps
\begin {align*} \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx\\ &=-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {a^2}{2}-2 a b \sec (c+d x)-\frac {1}{2} \left (a^2-4 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {a \left (a^2-7 b^2\right ) \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 \left (a^2+5 b^2\right )+a b \left (a^2+2 b^2\right ) \sec (c+d x)+\frac {1}{4} a^2 \left (a^2-7 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2}\\ &=-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {a \left (a^2-7 b^2\right ) \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^3 \left (a^2+5 b^2\right )-\left (-a^2 b \left (a^2+2 b^2\right )+\frac {1}{4} a^2 b \left (a^2+5 b^2\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{2 a^3 b \left (a^2-b^2\right )^2}+\frac {\left (\left (a^4-10 a^2 b^2-3 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 a b \left (a^2-b^2\right )^2}\\ &=-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {a \left (a^2-7 b^2\right ) \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a b \left (a^2-b^2\right )^2}+\frac {\left (3 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{8 a \left (a^2-b^2\right )^2}+\frac {\left (\left (a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 b \left (a^2-b^2\right )^2}\\ &=\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b (a+b)^3 d}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {a \left (a^2-7 b^2\right ) \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (3 \left (a^2+b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a \left (a^2-b^2\right )^2}+\frac {\left (a^2+5 b^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b \left (a^2-b^2\right )^2}\\ &=\frac {\left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b \left (a^2-b^2\right )^2 d}+\frac {3 \left (a^2+b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a \left (a^2-b^2\right )^2 d}+\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b (a+b)^3 d}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}+\frac {a \left (a^2-7 b^2\right ) \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 11.87, size = 289, normalized size = 1.14 \begin {gather*} \frac {-\frac {4 a \sqrt {\cos (c+d x)} \left (-a^2 b+7 b^3+a \left (a^2+5 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {6 \left (a^3-3 a b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 b \left (a^2+2 b^2\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac {2 \left (a^2+5 b^2\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 b d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1759\) vs.
\(2(317)=634\).
time = 0.62, size = 1760, normalized size = 6.96
method | result | size |
default | \(\text {Expression too large to display}\) | \(1760\) |
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 {1}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {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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