Optimal. Leaf size=184 \[ \frac {\left (2 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}-\frac {b \left (4 a^2-3 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2} \]
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Rubi [A] time = 0.48, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4264, 3847, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac {b \left (4 a^2-3 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {\left (2 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 3771
Rule 3787
Rule 3847
Rule 3849
Rule 4106
Rule 4264
Rubi steps
\begin {align*} \int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^2} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx\\ &=\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) 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 {-a^2+\frac {3 b^2}{2}+a b \sec (c+d x)-\frac {1}{2} b^2 \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) 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 {a \left (-a^2+\frac {3 b^2}{2}\right )-\left (-a^2 b+b \left (-a^2+\frac {3 b^2}{2}\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )}+\frac {\left (b^2 \left (5-\frac {3 b^2}{a^2}\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}{2 a \left (a^2-b^2\right )}\\ &=\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (b^2 \left (5 a^2-3 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac {\left (\left (2 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\left (b \left (4 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}+\frac {\left (2 a^2-3 b^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\left (b \left (4 a^2-3 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {b \left (4 a^2-3 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A] time = 2.08, size = 252, normalized size = 1.37 \[ \frac {\frac {4 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\frac {2 \left (2 a^2-b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {2 \left (2 a^2-3 b^2\right ) \sin (c+d x) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt {\sin ^2(c+d x)}}+8 b \left (\frac {b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}-F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{(a-b) (a+b)}}{4 a d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 10.57, size = 809, normalized size = 4.40 \[ \text {result too large to display} \]
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 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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