Optimal. Leaf size=270 \[ \frac {\left (a^2 C+A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b d \left (a^2-b^2\right )}+\frac {\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\left (a^4 (-C)-a^2 b^2 (5 A+C)+3 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 b d (a-b) (a+b)^2} \]
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Rubi [A] time = 1.07, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3056, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {\left (a^2 C+A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b d \left (a^2-b^2\right )}+\frac {\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {\left (-a^2 b^2 (5 A+C)+a^4 (-C)+3 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 b d (a-b) (a+b)^2}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 3002
Rule 3055
Rule 3056
Rule 3059
Rubi steps
\begin {align*} \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (-3 A b^2+2 a^2 \left (A-\frac {C}{2}\right )\right )-a b (A+C) \cos (c+d x)+\frac {1}{2} \left (A b^2+a^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {2 \int \frac {\frac {1}{4} b \left (3 A b^2-a^2 (4 A+C)\right )+\frac {1}{2} a \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x)+\frac {1}{4} b \left (3 A b^2-a^2 (2 A-C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {2 \int \frac {-\frac {1}{4} b^2 \left (3 A b^2-a^2 (4 A+C)\right )-\frac {1}{4} a b \left (A b^2+a^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^2 b \left (a^2-b^2\right )}+\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\left (A b^2+a^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a b \left (a^2-b^2\right )}+\frac {\left (3 A b^4-a^4 C-a^2 b^2 (5 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^2 b \left (a^2-b^2\right )}\\ &=\frac {\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}+\frac {\left (3 A b^4-a^4 C-a^2 b^2 (5 A+C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 (a-b) b (a+b)^2 d}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 4.11, size = 306, normalized size = 1.13 \[ \frac {4 \sqrt {\cos (c+d x)} \left (\frac {\left (a^2 b C+A b^3\right ) \sin (c+d x)}{\left (b^2-a^2\right ) (a+b \cos (c+d x))}+2 A \tan (c+d x)\right )-\frac {\frac {\left (8 a A b^2-4 a^3 (A-C)\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}-\frac {2 \left (a^2 b (10 A+C)-9 A b^3\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}-\frac {2 \left (a^2 (2 A-C)-3 A b^2\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}}{(b-a) (a+b)}}{4 a^2 d} \]
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 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 6.79, size = 899, normalized size = 3.33 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-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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