Optimal. Leaf size=256 \[ \frac {(A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac {\left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {\left (2 a^2 A+a b B-3 A b^2\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (a-b) (a+b)^2} \]
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Rubi [A] time = 0.92, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3000, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {(A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac {\left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}-\frac {\left (5 a^2 A b-3 a^3 B+a b^2 B-3 A b^3\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (a-b) (a+b)^2}+\frac {\left (2 a^2 A+a b B-3 A b^2\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 3000
Rule 3002
Rule 3055
Rule 3059
Rubi steps
\begin {align*} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=\frac {b (A b-a B) \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 (2 a^2 A-3 A b^2+a b B\right )-a (A b-a B) \cos (c+d x)+\frac {1}{2} b (A b-a B) \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 (2 a^2 A-3 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \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} \left (-4 a^2 A b+3 A b^3+2 a^3 B-a b^2 B\right )-\frac {1}{2} a \left (a^2 A-2 A b^2+a b B\right ) \cos (c+d x)-\frac {1}{4} b \left (2 a^2 A-3 A b^2+a b B\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 (2 a^2 A-3 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \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 (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right )-\frac {1}{4} a b^2 (A b-a B) \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 (2 a^2 A-3 A b^2+a b B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-3 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {(A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a \left (a^2-b^2\right )}-\frac {\left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {(A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}-\frac {\left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 (a-b) (a+b)^2 d}+\frac {\left (2 a^2 A-3 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \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 = 4.22, size = 316, normalized size = 1.23 \[ \frac {4 \sqrt {\cos (c+d x)} \left (\frac {b^2 (A b-a B) \sin (c+d x)}{\left (b^2-a^2\right ) (a+b \cos (c+d x))}+2 A \tan (c+d x)\right )-\frac {-\frac {8 a \left (a^2 A+a b B-2 A b^2\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}-\frac {2 \left (2 a^2 A+a b B-3 A b^2\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}+\frac {2 \left (4 a^3 B-10 a^2 A b-3 a b^2 B+9 A b^3\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{(b-a) (a+b)}}{4 a^2 d} \]
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 \[ \int \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 4.25, size = 883, normalized size = 3.45 \[ \text {result too large to display} \]
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 \[ \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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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