Optimal. Leaf size=370 \[ -\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d \left (a^2-b^2\right )}-\frac {a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {\left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac {a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^5 d \left (a^2-b^2\right )}-\frac {a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2} \]
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Rubi [A] time = 1.38, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3048, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac {a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^5 d \left (a^2-b^2\right )}+\frac {\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^2 b^2 (A-3 C)-7 a^4 C+5 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d \left (a^2-b^2\right )}-\frac {a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 3002
Rule 3048
Rule 3049
Rule 3059
Rubi steps
\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} \left (A b^2+a^2 C\right )-a b (A+C) \cos (c+d x)-\frac {1}{2} \left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac {\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {2 \int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{4} a \left (5 A b^2+7 a^2 C-2 b^2 C\right )+\frac {1}{2} b \left (5 A b^2+2 a^2 C+3 b^2 C\right ) \cos (c+d x)+\frac {5}{4} a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=-\frac {a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {4 \int \frac {\frac {5}{8} a^2 \left (3 A b^2+7 a^2 C-4 b^2 C\right )-\frac {1}{4} a b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \cos (c+d x)-\frac {3}{8} \left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=-\frac {a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {4 \int \frac {-\frac {5}{8} a^2 b \left (3 A b^2+7 a^2 C-4 b^2 C\right )-\frac {5}{8} a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^4 \left (a^2-b^2\right )}+\frac {\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 b^4 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}-\frac {a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (a^2 \left (5 A b^4-3 a^2 b^2 (A-3 C)-7 a^4 C\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^5 \left (a^2-b^2\right )}-\frac {\left (a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^5 \left (a^2-b^2\right )}\\ &=\frac {\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^5 \left (a^2-b^2\right ) d}-\frac {a^2 \left (5 A b^4-3 a^2 b^2 (A-3 C)-7 a^4 C\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{(a-b) b^5 (a+b)^2 d}-\frac {a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 4.16, size = 354, normalized size = 0.96 \[ \frac {4 \sqrt {\cos (c+d x)} \left (-\frac {15 a^2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-20 a C \sin (c+d x)+3 b C \sin (2 (c+d x))\right )+\frac {\frac {8 a \left (C \left (14 a^2+b^2\right )+15 A b^2\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac {2 \left (35 a^4 C+a^2 b^2 (15 A-32 C)-6 b^4 (5 A+3 C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{60 b^3 d} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 7.89, size = 1337, normalized size = 3.61 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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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