Optimal. Leaf size=445 \[ -\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b^2 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (-7 a^3 C+5 a^2 b B-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b^3 d \left (a^2-b^2\right )}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-35 a^4 C+25 a^3 b B-3 a^2 b^2 (5 A-8 C)-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac {a^2 \left (-7 a^4 C+5 a^3 b B-3 a^2 b^2 (A-3 C)-7 a b^3 B+5 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2}+\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-21 a^5 C+15 a^4 b B-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+4 a b^4 (3 A+C)-2 b^5 B\right )}{3 b^5 d \left (a^2-b^2\right )} \]
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Rubi [A] time = 1.59, antiderivative size = 445, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3047, 3049, 3059, 2639, 3002, 2641, 2805} \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-a^3 b^2 (9 A-20 C)-16 a^2 b^3 B+15 a^4 b B-21 a^5 C+4 a b^4 (3 A+C)-2 b^5 B\right )}{3 b^5 d \left (a^2-b^2\right )}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-3 a^2 b^2 (5 A-8 C)+25 a^3 b B-35 a^4 C-20 a b^3 B+2 b^4 (5 A+3 C)\right )}{5 b^4 d \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^2 b^2 (A-3 C)+5 a^3 b B-7 a^4 C-7 a b^3 B+5 A b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a-b) (a+b)^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (7 a^2 C-5 a b B+5 A b^2-2 b^2 C\right )}{5 b^2 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 b B-7 a^3 C-a b^2 (3 A-4 C)-2 b^3 B\right )}{3 b^3 d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 3002
Rule 3047
Rule 3049
Rule 3059
Rubi steps
\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=-\frac {\left (A b^2-a (b B-a 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 (b B-a C)\right )+b (b B-a (A+C)) \cos (c+d x)-\frac {1}{2} \left (5 A b^2-5 a b B+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^2-5 a b B+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 (b B-a 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^2-5 a b B+7 a^2 C-2 b^2 C\right )+\frac {1}{2} b \left (5 A b^2-5 a b B+2 a^2 C+3 b^2 C\right ) \cos (c+d x)-\frac {5}{4} \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 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 {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 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^2-5 a b B+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 (b B-a 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 \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right )+\frac {1}{4} b \left (10 a^2 b B+5 b^3 B-14 a^3 C-a b^2 (15 A+C)\right ) \cos (c+d x)+\frac {3}{8} \left (25 a^3 b B-20 a b^3 B-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 {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 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^2-5 a b B+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 (b B-a 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 b \left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 C\right )+\frac {5}{8} \left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a 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 (25 a^3 b B-20 a b^3 B-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 (25 a^3 b B-20 a b^3 B-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 {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 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^2-5 a b B+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 (b B-a 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+5 a^3 b B-7 a b^3 B-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 (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^5 \left (a^2-b^2\right )}\\ &=-\frac {\left (25 a^3 b B-20 a b^3 B-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 {\left (15 a^4 b B-16 a^2 b^3 B-2 b^5 B-a^3 b^2 (9 A-20 C)-21 a^5 C+4 a 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+5 a^3 b B-7 a b^3 B-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 {\left (5 a^2 b B-2 b^3 B-a b^2 (3 A-4 C)-7 a^3 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^2-5 a b B+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 (b B-a 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 = 5.74, size = 404, normalized size = 0.91 \[ \frac {4 \sqrt {\cos (c+d x)} \left (-\frac {15 a^2 \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+10 (b B-2 a C) \sin (c+d x)+3 b C \sin (2 (c+d x))\right )+\frac {\frac {8 \left (14 a^3 C-10 a^2 b B+a b^2 (15 A+C)-5 b^3 B\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac {2 \left (35 a^4 C-25 a^3 b B+a^2 b^2 (15 A-32 C)+40 a b^3 B-6 b^4 (5 A+3 C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {6 \sin (c+d x) \left (35 a^4 C-25 a^3 b B+3 a^2 b^2 (5 A-8 C)+20 a b^3 B-2 b^4 (5 A+3 C)\right ) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{60 b^3 d} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 10.17, size = 1382, normalized size = 3.11 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
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 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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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