Integrand size = 33, antiderivative size = 514 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (A b^4 m-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \sin (c+d x)}{b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (A b^4 m-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{a b \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (a^2 C (1+m)-b^2 (C-A m)\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (1+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (A b^2+a^2 C\right ) (1+m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt {\sin ^2(c+d x)}} \]
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Time = 1.01 (sec) , antiderivative size = 514, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3135, 3142, 2722, 2902, 3268, 440} \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\sin (c+d x) \left (a^2 C (m+1)-b^2 (C-A m)\right ) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{b^2 d (m+1) \left (a^2-b^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{a b d (m+2) \left (a^2-b^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\sin (c+d x) \left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )^2}-\frac {\sin (c+d x) \left (a^4 (-C) (m+1)+a^2 b^2 (A (-m)+A+C (m+2))+A b^4 m\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{a b d \left (a^2-b^2\right )^2} \]
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Rule 440
Rule 2722
Rule 2902
Rule 3135
Rule 3142
Rule 3268
Rubi steps \begin{align*} \text {integral}& = \frac {\left (A b^2+a^2 C\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\cos ^m(c+d x) \left (A b^2 m+a^2 (A+C+C m)-a b (A+C) \cos (c+d x)-\left (A b^2+a^2 C\right ) (1+m) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {\left (A b^2+a^2 C\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (\left (A b^2+a^2 C\right ) (1+m)\right ) \int \cos ^{1+m}(c+d x) \, dx}{a b \left (a^2-b^2\right )}+\frac {\left (A m-C \left (1-\frac {a^2 (1+m)}{b^2}\right )\right ) \int \cos ^m(c+d x) \, dx}{a^2-b^2}+\frac {\left (a^2 b^2 (A+C)-a^2 \left (A b^2+a^2 C\right ) (1+m)+b^2 \left (A b^2 m+a^2 (A+C+C m)\right )\right ) \int \frac {\cos ^m(c+d x)}{a+b \cos (c+d x)} \, dx}{a b^2 \left (a^2-b^2\right )} \\ & = \frac {\left (A b^2+a^2 C\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (A m-C \left (1-\frac {a^2 (1+m)}{b^2}\right )\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d (1+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (A b^2+a^2 C\right ) (1+m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a^2 b^2 (A+C)-a^2 \left (A b^2+a^2 C\right ) (1+m)+b^2 \left (A b^2 m+a^2 (A+C+C m)\right )\right ) \int \frac {\cos ^m(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}-\frac {\left (a^2 b^2 (A+C)-a^2 \left (A b^2+a^2 C\right ) (1+m)+b^2 \left (A b^2 m+a^2 (A+C+C m)\right )\right ) \int \frac {\cos ^{1+m}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx}{a b \left (a^2-b^2\right )} \\ & = \frac {\left (A b^2+a^2 C\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (A m-C \left (1-\frac {a^2 (1+m)}{b^2}\right )\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d (1+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (A b^2+a^2 C\right ) (1+m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (\left (a^2 b^2 (A+C)-a^2 \left (A b^2+a^2 C\right ) (1+m)+b^2 \left (A b^2 m+a^2 (A+C+C m)\right )\right ) \cos ^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )}(c+d x) \cos ^2(c+d x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (-1+m)}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (\left (a^2 b^2 (A+C)-a^2 \left (A b^2+a^2 C\right ) (1+m)+b^2 \left (A b^2 m+a^2 (A+C+C m)\right )\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{m/2}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{a b \left (a^2-b^2\right ) d} \\ & = \frac {\left (A b^4 m-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \sin (c+d x)}{b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (A b^4 m-a^4 C (1+m)+a^2 b^2 (A (1-m)+C (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{a b \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (A m-C \left (1-\frac {a^2 (1+m)}{b^2}\right )\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d (1+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (A b^2+a^2 C\right ) (1+m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(14082\) vs. \(2(514)=1028\).
Time = 51.73 (sec) , antiderivative size = 14082, normalized size of antiderivative = 27.40 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Result too large to show} \]
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\[\int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (a +\cos \left (d x +c \right ) b \right )^{2}}d x\]
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\[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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