Integrand size = 28, antiderivative size = 264 \[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {\left (a^2-2 b^2\right ) \cos ^{-1+m}(c+d x) \sin (c+d x)}{d m (2+m)}-\frac {2 a b \cos ^m(c+d x) \sin (c+d x)}{d \left (2+3 m+m^2\right )}-\frac {\cos ^{-1+m}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{d (2+m)}-\frac {\left (a^2 (1-m)-b^2 (2+m)\right ) \cos ^{-1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-m) m (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d m (1+m) \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.85 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4482, 2968, 3129, 3112, 3102, 2827, 2722} \[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {\left (a^2 (1-m)-b^2 (m+2)\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m-1}{2},\frac {m+1}{2},\cos ^2(c+d x)\right )}{d (1-m) m (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x)}{d m (m+2)}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {m+2}{2},\cos ^2(c+d x)\right )}{d m (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x)}{d \left (m^2+3 m+2\right )}-\frac {\sin (c+d x) \cos ^{m-1}(c+d x) (a \cos (c+d x)+b)^2}{d (m+2)} \]
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Rule 2722
Rule 2827
Rule 2968
Rule 3102
Rule 3112
Rule 3129
Rule 4482
Rubi steps \begin{align*} \text {integral}& = \int \cos ^{-2+m}(c+d x) (b+a \cos (c+d x))^2 \sin ^2(c+d x) \, dx \\ & = \int \cos ^{-2+m}(c+d x) (b+a \cos (c+d x))^2 \left (1-\cos ^2(c+d x)\right ) \, dx \\ & = -\frac {\cos ^{-1+m}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{d (2+m)}+\frac {\int \cos ^{-2+m}(c+d x) (b+a \cos (c+d x)) \left (3 b+a \cos (c+d x)-2 b \cos ^2(c+d x)\right ) \, dx}{2+m} \\ & = -\frac {2 a b \cos ^m(c+d x) \sin (c+d x)}{d \left (2+3 m+m^2\right )}-\frac {\cos ^{-1+m}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{d (2+m)}+\frac {\int \cos ^{-2+m}(c+d x) \left (3 b^2 (1+m)+2 a b (2+m) \cos (c+d x)+\left (a^2-2 b^2\right ) (1+m) \cos ^2(c+d x)\right ) \, dx}{2+3 m+m^2} \\ & = \frac {\left (a^2-2 b^2\right ) \cos ^{-1+m}(c+d x) \sin (c+d x)}{d m (2+m)}-\frac {2 a b \cos ^m(c+d x) \sin (c+d x)}{d \left (2+3 m+m^2\right )}-\frac {\cos ^{-1+m}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{d (2+m)}+\frac {\int \cos ^{-2+m}(c+d x) \left (-\left ((1+m) \left (a^2 (1-m)-b^2 (2+m)\right )\right )+2 a b m (2+m) \cos (c+d x)\right ) \, dx}{m \left (2+3 m+m^2\right )} \\ & = \frac {\left (a^2-2 b^2\right ) \cos ^{-1+m}(c+d x) \sin (c+d x)}{d m (2+m)}-\frac {2 a b \cos ^m(c+d x) \sin (c+d x)}{d \left (2+3 m+m^2\right )}-\frac {\cos ^{-1+m}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{d (2+m)}+\frac {(2 a b) \int \cos ^{-1+m}(c+d x) \, dx}{1+m}-\frac {\left (a^2 (1-m)-b^2 (2+m)\right ) \int \cos ^{-2+m}(c+d x) \, dx}{m (2+m)} \\ & = \frac {\left (a^2-2 b^2\right ) \cos ^{-1+m}(c+d x) \sin (c+d x)}{d m (2+m)}-\frac {2 a b \cos ^m(c+d x) \sin (c+d x)}{d \left (2+3 m+m^2\right )}-\frac {\cos ^{-1+m}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{d (2+m)}-\frac {\left (a^2 (1-m)-b^2 (2+m)\right ) \cos ^{-1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-m) m (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d m (1+m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 47.88 (sec) , antiderivative size = 6639, normalized size of antiderivative = 25.15 \[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\text {Result too large to show} \]
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\[\int \cos \left (d x +c \right )^{m} \left (\sin \left (d x +c \right ) a +b \tan \left (d x +c \right )\right )^{2}d x\]
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\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \]
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\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \cos ^{m}{\left (c + d x \right )}\, dx \]
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\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \]
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\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (a\,\sin \left (c+d\,x\right )+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \]
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