Optimal. Leaf size=264 \[ -\frac{\left (a^2 (1-m)-b^2 (m+2)\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m-1}{2},\frac{m+1}{2},\cos ^2(c+d x)\right )}{d (1-m) m (m+2) \sqrt{\sin ^2(c+d x)}}-\frac{2 a b \sin (c+d x) \cos ^m(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2},\frac{m+2}{2},\cos ^2(c+d x)\right )}{d m (m+1) \sqrt{\sin ^2(c+d x)}}+\frac{\left (a^2-2 b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x)}{d m (m+2)}-\frac{2 a b \sin (c+d x) \cos ^m(c+d x)}{d \left (m^2+3 m+2\right )}-\frac{\sin (c+d x) \cos ^{m-1}(c+d x) (a \cos (c+d x)+b)^2}{d (m+2)} \]
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Rubi [A] time = 0.764108, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4397, 2889, 3050, 3033, 3023, 2748, 2643} \[ -\frac{\left (a^2 (1-m)-b^2 (m+2)\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m-1}{2};\frac{m+1}{2};\cos ^2(c+d x)\right )}{d (1-m) m (m+2) \sqrt{\sin ^2(c+d x)}}+\frac{\left (a^2-2 b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x)}{d m (m+2)}-\frac{2 a b \sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};\cos ^2(c+d x)\right )}{d m (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{2 a b \sin (c+d x) \cos ^m(c+d x)}{d \left (m^2+3 m+2\right )}-\frac{\sin (c+d x) \cos ^{m-1}(c+d x) (a \cos (c+d x)+b)^2}{d (m+2)} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 2889
Rule 3050
Rule 3033
Rule 3023
Rule 2748
Rule 2643
Rubi steps
\begin{align*} \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx &=\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 (-(1+m) \left (a^2 (1-m)-b^2 (2+m)\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) \, _2F_1\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) \, _2F_1\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*}
Mathematica [A] time = 12.386, size = 401, normalized size = 1.52 \[ \frac{\sin (c+d x) \cos ^{m-1}(c+d x) \left (\frac{2048 a^2 \cos ^4(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+3}{2},\frac{m+5}{2},\cos ^2(c+d x)\right )}{m+3}-\frac{2048 a^2 \cos ^2(c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\cos ^2(c+d x)\right )}{m+1}-\frac{2048 a^2 \cos ^2(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\cos ^2(c+d x)\right )}{m+1}+\frac{8190 a b \cos ^3(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},\cos ^2(c+d x)\right )}{m+2}-\frac{2 a b \cos (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m}{2},\frac{m+2}{2},\cos ^2(c+d x)\right )}{m}-\frac{8190 a b \cos (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2},\frac{m+2}{2},\cos ^2(c+d x)\right )}{m}+\frac{4095 b^2 \cos ^2(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\cos ^2(c+d x)\right )}{m+1}-\frac{b^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m-1}{2},\frac{m+1}{2},\cos ^2(c+d x)\right )}{m-1}-\frac{4095 b^2 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m-1}{2},\frac{m+1}{2},\cos ^2(c+d x)\right )}{m-1}\right )}{4096 d \sqrt{\sin ^2(c+d x)}} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.85, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( a\sin \left ( dx+c \right ) +b\tan \left ( dx+c \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) \tan \left (d x + c\right ) - b^{2} \tan \left (d x + c\right )^{2} - a^{2}\right )} \cos \left (d x + c\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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