Optimal. Leaf size=185 \[ \frac{(a A+b B) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )}+\frac{b (A b-a B) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}-\frac{(A b-a B) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )} \]
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Rubi [A] time = 0.312891, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3613, 3538, 3476, 364, 3634, 64} \[ \frac{(a A+b B) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )}+\frac{b (A b-a B) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac{b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}-\frac{(A b-a B) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Rule 3613
Rule 3538
Rule 3476
Rule 364
Rule 3634
Rule 64
Rubi steps
\begin{align*} \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=\frac{\int \tan ^m(c+d x) (a A+b B-(A b-a B) \tan (c+d x)) \, dx}{a^2+b^2}+\frac{(b (A b-a B)) \int \frac{\tan ^m(c+d x) \left (1+\tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac{(A b-a B) \int \tan ^{1+m}(c+d x) \, dx}{a^2+b^2}+\frac{(a A+b B) \int \tan ^m(c+d x) \, dx}{a^2+b^2}+\frac{(b (A b-a B)) \operatorname{Subst}\left (\int \frac{x^m}{a+b x} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{b (A b-a B) \, _2F_1\left (1,1+m;2+m;-\frac{b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (1+m)}-\frac{(A b-a B) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}+\frac{(a A+b B) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{(a A+b B) \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{\left (a^2+b^2\right ) d (1+m)}+\frac{b (A b-a B) \, _2F_1\left (1,1+m;2+m;-\frac{b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (1+m)}-\frac{(A b-a B) \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{\left (a^2+b^2\right ) d (2+m)}\\ \end{align*}
Mathematica [A] time = 0.864652, size = 144, normalized size = 0.78 \[ \frac{\tan ^{m+1}(c+d x) \left ((a A+b B) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )+\frac{(A b-a B) \left (b (m+2) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{b \tan (c+d x)}{a}\right )-a (m+1) \tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )\right )}{a (m+2)}\right )}{d (m+1) \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.323, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ) }{a+b\tan \left ( dx+c \right ) }}\, 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 \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}\,{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 (\frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{a + b \tan{\left (c + d x \right )}}\, dx \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 \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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