Integrand size = 31, antiderivative size = 185 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {(a A+b B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{\left (a^2+b^2\right ) d (1+m)}+\frac {b (A b-a B) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (1+m)}-\frac {(A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{\left (a^2+b^2\right ) d (2+m)} \]
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Time = 0.36 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3694, 3619, 3557, 371, 3715, 66} \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {b (A b-a B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}+\frac {(a A+b B) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )}-\frac {(A b-a B) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )} \]
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Rule 66
Rule 371
Rule 3557
Rule 3619
Rule 3694
Rule 3715
Rubi steps \begin{align*} \text {integral}& = \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)) \text {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) \operatorname {Hypergeometric2F1}\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) \text {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) \text {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) \operatorname {Hypergeometric2F1}\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) \operatorname {Hypergeometric2F1}\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 {Hypergeometric2F1}\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*}
Time = 1.01 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\tan ^{1+m}(c+d x) \left ((a A+b B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )+\frac {(A b-a B) \left (b (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b \tan (c+d x)}{a}\right )-a (1+m) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)\right )}{a (2+m)}\right )}{\left (a^2+b^2\right ) d (1+m)} \]
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\[\int \frac {\tan \left (d x +c \right )^{m} \left (A +B \tan \left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}d x\]
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\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]
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\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
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