Integrand size = 33, antiderivative size = 187 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(A+i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}} \]
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Time = 0.49 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3684, 3683, 140, 138} \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(A+i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) \tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt {a+b \tan (c+d x)}} \]
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Rule 138
Rule 140
Rule 3683
Rule 3684
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} (A-i B) \int \frac {(1+i \tan (c+d x)) \tan ^m(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx+\frac {1}{2} (A+i B) \int \frac {(1-i \tan (c+d x)) \tan ^m(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {(A-i B) \text {Subst}\left (\int \frac {x^m}{(1-i x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(A+i B) \text {Subst}\left (\int \frac {x^m}{(1+i x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {\left ((A-i B) \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \text {Subst}\left (\int \frac {x^m}{(1-i x) \sqrt {1+\frac {b x}{a}}} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left ((A+i B) \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \text {Subst}\left (\int \frac {x^m}{(1+i x) \sqrt {1+\frac {b x}{a}}} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {a+b \tan (c+d x)}} \\ & = \frac {(A+i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}}+\frac {(A-i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}} \\ \end{align*}
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx \]
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\[\int \frac {\tan \left (d x +c \right )^{m} \left (A +B \tan \left (d x +c \right )\right )}{\sqrt {a +b \tan \left (d x +c \right )}}d x\]
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\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]
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\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]
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Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]
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