Integrand size = 33, antiderivative size = 193 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^2 (A+i B) \operatorname {AppellF1}\left (1+m,-\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {a^2 (A-i B) \operatorname {AppellF1}\left (1+m,-\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]
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Time = 0.73 (sec) , antiderivative size = 193, 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} \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^2 (A+i B) \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {5}{2},1,m+2,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}+\frac {a^2 (A-i B) \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {5}{2},1,m+2,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}} \]
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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 (1+i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} \, dx+\frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} \, dx \\ & = \frac {(A-i B) \text {Subst}\left (\int \frac {x^m (a+b x)^{5/2}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(A+i B) \text {Subst}\left (\int \frac {x^m (a+b x)^{5/2}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {\left (a^2 (A-i B) \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \left (1+\frac {b x}{a}\right )^{5/2}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {\left (a^2 (A+i B) \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \left (1+\frac {b x}{a}\right )^{5/2}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}} \\ & = \frac {a^2 (A+i B) \operatorname {AppellF1}\left (1+m,-\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {a^2 (A-i B) \operatorname {AppellF1}\left (1+m,-\frac {5}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \\ \end{align*}
\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx \]
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\[\int \tan \left (d x +c \right )^{m} \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}} \left (A +B \tan \left (d x +c \right )\right )d x\]
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\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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Timed out. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]
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