Integrand size = 33, antiderivative size = 187 \[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {(A+i B) \operatorname {AppellF1}\left (1+m,-\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {(A-i B) \operatorname {AppellF1}\left (1+m,-\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]
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Time = 0.50 (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} \[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {(A+i B) \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},1,m+2,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}+\frac {(A-i B) \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},1,m+2,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}} \]
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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) \sqrt {a+b \tan (c+d x)} \, dx+\frac {1}{2} (A+i B) \int (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 \sqrt {a+b x}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(A+i B) \text {Subst}\left (\int \frac {x^m \sqrt {a+b x}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {\left ((A-i B) \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \sqrt {1+\frac {b x}{a}}}{1-i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {\left ((A+i B) \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \sqrt {1+\frac {b x}{a}}}{1+i x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}} \\ & = \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 {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\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 {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \\ \end{align*}
\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx \]
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\[\int \tan \left (d x +c \right )^{m} \sqrt {a +b \tan \left (d x +c \right )}\, \left (A +B \tan \left (d x +c \right )\right )d x\]
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\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m} \,d x } \]
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\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{m}{\left (c + d x \right )}\, dx \]
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\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m} \,d x } \]
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\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
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