Optimal. Leaf size=158 \[ \frac{2 i B (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},-i \tan (c+d x)\right )}{d \sqrt{\cot (c+d x)}}+\frac{2 (A-i B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt{\cot (c+d x)}} \]
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Rubi [A] time = 0.391409, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4241, 3601, 3564, 130, 430, 429, 3599, 66, 64} \[ \frac{2 (A-i B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt{\cot (c+d x)}}+\frac{2 i B (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right )}{d \sqrt{\cot (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 4241
Rule 3601
Rule 3564
Rule 130
Rule 430
Rule 429
Rule 3599
Rule 66
Rule 64
Rubi steps
\begin{align*} \int \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\left (\left ((-A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx\right )+\frac{\left (i B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx}{a}\\ &=-\frac{\left (i a^2 (-A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+n}}{\sqrt{-\frac{i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac{\left (i a B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-1+n}}{\sqrt{x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2 a^3 (-A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\left (a+i a x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left (i B \sqrt{\cot (c+d x)} (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(1+i x)^{-1+n}}{\sqrt{x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{2 i B \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt{\cot (c+d x)}}+\frac{\left (2 a^2 (-A+i B) \sqrt{\cot (c+d x)} (1+i \tan (c+d x))^{-n} \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{\left (1+i x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{2 (A-i B) F_1\left (\frac{1}{2};1-n,1;\frac{3}{2};-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt{\cot (c+d x)}}+\frac{2 i B \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt{\cot (c+d x)}}\\ \end{align*}
Mathematica [F] time = 21.6215, size = 0, normalized size = 0. \[ \int \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.412, size = 0, normalized size = 0. \begin{align*} \int \sqrt{\cot \left ( dx+c \right ) } \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) \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{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sqrt{\cot \left (d x + c\right )}\,{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 ({\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sqrt{\cot \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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