Optimal. Leaf size=214 \[ \frac{(2 m+1) (-B+i A) \sqrt{a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \text{Hypergeometric2F1}\left (\frac{1}{2},-m,\frac{3}{2},1+i \tan (c+d x)\right )}{a d}+\frac{(A-i B) \sqrt{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{2 d (m+1) \sqrt{a+i a \tan (c+d x)}}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.625167, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3596, 3601, 3564, 135, 133, 3599, 67, 65} \[ \frac{(A-i B) \sqrt{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{2 d (m+1) \sqrt{a+i a \tan (c+d x)}}+\frac{(2 m+1) (-B+i A) \sqrt{a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};i \tan (c+d x)+1\right )}{a d}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3596
Rule 3601
Rule 3564
Rule 135
Rule 133
Rule 3599
Rule 67
Rule 65
Rubi steps
\begin{align*} \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-a (A m+i B (1+m))+\frac{1}{2} a (i A-B) (1+2 m) \tan (c+d x)\right ) \, dx}{a^2}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(A-i B) \int \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx}{2 a}-\frac{((A+i B) (1+2 m)) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx}{2 a^2}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(a (i A+B)) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a}\right )^m}{\sqrt{a+x} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{2 d}-\frac{((A+i B) (1+2 m)) \operatorname{Subst}\left (\int \frac{x^m}{\sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\left ((A+i B) (1+2 m) (-i \tan (c+d x))^{-m} \tan ^m(c+d x)\right ) \operatorname{Subst}\left (\int \frac{(-i x)^m}{\sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (a (i A+B) \sqrt{1+i \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a}\right )^m}{\sqrt{1+\frac{x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{2 d \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(A-i B) F_1\left (1+m;\frac{1}{2},1;2+m;-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt{1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{2 d (1+m) \sqrt{a+i a \tan (c+d x)}}+\frac{(i A-B) (1+2 m) \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d}\\ \end{align*}
Mathematica [F] time = 180.003, size = 0, normalized size = 0. \[ \text{\$Aborted} \]
Verification is Not applicable to the result.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.582, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ){\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2}{\left ({\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]