Optimal. Leaf size=226 \[ \frac{(1-m) (A (1-m)-i B (m+1)) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{4 a^2 d (m+1)}+\frac{m (B m+i A (2-m)) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{4 a^2 d (m+2)}+\frac{(A (2-m)-i B m) \tan ^{m+1}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.483971, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3596, 3538, 3476, 364} \[ \frac{(1-m) (A (1-m)-i B (m+1)) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{4 a^2 d (m+1)}+\frac{m (B m+i A (2-m)) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{4 a^2 d (m+2)}+\frac{(A (2-m)-i B m) \tan ^{m+1}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3596
Rule 3538
Rule 3476
Rule 364
Rubi steps
\begin{align*} \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\tan ^m(c+d x) (a (A (3-m)-i B (1+m))-a (i A-B) (1-m) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan ^m(c+d x) \left (2 a^2 (1-m) (A (1-m)-i B (1+m))+2 a^2 m (i A (2-m)+B m) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{(m (i A (2-m)+B m)) \int \tan ^{1+m}(c+d x) \, dx}{4 a^2}+\frac{((1-m) (A (1-m)-i B (1+m))) \int \tan ^m(c+d x) \, dx}{4 a^2}\\ &=\frac{(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{(m (i A (2-m)+B m)) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{4 a^2 d}+\frac{((1-m) (A (1-m)-i B (1+m))) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{4 a^2 d}\\ &=\frac{(1-m) (A (1-m)-i B (1+m)) \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{4 a^2 d (1+m)}+\frac{(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac{m (i A (2-m)+B m) \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{4 a^2 d (2+m)}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [B] time = 8.27824, size = 565, normalized size = 2.5 \[ -\frac{i e^{-2 i c} \left (-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-m} \sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (\frac{e^{4 i c} 2^{1-m} \left (A \left (2 m^2-4 m+1\right )+i B \left (2 m^2-1\right )\right ) \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (\left (1+e^{2 i (c+d x)}\right )^m \left ((m+1) \text{Hypergeometric2F1}\left (m,m,m+1,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+m \left (-1+e^{2 i (c+d x)}\right ) \text{Hypergeometric2F1}\left (m,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right )-2^m (m+1) \text{Hypergeometric2F1}\left (1,m,m+1,\frac{1-e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right )}{m (m+1)}+\frac{e^{4 i c} 2^{2-m} (A (2 m-3)+i (2 B m+B)) \left (-1+e^{2 i (c+d x)}\right )^{m+1} \text{Hypergeometric2F1}\left (m-1,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )}{m+1}+(A+i B) e^{-4 i d x} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{2-m}+(A (3-2 m)-i (2 B m+B)) e^{2 i (c-d x)} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{2-m}\right )}{16 d (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 1.386, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ) }{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{2}}}\, 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{{\left ({\left (A - i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A 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} e^{\left (-4 i \, d x - 4 i \, c\right )}}{4 \, a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]