Optimal. Leaf size=102 \[ \frac{4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (3 B+5 i A)}{3 d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.217649, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3593, 3591, 3533, 205} \[ \frac{4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (3 B+5 i A)}{3 d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3593
Rule 3591
Rule 3533
Rule 205
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{(a+i a \tan (c+d x)) \left (\frac{1}{2} a (5 i A+3 B)-\frac{1}{2} a (A-3 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (5 i A+3 B)}{3 d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{-3 a^2 (A-i B)-3 a^2 (i A+B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (5 i A+3 B)}{3 d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\left (12 a^4 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-3 a^2 (A-i B)+3 a^2 (i A+B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (5 i A+3 B)}{3 d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 3.18703, size = 96, normalized size = 0.94 \[ -\frac{2 a^2 \left (-6 i (A-i B) \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+A \cot (c+d x)+6 i A+3 B\right )}{3 d \sqrt{\tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.017, size = 504, normalized size = 4.9 \begin{align*} -{\frac{2\,{a}^{2}A}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,i{a}^{2}A}{d}{\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}}-2\,{\frac{{a}^{2}B}{d\sqrt{\tan \left ( dx+c \right ) }}}+{\frac{iB{a}^{2}\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{{\frac{i}{2}}{a}^{2}B\sqrt{2}}{d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{iB{a}^{2}\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{a}^{2}A\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{a}^{2}A\sqrt{2}}{2\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{{a}^{2}A\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{\frac{i}{2}}{a}^{2}A\sqrt{2}}{d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{iA{a}^{2}\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{iA{a}^{2}\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{a}^{2}B\sqrt{2}}{2\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{{a}^{2}B\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{a}^{2}B\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [B] time = 2.48213, size = 239, normalized size = 2.34 \begin{align*} \frac{3 \,{\left (2 \, \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2} + \frac{4 \,{\left (3 \,{\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] time = 1.80269, size = 1181, normalized size = 11.58 \begin{align*} -\frac{3 \, \sqrt{\frac{{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 3 \, \sqrt{\frac{{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 8 \,{\left ({\left (7 \, A - 3 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (5 \, A - 3 i \, B\right )} a^{2}\right )} \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}}}{12 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\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*} a^{2} \left (\int \frac{A}{\tan ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx + \int - \frac{A}{\sqrt{\tan{\left (c + d x \right )}}}\, dx + \int \frac{B}{\tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int - B \sqrt{\tan{\left (c + d x \right )}}\, dx + \int \frac{2 i A}{\tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{2 i B}{\sqrt{\tan{\left (c + d x \right )}}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.27096, size = 108, normalized size = 1.06 \begin{align*} -\frac{\left (2 i - 2\right ) \, \sqrt{2}{\left (-i \, A a^{2} - B a^{2}\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{d} - \frac{12 i \, A a^{2} \tan \left (d x + c\right ) + 6 \, B a^{2} \tan \left (d x + c\right ) + 2 \, A a^{2}}{3 \, d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]