Optimal. Leaf size=127 \[ \frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (5 B+7 i A)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.257492, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3593, 3591, 3529, 3533, 205} \[ \frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (5 B+7 i A)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3593
Rule 3591
Rule 3529
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{7}{2}}(c+d x)} \, dx &=-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+i a \tan (c+d x)) \left (\frac{1}{2} a (7 i A+5 B)-\frac{1}{2} a (3 A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{-5 a^2 (A-i B)-5 a^2 (i A+B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{-5 a^2 (i A+B)+5 a^2 (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{\left (20 a^4 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-5 a^2 (i A+B)-5 a^2 (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt [4]{-1} a^2 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [B] time = 4.94616, size = 272, normalized size = 2.14 \[ \frac{\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (-\frac{4 i e^{-2 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac{(\cos (2 c)-i \sin (2 c)) \csc ^2(c+d x) (5 (B+2 i A) \sin (2 (c+d x))+(33 A-30 i B) \cos (2 (c+d x))-27 A+30 i B)}{15 \sqrt{\tan (c+d x)}}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.018, size = 537, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A] time = 1.82405, size = 265, normalized size = 2.09 \begin{align*} -\frac{15 \,{\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 ({\left (30 \, A - 30 i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 5 \,{\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] time = 1.92264, size = 1359, normalized size = 10.7 \begin{align*} -\frac{15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) - 15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) -{\left ({\left (344 i \, A + 280 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-88 i \, A - 200 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-248 i \, A - 280 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (184 i \, A + 200 \, 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}}}{60 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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(-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.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.25522, size = 146, normalized size = 1.15 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{2}{\left (8 i \, A a^{2} + 8 \, 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 )}{4 \, d} + \frac{60 \, A a^{2} \tan \left (d x + c\right )^{2} - 60 i \, B a^{2} \tan \left (d x + c\right )^{2} - 20 i \, A a^{2} \tan \left (d x + c\right ) - 10 \, B a^{2} \tan \left (d x + c\right ) - 6 \, A a^{2}}{15 \, d \tan \left (d x + c\right )^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]