Optimal. Leaf size=236 \[ -\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}+\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.286259, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}+\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3595
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{\sqrt{\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac{\int \frac{\frac{1}{2} a (i A-B)-\frac{1}{2} a (A-3 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} a (i A-B)-\frac{1}{2} a (A-3 i B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^2 d}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac{\left (\left (\frac{1}{4}+\frac{i}{4}\right ) (A-(2+i) B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a d}--\frac{\left (\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a d}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}--\frac{\left (\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}--\frac{\left (\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}--\frac{\left (\left (\frac{1}{8}-\frac{i}{8}\right ) (A+(2-i) B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a d}--\frac{\left (\left (\frac{1}{8}-\frac{i}{8}\right ) (A+(2-i) B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a d}\\ &=\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt{2} a d}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt{2} a d}+\frac{(i A-B) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}--\frac{\left (\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}-\frac{\left (\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}\\ &=-\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}+\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt{2} a d}-\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-(2+i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt{2} a d}+\frac{(i A-B) \sqrt{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.58294, size = 198, normalized size = 0.84 \[ \frac{(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left (4 (A+i B) \sin (c+d x) (\sin (d x)+i \cos (d x))+(1+i) (-\sin (c)+i \cos (c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left ((A+(2-i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+i (A-(2+i) B) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.061, size = 192, normalized size = 0.8 \begin{align*}{\frac{{\frac{i}{2}}B}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }\sqrt{\tan \left ( dx+c \right ) }}+{\frac{A}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }\sqrt{\tan \left ( dx+c \right ) }}-{\frac{2\,iB}{ad \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }+{\frac{A}{ad \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }-{\frac{iB}{ad \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) } \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 [B] time = 1.80759, size = 1477, normalized size = 6.26 \begin{align*} \text{result too large to display} \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*} \frac{\int \frac{A \sqrt{\tan{\left (c + d x \right )}}}{i \tan{\left (c + d x \right )} + 1}\, dx + \int \frac{B \tan ^{\frac{3}{2}}{\left (c + d x \right )}}{i \tan{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.24527, size = 131, normalized size = 0.56 \begin{align*} -\frac{\left (i - 1\right ) \, \sqrt{2} B \arctan \left (\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{2 \, a d} - \frac{\left (i - 1\right ) \, \sqrt{2}{\left (A - i \, B\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{4 \, a d} + \frac{A \sqrt{\tan \left (d x + c\right )} + i \, B \sqrt{\tan \left (d x + c\right )}}{2 \, a d{\left (\tan \left (d x + c\right ) - i\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]