Optimal. Leaf size=296 \[ \frac{((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}-\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) ((6+i) A+(1+4 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (-B+i A)}{2 a d \sqrt{\tan (c+d x)}}+\frac{((7+5 i) A-(5-3 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{8 \sqrt{2} a d}+\frac{((5-3 i) B-(7+5 i) A) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{8 \sqrt{2} a d} \]
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Rubi [A] time = 0.401374, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}-\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) ((6+i) A+(1+4 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (-B+i A)}{2 a d \sqrt{\tan (c+d x)}}+\frac{((7+5 i) A-(5-3 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{8 \sqrt{2} a d}+\frac{((5-3 i) B-(7+5 i) A) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{8 \sqrt{2} a d} \]
Antiderivative was successfully verified.
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Rule 3596
Rule 3529
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx &=\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac{\int \frac{\frac{1}{2} a (7 A+3 i B)-\frac{5}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{2 a^2}\\ &=-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac{\int \frac{-\frac{5}{2} a (i A-B)-\frac{1}{2} a (7 A+3 i B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}\\ &=-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (i A-B)}{2 a d \sqrt{\tan (c+d x)}}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac{\int \frac{-\frac{1}{2} a (7 A+3 i B)+\frac{5}{2} a (i A-B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (i A-B)}{2 a d \sqrt{\tan (c+d x)}}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} a (7 A+3 i B)+\frac{5}{2} a (i A-B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^2 d}\\ &=-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (i A-B)}{2 a d \sqrt{\tan (c+d x)}}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}-\frac{((7+5 i) A-(5-3 i) B) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 a d}-\frac{((7-5 i) A+(5+3 i) B) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 a d}\\ &=-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (i A-B)}{2 a d \sqrt{\tan (c+d x)}}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac{((7+5 i) A-(5-3 i) B) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 \sqrt{2} a d}+\frac{((7+5 i) A-(5-3 i) B) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 \sqrt{2} a d}-\frac{((7-5 i) A+(5+3 i) B) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 a d}-\frac{((7-5 i) A+(5+3 i) B) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 a d}\\ &=\frac{((7+5 i) A-(5-3 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt{2} a d}-\frac{((7+5 i) A-(5-3 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt{2} a d}-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (i A-B)}{2 a d \sqrt{\tan (c+d x)}}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}-\frac{((7-5 i) A+(5+3 i) B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}+\frac{((7-5 i) A+(5+3 i) B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}\\ &=\frac{((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}-\frac{((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}+\frac{((7+5 i) A-(5-3 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt{2} a d}-\frac{((7+5 i) A-(5-3 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt{2} a d}-\frac{7 A+3 i B}{6 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{5 (i A-B)}{2 a d \sqrt{\tan (c+d x)}}+\frac{A+i B}{2 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.72783, size = 241, normalized size = 0.81 \[ \frac{(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left (\frac{2}{3} \csc (c+d x) (\cos (d x)-i \sin (d x)) ((-12 B+8 i A) \sin (2 (c+d x))+(11 A+15 i B) \cos (2 (c+d x))-19 A-15 i B)+(1-i) (\cos (c)+i \sin (c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left (((6+i) A+(1+4 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((4+i) B-(1+6 i) A) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.049, size = 289, normalized size = 1. \begin{align*}{\frac{{\frac{i}{2}}A}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }\sqrt{\tan \left ( dx+c \right ) }}-{\frac{B}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }\sqrt{\tan \left ( dx+c \right ) }}-4\,{\frac{B}{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{6\,iA}{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{iA}{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{B}{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{2\,iA}{ad}{\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}}-2\,{\frac{B}{ad\sqrt{\tan \left ( dx+c \right ) }}}-{\frac{2\,A}{3\,ad} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Fricas [B] time = 2.08418, size = 2156, normalized size = 7.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Giac [A] time = 1.25524, size = 190, normalized size = 0.64 \begin{align*} \frac{\left (i - 1\right ) \, \sqrt{2}{\left (i \, A + 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{\left (i - 1\right ) \, \sqrt{2}{\left (6 \, A + 4 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{-i \, A \sqrt{\tan \left (d x + c\right )} + B \sqrt{\tan \left (d x + c\right )}}{2 \, a d{\left (\tan \left (d x + c\right ) - i\right )}} + \frac{i \,{\left (6 \, A \tan \left (d x + c\right ) + 6 i \, B \tan \left (d x + c\right ) + 2 i \, A\right )}}{3 \, a d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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