Optimal. Leaf size=318 \[ \frac{((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}-\frac{\left (\frac{1}{16}-\frac{i}{16}\right ) ((2+23 i) A-(7+2 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{32}-\frac{i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{1}{32}-\frac{i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.578141, antiderivative size = 318, 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} \[ \frac{((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}-\frac{\left (\frac{1}{16}-\frac{i}{16}\right ) ((2+23 i) A-(7+2 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{32}-\frac{i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{1}{32}-\frac{i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2} \]
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{3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx &=\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a (9 A+i B)-\frac{5}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2}\\ &=\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac{\int \frac{\frac{5}{2} a^2 (5 A+i B)-\frac{3}{2} a^2 (7 i A-3 B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac{\int \frac{-\frac{3}{2} a^2 (7 i A-3 B)-\frac{5}{2} a^2 (5 A+i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{2} a^2 (7 i A-3 B)-\frac{5}{2} a^2 (5 A+i B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 a^4 d}\\ &=-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac{((25+21 i) A-(9-5 i) B) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^2 d}+\frac{((25-21 i) A+(9+5 i) B) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^2 d}\\ &=-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac{((25+21 i) A-(9-5 i) B) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^2 d}-\frac{((25+21 i) A-(9-5 i) B) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^2 d}-\frac{((25-21 i) A+(9+5 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 )}{32 \sqrt{2} a^2 d}-\frac{((25-21 i) A+(9+5 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 )}{32 \sqrt{2} a^2 d}\\ &=-\frac{((25-21 i) A+(9+5 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^2 d}+\frac{((25-21 i) A+(9+5 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^2 d}-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac{((25+21 i) A-(9-5 i) B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}+\frac{((25+21 i) A-(9-5 i) B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}\\ &=\frac{((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}-\frac{((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}-\frac{((25-21 i) A+(9+5 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^2 d}+\frac{((25-21 i) A+(9+5 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^2 d}-\frac{5 (5 A+i B)}{8 a^2 d \sqrt{\tan (c+d x)}}+\frac{7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt{\tan (c+d x)}}+\frac{A+i B}{4 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [A] time = 2.38228, size = 250, normalized size = 0.79 \[ \frac{\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left ((-2 \cos (2 d x)+2 i \sin (2 d x)) ((-7 B+43 i A) \sin (2 (c+d x))+(41 A+5 i B) \cos (2 (c+d x))-9 A-5 i B)+(-\sin (2 c)+i \cos (2 c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left (((21-25 i) A+(5+9 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((23+2 i) A+(2+7 i) B) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.055, size = 311, normalized size = 1. \begin{align*} -{\frac{9\,A}{8\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{\frac{5\,i}{8}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{11\,i}{8}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{7\,B}{8\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{{\frac{7\,i}{4}}B}{{a}^{2}d \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }-{\frac{23\,A}{4\,{a}^{2}d \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}{2\,{a}^{2}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }+{\frac{{\frac{i}{2}}B}{{a}^{2}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }-2\,{\frac{A}{{a}^{2}d\sqrt{\tan \left ( dx+c \right ) }}} \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 = 2.37269, size = 2009, normalized size = 6.32 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.22138, size = 193, normalized size = 0.61 \begin{align*} \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 )}{8 \, a^{2} d} + \frac{\left (i - 1\right ) \, \sqrt{2}{\left (-23 i \, A + 7 \, B\right )} \arctan \left (-\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} - \frac{2 \, A}{a^{2} d \sqrt{\tan \left (d x + c\right )}} - \frac{9 \, A \tan \left (d x + c\right )^{\frac{3}{2}} + 5 i \, B \tan \left (d x + c\right )^{\frac{3}{2}} - 11 i \, A \sqrt{\tan \left (d x + c\right )} + 7 \, B \sqrt{\tan \left (d x + c\right )}}{8 \, a^{2} d{\left (\tan \left (d x + c\right ) - i\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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