Optimal. Leaf size=277 \[ \frac{((1+3 i) A+(9+5 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}-\frac{((1+3 i) A+(9+5 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^2 d}+\frac{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{((1-3 i) A-(9-5 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}-\frac{((1-3 i) A-(9-5 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}+\frac{(-B+i A) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.500545, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ \frac{((1+3 i) A+(9+5 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^2 d}-\frac{((1+3 i) A+(9+5 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^2 d}+\frac{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{((1-3 i) A-(9-5 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}-\frac{((1-3 i) A-(9-5 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}+\frac{(-B+i A) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3595
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\sqrt{\tan (c+d x)} \left (\frac{3}{2} a (i A-B)-\frac{1}{2} a (A-7 i B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{-\frac{1}{2} a^2 (A+5 i B)-\frac{3}{2} a^2 (i A+3 B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} a^2 (A+5 i B)-\frac{3}{2} a^2 (i A+3 B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 a^4 d}\\ &=\frac{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\left (\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+2 i) A+(2-7 i) B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^2 d}-\frac{((1+3 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{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{((1-3 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{((1-3 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{((1+3 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{((1+3 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{((1-3 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{((1-3 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{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{((1+3 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{((1+3 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{((1+3 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{((1+3 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{((1-3 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{((1-3 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{(A+5 i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [A] time = 2.35207, size = 243, normalized size = 0.88 \[ \frac{\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (4 \sin (c+d x) (\sin (2 d x)+i \cos (2 d x)) ((3 A+7 i B) \sin (c+d x)+(5 B-i A) \cos (c+d x))-(1+i) (-\sin (2 c)+i \cos (2 c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left (((2+7 i) B-(1-2 i) A) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((7+2 i) B-(2-i) A) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.05, size = 294, normalized size = 1.1 \begin{align*}{\frac{7\,B}{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{3\,i}{8}}A}{{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}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{A}{8\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{7\,B}{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{{\frac{i}{4}}A}{{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{B}{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}}A}{{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 ) } \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 = 1.94814, size = 1754, normalized size = 6.33 \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 = 1.23344, size = 166, normalized size = 0.6 \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 (A - 7 i \, 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{3 i \, A \tan \left (d x + c\right )^{\frac{3}{2}} - 7 \, B \tan \left (d x + c\right )^{\frac{3}{2}} + A \sqrt{\tan \left (d x + c\right )} + 5 i \, 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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