Optimal. Leaf size=315 \[ -\frac{((7-5 i) A-2 i B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}+\frac{((7-5 i) A-2 i B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^3 d}-\frac{((7+5 i) A-2 i B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{((7+5 i) A-2 i B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3} \]
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Rubi [A] time = 0.640067, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{((7-5 i) A-2 i B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}+\frac{((7-5 i) A-2 i B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^3 d}-\frac{((7+5 i) A-2 i B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{((7+5 i) A-2 i B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3} \]
Antiderivative was successfully verified.
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Rule 3596
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)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx &=\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\frac{1}{2} a (11 A-i B)-\frac{5}{2} a (i A-B) \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{3 a^2 (6 A-i B)-3 a^2 (4 i A-B) \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{24 a^4}\\ &=\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\int \frac{3 a^3 (7 A-2 i B)-15 i a^3 A \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 a^3 (7 A-2 i B)-15 i a^3 A x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{((7-5 i) A-2 i B) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^3 d}+\frac{((7+5 i) A-2 i B) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^3 d}\\ &=\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{((7-5 i) A-2 i B) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^3 d}+\frac{((7-5 i) A-2 i B) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^3 d}-\frac{((7+5 i) A-2 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^3 d}-\frac{((7+5 i) A-2 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^3 d}\\ &=-\frac{((7+5 i) A-2 i B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{((7+5 i) A-2 i B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{((7-5 i) A-2 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^3 d}-\frac{((7-5 i) A-2 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^3 d}\\ &=-\frac{((7-5 i) A-2 i B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}+\frac{((7-5 i) A-2 i B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{((7+5 i) A-2 i B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{((7+5 i) A-2 i B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(A+i B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(4 A+i B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 A \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 2.9499, size = 258, normalized size = 0.82 \[ \frac{\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (\frac{4}{3} \sin (c+d x) (\cos (3 d x)-i \sin (3 d x)) ((-B+19 i A) \sin (2 (c+d x))+3 (7 A+i B) \cos (2 (c+d x))+6 A+3 i B)+(-\sin (3 c)+i \cos (3 c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left ((2 B+(5+7 i) A) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((1+6 i) A+(1-i) B) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 323, normalized size = 1. \begin{align*}{\frac{-{\frac{5\,i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{19\,A}{12\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{\frac{i}{12}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{9\,i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{B}{4\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{{\frac{3\,i}{2}}A}{{a}^{3}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}{4\,{a}^{3}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}^{3}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}{4\,{a}^{3}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 = 1.95843, size = 1820, normalized size = 5.78 \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 = 1.20348, size = 185, normalized size = 0.59 \begin{align*} -\frac{\left (i + 1\right ) \, \sqrt{2}{\left (6 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 )}{16 \, a^{3} d} + \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 )}{16 \, a^{3} d} - \frac{15 i \, A \tan \left (d x + c\right )^{\frac{5}{2}} + 38 \, A \tan \left (d x + c\right )^{\frac{3}{2}} + 2 i \, B \tan \left (d x + c\right )^{\frac{3}{2}} - 27 i \, A \sqrt{\tan \left (d x + c\right )} + 6 \, B \sqrt{\tan \left (d x + c\right )}}{24 \, a^{3} d{\left (\tan \left (d x + c\right ) - i\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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