Optimal. Leaf size=317 \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^3 d}-\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^3 d}+\frac{(2 i A+(1-i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}-\frac{(2 i A+(1-i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(2 B+i A) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.624298, antiderivative size = 317, 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 = {3595, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^3 d}-\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^3 d}+\frac{(2 i A+(1-i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}-\frac{(2 i A+(1-i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(-B+i A) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(2 B+i A) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3595
Rule 3596
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))^3} \, dx &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\frac{1}{2} a (i A-B)-\frac{1}{2} a (5 A-7 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(i A+2 B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}-\frac{\int \frac{3 i a^2 A-3 a^2 (A-2 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{24 a^4}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(i A+2 B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\int \frac{3 a^3 (2 i A+B)+3 i a^3 B \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(i A+2 B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{B \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 (2 i A+B)+3 i a^3 B x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(i A+2 B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\left (\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+i) A+B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^3 d}-\frac{(2 i A+(1-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{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(i A+2 B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\left (\left (\frac{1}{32}+\frac{i}{32}\right ) ((1+i) A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^3 d}-\frac{\left (\left (\frac{1}{32}+\frac{i}{32}\right ) ((1+i) A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^3 d}+\frac{(2 i A+(1-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{(2 i A+(1-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{(2 i A+(1-i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}-\frac{(2 i A+(1-i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(i A+2 B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}--\frac{\left (\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+i) A+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^3 d}-\frac{\left (\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+i) A+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^3 d}\\ &=\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+i) A+B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^3 d}-\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+i) A+B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^3 d}+\frac{(2 i A+(1-i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}-\frac{(2 i A+(1-i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(i A-B) \sqrt{\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac{(i A+2 B) \sqrt{\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac{B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 3.3741, size = 272, normalized size = 0.86 \[ \frac{e^{-4 i (c+d x)} \sec (c+d x) (\cos (3 (c+d x))-i \sin (3 (c+d x))) \left (\left (-2 e^{2 i (c+d x)}+e^{4 i (c+d x)}+2 e^{6 i (c+d x)}-1\right ) \left (A e^{2 i (c+d x)}+A-2 i B e^{2 i (c+d x)}+i B\right )-6 (A-i B) e^{6 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )-3 A e^{6 i (c+d x)} \sqrt{-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt{-1+e^{4 i (c+d x)}}\right )\right )}{96 a^3 d \sqrt{\tan (c+d x)}} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.066, size = 278, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{5\,B}{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}}A}{{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}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{A}{4\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{A}{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{A}{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}}B}{{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 = 1.61481, size = 1670, normalized size = 5.27 \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 = 1.21718, size = 177, normalized size = 0.56 \begin{align*} -\frac{\left (i + 1\right ) \, \sqrt{2} A \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 (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 )}{16 \, a^{3} d} - \frac{3 i \, B \tan \left (d x + c\right )^{\frac{5}{2}} + 2 i \, A \tan \left (d x + c\right )^{\frac{3}{2}} + 10 \, B \tan \left (d x + c\right )^{\frac{3}{2}} + 6 \, A \sqrt{\tan \left (d x + c\right )} - 3 i \, 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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