Optimal. Leaf size=393 \[ -\frac{3 (-5 B+2 i A) \tan ^{\frac{5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^3 d}-\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^3 d}+\frac{15 (-5 B+2 i A) \sqrt{\tan (c+d x)}}{8 a^3 d}-\frac{((28-30 i) A+(75+77 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}-\frac{\left (\frac{1}{32}+\frac{i}{32}\right ) ((1+29 i) A-(76+i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^3 d}+\frac{(-B+i A) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.82717, antiderivative size = 393, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3595, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{3 (-5 B+2 i A) \tan ^{\frac{5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^3 d}-\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^3 d}+\frac{15 (-5 B+2 i A) \sqrt{\tan (c+d x)}}{8 a^3 d}-\frac{((28-30 i) A+(75+77 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}-\frac{\left (\frac{1}{32}+\frac{i}{32}\right ) ((1+29 i) A-(76+i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^3 d}+\frac{(-B+i A) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3595
Rule 3528
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\tan ^{\frac{7}{2}}(c+d x) \left (\frac{9}{2} a (i A-B)+\frac{3}{2} a (A+5 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\tan ^{\frac{5}{2}}(c+d x) \left (-21 a^2 (A+2 i B)+3 a^2 (5 i A-16 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\int \tan ^{\frac{3}{2}}(c+d x) \left (-45 a^3 (2 i A-5 B)-21 a^3 (4 A+11 i B) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\int \sqrt{\tan (c+d x)} \left (21 a^3 (4 A+11 i B)-45 a^3 (2 i A-5 B) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac{15 (2 i A-5 B) \sqrt{\tan (c+d x)}}{8 a^3 d}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\int \frac{45 a^3 (2 i A-5 B)+21 a^3 (4 A+11 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac{15 (2 i A-5 B) \sqrt{\tan (c+d x)}}{8 a^3 d}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{45 a^3 (2 i A-5 B)+21 a^3 (4 A+11 i B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac{15 (2 i A-5 B) \sqrt{\tan (c+d x)}}{8 a^3 d}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(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 ) ((29+i) A+(1+76 i) 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{((28-30 i) A+(75+77 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{15 (2 i A-5 B) \sqrt{\tan (c+d x)}}{8 a^3 d}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(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 ) ((29+i) A+(1+76 i) 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 ) ((29+i) A+(1+76 i) 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{((28-30 i) A+(75+77 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{((28-30 i) A+(75+77 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{((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{((28-30 i) A+(75+77 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{15 (2 i A-5 B) \sqrt{\tan (c+d x)}}{8 a^3 d}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(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 ) ((29+i) A+(1+76 i) 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 ) ((29+i) A+(1+76 i) 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 ) ((29+i) A+(1+76 i) 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 ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^3 d}-\frac{((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{((28-30 i) A+(75+77 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{15 (2 i A-5 B) \sqrt{\tan (c+d x)}}{8 a^3 d}+\frac{7 (4 A+11 i B) \tan ^{\frac{3}{2}}(c+d x)}{24 a^3 d}+\frac{(i A-B) \tan ^{\frac{9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+2 i B) \tan ^{\frac{7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac{3 (2 i A-5 B) \tan ^{\frac{5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 4.99308, size = 300, normalized size = 0.76 \[ \frac{\sec ^3(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (3 (-\sin (3 c)+i \cos (3 c)) \sqrt{\sin (2 (c+d x))} \left (((30-28 i) A+(77+75 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+(1+i) ((1-76 i) B-(29-i) A) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )+\tan (c+d x) (\sin (3 d x)+i \cos (3 d x)) (2 (90 A+241 i B) \cos (2 (c+d x))+(147 A+349 i B) \cos (4 (c+d x))+194 i A \sin (2 (c+d x))+145 i A \sin (4 (c+d x))+33 A-502 B \sin (2 (c+d x))-347 B \sin (4 (c+d x))+69 i B)\right )}{96 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.057, size = 404, normalized size = 1. \begin{align*}{\frac{{\frac{2\,i}{3}}B}{{a}^{3}d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-6\,{\frac{B\sqrt{\tan \left ( dx+c \right ) }}{{a}^{3}d}}+{\frac{2\,iA}{{a}^{3}d}\sqrt{\tan \left ( dx+c \right ) }}+{\frac{{\frac{35\,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\,A}{2\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{91\,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{49\,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{7\,A}{4\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{{\frac{27\,i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{29\,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{19\,iB}{{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 = 1.98916, size = 2095, normalized size = 5.33 \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 = 1.34413, size = 284, normalized size = 0.72 \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 )}{16 \, a^{3} d} + \frac{\left (i - 1\right ) \, \sqrt{2}{\left (-29 i \, A + 76 \, 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{60 \, A \tan \left (d x + c\right )^{\frac{5}{2}} + 105 i \, B \tan \left (d x + c\right )^{\frac{5}{2}} - 98 i \, A \tan \left (d x + c\right )^{\frac{3}{2}} + 182 \, B \tan \left (d x + c\right )^{\frac{3}{2}} - 42 \, A \sqrt{\tan \left (d x + c\right )} - 81 i \, B \sqrt{\tan \left (d x + c\right )}}{24 \, a^{3} d{\left (\tan \left (d x + c\right ) - i\right )}^{3}} - \frac{-2 i \, B a^{6} d^{2} \tan \left (d x + c\right )^{\frac{3}{2}} - 6 i \, A a^{6} d^{2} \sqrt{\tan \left (d x + c\right )} + 18 \, B a^{6} d^{2} \sqrt{\tan \left (d x + c\right )}}{3 \, a^{9} d^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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