Optimal. Leaf size=238 \[ -\frac{5 c^3 (-11 B+3 i A) \sqrt{c-i c \tan (e+f x)}}{4 a^2 f}-\frac{5 c^2 (-11 B+3 i A) (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{5 c^{7/2} (-11 B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{2 \sqrt{2} a^2 f}-\frac{c (-11 B+3 i A) (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
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Rubi [A] time = 0.272166, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {3588, 78, 47, 50, 63, 208} \[ -\frac{5 c^3 (-11 B+3 i A) \sqrt{c-i c \tan (e+f x)}}{4 a^2 f}-\frac{5 c^2 (-11 B+3 i A) (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{5 c^{7/2} (-11 B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{2 \sqrt{2} a^2 f}-\frac{c (-11 B+3 i A) (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 78
Rule 47
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{5/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac{((3 A+11 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{5/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{\left (5 (3 A+11 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=-\frac{5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac{(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{\left (5 (3 A+11 i B) c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{8 a f}\\ &=-\frac{5 (3 i A-11 B) c^3 \sqrt{c-i c \tan (e+f x)}}{4 a^2 f}-\frac{5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac{(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{\left (5 (3 A+11 i B) c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=-\frac{5 (3 i A-11 B) c^3 \sqrt{c-i c \tan (e+f x)}}{4 a^2 f}-\frac{5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac{(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{\left (5 (3 i A-11 B) c^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{2 a f}\\ &=\frac{5 (3 i A-11 B) c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{2 \sqrt{2} a^2 f}-\frac{5 (3 i A-11 B) c^3 \sqrt{c-i c \tan (e+f x)}}{4 a^2 f}-\frac{5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac{(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2}\\ \end{align*}
Mathematica [F] time = 180.005, size = 0, normalized size = 0. \[ \text{\$Aborted} \]
Verification is Not applicable to the result.
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Maple [A] time = 0.112, size = 179, normalized size = 0.8 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ({\frac{i}{3}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+5\,iBc\sqrt{c-ic\tan \left ( fx+e \right ) }+Ac\sqrt{c-ic\tan \left ( fx+e \right ) }+2\,{c}^{2} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( -{\frac{17\,i}{8}}B-{\frac{9\,A}{8}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{3/2}+ \left ({\frac{15\,i}{4}}Bc+7/4\,Ac \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 55\,iB+15\,A \right ) \sqrt{2}}{16\,\sqrt{c}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c-ic\tan \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) } \right ) \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: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.64429, size = 1215, normalized size = 5.11 \begin{align*} \frac{3 \, \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt{-\frac{{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \log \left (\frac{{\left ({\left (15 i \, A - 55 \, B\right )} c^{4} + \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt{-\frac{{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) - 3 \, \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt{-\frac{{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \log \left (\frac{{\left ({\left (15 i \, A - 55 \, B\right )} c^{4} - \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt{-\frac{{\left (225 \, A^{2} + 1650 i \, A B - 3025 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) + \sqrt{2}{\left ({\left (-45 i \, A + 165 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-60 i \, A + 220 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-9 i \, A + 33 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (6 i \, A - 6 \, B\right )} c^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \,{\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \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: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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