Optimal. Leaf size=343 \[ \frac{7 c^{9/2} (-7 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac{7 c^4 (-7 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^3 f}+\frac{7 c^3 (-7 B+2 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac{14 c^2 (-7 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 c (-7 B+2 i A) (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.380989, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3588, 78, 47, 50, 63, 217, 203} \[ \frac{7 c^{9/2} (-7 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac{7 c^4 (-7 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^3 f}+\frac{7 c^3 (-7 B+2 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac{14 c^2 (-7 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 c (-7 B+2 i A) (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 78
Rule 47
Rule 50
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{7/2}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{((2 A+7 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{7/2}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{\left (7 (2 A+7 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{5/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a f}\\ &=\frac{14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left (7 (2 A+7 i B) c^3\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{3 a^2 f}\\ &=\frac{7 (2 i A-7 B) c^3 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac{14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left (7 (2 A+7 i B) c^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=\frac{7 (2 i A-7 B) c^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^3 f}+\frac{7 (2 i A-7 B) c^3 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac{14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left (7 (2 A+7 i B) c^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=\frac{7 (2 i A-7 B) c^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^3 f}+\frac{7 (2 i A-7 B) c^3 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac{14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{\left (7 (2 i A-7 B) c^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c-\frac{c x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{a^3 f}\\ &=\frac{7 (2 i A-7 B) c^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^3 f}+\frac{7 (2 i A-7 B) c^3 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac{14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{\left (7 (2 i A-7 B) c^5\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c-i c \tan (e+f x)}}\right )}{a^3 f}\\ &=\frac{7 (2 i A-7 B) c^{9/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac{7 (2 i A-7 B) c^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a^3 f}+\frac{7 (2 i A-7 B) c^3 \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac{14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 15.1775, size = 247, normalized size = 0.72 \[ -\frac{\sqrt{2} c^4 e^{-4 i (e+f x)} \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \left (105 (7 B-2 i A) e^{5 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2 \tan ^{-1}\left (e^{i (e+f x)}\right )-2 i A \left (-8 e^{2 i (e+f x)}+56 e^{4 i (e+f x)}+175 e^{6 i (e+f x)}+105 e^{8 i (e+f x)}+6\right )+B \left (-56 e^{2 i (e+f x)}+392 e^{4 i (e+f x)}+1225 e^{6 i (e+f x)}+735 e^{8 i (e+f x)}+12\right )\right )}{15 a^2 f \left (1+e^{2 i (e+f x)}\right )^2 \sqrt{a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.125, size = 899, normalized size = 2.6 \begin{align*} \text{result too large to display} \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 = 1.91753, size = 1820, normalized size = 5.31 \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 [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{9}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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