Optimal. Leaf size=219 \[ -\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]
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Rubi [A] time = 0.649914, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3565, 3628, 3539, 3537, 63, 208} \[ -\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3565
Rule 3628
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{\frac{1}{2} \left (2 b^3 c^2+3 a^3 c d-7 a b^2 c d+8 a^2 b d^2\right )+\frac{3}{2} d \left (3 a^2 b c-b^3 c-a^3 d+3 a b^2 d\right ) \tan (e+f x)+\frac{1}{2} b \left (2 a b c d-a^2 d^2+b^2 \left (2 c^2+3 d^2\right )\right ) \tan ^2(e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{\frac{3}{2} d \left (6 a^2 b c d-2 b^3 c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )\right )-\frac{3}{2} d \left (2 a^3 c d-6 a b^2 c d-3 a^2 b \left (c^2-d^2\right )+b^3 \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d \left (c^2+d^2\right )^2}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(a-i b)^3 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{(a+i b)^3 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{(i a+b)^3 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}+\frac{(i a-b)^3 \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{(a-i b)^3 \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c-i d)^2 d f}-\frac{(a+i b)^3 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c+i d)^2 d f}\\ &=\frac{(i a+b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2} f}-\frac{(i a-b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{5/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{4 (b c-a d)^2 \left (3 a c d+b \left (c^2+4 d^2\right )\right )}{3 d^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [C] time = 1.44385, size = 284, normalized size = 1.3 \[ -\frac{-3 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x)) \left (i (c+i d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )\right )-d \left (-3 a^2 b c+a^3 d-3 a b^2 d+b^3 c\right ) \left (i (c+i d) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )\right )+6 b^2 d (c-i d) (c+i d) (a+b \tan (e+f x))+4 b^3 c \left (c^2+d^2\right )}{3 d^2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.111, size = 26303, normalized size = 120.1 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{3}}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \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 )}^{3}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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