Optimal. Leaf size=276 \[ \frac{4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i (a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}+\frac{i (a+i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c+i d)^{5/2}} \]
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Rubi [A] time = 1.35211, antiderivative size = 276, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {3567, 3649, 3616, 3615, 93, 208} \[ \frac{4 \left (-3 a c d+b c^2-2 b d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i (a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}+\frac{i (a+i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (c+i d)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3567
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx &=\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \int \frac{\frac{1}{2} \left (-3 a^2 c+b^2 c-4 a b d\right )-\frac{3}{2} \left (2 a b c-a^2 d+b^2 d\right ) \tan (e+f x)-b (b c-a d) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (c^2+d^2\right )}\\ &=\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 \left (b c^2-3 a c d-2 b d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{4 \int \frac{-\frac{3}{4} (b c-a d) (a c+b c-a d+b d) (a c-b c+a d+b d)-\frac{3}{2} (b c-a d)^2 (a c+b d) \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{3 (b c-a d) \left (c^2+d^2\right )^2}\\ &=\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 \left (b c^2-3 a c d-2 b d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(a-i b)^2 \int \frac{1+i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{(a+i b)^2 \int \frac{1-i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 \left (b c^2-3 a c d-2 b d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(a-i b)^2 \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (c-i d)^2 f}+\frac{(a+i b)^2 \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 \left (b c^2-3 a c d-2 b d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(a-i b)^2 \operatorname{Subst}\left (\int \frac{1}{i a+b-(i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(c-i d)^2 f}+\frac{(a+i b)^2 \operatorname{Subst}\left (\int \frac{1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{(c+i d)^2 f}\\ &=-\frac{i (a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}+\frac{i (a+i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(c+i d)^{5/2} f}+\frac{2 (b c-a d) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 \left (b c^2-3 a c d-2 b d^2\right ) \sqrt{a+b \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 2.63845, size = 264, normalized size = 0.96 \[ \frac{-\frac{2 \sqrt{a+b \tan (e+f x)} \left (2 d \left (3 a c d-b c^2+2 b d^2\right ) \tan (e+f x)+7 a c^2 d+a d^3-3 b c^3+3 b c d^2\right )}{\left (c^2+d^2\right )^2 (c+d \tan (e+f x))^{3/2}}+\frac{3 i (-a+i b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(c-i d)^{5/2}}+\frac{3 i (a+i b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(-c-i d)^{5/2}}}{3 f} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \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 [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 )^{\frac{3}{2}}}{\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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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