Optimal. Leaf size=280 \[ -\frac{2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right )^2 (b c-a d) \sqrt{a+b \tan (e+f x)}}-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2}}+\frac{i \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2}} \]
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Rubi [A] time = 1.22366, antiderivative size = 280, 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 = {3568, 3649, 3616, 3615, 93, 208} \[ -\frac{2 b \left (-5 a^2 d+6 a b c+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right )^2 (b c-a d) \sqrt{a+b \tan (e+f x)}}-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{3/2}}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2}}+\frac{i \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3568
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{\sqrt{c+d \tan (e+f x)}}{(a+b \tan (e+f x))^{5/2}} \, dx &=-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac{2 \int \frac{\frac{1}{2} (-3 a c-b d)+\frac{3}{2} (b c-a d) \tan (e+f x)+b d \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right )}\\ &=-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac{2 b \left (6 a b c-5 a^2 d+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d) f \sqrt{a+b \tan (e+f x)}}+\frac{4 \int \frac{\frac{3}{4} (b c-a d) \left (a^2 c-b^2 c+2 a b d\right )-\frac{3}{4} (b c-a d) \left (2 a b c-a^2 d+b^2 d\right ) \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac{2 b \left (6 a b c-5 a^2 d+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d) f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d) \int \frac{1+i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2}+\frac{(c+i d) \int \frac{1-i \tan (e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac{2 b \left (6 a b c-5 a^2 d+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d) f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d) \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 (a-i b)^2 f}+\frac{(c+i d) \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 (a+i b)^2 f}\\ &=-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac{2 b \left (6 a b c-5 a^2 d+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d) f \sqrt{a+b \tan (e+f x)}}+\frac{(c-i d) \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 )}{(a-i b)^2 f}+\frac{(c+i d) \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 )}{(a+i b)^2 f}\\ &=-\frac{i \sqrt{c-i d} \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 )}{(a-i b)^{5/2} f}+\frac{i \sqrt{c+i d} \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 )}{(a+i b)^{5/2} f}-\frac{2 b \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{3/2}}-\frac{2 b \left (6 a b c-5 a^2 d+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right )^2 (b c-a d) f \sqrt{a+b \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 5.01646, size = 266, normalized size = 0.95 \[ \frac{\frac{2 b \sqrt{c+d \tan (e+f x)} \left (b \left (-5 a^2 d+6 a b c+b^2 d\right ) \tan (e+f x)+7 a^2 b c-6 a^3 d+b^3 c\right )}{\left (a^2+b^2\right )^2 (a d-b c) (a+b \tan (e+f x))^{3/2}}-\frac{3 i \sqrt{-c-i d} \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(a+i b)^{5/2}}+\frac{3 i \sqrt{c-i d} \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{(-a+i b)^{5/2}}}{3 f} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{\sqrt{c+d\tan \left ( fx+e \right ) } \left ( a+b\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{\sqrt{c + d \tan{\left (e + f x \right )}}}{\left (a + b \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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