Optimal. Leaf size=295 \[ -\frac{4 b^2 \left (-4 a^2 d+3 a b c-b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt{a+b \tan (e+f x)}}-\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2} \sqrt{c-i d}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2} \sqrt{c+i d}} \]
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Rubi [A] time = 1.26888, antiderivative size = 295, 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 = {3569, 3649, 3616, 3615, 93, 208} \[ -\frac{4 b^2 \left (-4 a^2 d+3 a b c-b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right )^2 (b c-a d)^2 \sqrt{a+b \tan (e+f x)}}-\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{3 f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a-i b)^{5/2} \sqrt{c-i d}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f (a+i b)^{5/2} \sqrt{c+i d}} \]
Antiderivative was successfully verified.
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Rule 3569
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{(a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \, dx &=-\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac{2 \int \frac{\frac{1}{2} \left (2 b^2 d-3 a (b c-a d)\right )+\frac{3}{2} b (b c-a d) \tan (e+f x)+b^2 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 ) (b c-a d)}\\ &=-\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac{4 b^2 \left (3 a b c-4 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)^2 f \sqrt{a+b \tan (e+f x)}}+\frac{4 \int \frac{\frac{3}{4} \left (a^2-b^2\right ) (b c-a d)^2-\frac{3}{2} a b (b c-a d)^2 \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)^2}\\ &=-\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac{4 b^2 \left (3 a b c-4 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)^2 f \sqrt{a+b \tan (e+f x)}}+\frac{\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{\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^2 \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac{4 b^2 \left (3 a b c-4 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)^2 f \sqrt{a+b \tan (e+f x)}}+\frac{\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{\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^2 \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac{4 b^2 \left (3 a b c-4 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)^2 f \sqrt{a+b \tan (e+f x)}}+\frac{\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{\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 \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} \sqrt{c-i d} f}+\frac{i \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} \sqrt{c+i d} f}-\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x))^{3/2}}-\frac{4 b^2 \left (3 a b c-4 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)^2 f \sqrt{a+b \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.98365, size = 308, normalized size = 1.04 \[ \frac{\frac{4 b^2 \left (4 a^2 d-3 a b c+b^2 d\right ) \sqrt{c+d \tan (e+f x)}}{(b c-a d)^2 \sqrt{a+b \tan (e+f x)}}+\frac{2 b^2 \left (a^2+b^2\right ) \sqrt{c+d \tan (e+f x)}}{(a d-b c) (a+b \tan (e+f x))^{3/2}}+3 i \left (\frac{(a-i b)^2 \tan ^{-1}\left (\frac{\sqrt{-c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+i b} \sqrt{-c-i d}}+\frac{(a+i b)^2 \tan ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{-a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{-a+i b} \sqrt{c-i d}}\right )}{3 f \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{d \tan \left (f x + e\right ) + c}}\,{d x} \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{1}{\left (a + b \tan{\left (e + f x \right )}\right )^{\frac{5}{2}} \sqrt{c + d \tan{\left (e + f x \right )}}}\, 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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