Optimal. Leaf size=119 \[ -\frac{i b \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac{2 b \sqrt{x} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac{x}{a+i b} \]
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Rubi [A] time = 0.174729, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3739, 3732, 2190, 2279, 2391} \[ -\frac{i b \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac{2 b \sqrt{x} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac{x}{a+i b} \]
Antiderivative was successfully verified.
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Rule 3739
Rule 3732
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{1}{a+b \tan \left (c+d \sqrt{x}\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{x}{a+b \tan (c+d x)} \, dx,x,\sqrt{x}\right )\\ &=\frac{x}{a+i b}+(4 i b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )\\ &=\frac{x}{a+i b}+\frac{2 b \sqrt{x} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{(2 b) \operatorname{Subst}\left (\int \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{x}{a+i b}+\frac{2 b \sqrt{x} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{\left (a^2+b^2\right ) d^2}\\ &=\frac{x}{a+i b}+\frac{2 b \sqrt{x} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{i b \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}\\ \end{align*}
Mathematica [A] time = 0.181958, size = 111, normalized size = 0.93 \[ \frac{i b \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )+2 b d \sqrt{x} \log \left (1+\frac{(a+i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )+d^2 x (a+i b)}{d^2 \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.154, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( c+d\sqrt{x} \right ) \right ) ^{-1}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.10001, size = 356, normalized size = 2.99 \begin{align*} \frac{{\left (a - i \, b\right )} d^{2} x - 2 i \, b d \sqrt{x} \arctan \left (\frac{2 \, a b \cos \left (2 \, d \sqrt{x} + 2 \, c\right ) -{\left (a^{2} - b^{2}\right )} \sin \left (2 \, d \sqrt{x} + 2 \, c\right )}{a^{2} + b^{2}}, \frac{2 \, a b \sin \left (2 \, d \sqrt{x} + 2 \, c\right ) + a^{2} + b^{2} +{\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt{x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + b d \sqrt{x} \log \left (\frac{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d \sqrt{x} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d \sqrt{x} + 2 \, c\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, d \sqrt{x} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt{x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - i \, b{\rm Li}_2\left (\frac{{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt{x} + 2 i \, c\right )}}{-i \, a + b}\right )}{{\left (a^{2} + b^{2}\right )} d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.57733, size = 1362, normalized size = 11.45 \begin{align*} \frac{2 \, a d^{2} x - 2 \, b c \log \left (\frac{{\left (i \, a b + b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} - a^{2} + i \, a b +{\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )}{\tan \left (d \sqrt{x} + c\right )^{2} + 1}\right ) - 2 \, b c \log \left (\frac{{\left (i \, a b - b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + a^{2} + i \, a b +{\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )}{\tan \left (d \sqrt{x} + c\right )^{2} + 1}\right ) - i \, b{\rm Li}_2\left (-\frac{{\left (2 i \, a b + 2 \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + 2 \, a^{2} - 2 i \, a b +{\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + i \, b{\rm Li}_2\left (-\frac{{\left (-2 i \, a b + 2 \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + 2 \, a^{2} + 2 i \, a b +{\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + 2 \,{\left (b d \sqrt{x} + b c\right )} \log \left (\frac{{\left (2 i \, a b + 2 \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + 2 \, a^{2} - 2 i \, a b +{\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + a^{2} + b^{2}}\right ) + 2 \,{\left (b d \sqrt{x} + b c\right )} \log \left (\frac{{\left (-2 i \, a b + 2 \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + 2 \, a^{2} + 2 i \, a b +{\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d \sqrt{x} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d \sqrt{x} + c\right )^{2} + a^{2} + b^{2}}\right )}{2 \,{\left (a^{2} + b^{2}\right )} d^{2}} \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}{a + b \tan{\left (c + d \sqrt{x} \right )}}\, 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{1}{b \tan \left (d \sqrt{x} + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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