3.851 \(\int \frac{\sqrt{d+e x}}{\sqrt{f+g x} (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=285 \[ \frac{2 \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{f+g x} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{2 \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \tanh ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{f+g x} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c f-g \left (b-\sqrt{b^2-4 a c}\right )}} \]

[Out]

(-2*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])/(S
qrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c]
)*g]) + (2*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e
*x])/(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*f - (b + Sqrt[b^2 -
4*a*c])*g])

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Rubi [A]  time = 0.52577, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {909, 93, 208} \[ \frac{2 \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{f+g x} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{2 \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \tanh ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{f+g x} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c f-g \left (b-\sqrt{b^2-4 a c}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]/(Sqrt[f + g*x]*(a + b*x + c*x^2)),x]

[Out]

(-2*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])/(S
qrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c]
)*g]) + (2*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e
*x])/(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*f - (b + Sqrt[b^2 -
4*a*c])*g])

Rule 909

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Int
[ExpandIntegrand[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b
, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{d+e x}}{\sqrt{f+g x} \left (a+b x+c x^2\right )} \, dx &=\int \left (\frac{e+\frac{2 c d-b e}{\sqrt{b^2-4 a c}}}{\left (b-\sqrt{b^2-4 a c}+2 c x\right ) \sqrt{d+e x} \sqrt{f+g x}}+\frac{e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}}{\left (b+\sqrt{b^2-4 a c}+2 c x\right ) \sqrt{d+e x} \sqrt{f+g x}}\right ) \, dx\\ &=\left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\left (b+\sqrt{b^2-4 a c}+2 c x\right ) \sqrt{d+e x} \sqrt{f+g x}} \, dx+\left (e+\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\left (b-\sqrt{b^2-4 a c}+2 c x\right ) \sqrt{d+e x} \sqrt{f+g x}} \, dx\\ &=\left (2 \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-2 c d+\left (b+\sqrt{b^2-4 a c}\right ) e-\left (-2 c f+\left (b+\sqrt{b^2-4 a c}\right ) g\right ) x^2} \, dx,x,\frac{\sqrt{d+e x}}{\sqrt{f+g x}}\right )+\left (2 \left (e+\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-2 c d+\left (b-\sqrt{b^2-4 a c}\right ) e-\left (-2 c f+\left (b-\sqrt{b^2-4 a c}\right ) g\right ) x^2} \, dx,x,\frac{\sqrt{d+e x}}{\sqrt{f+g x}}\right )\\ &=-\frac{2 \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} \tanh ^{-1}\left (\frac{\sqrt{2 c f-\left (b-\sqrt{b^2-4 a c}\right ) g} \sqrt{d+e x}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} \sqrt{f+g x}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c f-\left (b-\sqrt{b^2-4 a c}\right ) g}}+\frac{2 \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} \tanh ^{-1}\left (\frac{\sqrt{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g} \sqrt{d+e x}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} \sqrt{f+g x}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}}\\ \end{align*}

Mathematica [A]  time = 1.0404, size = 268, normalized size = 0.94 \[ \frac{2 \left (\frac{\sqrt{e \left (\sqrt{b^2-4 a c}+b\right )-2 c d} \tan ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{f+g x} \sqrt{e \left (\sqrt{b^2-4 a c}+b\right )-2 c d}}\right )}{\sqrt{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{\sqrt{e \left (b-\sqrt{b^2-4 a c}\right )-2 c d} \tan ^{-1}\left (\frac{\sqrt{d+e x} \sqrt{g \sqrt{b^2-4 a c}-b g+2 c f}}{\sqrt{f+g x} \sqrt{-e \sqrt{b^2-4 a c}+b e-2 c d}}\right )}{\sqrt{g \left (\sqrt{b^2-4 a c}-b\right )+2 c f}}\right )}{\sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d + e*x]/(Sqrt[f + g*x]*(a + b*x + c*x^2)),x]

[Out]

(2*(-((Sqrt[-2*c*d + (b - Sqrt[b^2 - 4*a*c])*e]*ArcTan[(Sqrt[2*c*f - b*g + Sqrt[b^2 - 4*a*c]*g]*Sqrt[d + e*x])
/(Sqrt[-2*c*d + b*e - Sqrt[b^2 - 4*a*c]*e]*Sqrt[f + g*x])])/Sqrt[2*c*f + (-b + Sqrt[b^2 - 4*a*c])*g]) + (Sqrt[
-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e]*ArcTan[(Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])/(Sqrt[-2*c*
d + (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]))/Sqrt[b^2 - 4*a*c]

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Maple [B]  time = 0.474, size = 5482, normalized size = 19.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)/(c*x^2+b*x+a)/(g*x+f)^(1/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e x + d}}{{\left (c x^{2} + b x + a\right )} \sqrt{g x + f}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(c*x^2+b*x+a)/(g*x+f)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(e*x + d)/((c*x^2 + b*x + a)*sqrt(g*x + f)), x)

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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.

[In]

integrate((e*x+d)^(1/2)/(c*x^2+b*x+a)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d + e x}}{\sqrt{f + g x} \left (a + b x + c x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(c*x**2+b*x+a)/(g*x+f)**(1/2),x)

[Out]

Integral(sqrt(d + e*x)/(sqrt(f + g*x)*(a + b*x + c*x**2)), x)

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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.

[In]

integrate((e*x+d)^(1/2)/(c*x^2+b*x+a)/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError