3.860 \(\int \frac{\sqrt{a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx\)

Optimal. Leaf size=673 \[ \frac{g \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{8 (e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}}+\frac{e \sqrt{a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g)^3}-\frac{e^2 \sqrt{a g^2-b f g+c f^2} \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{g (e f-d g)^3}+\frac{e^2 (2 c f-b g) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} g (e f-d g)^3}-\frac{g \sqrt{a+b x+c x^2} (-2 a g+x (2 c f-b g)+b f)}{4 (f+g x)^2 (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac{e (2 c f-b g) \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{2 g (e f-d g)^2 \sqrt{a g^2-b f g+c f^2}}+\frac{e \sqrt{a+b x+c x^2}}{(f+g x) (e f-d g)^2}-\frac{\sqrt{c} e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^2}-\frac{e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} (e f-d g)^3} \]

[Out]

(e*Sqrt[a + b*x + c*x^2])/((e*f - d*g)^2*(f + g*x)) - (g*(b*f - 2*a*g + (2*c*f - b*g)*x)*Sqrt[a + b*x + c*x^2]
)/(4*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*(f + g*x)^2) - (e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
+ b*x + c*x^2])])/(2*Sqrt[c]*(e*f - d*g)^3) + (e^2*(2*c*f - b*g)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x
+ c*x^2])])/(2*Sqrt[c]*g*(e*f - d*g)^3) - (Sqrt[c]*e*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(
g*(e*f - d*g)^2) + (e*Sqrt[c*d^2 - b*d*e + a*e^2]*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*
e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(e*f - d*g)^3 + ((b^2 - 4*a*c)*g*ArcTanh[(b*f - 2*a*g + (2*c*f - b*g)*x)/(
2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/(8*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)^(3/2)) + (e*(2*c
*f - b*g)*ArcTanh[(b*f - 2*a*g + (2*c*f - b*g)*x)/(2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/(2*g
*(e*f - d*g)^2*Sqrt[c*f^2 - b*f*g + a*g^2]) - (e^2*Sqrt[c*f^2 - b*f*g + a*g^2]*ArcTanh[(b*f - 2*a*g + (2*c*f -
b*g)*x)/(2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/(g*(e*f - d*g)^3)

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Rubi [A]  time = 0.861806, antiderivative size = 673, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {960, 734, 843, 621, 206, 724, 720, 732} \[ \frac{g \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{8 (e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}}+\frac{e \sqrt{a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g)^3}-\frac{e^2 \sqrt{a g^2-b f g+c f^2} \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{g (e f-d g)^3}+\frac{e^2 (2 c f-b g) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} g (e f-d g)^3}-\frac{g \sqrt{a+b x+c x^2} (-2 a g+x (2 c f-b g)+b f)}{4 (f+g x)^2 (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac{e (2 c f-b g) \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{2 g (e f-d g)^2 \sqrt{a g^2-b f g+c f^2}}+\frac{e \sqrt{a+b x+c x^2}}{(f+g x) (e f-d g)^2}-\frac{\sqrt{c} e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^2}-\frac{e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} (e f-d g)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x + c*x^2]/((d + e*x)*(f + g*x)^3),x]

[Out]

(e*Sqrt[a + b*x + c*x^2])/((e*f - d*g)^2*(f + g*x)) - (g*(b*f - 2*a*g + (2*c*f - b*g)*x)*Sqrt[a + b*x + c*x^2]
)/(4*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*(f + g*x)^2) - (e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
+ b*x + c*x^2])])/(2*Sqrt[c]*(e*f - d*g)^3) + (e^2*(2*c*f - b*g)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x
+ c*x^2])])/(2*Sqrt[c]*g*(e*f - d*g)^3) - (Sqrt[c]*e*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(
g*(e*f - d*g)^2) + (e*Sqrt[c*d^2 - b*d*e + a*e^2]*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*
e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(e*f - d*g)^3 + ((b^2 - 4*a*c)*g*ArcTanh[(b*f - 2*a*g + (2*c*f - b*g)*x)/(
2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/(8*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)^(3/2)) + (e*(2*c
*f - b*g)*ArcTanh[(b*f - 2*a*g + (2*c*f - b*g)*x)/(2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/(2*g
*(e*f - d*g)^2*Sqrt[c*f^2 - b*f*g + a*g^2]) - (e^2*Sqrt[c*f^2 - b*f*g + a*g^2]*ArcTanh[(b*f - 2*a*g + (2*c*f -
b*g)*x)/(2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqrt[a + b*x + c*x^2])])/(g*(e*f - d*g)^3)

Rule 960

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

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 843

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

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 0]

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx &=\int \left (\frac{e^3 \sqrt{a+b x+c x^2}}{(e f-d g)^3 (d+e x)}-\frac{g \sqrt{a+b x+c x^2}}{(e f-d g) (f+g x)^3}-\frac{e g \sqrt{a+b x+c x^2}}{(e f-d g)^2 (f+g x)^2}-\frac{e^2 g \sqrt{a+b x+c x^2}}{(e f-d g)^3 (f+g x)}\right ) \, dx\\ &=\frac{e^3 \int \frac{\sqrt{a+b x+c x^2}}{d+e x} \, dx}{(e f-d g)^3}-\frac{\left (e^2 g\right ) \int \frac{\sqrt{a+b x+c x^2}}{f+g x} \, dx}{(e f-d g)^3}-\frac{(e g) \int \frac{\sqrt{a+b x+c x^2}}{(f+g x)^2} \, dx}{(e f-d g)^2}-\frac{g \int \frac{\sqrt{a+b x+c x^2}}{(f+g x)^3} \, dx}{e f-d g}\\ &=\frac{e \sqrt{a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac{g (b f-2 a g+(2 c f-b g) x) \sqrt{a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}-\frac{e^2 \int \frac{b d-2 a e+(2 c d-b e) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{2 (e f-d g)^3}+\frac{e^2 \int \frac{b f-2 a g+(2 c f-b g) x}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{2 (e f-d g)^3}-\frac{e \int \frac{b+2 c x}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{2 (e f-d g)^2}+\frac{\left (\left (b^2-4 a c\right ) g\right ) \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )}\\ &=\frac{e \sqrt{a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac{g (b f-2 a g+(2 c f-b g) x) \sqrt{a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}-\frac{(e (2 c d-b e)) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2 (e f-d g)^3}+\frac{\left (e \left (c d^2-b d e+a e^2\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{(e f-d g)^3}+\frac{\left (e^2 (2 c f-b g)\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2 g (e f-d g)^3}-\frac{(c e) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{g (e f-d g)^2}+\frac{(e (2 c f-b g)) \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{2 g (e f-d g)^2}-\frac{\left (\left (b^2-4 a c\right ) g\right ) \operatorname{Subst}\left (\int \frac{1}{4 c f^2-4 b f g+4 a g^2-x^2} \, dx,x,\frac{-b f+2 a g-(2 c f-b g) x}{\sqrt{a+b x+c x^2}}\right )}{4 (e f-d g) \left (c f^2-b f g+a g^2\right )}-\frac{\left (e^2 \left (c f^2-b f g+a g^2\right )\right ) \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{g (e f-d g)^3}\\ &=\frac{e \sqrt{a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac{g (b f-2 a g+(2 c f-b g) x) \sqrt{a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac{\left (b^2-4 a c\right ) g \tanh ^{-1}\left (\frac{b f-2 a g+(2 c f-b g) x}{2 \sqrt{c f^2-b f g+a g^2} \sqrt{a+b x+c x^2}}\right )}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}}-\frac{(e (2 c d-b e)) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{(e f-d g)^3}-\frac{\left (2 e \left (c d^2-b d e+a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac{-b d+2 a e-(2 c d-b e) x}{\sqrt{a+b x+c x^2}}\right )}{(e f-d g)^3}+\frac{\left (e^2 (2 c f-b g)\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^3}-\frac{(2 c e) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^2}-\frac{(e (2 c f-b g)) \operatorname{Subst}\left (\int \frac{1}{4 c f^2-4 b f g+4 a g^2-x^2} \, dx,x,\frac{-b f+2 a g-(2 c f-b g) x}{\sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^2}+\frac{\left (2 e^2 \left (c f^2-b f g+a g^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c f^2-4 b f g+4 a g^2-x^2} \, dx,x,\frac{-b f+2 a g-(2 c f-b g) x}{\sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^3}\\ &=\frac{e \sqrt{a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac{g (b f-2 a g+(2 c f-b g) x) \sqrt{a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}-\frac{e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} (e f-d g)^3}+\frac{e^2 (2 c f-b g) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} g (e f-d g)^3}-\frac{\sqrt{c} e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^2}+\frac{e \sqrt{c d^2-b d e+a e^2} \tanh ^{-1}\left (\frac{b d-2 a e+(2 c d-b e) x}{2 \sqrt{c d^2-b d e+a e^2} \sqrt{a+b x+c x^2}}\right )}{(e f-d g)^3}+\frac{\left (b^2-4 a c\right ) g \tanh ^{-1}\left (\frac{b f-2 a g+(2 c f-b g) x}{2 \sqrt{c f^2-b f g+a g^2} \sqrt{a+b x+c x^2}}\right )}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}}+\frac{e (2 c f-b g) \tanh ^{-1}\left (\frac{b f-2 a g+(2 c f-b g) x}{2 \sqrt{c f^2-b f g+a g^2} \sqrt{a+b x+c x^2}}\right )}{2 g (e f-d g)^2 \sqrt{c f^2-b f g+a g^2}}-\frac{e^2 \sqrt{c f^2-b f g+a g^2} \tanh ^{-1}\left (\frac{b f-2 a g+(2 c f-b g) x}{2 \sqrt{c f^2-b f g+a g^2} \sqrt{a+b x+c x^2}}\right )}{g (e f-d g)^3}\\ \end{align*}

Mathematica [A]  time = 1.46861, size = 609, normalized size = 0.9 \[ \frac{\frac{g \left (b^2-4 a c\right ) (e f-d g)^2 \tanh ^{-1}\left (\frac{-2 a g+b (f-g x)+2 c f x}{2 \sqrt{a+x (b+c x)} \sqrt{g (a g-b f)+c f^2}}\right )}{\left (g (a g-b f)+c f^2\right )^{3/2}}+8 e \sqrt{e (a e-b d)+c d^2} \tanh ^{-1}\left (\frac{-2 a e+b (d-e x)+2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )+\frac{2 g \sqrt{a+x (b+c x)} (e f-d g)^2 (2 a g-b f+b g x-2 c f x)}{(f+g x)^2 \left (g (a g-b f)+c f^2\right )}-\frac{4 e (e f-d g) \left (2 \sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )-\frac{(2 c f-b g) \tanh ^{-1}\left (\frac{-2 a g+b (f-g x)+2 c f x}{2 \sqrt{a+x (b+c x)} \sqrt{g (a g-b f)+c f^2}}\right )}{\sqrt{g (a g-b f)+c f^2}}\right )}{g}+\frac{8 e \sqrt{a+x (b+c x)} (e f-d g)}{f+g x}+\frac{4 e (b e-2 c d) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{\sqrt{c}}+\frac{4 e^2 \left ((2 c f-b g) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{c} \sqrt{g (a g-b f)+c f^2} \tanh ^{-1}\left (\frac{-2 a g+b (f-g x)+2 c f x}{2 \sqrt{a+x (b+c x)} \sqrt{g (a g-b f)+c f^2}}\right )\right )}{\sqrt{c} g}}{8 (e f-d g)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x + c*x^2]/((d + e*x)*(f + g*x)^3),x]

[Out]

((8*e*(e*f - d*g)*Sqrt[a + x*(b + c*x)])/(f + g*x) + (2*g*(e*f - d*g)^2*(-(b*f) + 2*a*g - 2*c*f*x + b*g*x)*Sqr
t[a + x*(b + c*x)])/((c*f^2 + g*(-(b*f) + a*g))*(f + g*x)^2) + (4*e*(-2*c*d + b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt
[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c] + 8*e*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*ArcTanh[(-2*a*e + 2*c*d*x + b*(d - e
*x))/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])] + ((b^2 - 4*a*c)*g*(e*f - d*g)^2*ArcTanh[(-2*a*
g + 2*c*f*x + b*(f - g*x))/(2*Sqrt[c*f^2 + g*(-(b*f) + a*g)]*Sqrt[a + x*(b + c*x)])])/(c*f^2 + g*(-(b*f) + a*g
))^(3/2) - (4*e*(e*f - d*g)*(2*Sqrt[c]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] - ((2*c*f - b*g)
*ArcTanh[(-2*a*g + 2*c*f*x + b*(f - g*x))/(2*Sqrt[c*f^2 + g*(-(b*f) + a*g)]*Sqrt[a + x*(b + c*x)])])/Sqrt[c*f^
2 + g*(-(b*f) + a*g)]))/g + (4*e^2*((2*c*f - b*g)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] - 2*S
qrt[c]*Sqrt[c*f^2 + g*(-(b*f) + a*g)]*ArcTanh[(-2*a*g + 2*c*f*x + b*(f - g*x))/(2*Sqrt[c*f^2 + g*(-(b*f) + a*g
)]*Sqrt[a + x*(b + c*x)])]))/(Sqrt[c]*g))/(8*(e*f - d*g)^3)

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Maple [B]  time = 0.355, size = 6714, normalized size = 10. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(1/2)/(e*x+d)/(g*x+f)^3,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{c x^{2} + b x + a}}{{\left (e x + d\right )}{\left (g x + f\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(e*x+d)/(g*x+f)^3,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x + a)/((e*x + d)*(g*x + f)^3), 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((c*x^2+b*x+a)^(1/2)/(e*x+d)/(g*x+f)^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((c*x**2+b*x+a)**(1/2)/(e*x+d)/(g*x+f)**3,x)

[Out]

Timed out

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Giac [B]  time = 5.47724, size = 2489, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(e*x+d)/(g*x+f)^3,x, algorithm="giac")

[Out]

-1/4*(b^2*d^2*g^3 - 4*a*c*d^2*g^3 - 8*c^2*d*f^3*e + 12*b*c*d*f^2*g*e - 6*b^2*d*f*g^2*e + 4*a*b*d*g^3*e + 4*b*c
*f^3*e^2 - 3*b^2*f^2*g*e^2 - 12*a*c*f^2*g*e^2 + 12*a*b*f*g^2*e^2 - 8*a^2*g^3*e^2)*arctan(-((sqrt(c)*x - sqrt(c
*x^2 + b*x + a))*g + sqrt(c)*f)/sqrt(-c*f^2 + b*f*g - a*g^2))/((c*d^3*f^2*g^3 - b*d^3*f*g^4 + a*d^3*g^5 - 3*c*
d^2*f^3*g^2*e + 3*b*d^2*f^2*g^3*e - 3*a*d^2*f*g^4*e + 3*c*d*f^4*g*e^2 - 3*b*d*f^3*g^2*e^2 + 3*a*d*f^2*g^3*e^2
- c*f^5*e^3 + b*f^4*g*e^3 - a*f^3*g^2*e^3)*sqrt(-c*f^2 + b*f*g - a*g^2)) - 2*(c*d^2*e - b*d*e^2 + a*e^3)*arcta
n(-((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))/((d^3*g^3 - 3*d^2*f*g^2*e
+ 3*d*f^2*g*e^2 - f^3*e^3)*sqrt(-c*d^2 + b*d*e - a*e^2)) + 1/4*(8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*c^2*d
*f^2*g^2 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c*d*f*g^3 + (sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*d*g
^4 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c*d*g^4 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c*f^2*g^2*e
+ 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*f*g^3*e + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c*f*g^3*e -
4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*g^4*e + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(5/2)*d*f^3*g -
5*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^2*sqrt(c)*d*f*g^3 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*c^(3/
2)*d*f*g^3 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b*sqrt(c)*d*g^4 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
^2*c^(5/2)*f^4*e - 20*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^(3/2)*f^3*g*e + 9*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^2*b^2*sqrt(c)*f^2*g^2*e + 12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*c^(3/2)*f^2*g^2*e - 4*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^2*a*b*sqrt(c)*f*g^3*e - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^2*sqrt(c)*g^4*e + 8
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*c^2*d*f^3*g - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^2*c*d*f^2*g^2 - 8
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*c^2*d*f^2*g^2 - (sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^3*d*f*g^3 + 4*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))*a*b*c*d*f*g^3 + (sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^2*d*g^4 + 4*(sqrt(c)*
x - sqrt(c*x^2 + b*x + a))*a^2*c*d*g^4 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*c^2*f^4*e - 16*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))*b^2*c*f^3*g*e - 16*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*c^2*f^3*g*e + 5*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))*b^3*f^2*g^2*e + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b*c*f^2*g^2*e - 9*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))*a*b^2*f*g^3*e - 28*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*c*f*g^3*e + 4*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))*a^2*b*g^4*e + 2*b^2*c^(3/2)*d*f^3*g - b^3*sqrt(c)*d*f^2*g^2 - 4*a*b*c^(3/2)*d*f^2*g^2 + a
*b^2*sqrt(c)*d*f*g^3 + 4*a^2*c^(3/2)*d*f*g^3 + 2*b^2*c^(3/2)*f^4*e - 3*b^3*sqrt(c)*f^3*g*e - 8*a*b*c^(3/2)*f^3
*g*e + 15*a*b^2*sqrt(c)*f^2*g^2*e + 4*a^2*c^(3/2)*f^2*g^2*e - 20*a^2*b*sqrt(c)*f*g^3*e + 8*a^3*sqrt(c)*g^4*e)/
((c*d^2*f^2*g^3 - b*d^2*f*g^4 + a*d^2*g^5 - 2*c*d*f^3*g^2*e + 2*b*d*f^2*g^3*e - 2*a*d*f*g^4*e + c*f^4*g*e^2 -
b*f^3*g^2*e^2 + a*f^2*g^3*e^2)*((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*g + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a)
)*sqrt(c)*f + b*f - a*g)^2)