3.7.4 \(\int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^4} \, dx\)

Optimal. Leaf size=933 \[ \frac {(2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^3}{2 \sqrt {c} g (e f-d g)^4}-\frac {\sqrt {c f^2-b g f+a g^2} \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b g f+a g^2} \sqrt {c x^2+b x+a}}\right ) e^3}{g (e f-d g)^4}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^2}{g (e f-d g)^3}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^2}{2 \sqrt {c} (e f-d g)^4}+\frac {\sqrt {c d^2-b e d+a e^2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b e d+a e^2} \sqrt {c x^2+b x+a}}\right ) e^2}{(e f-d g)^4}+\frac {(2 c f-b g) \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b g f+a g^2} \sqrt {c x^2+b x+a}}\right ) e^2}{2 g (e f-d g)^3 \sqrt {c f^2-b g f+a g^2}}+\frac {\sqrt {c x^2+b x+a} e^2}{(e f-d g)^3 (f+g x)}+\frac {\left (b^2-4 a c\right ) g \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b g f+a g^2} \sqrt {c x^2+b x+a}}\right ) e}{8 (e f-d g)^2 \left (c f^2-b g f+a g^2\right )^{3/2}}-\frac {g (b f-2 a g+(2 c f-b g) x) \sqrt {c x^2+b x+a} e}{4 (e f-d g)^2 \left (c f^2-b g f+a g^2\right ) (f+g x)^2}+\frac {g^2 \left (c x^2+b x+a\right )^{3/2}}{3 (e f-d g) \left (c f^2-b g f+a g^2\right ) (f+g x)^3}+\frac {\left (b^2-4 a c\right ) g (2 c f-b g) \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b g f+a g^2} \sqrt {c x^2+b x+a}}\right )}{16 (e f-d g) \left (c f^2-b g f+a g^2\right )^{5/2}}-\frac {g (2 c f-b g) (b f-2 a g+(2 c f-b g) x) \sqrt {c x^2+b x+a}}{8 (e f-d g) \left (c f^2-b g f+a g^2\right )^2 (f+g x)^2} \]

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Rubi [A]  time = 1.22, antiderivative size = 933, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {960, 734, 843, 621, 206, 724, 730, 720, 732} \begin {gather*} \frac {(2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^3}{2 \sqrt {c} g (e f-d g)^4}-\frac {\sqrt {c f^2-b g f+a g^2} \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b g f+a g^2} \sqrt {c x^2+b x+a}}\right ) e^3}{g (e f-d g)^4}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^2}{g (e f-d g)^3}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^2}{2 \sqrt {c} (e f-d g)^4}+\frac {\sqrt {c d^2-b e d+a e^2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b e d+a e^2} \sqrt {c x^2+b x+a}}\right ) e^2}{(e f-d g)^4}+\frac {(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 g f+a g^2} \sqrt {c x^2+b x+a}}\right ) e^2}{2 g (e f-d g)^3 \sqrt {c f^2-b g f+a g^2}}+\frac {\sqrt {c x^2+b x+a} e^2}{(e f-d g)^3 (f+g x)}+\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 g f+a g^2} \sqrt {c x^2+b x+a}}\right ) e}{8 (e f-d g)^2 \left (c f^2-b g f+a g^2\right )^{3/2}}-\frac {g (b f-2 a g+(2 c f-b g) x) \sqrt {c x^2+b x+a} e}{4 (e f-d g)^2 \left (c f^2-b g f+a g^2\right ) (f+g x)^2}+\frac {g^2 \left (c x^2+b x+a\right )^{3/2}}{3 (e f-d g) \left (c f^2-b g f+a g^2\right ) (f+g x)^3}+\frac {\left (b^2-4 a c\right ) g (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 g f+a g^2} \sqrt {c x^2+b x+a}}\right )}{16 (e f-d g) \left (c f^2-b g f+a g^2\right )^{5/2}}-\frac {g (2 c f-b g) (b f-2 a g+(2 c f-b g) x) \sqrt {c x^2+b x+a}}{8 (e f-d g) \left (c f^2-b g f+a g^2\right )^2 (f+g x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*Sqrt[a + b*x + c*x^2])/((e*f - d*g)^3*(f + g*x)) - (g*(2*c*f - b*g)*(b*f - 2*a*g + (2*c*f - b*g)*x)*Sqrt[
a + b*x + c*x^2])/(8*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)^2*(f + g*x)^2) - (e*g*(b*f - 2*a*g + (2*c*f - b*g)*x)
*Sqrt[a + b*x + c*x^2])/(4*(e*f - d*g)^2*(c*f^2 - b*f*g + a*g^2)*(f + g*x)^2) + (g^2*(a + b*x + c*x^2)^(3/2))/
(3*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*(f + g*x)^3) - (e^2*(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)^4) + (e^3*(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)^4) - (Sqrt[c]*e^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])
/(g*(e*f - d*g)^3) + (e^2*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)^4 + ((b^2 - 4*a*c)*g*(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])])/(16*(e*f - d*g)*(c*f^2 - b*f*g + a*g^
2)^(5/2)) + ((b^2 - 4*a*c)*e*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)^2*(c*f^2 - b*f*g + a*g^2)^(3/2)) + (e^2*(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)^3*Sqrt[c*f^2 - b*f*g
+ a*g^2]) - (e^3*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)^4)

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 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 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 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 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, 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 + 3, 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]

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 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])

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^4} \, dx &=\int \left (\frac {e^4 \sqrt {a+b x+c x^2}}{(e f-d g)^4 (d+e x)}-\frac {g \sqrt {a+b x+c x^2}}{(e f-d g) (f+g x)^4}-\frac {e g \sqrt {a+b x+c x^2}}{(e f-d g)^2 (f+g x)^3}-\frac {e^2 g \sqrt {a+b x+c x^2}}{(e f-d g)^3 (f+g x)^2}-\frac {e^3 g \sqrt {a+b x+c x^2}}{(e f-d g)^4 (f+g x)}\right ) \, dx\\ &=\frac {e^4 \int \frac {\sqrt {a+b x+c x^2}}{d+e x} \, dx}{(e f-d g)^4}-\frac {\left (e^3 g\right ) \int \frac {\sqrt {a+b x+c x^2}}{f+g x} \, dx}{(e f-d g)^4}-\frac {\left (e^2 g\right ) \int \frac {\sqrt {a+b x+c x^2}}{(f+g x)^2} \, dx}{(e f-d g)^3}-\frac {(e g) \int \frac {\sqrt {a+b x+c x^2}}{(f+g x)^3} \, dx}{(e f-d g)^2}-\frac {g \int \frac {\sqrt {a+b x+c x^2}}{(f+g x)^4} \, dx}{e f-d g}\\ &=\frac {e^2 \sqrt {a+b x+c x^2}}{(e f-d g)^3 (f+g x)}-\frac {e g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^3}-\frac {e^3 \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)^4}+\frac {e^3 \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)^4}-\frac {e^2 \int \frac {b+2 c x}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g)^3}+\frac {\left (\left (b^2-4 a c\right ) e g\right ) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{8 (e f-d g)^2 \left (c f^2-b f g+a g^2\right )}-\frac {(g (2 c f-b g)) \int \frac {\sqrt {a+b x+c x^2}}{(f+g x)^3} \, dx}{2 (e f-d g) \left (c f^2-b f g+a g^2\right )}\\ &=\frac {e^2 \sqrt {a+b x+c x^2}}{(e f-d g)^3 (f+g x)}-\frac {g (2 c f-b g) (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)^2}-\frac {e g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^3}-\frac {\left (e^2 (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g)^4}+\frac {\left (e^2 \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)^4}+\frac {\left (e^3 (2 c f-b g)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 g (e f-d g)^4}-\frac {\left (c e^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{g (e f-d g)^3}+\frac {\left (e^2 (2 c f-b g)\right ) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{2 g (e f-d g)^3}+\frac {\left (\left (b^2-4 a c\right ) g (2 c f-b g)\right ) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{16 (e f-d g) \left (c f^2-b f g+a g^2\right )^2}-\frac {\left (\left (b^2-4 a c\right ) e 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)^2 \left (c f^2-b f g+a g^2\right )}-\frac {\left (e^3 \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)^4}\\ &=\frac {e^2 \sqrt {a+b x+c x^2}}{(e f-d g)^3 (f+g x)}-\frac {g (2 c f-b g) (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)^2}-\frac {e g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^3}+\frac {\left (b^2-4 a c\right ) e 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)^2 \left (c f^2-b f g+a g^2\right )^{3/2}}-\frac {\left (e^2 (2 c d-b e)\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 )}{(e f-d g)^4}-\frac {\left (2 e^2 \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)^4}+\frac {\left (e^3 (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)^4}-\frac {\left (2 c e^2\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 {\left (e^2 (2 c f-b 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 )}{g (e f-d g)^3}-\frac {\left (\left (b^2-4 a c\right ) g (2 c f-b 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 )}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^2}+\frac {\left (2 e^3 \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)^4}\\ &=\frac {e^2 \sqrt {a+b x+c x^2}}{(e f-d g)^3 (f+g x)}-\frac {g (2 c f-b g) (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)^2}-\frac {e g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^3}-\frac {e^2 (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)^4}+\frac {e^3 (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)^4}-\frac {\sqrt {c} e^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^3}+\frac {e^2 \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)^4}+\frac {\left (b^2-4 a c\right ) g (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 )}{16 (e f-d g) \left (c f^2-b f g+a g^2\right )^{5/2}}+\frac {\left (b^2-4 a c\right ) e 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)^2 \left (c f^2-b f g+a g^2\right )^{3/2}}+\frac {e^2 (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)^3 \sqrt {c f^2-b f g+a g^2}}-\frac {e^3 \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)^4}\\ \end {align*}

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Mathematica [A]  time = 4.16, size = 858, normalized size = 0.92 \begin {gather*} \frac {\frac {24 \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 {c f^2+g (a g-b f)} \tanh ^{-1}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (a g-b f)} \sqrt {a+x (b+c x)}}\right )\right ) e^3}{\sqrt {c} g}+24 \left (\frac {(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}}+2 \sqrt {c d^2+e (a e-b d)} \tanh ^{-1}\left (\frac {-2 a e+2 c d x+b (d-e x)}{2 \sqrt {c d^2+e (a e-b d)} \sqrt {a+x (b+c x)}}\right )\right ) e^2-\frac {24 (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+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (a g-b f)} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c f^2+g (a g-b f)}}\right ) e^2}{g}+\frac {48 (e f-d g) \sqrt {a+x (b+c x)} e^2}{f+g x}+\frac {6 \left (b^2-4 a c\right ) g (e f-d g)^2 \tanh ^{-1}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (a g-b f)} \sqrt {a+x (b+c x)}}\right ) e}{\left (c f^2+g (a g-b f)\right )^{3/2}}+\frac {12 g (e f-d g)^2 (-b f-2 c x f+2 a g+b g x) \sqrt {a+x (b+c x)} e}{\left (c f^2+g (a g-b f)\right ) (f+g x)^2}-\frac {16 g^2 (d g-e f)^3 (a+x (b+c x))^{3/2}}{\left (c f^2+g (a g-b f)\right ) (f+g x)^3}-\frac {3 g (2 c f-b g) (e f-d g)^3 \left (\frac {2 \sqrt {a+x (b+c x)} (-2 a g+2 c f x+b (f-g x))}{\left (c f^2+g (a g-b f)\right ) (f+g x)^2}+\frac {\left (4 a c-b^2\right ) \tanh ^{-1}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (a g-b f)} \sqrt {a+x (b+c x)}}\right )}{\left (c f^2+g (a g-b f)\right )^{3/2}}\right )}{c f^2+g (a g-b f)}}{48 (e f-d g)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((48*e^2*(e*f - d*g)*Sqrt[a + x*(b + c*x)])/(f + g*x) + (12*e*g*(e*f - d*g)^2*(-(b*f) + 2*a*g - 2*c*f*x + b*g*
x)*Sqrt[a + x*(b + c*x)])/((c*f^2 + g*(-(b*f) + a*g))*(f + g*x)^2) - (16*g^2*(-(e*f) + d*g)^3*(a + x*(b + c*x)
)^(3/2))/((c*f^2 + g*(-(b*f) + a*g))*(f + g*x)^3) + 24*e^2*(((-2*c*d + b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + x*(b + c*x)])])/Sqrt[c] + 2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*ArcTanh[(-2*a*e + 2*c*d*x + b*(d - e*x))/(2*S
qrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])]) + (6*(b^2 - 4*a*c)*e*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) - (3*g*(2*c*f - b*g)*(e*f - d*g)^3*((2*Sqrt[a + x*(b + c*x)]*(-2*a*g + 2*c*f*x + b*(f - g*x)))/((c*f^2 +
g*(-(b*f) + a*g))*(f + g*x)^2) + ((-b^2 + 4*a*c)*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)))/(c*f^2 + g*(-(b*f) + a*g)) - (24*e^2
*(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 + (24*e^3*((2*c*f - b*g)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] - 2*Sqrt[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))/(48*(e*f - d*g)^4)

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IntegrateAlgebraic [F]  time = 180.02, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)^4,x, algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)^4,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.02, size = 11995, normalized size = 12.86 \begin {gather*} \text {output too large to display} \end {gather*}

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)^4,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c x^{2} + b x + a}}{{\left (e x + d\right )} {\left (g x + f\right )}^{4}}\,{d x} \end {gather*}

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)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x + a)/((e*x + d)*(g*x + f)^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (f+g\,x\right )}^4\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(1/2)/((f + g*x)^4*(d + e*x)),x)

[Out]

int((a + b*x + c*x^2)^(1/2)/((f + g*x)^4*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x + c x^{2}}}{\left (d + e x\right ) \left (f + g x\right )^{4}}\, dx \end {gather*}

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)**4,x)

[Out]

Integral(sqrt(a + b*x + c*x**2)/((d + e*x)*(f + g*x)**4), x)

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