\(\int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx\) [860]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [F(-1)]
   Sympy [F]
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 29, antiderivative size = 673 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\frac {e \sqrt {a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac {g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}-\frac {e (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} (e f-d g)^3}+\frac {e^2 (2 c f-b g) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} g (e f-d g)^3}-\frac {\sqrt {c} e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^2}+\frac {e \sqrt {c d^2-b d e+a e^2} \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{(e f-d g)^3}+\frac {\left (b^2-4 a c\right ) g \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}}+\frac {e (2 c f-b g) \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{2 g (e f-d g)^2 \sqrt {c f^2-b f g+a g^2}}-\frac {e^2 \sqrt {c f^2-b f g+a g^2} \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^3} \]

[Out]

1/8*(-4*a*c+b^2)*g*arctanh(1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c*f^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/(-d*g
+e*f)/(a*g^2-b*f*g+c*f^2)^(3/2)-1/2*e*(-b*e+2*c*d)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/(-d*g+e*
f)^3/c^(1/2)+1/2*e^2*(-b*g+2*c*f)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/g/(-d*g+e*f)^3/c^(1/2)-e*
arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))*c^(1/2)/g/(-d*g+e*f)^2+e*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d
)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))*(a*e^2-b*d*e+c*d^2)^(1/2)/(-d*g+e*f)^3+1/2*e*(-b*g+2*c*f)*
arctanh(1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c*f^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/g/(-d*g+e*f)^2/(a*g^2-b*
f*g+c*f^2)^(1/2)-e^2*arctanh(1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(a*g^2-b*f*g+c*f^2)^(1/2)/(c*x^2+b*x+a)^(1/2))*(a*
g^2-b*f*g+c*f^2)^(1/2)/g/(-d*g+e*f)^3+e*(c*x^2+b*x+a)^(1/2)/(-d*g+e*f)^2/(g*x+f)-1/4*g*(b*f-2*a*g+(-b*g+2*c*f)
*x)*(c*x^2+b*x+a)^(1/2)/(-d*g+e*f)/(a*g^2-b*f*g+c*f^2)/(g*x+f)^2

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 673, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {974, 748, 857, 635, 212, 738, 734, 746} \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\frac {g \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{8 (e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}}+\frac {e \sqrt {a e^2-b d e+c d^2} \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^3}-\frac {e^2 \sqrt {a g^2-b f g+c f^2} \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{g (e f-d g)^3}+\frac {e^2 (2 c f-b g) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} g (e f-d g)^3}+\frac {e (2 c f-b g) \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 g (e f-d g)^2 \sqrt {a g^2-b f g+c f^2}}-\frac {\sqrt {c} e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^2}-\frac {e (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} (e f-d g)^3}-\frac {g \sqrt {a+b x+c x^2} (-2 a g+x (2 c f-b g)+b f)}{4 (f+g x)^2 (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac {e \sqrt {a+b x+c x^2}}{(f+g x) (e f-d g)^2} \]

[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 212

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

Rule 635

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 734

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] /; Free
Q[{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 738

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 746

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 748

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 857

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 974

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*} \text {integral}& = \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 ) \text {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)) \text {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 ) \text {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 ) \text {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) \text {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)) \text {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 ) \text {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] (verified)

Time = 10.90 (sec) , antiderivative size = 609, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\frac {\frac {8 e (e f-d g) \sqrt {a+x (b+c x)}}{f+g x}+\frac {2 g (e f-d g)^2 (-b f+2 a g-2 c f x+b g x) \sqrt {a+x (b+c x)}}{\left (c f^2+g (-b f+a g)\right ) (f+g x)^2}+\frac {4 e (-2 c d+b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}+8 e \sqrt {c d^2+e (-b d+a e)} \text {arctanh}\left (\frac {-2 a e+2 c d x+b (d-e x)}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )+\frac {\left (b^2-4 a c\right ) g (e f-d g)^2 \text {arctanh}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )}{\left (c f^2+g (-b f+a g)\right )^{3/2}}-\frac {4 e (e f-d g) \left (2 \sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-\frac {(2 c f-b g) \text {arctanh}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c f^2+g (-b f+a g)}}\right )}{g}+\frac {4 e^2 \left ((2 c f-b g) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} \sqrt {c f^2+g (-b f+a g)} \text {arctanh}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )\right )}{\sqrt {c} g}}{8 (e f-d g)^3} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2511\) vs. \(2(603)=1206\).

Time = 0.99 (sec) , antiderivative size = 2512, normalized size of antiderivative = 3.73

method result size
default \(\text {Expression too large to display}\) \(2512\)

[In]

int((c*x^2+b*x+a)^(1/2)/(e*x+d)/(g*x+f)^3,x,method=_RETURNVERBOSE)

[Out]

1/g^2/(d*g-e*f)*(-1/2/(a*g^2-b*f*g+c*f^2)*g^2/(x+f/g)^2*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)
/g^2)^(3/2)-1/4*(b*g-2*c*f)*g/(a*g^2-b*f*g+c*f^2)*(-1/(a*g^2-b*f*g+c*f^2)*g^2/(x+f/g)*((x+f/g)^2*c+(b*g-2*c*f)
/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(3/2)+1/2*(b*g-2*c*f)*g/(a*g^2-b*f*g+c*f^2)*(((x+f/g)^2*c+(b*g-2*c*f)/g*(x
+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)+1/2*(b*g-2*c*f)/g*ln((1/2*(b*g-2*c*f)/g+c*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(
b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/c^(1/2)-(a*g^2-b*f*g+c*f^2)/g^2/((a*g^2-b*f*g+c*f^2)/g^2)
^(1/2)*ln((2*(a*g^2-b*f*g+c*f^2)/g^2+(b*g-2*c*f)/g*(x+f/g)+2*((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g
-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/(x+f/g)))+2*c/(a*g^2-b*f*g+c*f^2)*g^2*(1/4*(2*c*(x+f/g)+(b*g
-2*c*f)/g)/c*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)+1/8*(4*c*(a*g^2-b*f*g+c*f^2)/g^
2-(b*g-2*c*f)^2/g^2)/c^(3/2)*ln((1/2*(b*g-2*c*f)/g+c*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^
2-b*f*g+c*f^2)/g^2)^(1/2))))+1/2*c/(a*g^2-b*f*g+c*f^2)*g^2*(((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*
f^2)/g^2)^(1/2)+1/2*(b*g-2*c*f)/g*ln((1/2*(b*g-2*c*f)/g+c*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+
(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/c^(1/2)-(a*g^2-b*f*g+c*f^2)/g^2/((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*ln((2*(a*g^2-b
*f*g+c*f^2)/g^2+(b*g-2*c*f)/g*(x+f/g)+2*((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*
g^2-b*f*g+c*f^2)/g^2)^(1/2))/(x+f/g))))-e^2/(d*g-e*f)^3*(((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2
)/e^2)^(1/2)+1/2*(b*e-2*c*d)/e*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*
e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+c*d^2)/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*
e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2
-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))+e^2/(d*g-e*f)^3*(((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^
2)^(1/2)+1/2*(b*g-2*c*f)/g*ln((1/2*(b*g-2*c*f)/g+c*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-
b*f*g+c*f^2)/g^2)^(1/2))/c^(1/2)-(a*g^2-b*f*g+c*f^2)/g^2/((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*ln((2*(a*g^2-b*f*g+c*
f^2)/g^2+(b*g-2*c*f)/g*(x+f/g)+2*((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f
*g+c*f^2)/g^2)^(1/2))/(x+f/g)))-1/g*e/(d*g-e*f)^2*(-1/(a*g^2-b*f*g+c*f^2)*g^2/(x+f/g)*((x+f/g)^2*c+(b*g-2*c*f)
/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(3/2)+1/2*(b*g-2*c*f)*g/(a*g^2-b*f*g+c*f^2)*(((x+f/g)^2*c+(b*g-2*c*f)/g*(x
+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)+1/2*(b*g-2*c*f)/g*ln((1/2*(b*g-2*c*f)/g+c*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(
b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/c^(1/2)-(a*g^2-b*f*g+c*f^2)/g^2/((a*g^2-b*f*g+c*f^2)/g^2)
^(1/2)*ln((2*(a*g^2-b*f*g+c*f^2)/g^2+(b*g-2*c*f)/g*(x+f/g)+2*((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g
-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/(x+f/g)))+2*c/(a*g^2-b*f*g+c*f^2)*g^2*(1/4*(2*c*(x+f/g)+(b*g
-2*c*f)/g)/c*((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)+1/8*(4*c*(a*g^2-b*f*g+c*f^2)/g^
2-(b*g-2*c*f)^2/g^2)/c^(3/2)*ln((1/2*(b*g-2*c*f)/g+c*(x+f/g))/c^(1/2)+((x+f/g)^2*c+(b*g-2*c*f)/g*(x+f/g)+(a*g^
2-b*f*g+c*f^2)/g^2)^(1/2))))

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\int \frac {\sqrt {a + b x + c x^{2}}}{\left (d + e x\right ) \left (f + g x\right )^{3}}\, dx \]

[In]

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

[Out]

Integral(sqrt(a + b*x + c*x**2)/((d + e*x)*(f + g*x)**3), x)

Maxima [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a}}{{\left (e x + d\right )} {\left (g x + f\right )}^{3}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1824 vs. \(2 (603) = 1206\).

Time = 1.07 (sec) , antiderivative size = 1824, normalized size of antiderivative = 2.71 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (f+g\,x\right )}^3\,\left (d+e\,x\right )} \,d x \]

[In]

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

[Out]

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