\(\int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx\) [877]

Optimal result

Integrand size = 29, antiderivative size = 587 \[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=\frac {g^2 \sqrt {a+b x+c x^2}}{2 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {3 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)}+\frac {e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)}+\frac {e^3 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c d^2-b d e+a e^2} (e f-d g)^3}-\frac {e g (2 c f-b g) \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{2 (e f-d g)^2 \left (c f^2-b f g+a g^2\right )^{3/2}}-\frac {e^2 g \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{(e f-d g)^3 \sqrt {c f^2-b f g+a g^2}}-\frac {g \left (8 c^2 f^2+3 b^2 g^2-4 c g (2 b f+a g)\right ) \text {arctanh}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^{5/2}} \]

[Out]

-1/2*e*g*(-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))/(-
d*g+e*f)^2/(a*g^2-b*f*g+c*f^2)^(3/2)-1/8*g*(8*c^2*f^2+3*b^2*g^2-4*c*g*(a*g+2*b*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))/(-d*g+e*f)/(a*g^2-b*f*g+c*f^2)^(5/2)+e^3*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))/(-d*g+e*f)^3/(a*e^2-b*d*e+c*d^2)^(
1/2)-e^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)^3/
(a*g^2-b*f*g+c*f^2)^(1/2)+1/2*g^2*(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+3/4*g^2*(-b*g+2
*c*f)*(c*x^2+b*x+a)^(1/2)/(-d*g+e*f)/(a*g^2-b*f*g+c*f^2)^2/(g*x+f)+e*g^2*(c*x^2+b*x+a)^(1/2)/(-d*g+e*f)^2/(a*g
^2-b*f*g+c*f^2)/(g*x+f)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 587, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {974, 738, 212, 758, 820, 744} \[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=-\frac {g \left (-4 c g (a g+2 b f)+3 b^2 g^2+8 c^2 f^2\right ) \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{8 (e f-d g) \left (a g^2-b f g+c f^2\right )^{5/2}}+\frac {e^3 \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^3 \sqrt {a e^2-b d e+c d^2}}-\frac {e^2 g \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{(e f-d g)^3 \sqrt {a g^2-b f g+c f^2}}-\frac {e g (2 c f-b g) \text {arctanh}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 (e f-d g)^2 \left (a g^2-b f g+c f^2\right )^{3/2}}+\frac {e g^2 \sqrt {a+b x+c x^2}}{(f+g x) (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}+\frac {3 g^2 \sqrt {a+b x+c x^2} (2 c f-b g)}{4 (f+g x) (e f-d g) \left (a g^2-b f g+c f^2\right )^2}+\frac {g^2 \sqrt {a+b x+c x^2}}{2 (f+g x)^2 (e f-d g) \left (a g^2-b f g+c f^2\right )} \]

[In]

Int[1/((d + e*x)*(f + g*x)^3*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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 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 744

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 758

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[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[S
implify[m + 2*p + 3], 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}{(e f-d g)^3 (d+e x) \sqrt {a+b x+c x^2}}-\frac {g}{(e f-d g) (f+g x)^3 \sqrt {a+b x+c x^2}}-\frac {e g}{(e f-d g)^2 (f+g x)^2 \sqrt {a+b x+c x^2}}-\frac {e^2 g}{(e f-d g)^3 (f+g x) \sqrt {a+b x+c x^2}}\right ) \, dx \\ & = \frac {e^3 \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^3}-\frac {\left (e^2 g\right ) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^3}-\frac {(e g) \int \frac {1}{(f+g x)^2 \sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^2}-\frac {g \int \frac {1}{(f+g x)^3 \sqrt {a+b x+c x^2}} \, dx}{e f-d g} \\ & = \frac {g^2 \sqrt {a+b x+c x^2}}{2 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)}-\frac {\left (2 e^3\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 (2 e^2 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 )}{(e f-d g)^3}-\frac {(e g (2 c f-b g)) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g)^2 \left (c f^2-b f g+a g^2\right )}+\frac {g \int \frac {\frac {1}{2} (-4 c f+3 b g)+c g x}{(f+g x)^2 \sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g) \left (c f^2-b f g+a g^2\right )} \\ & = \frac {g^2 \sqrt {a+b x+c x^2}}{2 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {3 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)}+\frac {e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)}+\frac {e^3 \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 )}{\sqrt {c d^2-b d e+a e^2} (e f-d g)^3}-\frac {e^2 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 )}{(e f-d g)^3 \sqrt {c f^2-b f g+a g^2}}+\frac {(e g (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 )}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right )}-\frac {\left (g \left (8 c^2 f^2+3 b^2 g^2-4 c g (2 b f+a g)\right )\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 )^2} \\ & = \frac {g^2 \sqrt {a+b x+c x^2}}{2 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {3 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)}+\frac {e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)}+\frac {e^3 \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 )}{\sqrt {c d^2-b d e+a e^2} (e f-d g)^3}-\frac {e 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 )}{2 (e f-d g)^2 \left (c f^2-b f g+a g^2\right )^{3/2}}-\frac {e^2 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 )}{(e f-d g)^3 \sqrt {c f^2-b f g+a g^2}}+\frac {\left (g \left (8 c^2 f^2+3 b^2 g^2-4 c g (2 b f+a g)\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 )}{4 (e f-d g) \left (c f^2-b f g+a g^2\right )^2} \\ & = \frac {g^2 \sqrt {a+b x+c x^2}}{2 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {3 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)}+\frac {e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) (f+g x)}+\frac {e^3 \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 )}{\sqrt {c d^2-b d e+a e^2} (e f-d g)^3}-\frac {e 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 )}{2 (e f-d g)^2 \left (c f^2-b f g+a g^2\right )^{3/2}}-\frac {e^2 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 )}{(e f-d g)^3 \sqrt {c f^2-b f g+a g^2}}-\frac {g \left (8 c^2 f^2+3 b^2 g^2-4 c g (2 b f+a g)\right ) \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 )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.31 (sec) , antiderivative size = 549, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=\frac {\frac {4 g^2 (e f-d g)^2 \sqrt {a+x (b+c x)}}{\left (c f^2+g (-b f+a g)\right ) (f+g x)^2}+\frac {8 e g^2 (e f-d g) \sqrt {a+x (b+c x)}}{\left (c f^2+g (-b f+a g)\right ) (f+g x)}+\frac {8 e^3 \text {arctanh}\left (\frac {-2 a e+2 c d x+b (d-e x)}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c d^2+e (-b d+a e)}}+\frac {4 e g (-2 c f+b g) (e f-d g) \text {arctanh}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )}{\left (c f^2+g (-b f+a g)\right )^{3/2}}-\frac {8 e^2 g \text {arctanh}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c f^2+g (-b f+a g)}}+g (e f-d g)^2 \left (\frac {6 g (2 c f-b g) \sqrt {a+x (b+c x)}}{\left (c f^2+g (-b f+a g)\right )^2 (f+g x)}-\frac {\left (8 c^2 f^2+3 b^2 g^2-4 c g (2 b f+a g)\right ) \text {arctanh}\left (\frac {-2 a g+2 c f x+b (f-g x)}{2 \sqrt {c f^2+g (-b f+a g)} \sqrt {a+x (b+c x)}}\right )}{\left (c f^2+g (-b f+a g)\right )^{5/2}}\right )}{8 (e f-d g)^3} \]

[In]

Integrate[1/((d + e*x)*(f + g*x)^3*Sqrt[a + b*x + c*x^2]),x]

[Out]

((4*g^2*(e*f - d*g)^2*Sqrt[a + x*(b + c*x)])/((c*f^2 + g*(-(b*f) + a*g))*(f + g*x)^2) + (8*e*g^2*(e*f - d*g)*S
qrt[a + x*(b + c*x)])/((c*f^2 + g*(-(b*f) + a*g))*(f + g*x)) + (8*e^3*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)])])/Sqrt[c*d^2 + e*(-(b*d) + a*e)] + (4*e*g*(-2*c*f +
b*g)*(e*f - d*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)])])/(c*f^2 + g*(-(b*f) + a*g))^(3/2) - (8*e^2*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*(e*f - d*g)^2*((6*g*(2*c*f - b*g)*
Sqrt[a + x*(b + c*x)])/((c*f^2 + g*(-(b*f) + a*g))^2*(f + g*x)) - ((8*c^2*f^2 + 3*b^2*g^2 - 4*c*g*(2*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)])])/(c*f^2
+ g*(-(b*f) + a*g))^(5/2)))/(8*(e*f - d*g)^3)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1184\) vs. \(2(541)=1082\).

Time = 0.91 (sec) , antiderivative size = 1185, normalized size of antiderivative = 2.02

method	result	size
default	\(\text {Expression too large to display}\)	\(1185\)

[In]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{\left (d + e x\right ) \left (f + g x\right )^{3} \sqrt {a + b x + c x^{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} {\left (g x + f\right )}^{3}} \,d x } \]

[In]

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

[Out]

integrate(1/(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. 2239 vs. \(2 (541) = 1082\).

Time = 0.77 (sec) , antiderivative size = 2239, normalized size of antiderivative = 3.81 \[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-2*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*(24*c^2*e^2*f^4*g - 24*c^2*d*e*f^3*
g^2 - 36*b*c*e^2*f^3*g^2 + 8*c^2*d^2*f^2*g^3 + 28*b*c*d*e*f^2*g^3 + 15*b^2*e^2*f^2*g^3 + 20*a*c*e^2*f^2*g^3 -
8*b*c*d^2*f*g^4 - 10*b^2*d*e*f*g^4 - 20*a*b*e^2*f*g^4 + 3*b^2*d^2*g^5 - 4*a*c*d^2*g^5 + 4*a*b*d*e*g^5 + 8*a^2*
e^2*g^5)*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^2*e^3*f
^7 - 3*c^2*d*e^2*f^6*g - 2*b*c*e^3*f^6*g + 3*c^2*d^2*e*f^5*g^2 + 6*b*c*d*e^2*f^5*g^2 + b^2*e^3*f^5*g^2 + 2*a*c
*e^3*f^5*g^2 - c^2*d^3*f^4*g^3 - 6*b*c*d^2*e*f^4*g^3 - 3*b^2*d*e^2*f^4*g^3 - 6*a*c*d*e^2*f^4*g^3 - 2*a*b*e^3*f
^4*g^3 + 2*b*c*d^3*f^3*g^4 + 3*b^2*d^2*e*f^3*g^4 + 6*a*c*d^2*e*f^3*g^4 + 6*a*b*d*e^2*f^3*g^4 + a^2*e^3*f^3*g^4
- b^2*d^3*f^2*g^5 - 2*a*c*d^3*f^2*g^5 - 6*a*b*d^2*e*f^2*g^5 - 3*a^2*d*e^2*f^2*g^5 + 2*a*b*d^3*f*g^6 + 3*a^2*d
^2*e*f*g^6 - a^2*d^3*g^7)*sqrt(-c*f^2 + b*f*g - a*g^2)) + 1/4*(16*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*c^2*e*
f^3*g^2 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*c^2*d*f^2*g^3 - 20*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c
*e*f^2*g^3 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c*d*f*g^4 + 7*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2
*e*f*g^4 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c*e*f*g^4 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*d
*g^5 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c*d*g^5 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*e*g^5 +
40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(5/2)*e*f^4*g - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(5/2)*d
*f^3*g^2 - 44*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^(3/2)*e*f^3*g^2 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a
))^2*b*c^(3/2)*d*f^2*g^3 + 13*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^2*sqrt(c)*e*f^2*g^3 - 4*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))^2*a*c^(3/2)*e*f^2*g^3 - 9*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^2*sqrt(c)*d*f*g^4 + 12*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*c^(3/2)*d*f*g^4 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b*sqrt(c)*e
*f*g^4 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^2*sqrt(c)*e*g^5 + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*
c^2*e*f^4*g - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*c^2*d*f^3*g^2 - 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*
b^2*c*e*f^3*g^2 - 64*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*c^2*e*f^3*g^2 + 20*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*b^2*c*d*f^2*g^3 + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*c^2*d*f^2*g^3 + 9*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*b^3*e*f^2*g^3 + 72*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b*c*e*f^2*g^3 - 5*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))*b^3*d*f*g^4 - 28*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b*c*d*f*g^4 - 13*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*a*b^2*e*f*g^4 - 28*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*c*e*f*g^4 + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*a*b^2*d*g^5 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*c*d*g^5 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a
^2*b*e*g^5 + 10*b^2*c^(3/2)*e*f^4*g - 6*b^2*c^(3/2)*d*f^3*g^2 - 7*b^3*sqrt(c)*e*f^3*g^2 - 32*a*b*c^(3/2)*e*f^3
*g^2 + 3*b^3*sqrt(c)*d*f^2*g^3 + 20*a*b*c^(3/2)*d*f^2*g^3 + 27*a*b^2*sqrt(c)*e*f^2*g^3 + 20*a^2*c^(3/2)*e*f^2*
g^3 - 11*a*b^2*sqrt(c)*d*f*g^4 - 12*a^2*c^(3/2)*d*f*g^4 - 28*a^2*b*sqrt(c)*e*f*g^4 + 8*a^2*b*sqrt(c)*d*g^5 + 8
*a^3*sqrt(c)*e*g^5)/((c^2*e^2*f^6 - 2*c^2*d*e*f^5*g - 2*b*c*e^2*f^5*g + c^2*d^2*f^4*g^2 + 4*b*c*d*e*f^4*g^2 +
b^2*e^2*f^4*g^2 + 2*a*c*e^2*f^4*g^2 - 2*b*c*d^2*f^3*g^3 - 2*b^2*d*e*f^3*g^3 - 4*a*c*d*e*f^3*g^3 - 2*a*b*e^2*f^
3*g^3 + b^2*d^2*f^2*g^4 + 2*a*c*d^2*f^2*g^4 + 4*a*b*d*e*f^2*g^4 + a^2*e^2*f^2*g^4 - 2*a*b*d^2*f*g^5 - 2*a^2*d*
e*f*g^5 + a^2*d^2*g^6)*((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. \[ \int \frac {1}{(d+e x) (f+g x)^3 \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{{\left (f+g\,x\right )}^3\,\left (d+e\,x\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

[In]

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

[Out]

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