3.884 \(\int \frac {1}{(d+e x) (f+g x)^2 (a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=642 \[ \frac {g^2 \sqrt {a+b x+c x^2} \left (-4 c g (2 a g+b f)+3 b^2 g^2+4 c^2 f^2\right )}{\left (b^2-4 a c\right ) (f+g x) (e f-d g) \left (a g^2-b f g+c f^2\right )^2}-\frac {2 e^2 \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (e f-d g)^2 \left (a e^2-b d e+c d^2\right )}+\frac {2 e g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}+\frac {2 g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) (f+g x) \sqrt {a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac {e^4 \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {e g^3 \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{(e f-d g)^2 \left (a g^2-b f g+c f^2\right )^{3/2}}-\frac {3 g^3 (2 c f-b g) \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 (e f-d g) \left (a g^2-b f g+c f^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.91, antiderivative size = 642, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {960, 740, 12, 724, 206, 806} \[ \frac {g^2 \sqrt {a+b x+c x^2} \left (-4 c g (2 a g+b f)+3 b^2 g^2+4 c^2 f^2\right )}{\left (b^2-4 a c\right ) (f+g x) (e f-d g) \left (a g^2-b f g+c f^2\right )^2}-\frac {2 e^2 \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (e f-d g)^2 \left (a e^2-b d e+c d^2\right )}+\frac {2 e g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}+\frac {2 g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) (f+g x) \sqrt {a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac {e^4 \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {e g^3 \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{(e f-d g)^2 \left (a g^2-b f g+c f^2\right )^{3/2}}-\frac {3 g^3 (2 c f-b g) \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 (e f-d g) \left (a g^2-b f g+c f^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*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] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 806

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

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Mathematica [A]  time = 5.08, size = 623, normalized size = 0.97 \[ \frac {g^2 \left (\frac {3 g (b g-2 c f) \tanh ^{-1}\left (\frac {2 a g-b f+b g x-2 c f x}{2 \sqrt {a+x (b+c x)} \sqrt {g (a g-b f)+c f^2}}\right )}{\left (g (a g-b f)+c f^2\right )^{5/2}}-\frac {2 \sqrt {a+x (b+c x)} \left (-4 c g (2 a g+b f)+3 b^2 g^2+4 c^2 f^2\right )}{\left (b^2-4 a c\right ) (f+g x) \left (g (a g-b f)+c f^2\right )^2}\right )}{2 (d g-e f)}-\frac {2 e^2 \left (-2 c (a e+c d x)+b^2 e+b c (e x-d)\right )}{\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} (e f-d g)^2 \left (e (b d-a e)-c d^2\right )}+\frac {2 e g \left (-2 c (a g+c f x)+b^2 g+b c (g x-f)\right )}{\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} (e f-d g)^2 \left (g (b f-a g)-c f^2\right )}-\frac {2 g \left (-2 c (a g+c f x)+b^2 g+b c (g x-f)\right )}{\left (b^2-4 a c\right ) (f+g x) \sqrt {a+x (b+c x)} (d g-e f) \left (g (b f-a g)-c f^2\right )}+\frac {e^4 \tanh ^{-1}\left (\frac {-2 a e+b (d-e x)+2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{(e f-d g)^2 \left (e (a e-b d)+c d^2\right )^{3/2}}-\frac {e g^3 \tanh ^{-1}\left (\frac {-2 a g+b (f-g x)+2 c f x}{2 \sqrt {a+x (b+c x)} \sqrt {g (a g-b f)+c f^2}}\right )}{(e f-d g)^2 \left (g (a g-b f)+c f^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.03, size = 2807, normalized size = 4.37 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(d*g-e*f)^2*e^3/(a*e^2-b*d*e+c*d^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/(d*g
-e*f)^2*e/(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(((b*g-2*c*f)*(x+f/g)/g+2*(a*g^2-b*f*g+c*f
^2)/g^2+2*((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/(
x+f/g))-1/(d*g-e*f)^2*e^3/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^
2)/e^2)^(1/2)*b^2+3/2*g^3/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^
2-b*f*g+c*f^2)/g^2)^(1/2)*b^3+3*g^2/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-
b*f*g+c*f^2)/g^2)^(1/2)*c*f-1/(d*g-e*f)^2*e^3/(a*e^2-b*d*e+c*d^2)/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c
*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(
a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))-g/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)/(x+f/g)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/
g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)+3*g^3/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*f)
*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*x*b^2*c+12*g/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/(4*a*c-b^2)/((x+f/g)^2*
c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*x*c^3*f^2-6*g^2/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/(4*a*c-
b^2)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*b^2*c*f+6*g/(d*g-e*f)/(a*g^2-b*f*g+c*f^
2)^2/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*b*c^2*f^2-2/(d*g-e*f)^2*e^3
/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b*c+4/(d*
g-e*f)^2*e^2/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)
*x*c^2*d+2/(d*g-e*f)^2*e^2/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d
^2)/e^2)^(1/2)*b*c*d-12*g^2/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*
g^2-b*f*g+c*f^2)/g^2)^(1/2)*x*b*c^2*f-2/(d*g-e*f)^2*e*g/(a*g^2-b*f*g+c*f^2)/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*
f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*b*c*f+2/(d*g-e*f)^2*e*g^2/(a*g^2-b*f*g+c*f^2)/(4*a*c-b^2)/((x+f/g)
^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*x*b*c-3/2*g^3/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/((x+f/
g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*b-1/(d*g-e*f)^2*e/(a*g^2-b*f*g+c*f^2)*g^2/((x+f/g)
^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)-4/(d*g-e*f)^2*e*g/(a*g^2-b*f*g+c*f^2)/(4*a*c-b^2)/((
x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*x*c^2*f-3*g^2/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/
((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*ln(((b*g-2*c*f)*(x+f/g)/g+2*(a*g^2-b*f*g+c*f^2)/g^2+2*((a*g^2-b*f*g+c*f^2)/g^2
)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/(x+f/g))*c*f-8*g/(d*g-e*f)*c^2/(a*g
^2-b*f*g+c*f^2)/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*x-4*g/(d*g-e*f)*
c/(a*g^2-b*f*g+c*f^2)/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*b+1/(d*g-e
*f)^2*e*g^2/(a*g^2-b*f*g+c*f^2)/(4*a*c-b^2)/((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*
b^2+3/2*g^3/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)^2/((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*ln(((b*g-2*c*f)*(x+f/g)/g+2*(a*g^2
-b*f*g+c*f^2)/g^2+2*((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2
)^(1/2))/(x+f/g))*b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (e x + d\right )} {\left (g x + f\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (f+g\,x\right )}^2\,\left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x\right ) \left (f + g x\right )^{2} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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