3.9.84 \(\int \frac {1}{(d+e x) (f+g x)^2 (a+b x+c x^2)^{3/2}} \, dx\) [884]

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

[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 = 0.57, 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 = {974, 754, 12, 738, 212, 820} \begin {gather*} \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}} \end {gather*}

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 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 754

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 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*} \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 ) \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 )}{\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 ) \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) \left (c f^2-b f g+a g^2\right )^2}+\frac {\left (2 e g^3\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)^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 = 13.24, size = 623, normalized size = 0.97 \begin {gather*} -\frac {2 e^2 \left (b^2 e-2 c (a e+c d x)+b c (-d+e x)\right )}{\left (b^2-4 a c\right ) \left (-c d^2+e (b d-a e)\right ) (e f-d g)^2 \sqrt {a+x (b+c x)}}+\frac {2 e g \left (b^2 g-2 c (a g+c f x)+b c (-f+g x)\right )}{\left (b^2-4 a c\right ) (e f-d g)^2 \left (-c f^2+g (b f-a g)\right ) \sqrt {a+x (b+c x)}}-\frac {2 g \left (b^2 g-2 c (a g+c f x)+b c (-f+g x)\right )}{\left (b^2-4 a c\right ) (-e f+d g) \left (-c f^2+g (b f-a g)\right ) (f+g x) \sqrt {a+x (b+c x)}}+\frac {g^2 \left (-\frac {2 \left (4 c^2 f^2+3 b^2 g^2-4 c g (b f+2 a g)\right ) \sqrt {a+x (b+c x)}}{\left (b^2-4 a c\right ) \left (c f^2+g (-b f+a g)\right )^2 (f+g x)}+\frac {3 g (-2 c f+b g) \tanh ^{-1}\left (\frac {-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)}}\right )}{\left (c f^2+g (-b f+a g)\right )^{5/2}}\right )}{2 (-e f+d g)}+\frac {e^4 \tanh ^{-1}\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 )}{\left (c d^2+e (-b d+a e)\right )^{3/2} (e f-d g)^2}-\frac {e g^3 \tanh ^{-1}\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 )}{(e f-d g)^2 \left (c f^2+g (-b f+a g)\right )^{3/2}} \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1480\) vs. \(2(608)=1216\).
time = 0.16, size = 1481, normalized size = 2.31

method	result	size
default	\(\text {Expression too large to display}\)	\(1481\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)*(g*x + f)^2*(x*e + d)), x)

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Fricas [F(-1)] Timed out
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(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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Sympy [F(-1)] Timed out
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(1/(e*x+d)/(g*x+f)**2/(c*x**2+b*x+a)**(3/2),x)

[Out]

Timed out

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

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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