3.9.83 \(\int \frac {1}{(d+e x) (f+g x) (a+b x+c x^2)^{3/2}} \, dx\) [883]

Optimal. Leaf size=352 \[ -\frac {2 e \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (e f-d g) \sqrt {a+b x+c x^2}}+\frac {2 g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {a+b x+c x^2}}+\frac {e^3 \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{\left (c d^2-b d e+a e^2\right )^{3/2} (e f-d g)}-\frac {g^3 \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{(e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.29, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {974, 754, 12, 738, 212} \begin {gather*} -\frac {2 e \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) \left (a e^2-b d e+c d^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 ) \sqrt {a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac {e^3 \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) \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {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) \left (a g^2-b f g+c f^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 3.12, size = 347, normalized size = 0.99 \begin {gather*} \frac {2 \left (-b^3 e g+b^2 c (d g+e (f-g x))-2 c^2 (a d g+c d f x+a e (f-g x))+b c (3 a e g+c (-d f+e f x+d g x))\right )}{\left (b^2-4 a c\right ) \left (-c d^2+e (b d-a e)\right ) \left (-c f^2+g (b f-a g)\right ) \sqrt {a+x (b+c x)}}-\frac {2 e^3 \sqrt {-c d^2+b d e-a e^2} \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{\left (c d^2+e (-b d+a e)\right )^2 (-e f+d g)}-\frac {2 g^3 \sqrt {-c f^2+b f g-a g^2} \tan ^{-1}\left (\frac {\sqrt {c} (f+g x)-g \sqrt {a+x (b+c x)}}{\sqrt {-c f^2+g (b f-a g)}}\right )}{(e f-d g) \left (c f^2+g (-b f+a g)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(814\) vs. \(2(332)=664\).
time = 0.16, size = 815, normalized size = 2.32

method	result	size
default	\(-\frac {\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\left (e b -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {e b -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (e b -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{d g -e f}+\frac {\frac {g^{2}}{\left (a \,g^{2}-b f g +c \,f^{2}\right ) \sqrt {\left (x +\frac {f}{g}\right )^{2} c +\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}-\frac {\left (b g -2 c f \right ) g \left (2 c \left (x +\frac {f}{g}\right )+\frac {b g -2 c f}{g}\right )}{\left (a \,g^{2}-b f g +c \,f^{2}\right ) \left (\frac {4 c \left (a \,g^{2}-b f g +c \,f^{2}\right )}{g^{2}}-\frac {\left (b g -2 c f \right )^{2}}{g^{2}}\right ) \sqrt {\left (x +\frac {f}{g}\right )^{2} c +\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}-\frac {g^{2} \ln \left (\frac {\frac {2 a \,g^{2}-2 b f g +2 c \,f^{2}}{g^{2}}+\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+2 \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}\, \sqrt {\left (x +\frac {f}{g}\right )^{2} c +\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}{x +\frac {f}{g}}\right )}{\left (a \,g^{2}-b f g +c \,f^{2}\right ) \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}}{d g -e f}\)	\(815\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d*g-e*f)*(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)))+1/(d*g-e*f)*(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)))

________________________________________________________________________________________

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)/(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)*(x*e + d)), x)

________________________________________________________________________________________

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)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________