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

Optimal. Leaf size=416 \[ -\frac {f \left (2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {c (x (c d-a f)+a e)}{a \sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.62, antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {976, 1034, 725, 206} \[ -\frac {f \left (2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {c (x (c d-a f)+a e)}{a \sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(a*e + (c*d - a*f)*x))/(a*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[a + c*x^2]) - (f*(2*a*f^2 + c*(e^2 - 2*d*f + e*Sqr
t[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt
[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e^2
- 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + (f*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e
+ Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqr
t[2]*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])])

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 976

Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((2*a*c^2*e + c*(2
*c^2*d - c*(2*a*f))*x)*(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1))/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p +
1)), x] - Dist[1/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Si
mp[2*c*((c*d - a*f)^2 - (-(a*e))*(c*e))*(p + 1) - (2*c^2*d - c*(2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(-2*a*
c^2*e)*(p + q + 2) + (2*f*(2*a*c^2*e)*(p + q + 2) - (2*c^2*d - c*(2*a*f))*(-(c*e*(2*p + q + 4))))*x + c*f*(2*c
^2*d - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && Lt
Q[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

Rule 1034

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=\frac {c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}-\frac {\int \frac {-2 a c \left (a f^2+c \left (e^2-d f\right )\right )-2 a c^2 e f x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 a c \left (a c e^2+(c d-a f)^2\right )}\\ &=\frac {c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}-\frac {\left (f \left (2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )\right )\right ) \int \frac {1}{\left (e+\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}+\frac {\left (f \left (2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )\right )\right ) \int \frac {1}{\left (e-\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}\\ &=\frac {c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {\left (f \left (2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a f^2+c \left (e+\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {a+c x^2}}\right )}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}-\frac {\left (f \left (2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {a+c x^2}}\right )}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}\\ &=\frac {c (a e+(c d-a f) x)}{a \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}-\frac {f \left (2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {f \left (2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}\\ \end {align*}

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Mathematica [A]  time = 2.06, size = 320, normalized size = 0.77 \[ \frac {c (a (e-f x)+c d x)}{a \sqrt {a+c x^2} \left (a^2 f^2+a c \left (e^2-2 d f\right )+c^2 d^2\right )}-\frac {2 \sqrt {2} f^3 \tanh ^{-1}\left (\frac {2 a f+c x \left (\sqrt {e^2-4 d f}-e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2-2 c \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt {e^2-4 d f} \left (2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac {2 \sqrt {2} f^3 \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2+2 c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {e^2-4 d f} \left (2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(c*d*x + a*(e - f*x)))/(a*(c^2*d^2 + a^2*f^2 + a*c*(e^2 - 2*d*f))*Sqrt[a + c*x^2]) - (2*Sqrt[2]*f^3*ArcTanh
[(2*a*f + c*(-e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x
^2])])/(Sqrt[e^2 - 4*d*f]*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]))^(3/2)) + (2*Sqrt[2]*f^3*ArcTanh[(2
*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])]
)/(Sqrt[e^2 - 4*d*f]*(2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]))^(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/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2

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maple [B]  time = 0.02, size = 1713, normalized size = 4.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` f
or more details)Is 4*d*f-e^2 positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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