3.1.23 \(\int \frac {(a+b x^2) \sqrt {c+d x^2}}{e+f x^2} \, dx\)

Optimal. Leaf size=128 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) (-2 a d f-b c f+2 b d e)}{2 \sqrt {d} f^2}+\frac {(b e-a f) \sqrt {d e-c f} \tanh ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {e} \sqrt {c+d x^2}}\right )}{\sqrt {e} f^2}+\frac {b x \sqrt {c+d x^2}}{2 f} \]

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Rubi [A]  time = 0.14, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {528, 523, 217, 206, 377, 208} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) (-2 a d f-b c f+2 b d e)}{2 \sqrt {d} f^2}+\frac {(b e-a f) \sqrt {d e-c f} \tanh ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {e} \sqrt {c+d x^2}}\right )}{\sqrt {e} f^2}+\frac {b x \sqrt {c+d x^2}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 528

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 124, normalized size = 0.97 \begin {gather*} \frac {\frac {\log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right ) (2 a d f+b c f-2 b d e)}{\sqrt {d}}-\frac {2 (b e-a f) \sqrt {c f-d e} \tan ^{-1}\left (\frac {x \sqrt {c f-d e}}{\sqrt {e} \sqrt {c+d x^2}}\right )}{\sqrt {e}}+b f x \sqrt {c+d x^2}}{2 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*f*x*Sqrt[c + d*x^2] - (2*(b*e - a*f)*Sqrt[-(d*e) + c*f]*ArcTan[(Sqrt[-(d*e) + c*f]*x)/(Sqrt[e]*Sqrt[c + d*x
^2])])/Sqrt[e] + ((-2*b*d*e + b*c*f + 2*a*d*f)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/(2*f^2)

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IntegrateAlgebraic [A]  time = 0.38, size = 151, normalized size = 1.18 \begin {gather*} \frac {\log \left (\sqrt {c+d x^2}-\sqrt {d} x\right ) (-2 a d f-b c f+2 b d e)}{2 \sqrt {d} f^2}+\frac {(b e-a f) \sqrt {c f-d e} \tan ^{-1}\left (\frac {-f x \sqrt {c+d x^2}+\sqrt {d} e+\sqrt {d} f x^2}{\sqrt {e} \sqrt {c f-d e}}\right )}{\sqrt {e} f^2}+\frac {b x \sqrt {c+d x^2}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)*Sqrt[c + d*x^2])/(e + f*x^2),x]

[Out]

(b*x*Sqrt[c + d*x^2])/(2*f) + ((b*e - a*f)*Sqrt[-(d*e) + c*f]*ArcTan[(Sqrt[d]*e + Sqrt[d]*f*x^2 - f*x*Sqrt[c +
d*x^2])/(Sqrt[e]*Sqrt[-(d*e) + c*f])])/(Sqrt[e]*f^2) + ((2*b*d*e - b*c*f - 2*a*d*f)*Log[-(Sqrt[d]*x) + Sqrt[c
+ d*x^2]])/(2*Sqrt[d]*f^2)

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fricas [A]  time = 4.88, size = 777, normalized size = 6.07 \begin {gather*} \left [\frac {2 \, \sqrt {d x^{2} + c} b d f x - {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (b d e - a d f\right )} \sqrt {\frac {d e - c f}{e}} \log \left (\frac {{\left (8 \, d^{2} e^{2} - 8 \, c d e f + c^{2} f^{2}\right )} x^{4} + c^{2} e^{2} + 2 \, {\left (4 \, c d e^{2} - 3 \, c^{2} e f\right )} x^{2} - 4 \, {\left (c e^{2} x + {\left (2 \, d e^{2} - c e f\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {d e - c f}{e}}}{f^{2} x^{4} + 2 \, e f x^{2} + e^{2}}\right )}{4 \, d f^{2}}, \frac {2 \, \sqrt {d x^{2} + c} b d f x + 2 \, {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (b d e - a d f\right )} \sqrt {\frac {d e - c f}{e}} \log \left (\frac {{\left (8 \, d^{2} e^{2} - 8 \, c d e f + c^{2} f^{2}\right )} x^{4} + c^{2} e^{2} + 2 \, {\left (4 \, c d e^{2} - 3 \, c^{2} e f\right )} x^{2} - 4 \, {\left (c e^{2} x + {\left (2 \, d e^{2} - c e f\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {d e - c f}{e}}}{f^{2} x^{4} + 2 \, e f x^{2} + e^{2}}\right )}{4 \, d f^{2}}, \frac {2 \, \sqrt {d x^{2} + c} b d f x - 2 \, {\left (b d e - a d f\right )} \sqrt {-\frac {d e - c f}{e}} \arctan \left (\frac {{\left ({\left (2 \, d e - c f\right )} x^{2} + c e\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {d e - c f}{e}}}{2 \, {\left ({\left (d^{2} e - c d f\right )} x^{3} + {\left (c d e - c^{2} f\right )} x\right )}}\right ) - {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{4 \, d f^{2}}, \frac {\sqrt {d x^{2} + c} b d f x + {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (b d e - a d f\right )} \sqrt {-\frac {d e - c f}{e}} \arctan \left (\frac {{\left ({\left (2 \, d e - c f\right )} x^{2} + c e\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {d e - c f}{e}}}{2 \, {\left ({\left (d^{2} e - c d f\right )} x^{3} + {\left (c d e - c^{2} f\right )} x\right )}}\right )}{2 \, d f^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(d*x^2 + c)*b*d*f*x - (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)
*x - c) - (b*d*e - a*d*f)*sqrt((d*e - c*f)/e)*log(((8*d^2*e^2 - 8*c*d*e*f + c^2*f^2)*x^4 + c^2*e^2 + 2*(4*c*d*
e^2 - 3*c^2*e*f)*x^2 - 4*(c*e^2*x + (2*d*e^2 - c*e*f)*x^3)*sqrt(d*x^2 + c)*sqrt((d*e - c*f)/e))/(f^2*x^4 + 2*e
*f*x^2 + e^2)))/(d*f^2), 1/4*(2*sqrt(d*x^2 + c)*b*d*f*x + 2*(2*b*d*e - (b*c + 2*a*d)*f)*sqrt(-d)*arctan(sqrt(-
d)*x/sqrt(d*x^2 + c)) - (b*d*e - a*d*f)*sqrt((d*e - c*f)/e)*log(((8*d^2*e^2 - 8*c*d*e*f + c^2*f^2)*x^4 + c^2*e
^2 + 2*(4*c*d*e^2 - 3*c^2*e*f)*x^2 - 4*(c*e^2*x + (2*d*e^2 - c*e*f)*x^3)*sqrt(d*x^2 + c)*sqrt((d*e - c*f)/e))/
(f^2*x^4 + 2*e*f*x^2 + e^2)))/(d*f^2), 1/4*(2*sqrt(d*x^2 + c)*b*d*f*x - 2*(b*d*e - a*d*f)*sqrt(-(d*e - c*f)/e)
*arctan(1/2*((2*d*e - c*f)*x^2 + c*e)*sqrt(d*x^2 + c)*sqrt(-(d*e - c*f)/e)/((d^2*e - c*d*f)*x^3 + (c*d*e - c^2
*f)*x)) - (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c))/(d*f^2), 1/2*(s
qrt(d*x^2 + c)*b*d*f*x + (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (b*d*e - a*
d*f)*sqrt(-(d*e - c*f)/e)*arctan(1/2*((2*d*e - c*f)*x^2 + c*e)*sqrt(d*x^2 + c)*sqrt(-(d*e - c*f)/e)/((d^2*e -
c*d*f)*x^3 + (c*d*e - c^2*f)*x)))/(d*f^2)]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:

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maple [B]  time = 0.03, size = 1942, normalized size = 15.17

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (b\,x^2+a\right )\,\sqrt {d\,x^2+c}}{f\,x^2+e} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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