3.69 \(\int \frac{\sqrt{e+f x^2}}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=401 \[ \frac{d e^{3/2} \sqrt{f} \sqrt{c+d x^2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{3 c^2 \sqrt{e+f x^2} (b c-a d) (d e-c f) \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{b^2 e^{3/2} \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a c \sqrt{f} \sqrt{e+f x^2} (b c-a d)^2 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{\sqrt{d} \sqrt{e+f x^2} (b c (5 d e-4 c f)-a d (2 d e-c f)) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{c f}{d e}\right )}{3 c^{3/2} \sqrt{c+d x^2} (b c-a d)^2 (d e-c f) \sqrt{\frac{c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac{d x \sqrt{e+f x^2}}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)} \]

[Out]

-(d*x*Sqrt[e + f*x^2])/(3*c*(b*c - a*d)*(c + d*x^2)^(3/2)) - (Sqrt[d]*(b*c*(5*d*e - 4*c*f) - a*d*(2*d*e - c*f)
)*Sqrt[e + f*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(3*c^(3/2)*(b*c - a*d)^2*(d*e - c*f
)*Sqrt[c + d*x^2]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))]) + (d*e^(3/2)*Sqrt[f]*Sqrt[c + d*x^2]*EllipticF[ArcTan
[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*(b*c - a*d)*(d*e - c*f)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*
Sqrt[e + f*x^2]) + (b^2*e^(3/2)*Sqrt[c + d*x^2]*EllipticPi[1 - (b*e)/(a*f), ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (
d*e)/(c*f)])/(a*c*(b*c - a*d)^2*Sqrt[f]*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.366965, antiderivative size = 401, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {546, 539, 526, 525, 418, 411} \[ \frac{b^2 e^{3/2} \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a c \sqrt{f} \sqrt{e+f x^2} (b c-a d)^2 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{d e^{3/2} \sqrt{f} \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 c^2 \sqrt{e+f x^2} (b c-a d) (d e-c f) \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{\sqrt{d} \sqrt{e+f x^2} (b c (5 d e-4 c f)-a d (2 d e-c f)) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{c f}{d e}\right )}{3 c^{3/2} \sqrt{c+d x^2} (b c-a d)^2 (d e-c f) \sqrt{\frac{c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac{d x \sqrt{e+f x^2}}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e + f*x^2]/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

-(d*x*Sqrt[e + f*x^2])/(3*c*(b*c - a*d)*(c + d*x^2)^(3/2)) - (Sqrt[d]*(b*c*(5*d*e - 4*c*f) - a*d*(2*d*e - c*f)
)*Sqrt[e + f*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(3*c^(3/2)*(b*c - a*d)^2*(d*e - c*f
)*Sqrt[c + d*x^2]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))]) + (d*e^(3/2)*Sqrt[f]*Sqrt[c + d*x^2]*EllipticF[ArcTan
[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*(b*c - a*d)*(d*e - c*f)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*
Sqrt[e + f*x^2]) + (b^2*e^(3/2)*Sqrt[c + d*x^2]*EllipticPi[1 - (b*e)/(a*f), ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (
d*e)/(c*f)])/(a*c*(b*c - a*d)^2*Sqrt[f]*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

Rule 546

Int[(((c_) + (d_.)*(x_)^2)^(q_)*((e_) + (f_.)*(x_)^2)^(r_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[b^2/(b*c
- a*d)^2, Int[((c + d*x^2)^(q + 2)*(e + f*x^2)^r)/(a + b*x^2), x], x] - Dist[d/(b*c - a*d)^2, Int[(c + d*x^2)^
q*(e + f*x^2)^r*(2*b*c - a*d + b*d*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, r}, x] && LtQ[q, -1]

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 526

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

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\sqrt{e+f x^2}}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx &=\frac{b^2 \int \frac{\sqrt{e+f x^2}}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{(b c-a d)^2}-\frac{d \int \frac{\left (2 b c-a d+b d x^2\right ) \sqrt{e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx}{(b c-a d)^2}\\ &=-\frac{d x \sqrt{e+f x^2}}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{b^2 e^{3/2} \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a c (b c-a d)^2 \sqrt{f} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}+\frac{\int \frac{-d (5 b c-2 a d) e-d (4 b c-a d) f x^2}{\left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}} \, dx}{3 c (b c-a d)^2}\\ &=-\frac{d x \sqrt{e+f x^2}}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{b^2 e^{3/2} \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a c (b c-a d)^2 \sqrt{f} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}+\frac{(d e f) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{3 c (b c-a d) (d e-c f)}-\frac{(d (b c (5 d e-4 c f)-a d (2 d e-c f))) \int \frac{\sqrt{e+f x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c (b c-a d)^2 (d e-c f)}\\ &=-\frac{d x \sqrt{e+f x^2}}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{\sqrt{d} (b c (5 d e-4 c f)-a d (2 d e-c f)) \sqrt{e+f x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{c f}{d e}\right )}{3 c^{3/2} (b c-a d)^2 (d e-c f) \sqrt{c+d x^2} \sqrt{\frac{c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}+\frac{d e^{3/2} \sqrt{f} \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 c^2 (b c-a d) (d e-c f) \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}+\frac{b^2 e^{3/2} \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a c (b c-a d)^2 \sqrt{f} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}\\ \end{align*}

Mathematica [C]  time = 3.43655, size = 427, normalized size = 1.06 \[ \frac{-i a \left (c+d x^2\right ) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (c f-d e) \left (2 a d^2 e+b c (3 c f-5 d e)\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )+a c x \left (\frac{d}{c}\right )^{3/2} \left (e+f x^2\right ) \left (a d \left (2 c^2 f-3 c d e+c d f x^2-2 d^2 e x^2\right )+b c \left (-5 c^2 f+6 c d e-4 c d f x^2+5 d^2 e x^2\right )\right )-3 i b c^2 \left (c+d x^2\right ) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (b e-a f) (c f-d e) \Pi \left (\frac{b c}{a d};i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )-i a d e \left (c+d x^2\right ) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (a d (2 d e-c f)+b c (4 c f-5 d e)) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )}{3 a c^2 \sqrt{\frac{d}{c}} \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2} (b c-a d)^2 (c f-d e)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e + f*x^2]/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

(a*c*(d/c)^(3/2)*x*(e + f*x^2)*(b*c*(6*c*d*e - 5*c^2*f + 5*d^2*e*x^2 - 4*c*d*f*x^2) + a*d*(-3*c*d*e + 2*c^2*f
- 2*d^2*e*x^2 + c*d*f*x^2)) - I*a*d*e*(a*d*(2*d*e - c*f) + b*c*(-5*d*e + 4*c*f))*(c + d*x^2)*Sqrt[1 + (d*x^2)/
c]*Sqrt[1 + (f*x^2)/e]*EllipticE[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] - I*a*(-(d*e) + c*f)*(2*a*d^2*e + b*c*(-
5*d*e + 3*c*f))*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d
*e)] - (3*I)*b*c^2*(b*e - a*f)*(-(d*e) + c*f)*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticPi[(
b*c)/(a*d), I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(3*a*c^2*Sqrt[d/c]*(b*c - a*d)^2*(-(d*e) + c*f)*(c + d*x^2)^
(3/2)*Sqrt[e + f*x^2])

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Maple [B]  time = 0.036, size = 2068, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^2+e)^(1/2)/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

1/3*(x^5*a^2*c*d^3*f^2*(-d/c)^(1/2)+2*x^3*a*b*c^2*d^2*e*f*(-d/c)^(1/2)-5*x*a*b*c^3*d*e*f*(-d/c)^(1/2)-3*Ellipt
icPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*x^2*b^2*c^2*d^2*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(
1/2)+2*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a^2*c^2*d^2*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+5*Ell
ipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*b*c^2*d^2*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-EllipticE(x*(-d
/c)^(1/2),(c*f/d/e)^(1/2))*a^2*c^2*d^2*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-5*EllipticE(x*(-d/c)^(1/2),
(c*f/d/e)^(1/2))*a*b*c^2*d^2*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-2*x^3*a^2*d^4*e^2*(-d/c)^(1/2)-5*Elli
pticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*b*c*d^3*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+3*EllipticF(x*
(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*b*c^3*d*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+5*EllipticF(x*(-d/c)^(
1/2),(c*f/d/e)^(1/2))*x^2*a*b*c*d^3*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*EllipticPi(x*(-d/c)^(1/2),b*
c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*x^2*a*b*c^3*d*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-2*x^5*a^2*d^4*e*f*(
-d/c)^(1/2)+2*x^3*a^2*c^2*d^2*f^2*(-d/c)^(1/2)-3*x*a^2*c*d^3*e^2*(-d/c)^(1/2)-4*x^5*a*b*c^2*d^2*f^2*(-d/c)^(1/
2)-5*x^3*a*b*c^3*d*f^2*(-d/c)^(1/2)+5*x^3*a*b*c*d^3*e^2*(-d/c)^(1/2)+2*x*a^2*c^2*d^2*e*f*(-d/c)^(1/2)+6*x*a*b*
c^2*d^2*e^2*(-d/c)^(1/2)-2*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a^2*d^4*e^2*((d*x^2+c)/c)^(1/2)*((f*x
^2+e)/e)^(1/2)+2*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a^2*d^4*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(
1/2)-2*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a^2*c*d^3*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+3*Ellip
ticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*b*c^4*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+2*EllipticE(x*(-d/c)^
(1/2),(c*f/d/e)^(1/2))*a^2*c*d^3*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*EllipticPi(x*(-d/c)^(1/2),b*c/a
/d,(-f/e)^(1/2)/(-d/c)^(1/2))*a*b*c^4*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+3*EllipticPi(x*(-d/c)^(1/2),
b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*b^2*c^4*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*EllipticPi(x*(-d/c)^(
1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*b^2*c^3*d*e^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-2*x^3*a^2*c*d^3*
e*f*(-d/c)^(1/2)+5*x^5*a*b*c*d^3*e*f*(-d/c)^(1/2)+3*EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2
))*x^2*b^2*c^3*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-8*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*b*c
^3*d*e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+4*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*b*c^3*d*e*f*((d
*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+3*EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*a*b*c^3*d*
e*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+2*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a^2*c*d^3*e*f*((d*
x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a^2*c*d^3*e*f*((d*x^2+c)/c)^
(1/2)*((f*x^2+e)/e)^(1/2)+4*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*b*c^2*d^2*e*f*((d*x^2+c)/c)^(1/2)*
((f*x^2+e)/e)^(1/2)+3*EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*x^2*a*b*c^2*d^2*e*f*((d*x^2
+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-8*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*b*c^2*d^2*e*f*((d*x^2+c)/c)
^(1/2)*((f*x^2+e)/e)^(1/2))/(f*x^2+e)^(1/2)/(a*d-b*c)^2/(-d/c)^(1/2)/c^2/(c*f-d*e)/a/(d*x^2+c)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{f x^{2} + e}}{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e)^(1/2)/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(f*x^2 + e)/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e)^(1/2)/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e + f x^{2}}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**2+e)**(1/2)/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Integral(sqrt(e + f*x**2)/((a + b*x**2)*(c + d*x**2)**(5/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{f x^{2} + e}}{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e)^(1/2)/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(f*x^2 + e)/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)