3.42 \(\int \frac{(a+b x^2) (c+d x^2)^{5/2}}{(e+f x^2)^{3/2}} \, dx\)

Optimal. Leaf size=501 \[ -\frac{\sqrt{e} \sqrt{c+d x^2} \left (10 a d f (2 d e-3 c f)-b \left (15 c^2 f^2-41 c d e f+24 d^2 e^2\right )\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{15 f^{7/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{x \sqrt{c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right )}{15 e f^3 \sqrt{e+f x^2}}+\frac{\sqrt{c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 \sqrt{e} f^{7/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{d x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2} (6 b e-5 a f)}{5 e f^2}-\frac{d x \sqrt{c+d x^2} \sqrt{e+f x^2} (b e (24 d e-23 c f)-5 a f (4 d e-3 c f))}{15 e f^3}-\frac{x \left (c+d x^2\right )^{5/2} (b e-a f)}{e f \sqrt{e+f x^2}} \]

[Out]

-((5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) - 2*b*e*(24*d^2*e^2 - 44*c*d*e*f + 19*c^2*f^2))*x*Sqrt[c + d*x^2
])/(15*e*f^3*Sqrt[e + f*x^2]) - ((b*e - a*f)*x*(c + d*x^2)^(5/2))/(e*f*Sqrt[e + f*x^2]) - (d*(b*e*(24*d*e - 23
*c*f) - 5*a*f*(4*d*e - 3*c*f))*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*e*f^3) + (d*(6*b*e - 5*a*f)*x*(c + d*x^2
)^(3/2)*Sqrt[e + f*x^2])/(5*e*f^2) + ((5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) - 2*b*e*(24*d^2*e^2 - 44*c*d
*e*f + 19*c^2*f^2))*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*Sqrt[e]*f^(7/
2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - (Sqrt[e]*(10*a*d*f*(2*d*e - 3*c*f) - b*(24*d^2*e^2
- 41*c*d*e*f + 15*c^2*f^2))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*f^(7
/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.598706, antiderivative size = 501, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {526, 528, 531, 418, 492, 411} \[ -\frac{x \sqrt{c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right )}{15 e f^3 \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} \left (10 a d f (2 d e-3 c f)-b \left (15 c^2 f^2-41 c d e f+24 d^2 e^2\right )\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 f^{7/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{\sqrt{c+d x^2} \left (5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )-2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 \sqrt{e} f^{7/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{d x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2} (6 b e-5 a f)}{5 e f^2}-\frac{d x \sqrt{c+d x^2} \sqrt{e+f x^2} (b e (24 d e-23 c f)-5 a f (4 d e-3 c f))}{15 e f^3}-\frac{x \left (c+d x^2\right )^{5/2} (b e-a f)}{e f \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(c + d*x^2)^(5/2))/(e + f*x^2)^(3/2),x]

[Out]

-((5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) - 2*b*e*(24*d^2*e^2 - 44*c*d*e*f + 19*c^2*f^2))*x*Sqrt[c + d*x^2
])/(15*e*f^3*Sqrt[e + f*x^2]) - ((b*e - a*f)*x*(c + d*x^2)^(5/2))/(e*f*Sqrt[e + f*x^2]) - (d*(b*e*(24*d*e - 23
*c*f) - 5*a*f*(4*d*e - 3*c*f))*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*e*f^3) + (d*(6*b*e - 5*a*f)*x*(c + d*x^2
)^(3/2)*Sqrt[e + f*x^2])/(5*e*f^2) + ((5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) - 2*b*e*(24*d^2*e^2 - 44*c*d
*e*f + 19*c^2*f^2))*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*Sqrt[e]*f^(7/
2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - (Sqrt[e]*(10*a*d*f*(2*d*e - 3*c*f) - b*(24*d^2*e^2
- 41*c*d*e*f + 15*c^2*f^2))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*f^(7
/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

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 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]

Rule 531

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

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 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !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{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{\left (e+f x^2\right )^{3/2}} \, dx &=-\frac{(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt{e+f x^2}}-\frac{\int \frac{\left (c+d x^2\right )^{3/2} \left (-b c e-d (6 b e-5 a f) x^2\right )}{\sqrt{e+f x^2}} \, dx}{e f}\\ &=-\frac{(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt{e+f x^2}}+\frac{d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 e f^2}-\frac{\int \frac{\sqrt{c+d x^2} \left (c e (6 b d e-5 b c f-5 a d f)+d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x^2\right )}{\sqrt{e+f x^2}} \, dx}{5 e f^2}\\ &=-\frac{(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt{e+f x^2}}-\frac{d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 e f^3}+\frac{d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 e f^2}-\frac{\int \frac{c e \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right )+d \left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 e f^3}\\ &=-\frac{(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt{e+f x^2}}-\frac{d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 e f^3}+\frac{d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 e f^2}-\frac{\left (c \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right )\right ) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 f^3}-\frac{\left (d \left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right )\right ) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 e f^3}\\ &=-\frac{\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 e f^3 \sqrt{e+f x^2}}-\frac{(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt{e+f x^2}}-\frac{d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 e f^3}+\frac{d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 e f^2}-\frac{\sqrt{e} \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right ) \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 f^{7/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}+\frac{\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) \int \frac{\sqrt{c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{15 f^3}\\ &=-\frac{\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 e f^3 \sqrt{e+f x^2}}-\frac{(b e-a f) x \left (c+d x^2\right )^{5/2}}{e f \sqrt{e+f x^2}}-\frac{d (b e (24 d e-23 c f)-5 a f (4 d e-3 c f)) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 e f^3}+\frac{d (6 b e-5 a f) x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 e f^2}+\frac{\left (5 a f \left (8 d^2 e^2-13 c d e f+3 c^2 f^2\right )-2 b e \left (24 d^2 e^2-44 c d e f+19 c^2 f^2\right )\right ) \sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 \sqrt{e} f^{7/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{\sqrt{e} \left (10 a d f (2 d e-3 c f)-b \left (24 d^2 e^2-41 c d e f+15 c^2 f^2\right )\right ) \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 f^{7/2} \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 = 1.16102, size = 369, normalized size = 0.74 \[ \frac{-i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (c f-d e) \left (5 a d f (9 c f-8 d e)+b \left (15 c^2 f^2-64 c d e f+48 d^2 e^2\right )\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )+f x \sqrt{\frac{d}{c}} \left (c+d x^2\right ) \left (5 a f \left (3 c^2 f^2-6 c d e f+d^2 e \left (4 e+f x^2\right )\right )+b e \left (-15 c^2 f^2+c d f \left (41 e+11 f x^2\right )-3 d^2 \left (8 e^2+2 e f x^2-f^2 x^4\right )\right )\right )-i d e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \left (2 b e \left (19 c^2 f^2-44 c d e f+24 d^2 e^2\right )-5 a f \left (3 c^2 f^2-13 c d e f+8 d^2 e^2\right )\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )}{15 e f^4 \sqrt{\frac{d}{c}} \sqrt{c+d x^2} \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(c + d*x^2)^(5/2))/(e + f*x^2)^(3/2),x]

[Out]

(Sqrt[d/c]*f*x*(c + d*x^2)*(5*a*f*(-6*c*d*e*f + 3*c^2*f^2 + d^2*e*(4*e + f*x^2)) + b*e*(-15*c^2*f^2 + c*d*f*(4
1*e + 11*f*x^2) - 3*d^2*(8*e^2 + 2*e*f*x^2 - f^2*x^4))) - I*d*e*(-5*a*f*(8*d^2*e^2 - 13*c*d*e*f + 3*c^2*f^2) +
2*b*e*(24*d^2*e^2 - 44*c*d*e*f + 19*c^2*f^2))*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticE[I*ArcSinh[Sqr
t[d/c]*x], (c*f)/(d*e)] - I*e*(-(d*e) + c*f)*(5*a*d*f*(-8*d*e + 9*c*f) + b*(48*d^2*e^2 - 64*c*d*e*f + 15*c^2*f
^2))*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(15*Sqrt[d/c]*e*f
^4*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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Maple [B]  time = 0.046, size = 1169, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x^2+c)^(5/2)/(f*x^2+e)^(3/2),x)

[Out]

-1/15*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)*(-40*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c
*f/d/e)^(1/2))*a*d^3*e^3*f+24*(-d/c)^(1/2)*x*b*c*d^2*e^3*f-38*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*Elliptic
E(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^2*d*e^2*f^2-15*x*a*c^3*f^4*(-d/c)^(1/2)+79*((d*x^2+c)/c)^(1/2)*((f*x^2+e
)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^2*d*e^2*f^2-65*((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*c*d^2*e^2*f^2-112*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*Ellipt
icF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c*d^2*e^3*f+15*((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*c^2*d*e*f^3-45*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*
f/d/e)^(1/2))*a*c^2*d*e*f^3-15*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2
))*b*c^3*e*f^3+40*((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*d^3*e^3*
f-5*(-d/c)^(1/2)*x^5*a*d^3*e*f^3+48*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)
^(1/2))*b*d^3*e^4-48*((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))*b*d^3*e
^4-14*(-d/c)^(1/2)*x^5*b*c*d^2*e*f^3+25*(-d/c)^(1/2)*x^3*a*c*d^2*e*f^3+4*(-d/c)^(1/2)*x^3*b*c^2*d*e*f^3+88*((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))*b*c*d^2*e^3*f-41*(-d/c)^(1/2)*x
*b*c^2*d*e^2*f^2-35*(-d/c)^(1/2)*x^3*b*c*d^2*e^2*f^2+30*(-d/c)^(1/2)*x*a*c^2*d*e*f^3-20*(-d/c)^(1/2)*x*a*c*d^2
*e^2*f^2+6*(-d/c)^(1/2)*x^5*b*d^3*e^2*f^2-15*(-d/c)^(1/2)*x^3*a*c^2*d*f^4-20*(-d/c)^(1/2)*x^3*a*d^3*e^2*f^2+24
*(-d/c)^(1/2)*x^3*b*d^3*e^3*f+15*(-d/c)^(1/2)*x*b*c^3*e*f^3-3*(-d/c)^(1/2)*x^7*b*d^3*e*f^3+85*((d*x^2+c)/c)^(1
/2)*((f*x^2+e)/e)^(1/2)*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c*d^2*e^2*f^2)/f^4/(d*f*x^4+c*f*x^2+d*e*x^
2+c*e)/(-d/c)^(1/2)/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{\frac{5}{2}}}{{\left (f x^{2} + e\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)^(5/2)/(f*x^2+e)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)*(d*x^2 + c)^(5/2)/(f*x^2 + e)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b d^{2} x^{6} +{\left (2 \, b c d + a d^{2}\right )} x^{4} + a c^{2} +{\left (b c^{2} + 2 \, a c d\right )} x^{2}\right )} \sqrt{d x^{2} + c} \sqrt{f x^{2} + e}}{f^{2} x^{4} + 2 \, e f x^{2} + e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)^(5/2)/(f*x^2+e)^(3/2),x, algorithm="fricas")

[Out]

integral((b*d^2*x^6 + (2*b*c*d + a*d^2)*x^4 + a*c^2 + (b*c^2 + 2*a*c*d)*x^2)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(
f^2*x^4 + 2*e*f*x^2 + e^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{5}{2}}}{\left (e + f x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(d*x**2+c)**(5/2)/(f*x**2+e)**(3/2),x)

[Out]

Integral((a + b*x**2)*(c + d*x**2)**(5/2)/(e + f*x**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{\frac{5}{2}}}{{\left (f x^{2} + e\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)^(5/2)/(f*x^2+e)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)*(d*x^2 + c)^(5/2)/(f*x^2 + e)^(3/2), x)