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

Optimal. Leaf size=375 \[ -\frac{\sqrt{e} \sqrt{f} \sqrt{c+d x^2} (a d (d e-9 c f)+b c (3 c f+5 d e)) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{3 c^2 \sqrt{e+f x^2} (d e-c f)^3 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{\sqrt{f} \sqrt{c+d x^2} \left (a \left (-3 c^2 f^2-7 c d e f+2 d^2 e^2\right )+b c e (7 c f+d e)\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 c^2 \sqrt{e} \sqrt{e+f x^2} (d e-c f)^3 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x (2 a d (d e-3 c f)+b c (3 c f+d e))}{3 c^2 \sqrt{c+d x^2} \sqrt{e+f x^2} (d e-c f)^2}-\frac{x (b c-a d)}{3 c \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2} (d e-c f)} \]

[Out]

-((b*c - a*d)*x)/(3*c*(d*e - c*f)*(c + d*x^2)^(3/2)*Sqrt[e + f*x^2]) + ((2*a*d*(d*e - 3*c*f) + b*c*(d*e + 3*c*
f))*x)/(3*c^2*(d*e - c*f)^2*Sqrt[c + d*x^2]*Sqrt[e + f*x^2]) + (Sqrt[f]*(b*c*e*(d*e + 7*c*f) + a*(2*d^2*e^2 -
7*c*d*e*f - 3*c^2*f^2))*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*Sqrt[e
]*(d*e - c*f)^3*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - (Sqrt[e]*Sqrt[f]*(a*d*(d*e - 9*c*f) +
b*c*(5*d*e + 3*c*f))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*(d*e - c
*f)^3*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.3925, antiderivative size = 375, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {527, 525, 418, 411} \[ \frac{\sqrt{f} \sqrt{c+d x^2} \left (a \left (-3 c^2 f^2-7 c d e f+2 d^2 e^2\right )+b c e (7 c f+d e)\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 c^2 \sqrt{e} \sqrt{e+f x^2} (d e-c f)^3 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x (2 a d (d e-3 c f)+b c (3 c f+d e))}{3 c^2 \sqrt{c+d x^2} \sqrt{e+f x^2} (d e-c f)^2}-\frac{\sqrt{e} \sqrt{f} \sqrt{c+d x^2} (a d (d e-9 c f)+b c (3 c f+5 d e)) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 c^2 \sqrt{e+f x^2} (d e-c f)^3 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{x (b c-a d)}{3 c \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2} (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c - a*d)*x)/(3*c*(d*e - c*f)*(c + d*x^2)^(3/2)*Sqrt[e + f*x^2]) + ((2*a*d*(d*e - 3*c*f) + b*c*(d*e + 3*c*
f))*x)/(3*c^2*(d*e - c*f)^2*Sqrt[c + d*x^2]*Sqrt[e + f*x^2]) + (Sqrt[f]*(b*c*e*(d*e + 7*c*f) + a*(2*d^2*e^2 -
7*c*d*e*f - 3*c^2*f^2))*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*Sqrt[e
]*(d*e - c*f)^3*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - (Sqrt[e]*Sqrt[f]*(a*d*(d*e - 9*c*f) +
b*c*(5*d*e + 3*c*f))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*c^2*(d*e - c
*f)^3*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

Rule 527

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 + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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{a+b x^2}{\left (c+d x^2\right )^{5/2} \left (e+f x^2\right )^{3/2}} \, dx &=-\frac{(b c-a d) x}{3 c (d e-c f) \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}-\frac{\int \frac{-b c e-2 a d e+3 a c f+3 (b c-a d) f x^2}{\left (c+d x^2\right )^{3/2} \left (e+f x^2\right )^{3/2}} \, dx}{3 c (d e-c f)}\\ &=-\frac{(b c-a d) x}{3 c (d e-c f) \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}+\frac{(2 a d (d e-3 c f)+b c (d e+3 c f)) x}{3 c^2 (d e-c f)^2 \sqrt{c+d x^2} \sqrt{e+f x^2}}+\frac{\int \frac{-c f (4 b c e-a d e-3 a c f)+f (2 a d (d e-3 c f)+b c (d e+3 c f)) x^2}{\sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}} \, dx}{3 c^2 (d e-c f)^2}\\ &=-\frac{(b c-a d) x}{3 c (d e-c f) \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}+\frac{(2 a d (d e-3 c f)+b c (d e+3 c f)) x}{3 c^2 (d e-c f)^2 \sqrt{c+d x^2} \sqrt{e+f x^2}}-\frac{(f (a d (d e-9 c f)+b c (5 d e+3 c f))) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{3 c (d e-c f)^3}+\frac{\left (f \left (b c e (d e+7 c f)+a \left (2 d^2 e^2-7 c d e f-3 c^2 f^2\right )\right )\right ) \int \frac{\sqrt{c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{3 c^2 (d e-c f)^3}\\ &=-\frac{(b c-a d) x}{3 c (d e-c f) \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}+\frac{(2 a d (d e-3 c f)+b c (d e+3 c f)) x}{3 c^2 (d e-c f)^2 \sqrt{c+d x^2} \sqrt{e+f x^2}}+\frac{\sqrt{f} \left (b c e (d e+7 c f)+a \left (2 d^2 e^2-7 c d e f-3 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 )}{3 c^2 \sqrt{e} (d e-c f)^3 \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{f} (a d (d e-9 c f)+b c (5 d e+3 c 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 (d e-c f)^3 \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 = 2.06465, size = 428, normalized size = 1.14 \[ \frac{-i e \left (c+d x^2\right ) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (c f-d e) (2 a d (d e-3 c f)+b c (3 c f+d e)) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )+x \sqrt{\frac{d}{c}} \left (a \left (c^2 d^2 f \left (8 e^2+8 e f x^2+3 f^2 x^4\right )+6 c^3 d f^3 x^2+3 c^4 f^3+c d^3 e \left (-3 e^2+4 e f x^2+7 f^2 x^4\right )-2 d^4 e^2 x^2 \left (e+f x^2\right )\right )-b c e \left (c^2 d f \left (5 e+11 f x^2\right )+3 c^3 f^2+c d^2 f x^2 \left (4 e+7 f x^2\right )+d^3 e x^2 \left (e+f x^2\right )\right )\right )-i d e \left (c+d x^2\right ) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \left (a \left (-3 c^2 f^2-7 c d e f+2 d^2 e^2\right )+b c e (7 c f+d e)\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )}{3 c^2 e \sqrt{\frac{d}{c}} \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2} (c f-d e)^3} \]

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]*x*(-(b*c*e*(3*c^3*f^2 + d^3*e*x^2*(e + f*x^2) + c*d^2*f*x^2*(4*e + 7*f*x^2) + c^2*d*f*(5*e + 11*f*x
^2))) + a*(3*c^4*f^3 + 6*c^3*d*f^3*x^2 - 2*d^4*e^2*x^2*(e + f*x^2) + c^2*d^2*f*(8*e^2 + 8*e*f*x^2 + 3*f^2*x^4)
+ c*d^3*e*(-3*e^2 + 4*e*f*x^2 + 7*f^2*x^4))) - I*d*e*(b*c*e*(d*e + 7*c*f) + a*(2*d^2*e^2 - 7*c*d*e*f - 3*c^2*
f^2))*(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*e
*(-(d*e) + c*f)*(2*a*d*(d*e - 3*c*f) + b*c*(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*c^2*Sqrt[d/c]*e*(-(d*e) + c*f)^3*(c + d*x^2)^(3/2)*Sqrt[e +
f*x^2])

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Maple [B]  time = 0.038, size = 1742, normalized size = 4.7 \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]

-1/3*(3*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c^3*d*e*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+7*Elli
pticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c^2*d^2*e^2*f*((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))*x^2*b*c*d^3*e^3*((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*b*c*d^3*e^3*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+6*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/
2))*a*c^3*d*e*f^2*((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*c^2*d^
2*e^2*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))*b*c^3*d*e^2*f*((d*
x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*x*a*c^4*f^3*(-d/c)^(1/2)-7*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c
^3*d*e^2*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+2*x^3*a*d^4*e^3*(-d/c)^(1/2)-7*x^5*a*c*d^3*e*f^2*(-d/c)^(1/
2)-2*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*d^4*e^3*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+2*Ellipti
cF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*c*d^3*e^3*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-3*x^5*a*c^2*d^2*f^3*(-d
/c)^(1/2)+5*x*b*c^3*d*e^2*f*(-d/c)^(1/2)+2*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*d^4*e^3*((d*x^2+c)/
c)^(1/2)*((f*x^2+e)/e)^(1/2)-8*x^3*a*c^2*d^2*e*f^2*(-d/c)^(1/2)-4*x^3*a*c*d^3*e^2*f*(-d/c)^(1/2)+11*x^3*b*c^3*
d*e*f^2*(-d/c)^(1/2)+4*x^3*b*c^2*d^2*e^2*f*(-d/c)^(1/2)+7*x^5*b*c^2*d^2*e*f^2*(-d/c)^(1/2)+x^5*b*c*d^3*e^2*f*(
-d/c)^(1/2)-3*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^4*e*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+El
lipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*c^2*d^2*e^3*((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*c*d^3*e^3*((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^2*d^2*e^3*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-8*x*a*c^2*d^2*e^2*f*(-d/c)^(1/2)+2*x^5*a*d^4
*e^2*f*(-d/c)^(1/2)-6*x^3*a*c^3*d*f^3*(-d/c)^(1/2)+x^3*b*c*d^3*e^3*(-d/c)^(1/2)+3*x*a*c*d^3*e^3*(-d/c)^(1/2)+3
*x*b*c^4*e*f^2*(-d/c)^(1/2)-7*EllipticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*b*c^2*d^2*e^2*f*((d*x^2+c)/c)^(1/2
)*((f*x^2+e)/e)^(1/2)+6*EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*c^2*d^2*e*f^2*((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*c*d^3*e^2*f*((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*b*c^3*d*e*f^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))*x^2*b*c^2*d^2*e^2*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+3*Elli
pticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*c^2*d^2*e*f^2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+7*EllipticE(
x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*x^2*a*c*d^3*e^2*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2))/(f*x^2+e)^(1/2)/(c*
f-d*e)^3/(-d/c)^(1/2)/e/c^2/(d*x^2+c)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b x^{2} + a}{{\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 x^{2} + a\right )} \sqrt{d x^{2} + c} \sqrt{f x^{2} + e}}{d^{3} f^{2} x^{10} +{\left (2 \, d^{3} e f + 3 \, c d^{2} f^{2}\right )} x^{8} +{\left (d^{3} e^{2} + 6 \, c d^{2} e f + 3 \, c^{2} d f^{2}\right )} x^{6} + c^{3} e^{2} +{\left (3 \, c d^{2} e^{2} + 6 \, c^{2} d e f + c^{3} f^{2}\right )} x^{4} +{\left (3 \, c^{2} d e^{2} + 2 \, c^{3} e f\right )} x^{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*x^2 + a)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(d^3*f^2*x^10 + (2*d^3*e*f + 3*c*d^2*f^2)*x^8 + (d^3*e^2
+ 6*c*d^2*e*f + 3*c^2*d*f^2)*x^6 + c^3*e^2 + (3*c*d^2*e^2 + 6*c^2*d*e*f + c^3*f^2)*x^4 + (3*c^2*d*e^2 + 2*c^3*
e*f)*x^2), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b x^{2} + a}{{\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)