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

Optimal. Leaf size=358 \[ -\frac{\sqrt{e} \sqrt{c+d x^2} (-3 a d f-3 b c f+4 b d e) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{3 f^{5/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{d x \sqrt{c+d x^2} \sqrt{e+f x^2} (4 b e-3 a f)}{3 e f^2}-\frac{x \sqrt{c+d x^2} (b e (8 d e-7 c f)-3 a f (2 d e-c f))}{3 e f^2 \sqrt{e+f x^2}}+\frac{\sqrt{c+d x^2} (b e (8 d e-7 c f)-3 a f (2 d e-c f)) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 \sqrt{e} f^{5/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{x \left (c+d x^2\right )^{3/2} (b e-a f)}{e f \sqrt{e+f x^2}} \]

[Out]

-((b*e*(8*d*e - 7*c*f) - 3*a*f*(2*d*e - c*f))*x*Sqrt[c + d*x^2])/(3*e*f^2*Sqrt[e + f*x^2]) - ((b*e - a*f)*x*(c
+ d*x^2)^(3/2))/(e*f*Sqrt[e + f*x^2]) + (d*(4*b*e - 3*a*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(3*e*f^2) + ((b
*e*(8*d*e - 7*c*f) - 3*a*f*(2*d*e - c*f))*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*
f)])/(3*Sqrt[e]*f^(5/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - (Sqrt[e]*(4*b*d*e - 3*b*c*f -
3*a*d*f)*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*f^(5/2)*Sqrt[(e*(c + d*x
^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

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Rubi [A]  time = 0.36314, antiderivative size = 358, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{d x \sqrt{c+d x^2} \sqrt{e+f x^2} (4 b e-3 a f)}{3 e f^2}-\frac{x \sqrt{c+d x^2} (b e (8 d e-7 c f)-3 a f (2 d e-c f))}{3 e f^2 \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} (-3 a d f-3 b c f+4 b d e) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 f^{5/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{\sqrt{c+d x^2} (b e (8 d e-7 c f)-3 a f (2 d e-c f)) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{3 \sqrt{e} f^{5/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{x \left (c+d x^2\right )^{3/2} (b e-a f)}{e f \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*(8*d*e - 7*c*f) - 3*a*f*(2*d*e - c*f))*x*Sqrt[c + d*x^2])/(3*e*f^2*Sqrt[e + f*x^2]) - ((b*e - a*f)*x*(c
+ d*x^2)^(3/2))/(e*f*Sqrt[e + f*x^2]) + (d*(4*b*e - 3*a*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(3*e*f^2) + ((b
*e*(8*d*e - 7*c*f) - 3*a*f*(2*d*e - c*f))*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*
f)])/(3*Sqrt[e]*f^(5/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) - (Sqrt[e]*(4*b*d*e - 3*b*c*f -
3*a*d*f)*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(3*f^(5/2)*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 )^{3/2}}{\left (e+f x^2\right )^{3/2}} \, dx &=-\frac{(b e-a f) x \left (c+d x^2\right )^{3/2}}{e f \sqrt{e+f x^2}}-\frac{\int \frac{\sqrt{c+d x^2} \left (-b c e-d (4 b e-3 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 )^{3/2}}{e f \sqrt{e+f x^2}}+\frac{d (4 b e-3 a f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{3 e f^2}-\frac{\int \frac{c e (4 b d e-3 b c f-3 a d f)+d (b e (8 d e-7 c f)-3 a f (2 d e-c f)) x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{3 e f^2}\\ &=-\frac{(b e-a f) x \left (c+d x^2\right )^{3/2}}{e f \sqrt{e+f x^2}}+\frac{d (4 b e-3 a f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{3 e f^2}-\frac{(c (4 b d e-3 b c f-3 a d f)) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{3 f^2}-\frac{(d (b e (8 d e-7 c f)-3 a f (2 d e-c f))) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{3 e f^2}\\ &=-\frac{(b e (8 d e-7 c f)-3 a f (2 d e-c f)) x \sqrt{c+d x^2}}{3 e f^2 \sqrt{e+f x^2}}-\frac{(b e-a f) x \left (c+d x^2\right )^{3/2}}{e f \sqrt{e+f x^2}}+\frac{d (4 b e-3 a f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{3 e f^2}-\frac{\sqrt{e} (4 b d e-3 b c f-3 a d 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 f^{5/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}+\frac{(b e (8 d e-7 c f)-3 a f (2 d e-c f)) \int \frac{\sqrt{c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{3 f^2}\\ &=-\frac{(b e (8 d e-7 c f)-3 a f (2 d e-c f)) x \sqrt{c+d x^2}}{3 e f^2 \sqrt{e+f x^2}}-\frac{(b e-a f) x \left (c+d x^2\right )^{3/2}}{e f \sqrt{e+f x^2}}+\frac{d (4 b e-3 a f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{3 e f^2}+\frac{(b e (8 d e-7 c f)-3 a f (2 d e-c f)) \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 \sqrt{e} f^{5/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{\sqrt{e} (4 b d e-3 b c f-3 a d 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 f^{5/2} \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 = 0.765159, size = 260, normalized size = 0.73 \[ \frac{-i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (c f-d e) (6 a d f+3 b c f-8 b d e) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )+f x \sqrt{\frac{d}{c}} \left (c+d x^2\right ) \left (3 a f (c f-d e)+b e \left (-3 c f+4 d e+d f x^2\right )\right )-i d e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (b e (7 c f-8 d e)-3 a f (c f-2 d e)) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )}{3 e f^3 \sqrt{\frac{d}{c}} \sqrt{c+d x^2} \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/c]*f*x*(c + d*x^2)*(3*a*f*(-(d*e) + c*f) + b*e*(4*d*e - 3*c*f + d*f*x^2)) - I*d*e*(-3*a*f*(-2*d*e + c*
f) + b*e*(-8*d*e + 7*c*f))*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)*(-8*b*d*e + 3*b*c*f + 6*a*d*f)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticF[I*Ar
cSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(3*Sqrt[d/c]*e*f^3*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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Maple [A]  time = 0.025, size = 750, normalized size = 2.1 \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)^(3/2)/(f*x^2+e)^(3/2),x)

[Out]

1/3*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)*((-d/c)^(1/2)*x^5*b*d^2*e*f^2+3*(-d/c)^(1/2)*x^3*a*c*d*f^3-3*(-d/c)^(1/2)*
x^3*a*d^2*e*f^2-2*(-d/c)^(1/2)*x^3*b*c*d*e*f^2+4*(-d/c)^(1/2)*x^3*b*d^2*e^2*f+6*((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*e*f^2-6*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*Elli
pticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*d^2*e^2*f+3*((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*e*f^2-11*((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*d*e^2*f+8*((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
^2*e^3-3*((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))*a*c*d*e*f^2+6*((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^2*e^2*f+7*((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*e^2*f-8*((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^2*e^3+3*x*a*c^2*f^3*(-d/c)^(1/2)-3*(-d/c)^(1/2)*x*a*c*d*e
*f^2-3*(-d/c)^(1/2)*x*b*c^2*e*f^2+4*(-d/c)^(1/2)*x*b*c*d*e^2*f)/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)/f^3/e/(-d/c)^(1/
2)

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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{3}{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)^(3/2)/(f*x^2+e)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)*(d*x^2 + c)^(3/2)/(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 x^{4} +{\left (b c + a d\right )} x^{2} + a c\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)^(3/2)/(f*x^2+e)^(3/2),x, algorithm="fricas")

[Out]

integral((b*d*x^4 + (b*c + a*d)*x^2 + a*c)*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{3}{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)**(3/2)/(f*x**2+e)**(3/2),x)

[Out]

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

[Out]

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