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

Optimal. Leaf size=381 \[ -\frac{e^{3/2} \sqrt{c+d x^2} (-10 a d f+b c f+b d e) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{15 d f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right )}{15 d^2 f \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d^2 f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \sqrt{e+f x^2} (5 a d f-2 b c f+b d e)}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d} \]

[Out]

((5*a*d*f*(d*e + c*f) - 2*b*(d^2*e^2 - c*d*e*f + c^2*f^2))*x*Sqrt[c + d*x^2])/(15*d^2*f*Sqrt[e + f*x^2]) + ((b
*d*e - 2*b*c*f + 5*a*d*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*d*f) + (b*x*(c + d*x^2)^(3/2)*Sqrt[e + f*x^2]
)/(5*d) - (Sqrt[e]*(5*a*d*f*(d*e + c*f) - 2*b*(d^2*e^2 - c*d*e*f + c^2*f^2))*Sqrt[c + d*x^2]*EllipticE[ArcTan[
(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*d^2*f^(3/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])
- (e^(3/2)*(b*d*e + b*c*f - 10*a*d*f)*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)]
)/(15*d*f^(3/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

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Rubi [A]  time = 0.377796, antiderivative size = 381, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {528, 531, 418, 492, 411} \[ \frac{x \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right )}{15 d^2 f \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d^2 f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{e^{3/2} \sqrt{c+d x^2} (-10 a d f+b c f+b d e) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \sqrt{e+f x^2} (5 a d f-2 b c f+b d e)}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a*d*f*(d*e + c*f) - 2*b*(d^2*e^2 - c*d*e*f + c^2*f^2))*x*Sqrt[c + d*x^2])/(15*d^2*f*Sqrt[e + f*x^2]) + ((b
*d*e - 2*b*c*f + 5*a*d*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*d*f) + (b*x*(c + d*x^2)^(3/2)*Sqrt[e + f*x^2]
)/(5*d) - (Sqrt[e]*(5*a*d*f*(d*e + c*f) - 2*b*(d^2*e^2 - c*d*e*f + c^2*f^2))*Sqrt[c + d*x^2]*EllipticE[ArcTan[
(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*d^2*f^(3/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])
- (e^(3/2)*(b*d*e + b*c*f - 10*a*d*f)*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)]
)/(15*d*f^(3/2)*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2])

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 \left (a+b x^2\right ) \sqrt{c+d x^2} \sqrt{e+f x^2} \, dx &=\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}+\frac{\int \frac{\sqrt{c+d x^2} \left (-(b c-5 a d) e+(b d e-2 b c f+5 a d f) x^2\right )}{\sqrt{e+f x^2}} \, dx}{5 d}\\ &=\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}+\frac{\int \frac{-c e (b d e+b c f-10 a d f)+\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d f}\\ &=\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}-\frac{(c e (b d e+b c f-10 a d f)) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d f}+\frac{\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d f}\\ &=\frac{\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 d^2 f \sqrt{e+f x^2}}+\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}-\frac{e^{3/2} (b d e+b c f-10 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 )}{15 d f^{3/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{\left (e \left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right )\right ) \int \frac{\sqrt{c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{15 d^2 f}\\ &=\frac{\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 d^2 f \sqrt{e+f x^2}}+\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}-\frac{\sqrt{e} \left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+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 d^2 f^{3/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{e^{3/2} (b d e+b c f-10 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 )}{15 d f^{3/2} \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 = 0.75238, size = 267, normalized size = 0.7 \[ \frac{-i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (c f-d e) (5 a d f+b c f-2 b d e) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )+i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \left (2 b \left (c^2 f^2-c d e f+d^2 e^2\right )-5 a d f (c f+d e)\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )+f x \sqrt{\frac{d}{c}} \left (c+d x^2\right ) \left (e+f x^2\right ) \left (5 a d f+b c f+b d \left (e+3 f x^2\right )\right )}{15 d f^2 \sqrt{\frac{d}{c}} \sqrt{c+d x^2} \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/c]*f*x*(c + d*x^2)*(e + f*x^2)*(b*c*f + 5*a*d*f + b*d*(e + 3*f*x^2)) + I*e*(-5*a*d*f*(d*e + c*f) + 2*b
*(d^2*e^2 - c*d*e*f + c^2*f^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*b*d*e + b*c*f + 5*a*d*f)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticF[I
*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)])/(15*d*Sqrt[d/c]*f^2*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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Maple [B]  time = 0.015, size = 865, 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)^(1/2)*(f*x^2+e)^(1/2),x)

[Out]

1/15*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)*(3*(-d/c)^(1/2)*x^7*b*d^2*f^3+5*(-d/c)^(1/2)*x^5*a*d^2*f^3+4*(-d/c)^(1/2)
*x^5*b*c*d*f^3+4*(-d/c)^(1/2)*x^5*b*d^2*e*f^2+5*(-d/c)^(1/2)*x^3*a*c*d*f^3+5*(-d/c)^(1/2)*x^3*a*d^2*e*f^2+(-d/
c)^(1/2)*x^3*b*c^2*f^3+5*(-d/c)^(1/2)*x^3*b*c*d*e*f^2+(-d/c)^(1/2)*x^3*b*d^2*e^2*f+5*((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^2-5*((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^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))*b*c^2*e*f^2-3*((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*d*e^2*f+2*((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+5*((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+5*((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-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))*b*c^2*e*f^2+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))*b*c*d*e^2*f-2*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*Ellip
ticE(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*b*d^2*e^3+5*(-d/c)^(1/2)*x*a*c*d*e*f^2+(-d/c)^(1/2)*x*b*c^2*e*f^2+(-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)/d/f^2/(-d/c)^(1/2)

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

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

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)^(1/2),x, algorithm="fricas")

[Out]

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

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

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)**(1/2),x)

[Out]

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

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

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)^(1/2),x, algorithm="giac")

[Out]

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