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

Optimal. Leaf size=271 \[ \frac{b e^{3/2} \sqrt{c+d x^2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{c d \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{f x \sqrt{c+d x^2} (2 b c-a d)}{c d^2 \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{f} \sqrt{c+d x^2} (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{c d^2 \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{x \sqrt{e+f x^2} (b c-a d)}{c d \sqrt{c+d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.17057, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {526, 531, 418, 492, 411} \[ \frac{f x \sqrt{c+d x^2} (2 b c-a d)}{c d^2 \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{f} \sqrt{c+d x^2} (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{c d^2 \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{x \sqrt{e+f x^2} (b c-a d)}{c d \sqrt{c+d x^2}}+\frac{b e^{3/2} \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{c d \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c - a*d)*f*x*Sqrt[c + d*x^2])/(c*d^2*Sqrt[e + f*x^2]) - ((b*c - a*d)*x*Sqrt[e + f*x^2])/(c*d*Sqrt[c + d*
x^2]) - ((2*b*c - a*d)*Sqrt[e]*Sqrt[f]*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)]
)/(c*d^2*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) + (b*e^(3/2)*Sqrt[c + d*x^2]*EllipticF[ArcTan[
(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(c*d*Sqrt[f]*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 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 ) \sqrt{e+f x^2}}{\left (c+d x^2\right )^{3/2}} \, dx &=-\frac{(b c-a d) x \sqrt{e+f x^2}}{c d \sqrt{c+d x^2}}-\frac{\int \frac{-b c e-(2 b c-a d) f x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{c d}\\ &=-\frac{(b c-a d) x \sqrt{e+f x^2}}{c d \sqrt{c+d x^2}}+\frac{(b e) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{d}+\frac{((2 b c-a d) f) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{c d}\\ &=\frac{(2 b c-a d) f x \sqrt{c+d x^2}}{c d^2 \sqrt{e+f x^2}}-\frac{(b c-a d) x \sqrt{e+f x^2}}{c d \sqrt{c+d x^2}}+\frac{b e^{3/2} \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{c d \sqrt{f} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{((2 b c-a d) e f) \int \frac{\sqrt{c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{c d^2}\\ &=\frac{(2 b c-a d) f x \sqrt{c+d x^2}}{c d^2 \sqrt{e+f x^2}}-\frac{(b c-a d) x \sqrt{e+f x^2}}{c d \sqrt{c+d x^2}}-\frac{(2 b c-a d) \sqrt{e} \sqrt{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 )}{c d^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^{3/2} \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{c d \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 = 0.348876, size = 192, normalized size = 0.71 \[ \frac{-(b c-a d) \left (x \sqrt{\frac{d}{c}} \left (e+f x^2\right )-i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )\right )-i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (2 b c-a d) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )}{c^2 \left (\frac{d}{c}\right )^{3/2} \sqrt{c+d x^2} \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.038, size = 328, normalized size = 1.2 \begin{align*}{\frac{1}{d \left ( df{x}^{4}+cf{x}^{2}+de{x}^{2}+ce \right ) c}\sqrt{f{x}^{2}+e}\sqrt{d{x}^{2}+c} \left ({x}^{3}adf\sqrt{-{\frac{d}{c}}}-{x}^{3}bcf\sqrt{-{\frac{d}{c}}}+{\it EllipticF} \left ( x\sqrt{-{\frac{d}{c}}},\sqrt{{\frac{cf}{de}}} \right ) ade\sqrt{{\frac{d{x}^{2}+c}{c}}}\sqrt{{\frac{f{x}^{2}+e}{e}}}-{\it EllipticF} \left ( x\sqrt{-{\frac{d}{c}}},\sqrt{{\frac{cf}{de}}} \right ) bce\sqrt{{\frac{d{x}^{2}+c}{c}}}\sqrt{{\frac{f{x}^{2}+e}{e}}}-{\it EllipticE} \left ( x\sqrt{-{\frac{d}{c}}},\sqrt{{\frac{cf}{de}}} \right ) ade\sqrt{{\frac{d{x}^{2}+c}{c}}}\sqrt{{\frac{f{x}^{2}+e}{e}}}+2\,{\it EllipticE} \left ( x\sqrt{-{\frac{d}{c}}},\sqrt{{\frac{cf}{de}}} \right ) bce\sqrt{{\frac{d{x}^{2}+c}{c}}}\sqrt{{\frac{f{x}^{2}+e}{e}}}+xade\sqrt{-{\frac{d}{c}}}-xbce\sqrt{-{\frac{d}{c}}} \right ){\frac{1}{\sqrt{-{\frac{d}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(3/2),x)

[Out]

(f*x^2+e)^(1/2)*(d*x^2+c)^(1/2)*(x^3*a*d*f*(-d/c)^(1/2)-x^3*b*c*f*(-d/c)^(1/2)+EllipticF(x*(-d/c)^(1/2),(c*f/d
/e)^(1/2))*a*d*e*((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*e*((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*e*((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))*b*c*e*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+x*a*d*e
*(-d/c)^(1/2)-x*b*c*e*(-d/c)^(1/2))/d/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)/c/(-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 )} \sqrt{f x^{2} + e}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)*sqrt(f*x^2 + e)/(d*x^2 + c)^(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^{2} x^{4} + 2 \, c d x^{2} + c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(f*x^2 + e)/(d^2*x^4 + 2*c*d*x^2 + c^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 ) \sqrt{e + f x^{2}}}{\left (c + d 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)*(f*x**2+e)**(1/2)/(d*x**2+c)**(3/2),x)

[Out]

Integral((a + b*x**2)*sqrt(e + f*x**2)/(c + d*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 )} \sqrt{f x^{2} + e}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(f*x^2+e)^(1/2)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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