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

Optimal. Leaf size=328 \[ \frac{e^{3/2} \sqrt{f} \sqrt{c+d x^2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{b c \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{e^{3/2} \sqrt{c+d x^2} (b e-a f) \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a b c \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{f^2 x \sqrt{c+d x^2}}{b d \sqrt{e+f x^2}}-\frac{\sqrt{e} f^{3/2} \sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{b d \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

[Out]

(f^2*x*Sqrt[c + d*x^2])/(b*d*Sqrt[e + f*x^2]) - (Sqrt[e]*f^(3/2)*Sqrt[c + d*x^2]*EllipticE[ArcTan[(Sqrt[f]*x)/
Sqrt[e]], 1 - (d*e)/(c*f)])/(b*d*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) + (e^(3/2)*Sqrt[f]*Sqr
t[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(b*c*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2)
)]*Sqrt[e + f*x^2]) + (e^(3/2)*(b*e - a*f)*Sqrt[c + d*x^2]*EllipticPi[1 - (b*e)/(a*f), ArcTan[(Sqrt[f]*x)/Sqrt
[e]], 1 - (d*e)/(c*f)])/(a*b*c*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 = 0.191238, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {545, 422, 418, 492, 411, 539} \[ \frac{e^{3/2} \sqrt{c+d x^2} (b e-a f) \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a b c \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{e^{3/2} \sqrt{f} \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{b c \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{f^2 x \sqrt{c+d x^2}}{b d \sqrt{e+f x^2}}-\frac{\sqrt{e} f^{3/2} \sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{b d \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 545

Int[(((c_) + (d_.)*(x_)^2)^(q_)*((e_) + (f_.)*(x_)^2)^(r_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/b, Int[
(c + d*x^2)^(q - 1)*(e + f*x^2)^r, x], x] + Dist[(b*c - a*d)/b, Int[((c + d*x^2)^(q - 1)*(e + f*x^2)^r)/(a + b
*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, r}, x] && GtQ[q, 1]

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[a, Int[1/(Sqrt[a + b*x^2]*Sqrt[c +
d*x^2]), x], x] + Dist[b, Int[x^2/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[
d/c] && PosQ[b/a]

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]

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\left (e+f x^2\right )^{3/2}}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx &=\frac{f \int \frac{\sqrt{e+f x^2}}{\sqrt{c+d x^2}} \, dx}{b}+\frac{(b e-a f) \int \frac{\sqrt{e+f x^2}}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{b}\\ &=\frac{e^{3/2} (b e-a f) \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a b c \sqrt{f} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}+\frac{(e f) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{b}+\frac{f^2 \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{b}\\ &=\frac{f^2 x \sqrt{c+d x^2}}{b d \sqrt{e+f x^2}}+\frac{e^{3/2} \sqrt{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 )}{b c \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 e-a f) \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a b c \sqrt{f} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{\left (e f^2\right ) \int \frac{\sqrt{c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{b d}\\ &=\frac{f^2 x \sqrt{c+d x^2}}{b d \sqrt{e+f x^2}}-\frac{\sqrt{e} f^{3/2} \sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{b d \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} \sqrt{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 )}{b c \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 e-a f) \sqrt{c+d x^2} \Pi \left (1-\frac{b e}{a f};\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{a b c \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 = 0.355811, size = 184, normalized size = 0.56 \[ -\frac{i \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \left ((b e-a f) \left (a f \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )+(b e-a f) \Pi \left (\frac{b c}{a d};i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )\right )+a b e f E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )\right )}{a b^2 \sqrt{\frac{d}{c}} \sqrt{c+d x^2} \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.018, size = 300, normalized size = 0.9 \begin{align*}{\frac{1}{a{b}^{2} \left ( df{x}^{4}+cf{x}^{2}+de{x}^{2}+ce \right ) } \left ( -{\it EllipticF} \left ( x\sqrt{-{\frac{d}{c}}},\sqrt{{\frac{cf}{de}}} \right ){a}^{2}{f}^{2}+{\it EllipticF} \left ( x\sqrt{-{\frac{d}{c}}},\sqrt{{\frac{cf}{de}}} \right ) abef+{\it EllipticE} \left ( x\sqrt{-{\frac{d}{c}}},\sqrt{{\frac{cf}{de}}} \right ) abef+{\it EllipticPi} \left ( x\sqrt{-{\frac{d}{c}}},{\frac{bc}{ad}},{\sqrt{-{\frac{f}{e}}}{\frac{1}{\sqrt{-{\frac{d}{c}}}}}} \right ){a}^{2}{f}^{2}-2\,{\it EllipticPi} \left ( x\sqrt{-{\frac{d}{c}}},{\frac{bc}{ad}},{\sqrt{-{\frac{f}{e}}}{\frac{1}{\sqrt{-{\frac{d}{c}}}}}} \right ) abef+{\it EllipticPi} \left ( x\sqrt{-{\frac{d}{c}}},{\frac{bc}{ad}},{\sqrt{-{\frac{f}{e}}}{\frac{1}{\sqrt{-{\frac{d}{c}}}}}} \right ){b}^{2}{e}^{2} \right ) \sqrt{{\frac{f{x}^{2}+e}{e}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}\sqrt{d{x}^{2}+c}\sqrt{f{x}^{2}+e}{\frac{1}{\sqrt{-{\frac{d}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a^2*f^2+EllipticF(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*b*e*f+Elliptic
E(x*(-d/c)^(1/2),(c*f/d/e)^(1/2))*a*b*e*f+EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*a^2*f^2
-2*EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/e)^(1/2)/(-d/c)^(1/2))*a*b*e*f+EllipticPi(x*(-d/c)^(1/2),b*c/a/d,(-f/
e)^(1/2)/(-d/c)^(1/2))*b^2*e^2)*((f*x^2+e)/e)^(1/2)*((d*x^2+c)/c)^(1/2)*(d*x^2+c)^(1/2)*(f*x^2+e)^(1/2)/a/(-d/
c)^(1/2)/b^2/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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