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

Optimal. Leaf size=400 \[ \frac{e^{3/2} \sqrt{c+d x^2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\right )\right ) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{15 c d^2 \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right )}{15 d^3 \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d^3 \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \sqrt{e+f x^2} (5 a d f-4 b c f+3 b d e)}{15 d^2}+\frac{b x \sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d} \]

[Out]

((10*a*d*f*(2*d*e - c*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*c^2*f^2))*x*Sqrt[c + d*x^2])/(15*d^3*Sqrt[e + f*x^2])
+ ((3*b*d*e - 4*b*c*f + 5*a*d*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*d^2) + (b*x*Sqrt[c + d*x^2]*(e + f*x^
2)^(3/2))/(5*d) - (Sqrt[e]*(10*a*d*f*(2*d*e - c*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*c^2*f^2))*Sqrt[c + d*x^2]*E
llipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*d^3*Sqrt[f]*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*
Sqrt[e + f*x^2]) + (e^(3/2)*(5*a*d*(3*d*e - c*f) - b*(6*c*d*e - 4*c^2*f))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sq
rt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*c*d^2*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 = 0.43864, antiderivative size = 400, 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} \[ \frac{x \sqrt{c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right )}{15 d^3 \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d^3 \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{e^{3/2} \sqrt{c+d x^2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\right )\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 c d^2 \sqrt{f} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \sqrt{e+f x^2} (5 a d f-4 b c f+3 b d e)}{15 d^2}+\frac{b x \sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((10*a*d*f*(2*d*e - c*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*c^2*f^2))*x*Sqrt[c + d*x^2])/(15*d^3*Sqrt[e + f*x^2])
+ ((3*b*d*e - 4*b*c*f + 5*a*d*f)*x*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])/(15*d^2) + (b*x*Sqrt[c + d*x^2]*(e + f*x^
2)^(3/2))/(5*d) - (Sqrt[e]*(10*a*d*f*(2*d*e - c*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*c^2*f^2))*Sqrt[c + d*x^2]*E
llipticE[ArcTan[(Sqrt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*d^3*Sqrt[f]*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*
Sqrt[e + f*x^2]) + (e^(3/2)*(5*a*d*(3*d*e - c*f) - b*(6*c*d*e - 4*c^2*f))*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sq
rt[f]*x)/Sqrt[e]], 1 - (d*e)/(c*f)])/(15*c*d^2*Sqrt[f]*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 \frac{\left (a+b x^2\right ) \left (e+f x^2\right )^{3/2}}{\sqrt{c+d x^2}} \, dx &=\frac{b x \sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}+\frac{\int \frac{\sqrt{e+f x^2} \left (-(b c-5 a d) e+(3 b d e-4 b c f+5 a d f) x^2\right )}{\sqrt{c+d x^2}} \, dx}{5 d}\\ &=\frac{(3 b d e-4 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d^2}+\frac{b x \sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}+\frac{\int \frac{-e (2 b c (3 d e-2 c f)-5 a d (3 d e-c f))+\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d^2}\\ &=\frac{(3 b d e-4 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d^2}+\frac{b x \sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}-\frac{(e (2 b c (3 d e-2 c f)-5 a d (3 d e-c f))) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d^2}+\frac{\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d^2}\\ &=\frac{\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 d^3 \sqrt{e+f x^2}}+\frac{(3 b d e-4 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d^2}+\frac{b x \sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}+\frac{e^{3/2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\right )\right ) \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 c d^2 \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 \left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 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^3}\\ &=\frac{\left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 d^3 \sqrt{e+f x^2}}+\frac{(3 b d e-4 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d^2}+\frac{b x \sqrt{c+d x^2} \left (e+f x^2\right )^{3/2}}{5 d}-\frac{\sqrt{e} \left (10 a d f (2 d e-c f)+b \left (3 d^2 e^2-13 c d e f+8 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^3 \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^{3/2} \left (5 a d (3 d e-c f)-b \left (6 c d e-4 c^2 f\right )\right ) \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 c d^2 \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 = 0.84414, size = 275, normalized size = 0.69 \[ \frac{i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (c f-d e) (-5 a d f+4 b c f-3 b d e) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )-i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \left (10 a d f (2 d e-c f)+b \left (8 c^2 f^2-13 c d e f+3 d^2 e^2\right )\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )+f x \left (-\sqrt{\frac{d}{c}}\right ) \left (c+d x^2\right ) \left (e+f x^2\right ) \left (-5 a d f+4 b c f-3 b d \left (2 e+f x^2\right )\right )}{15 c^2 f \left (\frac{d}{c}\right )^{5/2} \sqrt{c+d x^2} \sqrt{e+f x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d/c]*f*x*(c + d*x^2)*(e + f*x^2)*(4*b*c*f - 5*a*d*f - 3*b*d*(2*e + f*x^2))) - I*e*(10*a*d*f*(2*d*e - c
*f) + b*(3*d^2*e^2 - 13*c*d*e*f + 8*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)*(-3*b*d*e + 4*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*c^2*(d/c)^(5/2)*f*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])

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Maple [B]  time = 0.018, size = 870, normalized size = 2.2 \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)*(f*x^2+e)^(3/2)/(d*x^2+c)^(1/2),x)

[Out]

1/15*(f*x^2+e)^(1/2)*(d*x^2+c)^(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-(-d/c)^(1/2)*x
^5*b*c*d*f^3+9*(-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^2-4*(-d/
c)^(1/2)*x^3*b*c^2*f^3+5*(-d/c)^(1/2)*x^3*b*c*d*e*f^2+6*(-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-4*((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+7*((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-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*d^2*e^3-10*((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+
20*((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+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*c^2*e*f^2-13*((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+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*d^2*e^3+5*(-d/c)^(1/2)*x*a*c*d*e*f^2-4*(-d/c)^(1/2)*x*b*c^2*e*f^
2+6*(-d/c)^(1/2)*x*b*c*d*e^2*f)/d^2/f/(d*f*x^4+c*f*x^2+d*e*x^2+c*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 (f x^{2} + e\right )}^{\frac{3}{2}}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b f x^{4} +{\left (b e + a f\right )} x^{2} + a e\right )} \sqrt{f x^{2} + e}}{\sqrt{d x^{2} + c}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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