3.86 \(\int \frac {1}{(a+b x^2) \sqrt {c+d x^2} (e+f x^2)^{3/2}} \, dx\)

Optimal. Leaf size=344 \[ \frac {b^2 e^{3/2} \sqrt {c+d x^2} \Pi \left (1-\frac {b e}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a c \sqrt {f} \sqrt {e+f x^2} (b e-a f)^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {f^{3/2} \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{\sqrt {e} \sqrt {e+f x^2} (b e-a f) (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {\sqrt {e} \sqrt {f} \sqrt {c+d x^2} (-a d f-b c f+2 b d e) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{c \sqrt {e+f x^2} (b e-a f)^2 (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {546, 539, 525, 418, 411} \[ \frac {b^2 e^{3/2} \sqrt {c+d x^2} \Pi \left (1-\frac {b e}{a f};\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{a c \sqrt {f} \sqrt {e+f x^2} (b e-a f)^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {f^{3/2} \sqrt {c+d x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{\sqrt {e} \sqrt {e+f x^2} (b e-a f) (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac {\sqrt {e} \sqrt {f} \sqrt {c+d x^2} (-a d f-b c f+2 b d e) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{c \sqrt {e+f x^2} (b e-a f)^2 (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, 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]

Rule 546

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

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}} \, dx &=\frac {b^2 \int \frac {\sqrt {e+f x^2}}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{(b e-a f)^2}-\frac {f \int \frac {2 b e-a f+b f x^2}{\sqrt {c+d x^2} \left (e+f x^2\right )^{3/2}} \, dx}{(b e-a f)^2}\\ &=\frac {b^2 e^{3/2} \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 c \sqrt {f} (b e-a f)^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {f^2 \int \frac {\sqrt {c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{(b e-a f) (d e-c f)}-\frac {(f (2 b d e-b c f-a d f)) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{(b e-a f)^2 (d e-c f)}\\ &=\frac {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 )}{\sqrt {e} (b e-a f) (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}-\frac {\sqrt {e} \sqrt {f} (2 b d e-b c f-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 )}{c (b e-a f)^2 (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}+\frac {b^2 e^{3/2} \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 c \sqrt {f} (b e-a f)^2 \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.64, size = 221, normalized size = 0.64 \[ \frac {-i b e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (c f-d e) \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-i a d e f \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )-a f^2 x \sqrt {\frac {d}{c}} \left (c+d x^2\right )}{a e \sqrt {\frac {d}{c}} \sqrt {c+d x^2} \sqrt {e+f x^2} (a f-b e) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} {\left (f x^{2} + e\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 303, normalized size = 0.88 \[ \frac {\left (\sqrt {-\frac {d}{c}}\, a d \,f^{2} x^{3}+\sqrt {-\frac {d}{c}}\, a c \,f^{2} x -\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a d e f \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b c e f \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )+\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b d \,e^{2} \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )\right ) \sqrt {f \,x^{2}+e}\, \sqrt {d \,x^{2}+c}}{\left (c f -d e \right ) \sqrt {-\frac {d}{c}}\, \left (a f -b e \right ) \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) a e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-1/c*d)^(1/2)*a*d*f^2*x^3-((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)*EllipticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2)
)*a*d*e*f-EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/(-1/c*d)^(1/2))*b*c*e*f*((d*x^2+c)/c)^(1/2)*((f
*x^2+e)/e)^(1/2)+EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/(-1/c*d)^(1/2))*b*d*e^2*((d*x^2+c)/c)^(1
/2)*((f*x^2+e)/e)^(1/2)+(-1/c*d)^(1/2)*a*c*f^2*x)*(f*x^2+e)^(1/2)*(d*x^2+c)^(1/2)/e/(c*f-d*e)/a/(-1/c*d)^(1/2)
/(a*f-b*e)/(d*f*x^4+c*f*x^2+d*e*x^2+c*e)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} {\left (f x^{2} + e\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\left (b\,x^2+a\right )\,\sqrt {d\,x^2+c}\,{\left (f\,x^2+e\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{2}\right ) \sqrt {c + d x^{2}} \left (e + f x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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