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

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

[Out]

-d*(a*d*f-2*b*c*f+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)*(d*x^2+c)^(1/2)/c/(-a*d+b*c)^2/(-c*f+d*e)/f^(1/2)/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)/
(f*x^2+e)^(1/2)-d^(3/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),
(1-c*f/d/e)^(1/2))*(f*x^2+e)^(1/2)/(-a*d+b*c)/(-c*f+d*e)/c^(1/2)/(d*x^2+c)^(1/2)/(c*(f*x^2+e)/e/(d*x^2+c))^(1/
2)+b^2*c^(3/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticPi(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),1-b*c/a/
d,(1-c*f/d/e)^(1/2))*(f*x^2+e)^(1/2)/a/(-a*d+b*c)^2/e/d^(1/2)/(d*x^2+c)^(1/2)/(c*(f*x^2+e)/e/(d*x^2+c))^(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} \[ \frac {b^2 c^{3/2} \sqrt {e+f x^2} \Pi \left (1-\frac {b c}{a d};\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{a \sqrt {d} e \sqrt {c+d x^2} (b c-a d)^2 \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac {d^{3/2} \sqrt {e+f x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{\sqrt {c} \sqrt {c+d x^2} (b c-a d) (d e-c f) \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac {d \sqrt {e} \sqrt {c+d x^2} (a d f-2 b c f+b d e) F\left (\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )|1-\frac {d e}{c f}\right )}{c \sqrt {f} \sqrt {e+f x^2} (b c-a d)^2 (d e-c f) \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^(3/2)*Sqrt[e + f*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(Sqrt[c]*(b*c - a*d)*(d*e
- c*f)*Sqrt[c + d*x^2]*Sqrt[(c*(e + f*x^2))/(e*(c + d*x^2))])) - (d*Sqrt[e]*(b*d*e - 2*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*c - a*d)^2*Sqrt[f]*(d*e - c*f)*Sqrt[(e*
(c + d*x^2))/(c*(e + f*x^2))]*Sqrt[e + f*x^2]) + (b^2*c^(3/2)*Sqrt[e + f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcT
an[(Sqrt[d]*x)/Sqrt[c]], 1 - (c*f)/(d*e)])/(a*Sqrt[d]*(b*c - a*d)^2*e*Sqrt[c + d*x^2]*Sqrt[(c*(e + f*x^2))/(e*
(c + d*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 ) \left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}} \, dx &=\frac {b^2 \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right ) \sqrt {e+f x^2}} \, dx}{(b c-a d)^2}-\frac {d \int \frac {2 b c-a d+b d x^2}{\left (c+d x^2\right )^{3/2} \sqrt {e+f x^2}} \, dx}{(b c-a d)^2}\\ &=\frac {b^2 c^{3/2} \sqrt {e+f x^2} \Pi \left (1-\frac {b c}{a d};\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{a \sqrt {d} (b c-a d)^2 e \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac {d^2 \int \frac {\sqrt {e+f x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{(b c-a d) (d e-c f)}-\frac {(d (b d e-2 b c f+a d f)) \int \frac {1}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx}{(b c-a d)^2 (d e-c f)}\\ &=-\frac {d^{3/2} \sqrt {e+f x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{\sqrt {c} (b c-a d) (d e-c f) \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}-\frac {d \sqrt {e} (b d e-2 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 c-a d)^2 \sqrt {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 {b^2 c^{3/2} \sqrt {e+f x^2} \Pi \left (1-\frac {b c}{a d};\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {c f}{d e}\right )}{a \sqrt {d} (b c-a d)^2 e \sqrt {c+d x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}}}\\ \end {align*}

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Mathematica [C]  time = 0.95, size = 365, normalized size = 1.06 \[ \frac {\sqrt {\frac {d}{c}} \left (-i b c^2 f \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+i b c d e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+i a d^2 e \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+i a d \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} (c f-d e) F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+a c d e x \left (\frac {d}{c}\right )^{3/2}+a c d f x^3 \left (\frac {d}{c}\right )^{3/2}\right )}{a d \sqrt {c+d x^2} \sqrt {e+f x^2} (a d-b c) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/c]*(a*c*d*(d/c)^(3/2)*e*x + a*c*d*(d/c)^(3/2)*f*x^3 + I*a*d^2*e*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e
]*EllipticE[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] + I*a*d*(-(d*e) + c*f)*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e
]*EllipticF[I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] + I*b*c*d*e*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticP
i[(b*c)/(a*d), I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)] - I*b*c^2*f*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*Ellipt
icPi[(b*c)/(a*d), I*ArcSinh[Sqrt[d/c]*x], (c*f)/(d*e)]))/(a*d*(-(b*c) + a*d)*(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)^(3/2)/(f*x^2+e)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 413, normalized size = 1.20 \[ \frac {\left (-\sqrt {-\frac {d}{c}}\, a \,d^{2} f \,x^{3}+\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a c d f \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-\sqrt {-\frac {d}{c}}\, a \,d^{2} e x +\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a \,d^{2} e \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, a \,d^{2} e \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b \,c^{2} f \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )+\sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, b c d e \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )\right ) \sqrt {f \,x^{2}+e}\, \sqrt {d \,x^{2}+c}}{\left (c f -d e \right ) \sqrt {-\frac {d}{c}}\, \left (a d -b c \right ) \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x^3*a*d^2*f*(-1/c*d)^(1/2)+EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*c*d*f*((d*x^2+c)/c)^(1/2)*((f*x^2+e
)/e)^(1/2)-EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*d^2*e*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)+Ellipti
cE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*d^2*e*((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*c^2*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*c*d*e*((d*x^2+c)/c)^(1/2)*((f*x^2+e)/e)^(1/2)-x*a*d^2*e*(
-1/c*d)^(1/2))*(f*x^2+e)^(1/2)*(d*x^2+c)^(1/2)/c/(c*f-d*e)/a/(-1/c*d)^(1/2)/(a*d-b*c)/(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 \[ \int \frac {1}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}} \sqrt {f x^{2} + e}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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