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

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

[Out]

f^2*x*(d*x^2+c)^(1/2)/b/d/(f*x^2+e)^(1/2)-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))*e^(1/2)*(d*x^2+c)^(1/2)/b/d/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)/(f*x^
2+e)^(1/2)+e^(3/2)*(-a*f+b*e)*(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/b/c/f^(1/2)/(e*(d*x^2+c)/c/(f*x^2+e))^(1/2)/(f*x^2+e)^(1
/2)+e^(3/2)*(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))*f^(1/2)*(d*x^2+c)^(1/2)/b/c/(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 = 0.19, antiderivative size = 328, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.36, size = 184, normalized size = 0.56 \[ -\frac {i \sqrt {\frac {d x^2}{c}+1} \sqrt {\frac {f x^2}{e}+1} \left (a b e f E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+(b e-a f) \left ((b e-a f) \Pi \left (\frac {b c}{a d};i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )+a f F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c}} x\right )|\frac {c f}{d e}\right )\right )\right )}{a b^2 \sqrt {\frac {d}{c}} \sqrt {c+d x^2} \sqrt {e+f x^2}} \]

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

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)

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maple [A]  time = 0.02, size = 300, normalized size = 0.91 \[ \frac {\left (-a^{2} f^{2} \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+a^{2} f^{2} \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )+a b e f \EllipticE \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )+a b e f \EllipticF \left (\sqrt {-\frac {d}{c}}\, x , \sqrt {\frac {c f}{d e}}\right )-2 a b e f \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )+b^{2} e^{2} \EllipticPi \left (\sqrt {-\frac {d}{c}}\, x , \frac {b c}{a d}, \frac {\sqrt {-\frac {f}{e}}}{\sqrt {-\frac {d}{c}}}\right )\right ) \sqrt {\frac {f \,x^{2}+e}{e}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}{\sqrt {-\frac {d}{c}}\, \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right ) a \,b^{2}} \]

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((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a^2*f^2+EllipticF((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*b*e*f+Elli
pticE((-1/c*d)^(1/2)*x,(c/d/e*f)^(1/2))*a*b*e*f+EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/(-1/c*d)^
(1/2))*a^2*f^2-2*EllipticPi((-1/c*d)^(1/2)*x,1/a*b*c/d,(-1/e*f)^(1/2)/(-1/c*d)^(1/2))*a*b*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^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/(-1/c*d)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x^{2} + e\right )}^{\frac {3}{2}}}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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