3.273 \(\int \frac {\sqrt {\frac {e (a+b x^2)}{c+d x^2}}}{x^2} \, dx\)

Optimal. Leaf size=239 \[ \frac {d x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c}-\frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c x}+\frac {b \sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {6719, 475, 21, 422, 418, 492, 411} \[ \frac {d x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c}-\frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c x}+\frac {b \sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 475

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 111, normalized size = 0.46 \[ \frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\frac {b \sqrt {\frac {b x^2}{a}+1} E\left (\sin ^{-1}\left (\sqrt {-\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{\sqrt {-\frac {b}{a}} \left (a+b x^2\right ) \sqrt {\frac {d x^2}{c}+1}}-\frac {1}{x}\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)*(-x^(-1) + (b*Sqrt[1 + (b*x^2)/a]*EllipticE[ArcSin[Sqrt[-(b/a)]
*x], (a*d)/(b*c)])/(Sqrt[-(b/a)]*(a + b*x^2)*Sqrt[1 + (d*x^2)/c])))/c

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 192, normalized size = 0.80 \[ -\frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (\sqrt {-\frac {b}{a}}\, b d \,x^{4}+\sqrt {-\frac {b}{a}}\, a d \,x^{2}+\sqrt {-\frac {b}{a}}\, b c \,x^{2}-\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b c x \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {-\frac {b}{a}}\, a c \right )}{\sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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((e*(b*x**2+a)/(d*x**2+c))**(1/2)/x**2,x)

[Out]

Timed out

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