3.274 \(\int \frac {\sqrt {\frac {e (a+b x^2)}{c+d x^2}}}{x^4} \, dx\)

Optimal. Leaf size=321 \[ -\frac {\sqrt {d} (b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a c^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {d x (b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 a c^2}-\frac {\left (c+d x^2\right ) (b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 a c^2 x}-\frac {b \sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a \sqrt {c} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 c x^3} \]

[Out]

1/3*d*(-2*a*d+b*c)*x*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a/c^2-1/3*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/c/x^3-1/3
*(-2*a*d+b*c)*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a/c^2/x-1/3*(-2*a*d+b*c)*(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
)/a/c^(3/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)-1/3*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))*d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a/c^(1/2)/(c*(b*x^2+a)/a/(d
*x^2+c))^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 321, 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, 583, 531, 418, 492, 411} \[ \frac {d x (b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 a c^2}-\frac {\left (c+d x^2\right ) (b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 a c^2 x}-\frac {\sqrt {d} (b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a c^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 c x^3}-\frac {b \sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 a \sqrt {c} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/x^4,x]

[Out]

(d*(b*c - 2*a*d)*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(3*a*c^2) - (Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^
2))/(3*c*x^3) - ((b*c - 2*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(3*a*c^2*x) - (Sqrt[d]*(b*c - 2*
a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*a*c^(3/2)*S
qrt[(c*(a + b*x^2))/(a*(c + d*x^2))]) - (b*Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]
*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*a*Sqrt[c]*Sqrt[(c*(a + b*x^2))/(a*(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 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 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 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

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^4} \, 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^4 \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 )}{3 c x^3}+\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-2 a d-b d x^2}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 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 )}{3 c x^3}-\frac {(b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 a c^2 x}-\frac {\left (\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {a b c d-b d (b c-2 a d) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 a c^2 \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 )}{3 c x^3}-\frac {(b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 a c^2 x}-\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 {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 c \sqrt {a+b x^2}}+\frac {\left (b d (b c-2 a 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}{3 a c^2 \sqrt {a+b x^2}}\\ &=\frac {d (b c-2 a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 a c^2}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 c x^3}-\frac {(b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 a c^2 x}-\frac {b \sqrt {d} \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 )}{3 a \sqrt {c} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\left (d (b c-2 a 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}{3 a c \sqrt {a+b x^2}}\\ &=\frac {d (b c-2 a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 a c^2}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 c x^3}-\frac {(b c-2 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 a c^2 x}-\frac {\sqrt {d} (b c-2 a 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 )}{3 a c^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {b \sqrt {d} \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 )}{3 a \sqrt {c} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\\ \end {align*}

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Mathematica [C]  time = 0.66, size = 238, normalized size = 0.74 \[ -\frac {\sqrt {\frac {b}{a}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {\frac {b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (a \left (c-2 d x^2\right )+b c x^2\right )+i b c x^3 \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i b c x^3 \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (2 a d-b c) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )\right )}{3 b c^2 x^3 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/x^4,x]

[Out]

-1/3*(Sqrt[b/a]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[b/a]*(a + b*x^2)*(c + d*x^2)*(b*c*x^2 + a*(c - 2*d*x^2
)) - I*b*c*(-(b*c) + 2*a*d)*x^3*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d
)/(b*c)] + I*b*c*(-(b*c) + a*d)*x^3*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x],
(a*d)/(b*c)]))/(b*c^2*x^3*(a + b*x^2))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^4,x, algorithm="fricas")

[Out]

integral(sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate(sqrt((b*x^2 + a)*e/(d*x^2 + c))/x^4, x)

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maple [A]  time = 0.04, size = 444, normalized size = 1.38 \[ \frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (2 \sqrt {-\frac {b}{a}}\, a b \,d^{2} x^{6}-\sqrt {-\frac {b}{a}}\, b^{2} c d \,x^{6}+2 \sqrt {-\frac {b}{a}}\, a^{2} d^{2} x^{4}-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c d \,x^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b c d \,x^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-\sqrt {-\frac {b}{a}}\, b^{2} c^{2} x^{4}+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{2} x^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{2} x^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {-\frac {b}{a}}\, a^{2} c d \,x^{2}-2 \sqrt {-\frac {b}{a}}\, a b \,c^{2} x^{2}-\sqrt {-\frac {b}{a}}\, a^{2} c^{2}\right )}{3 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a \,c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((b*x^2+a)/(d*x^2+c)*e)^(1/2)*(d*x^2+c)*(2*(-1/a*b)^(1/2)*x^6*a*b*d^2-(-1/a*b)^(1/2)*x^6*b^2*c*d+b*d*((b*x
^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^3*a*c-((b*x^2+a)/a)^(1/2)*((d
*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^3*b^2*c^2-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(
1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^3*a*b*c*d+((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE
((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*x^3*b^2*c^2+2*(-1/a*b)^(1/2)*x^4*a^2*d^2-(-1/a*b)^(1/2)*x^4*b^2*c^2+(-1/a*b
)^(1/2)*x^2*a^2*c*d-2*(-1/a*b)^(1/2)*x^2*a*b*c^2-(-1/a*b)^(1/2)*a^2*c^2)/((d*x^2+c)*(b*x^2+a))^(1/2)/c^2/x^3/(
-1/a*b)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt((b*x^2 + a)*e/(d*x^2 + c))/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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