3.306 \(\int \frac{1}{x^4 \sqrt{\frac{e (a+b x^2)}{c+d x^2}}} \, dx\)

Optimal. Leaf size=372 \[ \frac{\left (a+b x^2\right ) (2 b c-a d)}{3 a^2 c x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{d x \left (a+b x^2\right ) (2 b c-a d)}{3 a^2 c \left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{b \sqrt{c} \sqrt{d} \left (a+b x^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a^2 \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\sqrt{d} \left (a+b x^2\right ) (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a^2 \sqrt{c} \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{a+b x^2}{3 a x^3 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.453529, antiderivative size = 372, normalized size of antiderivative = 1., 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} \[ \frac{\left (a+b x^2\right ) (2 b c-a d)}{3 a^2 c x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{d x \left (a+b x^2\right ) (2 b c-a d)}{3 a^2 c \left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{b \sqrt{c} \sqrt{d} \left (a+b x^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a^2 \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\sqrt{d} \left (a+b x^2\right ) (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{3 a^2 \sqrt{c} \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{a+b x^2}{3 a x^3 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]),x]

[Out]

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

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]

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 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.403667, size = 238, normalized size = 0.64 \[ \frac{-\sqrt{\frac{b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (a \left (c+d x^2\right )-2 b c x^2\right )+2 i b c x^3 \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )-i b c x^3 \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-2 b c) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{3 a^2 c x^3 \sqrt{\frac{b}{a}} \left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]),x]

[Out]

(-(Sqrt[b/a]*(a + b*x^2)*(c + d*x^2)*(-2*b*c*x^2 + a*(c + d*x^2))) - I*b*c*(-2*b*c + 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)] + (2*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)])/(3*a^2*Sqrt[b/a]*c*x^3*Sqrt[(e
*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))

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Maple [A]  time = 0.01, size = 444, normalized size = 1.2 \begin{align*} -{\frac{b{x}^{2}+a}{3\,{a}^{2}{x}^{3}c} \left ( \sqrt{-{\frac{b}{a}}}{x}^{6}ab{d}^{2}-2\,\sqrt{-{\frac{b}{a}}}{x}^{6}{b}^{2}cd+2\,bd\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){x}^{3}ac-2\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){x}^{3}{b}^{2}{c}^{2}-\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){x}^{3}abcd+2\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ){x}^{3}{b}^{2}{c}^{2}+\sqrt{-{\frac{b}{a}}}{x}^{4}{a}^{2}{d}^{2}-2\,\sqrt{-{\frac{b}{a}}}{x}^{4}{b}^{2}{c}^{2}+2\,\sqrt{-{\frac{b}{a}}}{x}^{2}{a}^{2}cd-\sqrt{-{\frac{b}{a}}}{x}^{2}ab{c}^{2}+\sqrt{-{\frac{b}{a}}}{a}^{2}{c}^{2} \right ){\frac{1}{\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}{\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b*x^2+a)*((-b/a)^(1/2)*x^6*a*b*d^2-2*(-b/a)^(1/2)*x^6*b^2*c*d+2*b*d*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1
/2)*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^3*a*c-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-
b/a)^(1/2),(a*d/b/c)^(1/2))*x^3*b^2*c^2-((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/
b/c)^(1/2))*x^3*a*b*c*d+2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*x^
3*b^2*c^2+(-b/a)^(1/2)*x^4*a^2*d^2-2*(-b/a)^(1/2)*x^4*b^2*c^2+2*(-b/a)^(1/2)*x^2*a^2*c*d-(-b/a)^(1/2)*x^2*a*b*
c^2+(-b/a)^(1/2)*a^2*c^2)/(e*(b*x^2+a)/(d*x^2+c))^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/a^2/x^3/c/(-b/a)^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d x^{2} + c\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{b e x^{6} + a e x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x^2 + c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b*e*x^6 + a*e*x^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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