3.389 \(\int \frac{1}{x^{9/2} \sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=326 \[ \frac{7 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{14 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}-\frac{14 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}} \]

[Out]

(14*c^(5/2)*x^(3/2)*(b + c*x^2))/(15*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (2*Sqrt[b*x^2 + c*x^4])/
(9*b*x^(11/2)) + (14*c*Sqrt[b*x^2 + c*x^4])/(45*b^2*x^(7/2)) - (14*c^2*Sqrt[b*x^2 + c*x^4])/(15*b^3*x^(3/2)) -
(14*c^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqr
t[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4]) + (7*c^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/
(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4]
)

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Rubi [A]  time = 0.358624, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2025, 2032, 329, 305, 220, 1196} \[ \frac{14 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{7 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{14 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(14*c^(5/2)*x^(3/2)*(b + c*x^2))/(15*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (2*Sqrt[b*x^2 + c*x^4])/
(9*b*x^(11/2)) + (14*c*Sqrt[b*x^2 + c*x^4])/(45*b^2*x^(7/2)) - (14*c^2*Sqrt[b*x^2 + c*x^4])/(15*b^3*x^(3/2)) -
(14*c^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqr
t[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4]) + (7*c^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/
(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4]
)

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{1}{x^{9/2} \sqrt{b x^2+c x^4}} \, dx &=-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{(7 c) \int \frac{1}{x^{5/2} \sqrt{b x^2+c x^4}} \, dx}{9 b}\\ &=-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}+\frac{\left (7 c^2\right ) \int \frac{1}{\sqrt{x} \sqrt{b x^2+c x^4}} \, dx}{15 b^2}\\ &=-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{\left (7 c^3\right ) \int \frac{x^{3/2}}{\sqrt{b x^2+c x^4}} \, dx}{15 b^3}\\ &=-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{\left (7 c^3 x \sqrt{b+c x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{b+c x^2}} \, dx}{15 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{\left (14 c^3 x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{\left (14 c^{5/2} x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 b^{5/2} \sqrt{b x^2+c x^4}}-\frac{\left (14 c^{5/2} x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{b}}}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 b^{5/2} \sqrt{b x^2+c x^4}}\\ &=\frac{14 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}-\frac{14 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{7 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0132543, size = 57, normalized size = 0.17 \[ -\frac{2 \sqrt{\frac{c x^2}{b}+1} \, _2F_1\left (-\frac{9}{4},\frac{1}{2};-\frac{5}{4};-\frac{c x^2}{b}\right )}{9 x^{7/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(-2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-9/4, 1/2, -5/4, -((c*x^2)/b)])/(9*x^(7/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.188, size = 230, normalized size = 0.7 \begin{align*}{\frac{1}{45\,{b}^{3}} \left ( 42\,\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{4}b{c}^{2}-21\,\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{4}b{c}^{2}-42\,{c}^{3}{x}^{6}-28\,b{c}^{2}{x}^{4}+4\,{b}^{2}c{x}^{2}-10\,{b}^{3} \right ){x}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(9/2)/(c*x^4+b*x^2)^(1/2),x)

[Out]

1/45/(c*x^4+b*x^2)^(1/2)/x^(7/2)*(42*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*
c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b
*c^2-21*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^
(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b*c^2-42*c^3*x^6-28*b*c^2*x^4+
4*b^2*c*x^2-10*b^3)/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{4} + b x^{2}} x^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{x}}{c x^{9} + b x^{7}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*sqrt(x)/(c*x^9 + b*x^7), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{9}{2}} \sqrt{x^{2} \left (b + c x^{2}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(9/2)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(1/(x**(9/2)*sqrt(x**2*(b + c*x**2))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{4} + b x^{2}} x^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)