3.401 \(\int \frac{1}{x^{3/2} (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=173 \[ \frac{15 c^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{14 b^{13/4} \sqrt{b x^2+c x^4}}+\frac{15 c \sqrt{b x^2+c x^4}}{7 b^3 x^{5/2}}-\frac{9 \sqrt{b x^2+c x^4}}{7 b^2 x^{9/2}}+\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}} \]

[Out]

1/(b*x^(5/2)*Sqrt[b*x^2 + c*x^4]) - (9*Sqrt[b*x^2 + c*x^4])/(7*b^2*x^(9/2)) + (15*c*Sqrt[b*x^2 + c*x^4])/(7*b^
3*x^(5/2)) + (15*c^(7/4)*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])/(14*b^(13/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.233287, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2023, 2025, 2032, 329, 220} \[ \frac{15 c^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{14 b^{13/4} \sqrt{b x^2+c x^4}}+\frac{15 c \sqrt{b x^2+c x^4}}{7 b^3 x^{5/2}}-\frac{9 \sqrt{b x^2+c x^4}}{7 b^2 x^{9/2}}+\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^(5/2)*Sqrt[b*x^2 + c*x^4]) - (9*Sqrt[b*x^2 + c*x^4])/(7*b^2*x^(9/2)) + (15*c*Sqrt[b*x^2 + c*x^4])/(7*b^
3*x^(5/2)) + (15*c^(7/4)*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])/(14*b^(13/4)*Sqrt[b*x^2 + c*x^4])

Rule 2023

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*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}}+\frac{9 \int \frac{1}{x^{7/2} \sqrt{b x^2+c x^4}} \, dx}{2 b}\\ &=\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}}-\frac{9 \sqrt{b x^2+c x^4}}{7 b^2 x^{9/2}}-\frac{(45 c) \int \frac{1}{x^{3/2} \sqrt{b x^2+c x^4}} \, dx}{14 b^2}\\ &=\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}}-\frac{9 \sqrt{b x^2+c x^4}}{7 b^2 x^{9/2}}+\frac{15 c \sqrt{b x^2+c x^4}}{7 b^3 x^{5/2}}+\frac{\left (15 c^2\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{14 b^3}\\ &=\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}}-\frac{9 \sqrt{b x^2+c x^4}}{7 b^2 x^{9/2}}+\frac{15 c \sqrt{b x^2+c x^4}}{7 b^3 x^{5/2}}+\frac{\left (15 c^2 x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{14 b^3 \sqrt{b x^2+c x^4}}\\ &=\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}}-\frac{9 \sqrt{b x^2+c x^4}}{7 b^2 x^{9/2}}+\frac{15 c \sqrt{b x^2+c x^4}}{7 b^3 x^{5/2}}+\frac{\left (15 c^2 x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{7 b^3 \sqrt{b x^2+c x^4}}\\ &=\frac{1}{b x^{5/2} \sqrt{b x^2+c x^4}}-\frac{9 \sqrt{b x^2+c x^4}}{7 b^2 x^{9/2}}+\frac{15 c \sqrt{b x^2+c x^4}}{7 b^3 x^{5/2}}+\frac{15 c^{7/4} 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 )}{14 b^{13/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0171167, size = 60, normalized size = 0.35 \[ -\frac{2 \sqrt{\frac{c x^2}{b}+1} \, _2F_1\left (-\frac{7}{4},\frac{3}{2};-\frac{3}{4};-\frac{c x^2}{b}\right )}{7 b x^{5/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-7/4, 3/2, -3/4, -((c*x^2)/b)])/(7*b*x^(5/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.217, size = 141, normalized size = 0.8 \begin{align*}{\frac{c{x}^{2}+b}{14\,{b}^{3}} \left ( 15\,\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 ) \sqrt{-bc}{x}^{3}c+30\,{c}^{2}{x}^{4}+12\,bc{x}^{2}-4\,{b}^{2} \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x)

[Out]

1/14/(c*x^4+b*x^2)^(3/2)/x^(1/2)*(c*x^2+b)*(15*((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))*(-b*c)^(1/2)*x^3*c+30*c^2*x^4+12*b*c*x^2-4*b^2)/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^(3/2)), 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^{2} x^{10} + 2 \, b c x^{8} + b^{2} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*sqrt(x)/(c^2*x^10 + 2*b*c*x^8 + b^2*x^6), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{3}{2}} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(1/(x**(3/2)*(x**2*(b + c*x**2))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^(3/2)), x)