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

Optimal. Leaf size=118 \[ \frac{2 b^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{3 \sqrt [4]{c} \sqrt{b x^2+c x^4}}+\frac{2 \sqrt{b x^2+c x^4}}{3 \sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.133109, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2021, 2032, 329, 220} \[ \frac{2 b^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{c} \sqrt{b x^2+c x^4}}+\frac{2 \sqrt{b x^2+c x^4}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^(3/2),x]

[Out]

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

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*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]) && G
tQ[p, 0] && NeQ[m + n*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{\sqrt{b x^2+c x^4}}{x^{3/2}} \, dx &=\frac{2 \sqrt{b x^2+c x^4}}{3 \sqrt{x}}+\frac{1}{3} (2 b) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx\\ &=\frac{2 \sqrt{b x^2+c x^4}}{3 \sqrt{x}}+\frac{\left (2 b x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{3 \sqrt{b x^2+c x^4}}\\ &=\frac{2 \sqrt{b x^2+c x^4}}{3 \sqrt{x}}+\frac{\left (4 b x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{3 \sqrt{b x^2+c x^4}}\\ &=\frac{2 \sqrt{b x^2+c x^4}}{3 \sqrt{x}}+\frac{2 b^{3/4} 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 )}{3 \sqrt [4]{c} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.013025, size = 55, normalized size = 0.47 \[ \frac{2 \sqrt{x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{b}\right )}{\sqrt{x} \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.186, size = 130, normalized size = 1.1 \begin{align*}{\frac{2}{ \left ( 3\,c{x}^{2}+3\,b \right ) c}\sqrt{c{x}^{4}+b{x}^{2}} \left ( b\sqrt{-bc}\sqrt{{ \left ( cx+\sqrt{-bc} \right ){\frac{1}{\sqrt{-bc}}}}}\sqrt{2}\sqrt{{ \left ( -cx+\sqrt{-bc} \right ){\frac{1}{\sqrt{-bc}}}}}\sqrt{-{cx{\frac{1}{\sqrt{-bc}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( cx+\sqrt{-bc} \right ){\frac{1}{\sqrt{-bc}}}}},{\frac{\sqrt{2}}{2}} \right ) +{c}^{2}{x}^{3}+bcx \right ){x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^4 + b*x^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}}}{x^{\frac{3}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + b*x^2)/x^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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