3.45 \(\int \frac{\sqrt{a x+b x^3}}{x^4} \, dx\)

Optimal. Leaf size=283 \[ \frac{2 b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a x+b x^3}}-\frac{4 b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a x+b x^3}}+\frac{4 b^{3/2} x \left (a+b x^2\right )}{5 a \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}-\frac{4 b \sqrt{a x+b x^3}}{5 a x}-\frac{2 \sqrt{a x+b x^3}}{5 x^3} \]

[Out]

(4*b^(3/2)*x*(a + b*x^2))/(5*a*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) - (2*Sqrt[a*x + b*x^3])/(5*x^3) - (4*b
*Sqrt[a*x + b*x^3])/(5*a*x) - (4*b^(5/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^
2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*a^(3/4)*Sqrt[a*x + b*x^3]) + (2*b^(5/4)*Sqrt[x]*(Sq
rt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/
2])/(5*a^(3/4)*Sqrt[a*x + b*x^3])

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Rubi [A]  time = 0.250931, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {2020, 2025, 2032, 329, 305, 220, 1196} \[ \frac{2 b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a x+b x^3}}-\frac{4 b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a x+b x^3}}+\frac{4 b^{3/2} x \left (a+b x^2\right )}{5 a \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}-\frac{4 b \sqrt{a x+b x^3}}{5 a x}-\frac{2 \sqrt{a x+b x^3}}{5 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x + b*x^3]/x^4,x]

[Out]

(4*b^(3/2)*x*(a + b*x^2))/(5*a*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) - (2*Sqrt[a*x + b*x^3])/(5*x^3) - (4*b
*Sqrt[a*x + b*x^3])/(5*a*x) - (4*b^(5/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^
2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*a^(3/4)*Sqrt[a*x + b*x^3]) + (2*b^(5/4)*Sqrt[x]*(Sq
rt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/
2])/(5*a^(3/4)*Sqrt[a*x + b*x^3])

Rule 2020

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

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

Mathematica [C]  time = 0.0125866, size = 53, normalized size = 0.19 \[ -\frac{2 \sqrt{x \left (a+b x^2\right )} \, _2F_1\left (-\frac{5}{4},-\frac{1}{2};-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 x^3 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x + b*x^3]/x^4,x]

[Out]

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

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Maple [A]  time = 0.006, size = 201, normalized size = 0.7 \begin{align*} -{\frac{2}{5\,{x}^{3}}\sqrt{b{x}^{3}+ax}}-{\frac{ \left ( 4\,b{x}^{2}+4\,a \right ) b}{5\,a}{\frac{1}{\sqrt{x \left ( b{x}^{2}+a \right ) }}}}+{\frac{2\,b}{5\,a}\sqrt{-ab}\sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-2\,{\frac{b}{\sqrt{-ab}} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}} \left ( -2\,{\frac{\sqrt{-ab}}{b}{\it EllipticE} \left ( \sqrt{{\frac{b}{\sqrt{-ab}} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }},1/2\,\sqrt{2} \right ) }+{\frac{1}{b}\sqrt{-ab}{\it EllipticF} \left ( \sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) } \right ){\frac{1}{\sqrt{b{x}^{3}+ax}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(b*x^3+a*x)^(1/2)/x^3-4/5*(b*x^2+a)*b/a/(x*(b*x^2+a))^(1/2)+2/5/a*b*(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))*b/
(-a*b)^(1/2))^(1/2)*(-2*(x-1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)/(b*x^3+a*x)^(1/2)
*(-2/b*(-a*b)^(1/2)*EllipticE(((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))+1/b*(-a*b)^(1/2)*Ellipt
icF(((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x)/x^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{3} + a x}}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^3 + a*x)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(a + b*x**2))/x**4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x)/x^4, x)