3.49 \(\int \frac{(a x+b x^3)^{3/2}}{x} \, dx\)

Optimal. Leaf size=275 \[ \frac{4 a^{9/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a x+b x^3}}-\frac{8 a^{9/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a x+b x^3}}+\frac{8 a^2 x \left (a+b x^2\right )}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}+\frac{4}{15} a x \sqrt{a x+b x^3}+\frac{2}{9} \left (a x+b x^3\right )^{3/2} \]

[Out]

(8*a^2*x*(a + b*x^2))/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) + (4*a*x*Sqrt[a*x + b*x^3])/15 + (2
*(a*x + b*x^3)^(3/2))/9 - (8*a^(9/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*E
llipticE[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a*x + b*x^3]) + (4*a^(9/4)*Sqrt[x]*(Sqrt[
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])
/(15*b^(3/4)*Sqrt[a*x + b*x^3])

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Rubi [A]  time = 0.224599, antiderivative size = 275, 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 = {2021, 2004, 2032, 329, 305, 220, 1196} \[ \frac{4 a^{9/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a x+b x^3}}-\frac{8 a^{9/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a x+b x^3}}+\frac{8 a^2 x \left (a+b x^2\right )}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}+\frac{4}{15} a x \sqrt{a x+b x^3}+\frac{2}{9} \left (a x+b x^3\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^3)^(3/2)/x,x]

[Out]

(8*a^2*x*(a + b*x^2))/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) + (4*a*x*Sqrt[a*x + b*x^3])/15 + (2
*(a*x + b*x^3)^(3/2))/9 - (8*a^(9/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*E
llipticE[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a*x + b*x^3]) + (4*a^(9/4)*Sqrt[x]*(Sqrt[
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])
/(15*b^(3/4)*Sqrt[a*x + b*x^3])

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 2004

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(n*p + 1), x] + Dist[(
a*(n - j)*p)/(n*p + 1), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[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 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{\left (a x+b x^3\right )^{3/2}}{x} \, dx &=\frac{2}{9} \left (a x+b x^3\right )^{3/2}+\frac{1}{3} (2 a) \int \sqrt{a x+b x^3} \, dx\\ &=\frac{4}{15} a x \sqrt{a x+b x^3}+\frac{2}{9} \left (a x+b x^3\right )^{3/2}+\frac{1}{15} \left (4 a^2\right ) \int \frac{x}{\sqrt{a x+b x^3}} \, dx\\ &=\frac{4}{15} a x \sqrt{a x+b x^3}+\frac{2}{9} \left (a x+b x^3\right )^{3/2}+\frac{\left (4 a^2 \sqrt{x} \sqrt{a+b x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x^2}} \, dx}{15 \sqrt{a x+b x^3}}\\ &=\frac{4}{15} a x \sqrt{a x+b x^3}+\frac{2}{9} \left (a x+b x^3\right )^{3/2}+\frac{\left (8 a^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 )}{15 \sqrt{a x+b x^3}}\\ &=\frac{4}{15} a x \sqrt{a x+b x^3}+\frac{2}{9} \left (a x+b x^3\right )^{3/2}+\frac{\left (8 a^{5/2} \sqrt{x} \sqrt{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{b} \sqrt{a x+b x^3}}-\frac{\left (8 a^{5/2} \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 )}{15 \sqrt{b} \sqrt{a x+b x^3}}\\ &=\frac{8 a^2 x \left (a+b x^2\right )}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}+\frac{4}{15} a x \sqrt{a x+b x^3}+\frac{2}{9} \left (a x+b x^3\right )^{3/2}-\frac{8 a^{9/4} \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 )}{15 b^{3/4} \sqrt{a x+b x^3}}+\frac{4 a^{9/4} \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 )}{15 b^{3/4} \sqrt{a x+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0122047, size = 52, normalized size = 0.19 \[ \frac{2 a x \sqrt{x \left (a+b x^2\right )} \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )}{3 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^3)^(3/2)/x,x]

[Out]

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

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Maple [A]  time = 0.011, size = 195, normalized size = 0.7 \begin{align*}{\frac{2\,b{x}^{3}}{9}\sqrt{b{x}^{3}+ax}}+{\frac{22\,ax}{45}\sqrt{b{x}^{3}+ax}}+{\frac{4\,{a}^{2}}{15\,b}\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)^(3/2)/x,x)

[Out]

2/9*b*x^3*(b*x^3+a*x)^(1/2)+22/45*a*x*(b*x^3+a*x)^(1/2)+4/15*a^2/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)*EllipticF(((
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{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b x^{3} + a x}{\left (b x^{2} + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^3 + a*x)*(b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(a + b*x**2))**(3/2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x, x)