3.2.33 \(\int x \sqrt {b \sqrt [3]{x}+a x} \, dx\) [133]

Optimal. Leaf size=213 \[ \frac {12 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^3}-\frac {36 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^2}+\frac {4 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a}+\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}-\frac {6 b^{15/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 a^{13/4} \sqrt {b \sqrt [3]{x}+a x}} \]

[Out]

12/77*b^3*(b*x^(1/3)+a*x)^(1/2)/a^3-36/385*b^2*x^(2/3)*(b*x^(1/3)+a*x)^(1/2)/a^2+4/55*b*x^(4/3)*(b*x^(1/3)+a*x
)^(1/2)/a+2/5*x^2*(b*x^(1/3)+a*x)^(1/2)-6/77*b^(15/4)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/2)
/cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))*EllipticF(sin(2*arctan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*2^(1/2))*(x^(1/3)
*a^(1/2)+b^(1/2))*((b+a*x^(2/3))/(x^(1/3)*a^(1/2)+b^(1/2))^2)^(1/2)/a^(13/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2043, 2046, 2049, 2036, 335, 226} \begin {gather*} -\frac {6 b^{15/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 a^{13/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {12 b^3 \sqrt {a x+b \sqrt [3]{x}}}{77 a^3}-\frac {36 b^2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{385 a^2}+\frac {4 b x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{55 a}+\frac {2}{5} x^2 \sqrt {a x+b \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[b*x^(1/3) + a*x],x]

[Out]

(12*b^3*Sqrt[b*x^(1/3) + a*x])/(77*a^3) - (36*b^2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(385*a^2) + (4*b*x^(4/3)*Sqrt
[b*x^(1/3) + a*x])/(55*a) + (2*x^2*Sqrt[b*x^(1/3) + a*x])/5 - (6*b^(15/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[(b
+ a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(77*a
^(13/4)*Sqrt[b*x^(1/3) + a*x])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

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 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 2046

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 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rubi steps

\begin {align*} \int x \sqrt {b \sqrt [3]{x}+a x} \, dx &=3 \text {Subst}\left (\int x^5 \sqrt {b x+a x^3} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}+\frac {1}{5} (2 b) \text {Subst}\left (\int \frac {x^6}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {4 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a}+\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (18 b^2\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{55 a}\\ &=-\frac {36 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^2}+\frac {4 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a}+\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}+\frac {\left (18 b^3\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a^2}\\ &=\frac {12 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^3}-\frac {36 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^2}+\frac {4 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a}+\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (6 b^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a^3}\\ &=\frac {12 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^3}-\frac {36 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^2}+\frac {4 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a}+\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (6 b^4 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{77 a^3 \sqrt {b \sqrt [3]{x}+a x}}\\ &=\frac {12 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^3}-\frac {36 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^2}+\frac {4 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a}+\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (12 b^4 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{77 a^3 \sqrt {b \sqrt [3]{x}+a x}}\\ &=\frac {12 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^3}-\frac {36 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^2}+\frac {4 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a}+\frac {2}{5} x^2 \sqrt {b \sqrt [3]{x}+a x}-\frac {6 b^{15/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 a^{13/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.08, size = 118, normalized size = 0.55 \begin {gather*} \frac {2 \sqrt {b \sqrt [3]{x}+a x} \left (\sqrt {1+\frac {a x^{2/3}}{b}} \left (45 b^3-18 a b^2 x^{2/3}+14 a^2 b x^{4/3}+77 a^3 x^2\right )-45 b^3 \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {a x^{2/3}}{b}\right )\right )}{385 a^3 \sqrt {1+\frac {a x^{2/3}}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[b*x^(1/3) + a*x],x]

[Out]

(2*Sqrt[b*x^(1/3) + a*x]*(Sqrt[1 + (a*x^(2/3))/b]*(45*b^3 - 18*a*b^2*x^(2/3) + 14*a^2*b*x^(4/3) + 77*a^3*x^2)
- 45*b^3*Hypergeometric2F1[-1/2, 1/4, 5/4, -((a*x^(2/3))/b)]))/(385*a^3*Sqrt[1 + (a*x^(2/3))/b])

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Maple [A]
time = 0.34, size = 198, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {2 x^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{5}+\frac {4 b \,x^{\frac {4}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{55 a}-\frac {36 b^{2} x^{\frac {2}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{385 a^{2}}+\frac {12 b^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 a^{3}}-\frac {6 b^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a^{4} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(198\)
default	\(\frac {2 x^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{5}+\frac {4 b \,x^{\frac {4}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{55 a}-\frac {36 b^{2} x^{\frac {2}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{385 a^{2}}+\frac {12 b^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 a^{3}}-\frac {6 b^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a^{4} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(198\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^(1/3)+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/5*x^2*(b*x^(1/3)+a*x)^(1/2)+4/55*b*x^(4/3)*(b*x^(1/3)+a*x)^(1/2)/a-36/385*b^2*x^(2/3)*(b*x^(1/3)+a*x)^(1/2)/
a^2+12/77*b^3*(b*x^(1/3)+a*x)^(1/2)/a^3-6/77*b^4/a^4*(-a*b)^(1/2)*((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^
(1/2)*(-2*(x^(1/3)-1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)/(b*x^(1/3)+a*x)^(1/
2)*EllipticF(((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x + b*x^(1/3))*x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*x + b*x^(1/3))*x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {a x + b \sqrt [3]{x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(a*x + b*x**(1/3)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x + b*x^(1/3))*x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x\,\sqrt {a\,x+b\,x^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a*x + b*x^(1/3))^(1/2),x)

[Out]

int(x*(a*x + b*x^(1/3))^(1/2), x)

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