3.137 \(\int \frac {\sqrt {b \sqrt [3]{x}+a x}}{x^3} \, dx\)

Optimal. Leaf size=188 \[ \frac {10 a^{11/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 b^{9/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {20 a^2 \sqrt {a x+b \sqrt [3]{x}}}{77 b^2 x^{2/3}}-\frac {12 a \sqrt {a x+b \sqrt [3]{x}}}{77 b x^{4/3}}-\frac {6 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2} \]

[Out]

-6/11*(b*x^(1/3)+a*x)^(1/2)/x^2-12/77*a*(b*x^(1/3)+a*x)^(1/2)/b/x^(4/3)+20/77*a^2*(b*x^(1/3)+a*x)^(1/2)/b^2/x^
(2/3)+10/77*a^(11/4)*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)/b^(9/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2018, 2020, 2025, 2011, 329, 220} \[ \frac {20 a^2 \sqrt {a x+b \sqrt [3]{x}}}{77 b^2 x^{2/3}}+\frac {10 a^{11/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 b^{9/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {12 a \sqrt {a x+b \sqrt [3]{x}}}{77 b x^{4/3}}-\frac {6 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^(1/3) + a*x]/x^3,x]

[Out]

(-6*Sqrt[b*x^(1/3) + a*x])/(11*x^2) - (12*a*Sqrt[b*x^(1/3) + a*x])/(77*b*x^(4/3)) + (20*a^2*Sqrt[b*x^(1/3) + a
*x])/(77*b^2*x^(2/3)) + (10*a^(11/4)*(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*b^(9/4)*Sqrt[b*x^(1/3) + a*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]

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 2011

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 2018

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {b \sqrt [3]{x}+a x}}{x^3} \, dx &=3 \operatorname {Subst}\left (\int \frac {\sqrt {b x+a x^3}}{x^7} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{11 x^2}+\frac {1}{11} (6 a) \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{11 x^2}-\frac {12 a \sqrt {b \sqrt [3]{x}+a x}}{77 b x^{4/3}}-\frac {\left (30 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 b}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{11 x^2}-\frac {12 a \sqrt {b \sqrt [3]{x}+a x}}{77 b x^{4/3}}+\frac {20 a^2 \sqrt {b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}+\frac {\left (10 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 b^2}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{11 x^2}-\frac {12 a \sqrt {b \sqrt [3]{x}+a x}}{77 b x^{4/3}}+\frac {20 a^2 \sqrt {b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}+\frac {\left (10 a^3 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{77 b^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{11 x^2}-\frac {12 a \sqrt {b \sqrt [3]{x}+a x}}{77 b x^{4/3}}+\frac {20 a^2 \sqrt {b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}+\frac {\left (20 a^3 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{77 b^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{11 x^2}-\frac {12 a \sqrt {b \sqrt [3]{x}+a x}}{77 b x^{4/3}}+\frac {20 a^2 \sqrt {b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}+\frac {10 a^{11/4} \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 b^{9/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 59, normalized size = 0.31 \[ -\frac {6 \sqrt {a x+b \sqrt [3]{x}} \, _2F_1\left (-\frac {11}{4},-\frac {1}{2};-\frac {7}{4};-\frac {a x^{2/3}}{b}\right )}{11 x^2 \sqrt {\frac {a x^{2/3}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^(1/3) + a*x]/x^3,x]

[Out]

(-6*Sqrt[b*x^(1/3) + a*x]*Hypergeometric2F1[-11/4, -1/2, -7/4, -((a*x^(2/3))/b)])/(11*Sqrt[1 + (a*x^(2/3))/b]*
x^2)

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fricas [F]  time = 1.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a x + b x^{\frac {1}{3}}}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*x + b*x^(1/3))/x^3, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(x)]sym2poly
/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueEvaluation time: 3.256*((-4
20*b^3/4620/b^3/x^(1/3)/x^(1/3)-120*b^2*a/4620/b^3)/x^(1/3)/x^(1/3)+200*b*a^2/4620/b^3)*sqrt(a/x^(1/3)+b/x)+in
tegrate(600*b*a^3/4620/b^3/3/((x^(1/6))^5*sqrt(a*(x^(1/3))^2+b)*sign(x)),x)

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maple [A]  time = 0.08, size = 179, normalized size = 0.95 \[ \frac {10 \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, a^{2} \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 \sqrt {a x +b \,x^{\frac {1}{3}}}\, b^{2}}+\frac {20 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a^{2}}{77 b^{2} x^{\frac {2}{3}}}-\frac {12 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a}{77 b \,x^{\frac {4}{3}}}-\frac {6 \sqrt {a x +b \,x^{\frac {1}{3}}}}{11 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/11*(a*x+b*x^(1/3))^(1/2)/x^2-12/77*a*(a*x+b*x^(1/3))^(1/2)/b/x^(4/3)+20/77*a^2*(a*x+b*x^(1/3))^(1/2)/b^2/x^
(2/3)+10/77*a^2/b^2*(-a*b)^(1/2)*((x^(1/3)+(-a*b)^(1/2)/a)/(-a*b)^(1/2)*a)^(1/2)*(-2*(x^(1/3)-(-a*b)^(1/2)/a)/
(-a*b)^(1/2)*a)^(1/2)*(-1/(-a*b)^(1/2)*a*x^(1/3))^(1/2)/(a*x+b*x^(1/3))^(1/2)*EllipticF(((x^(1/3)+(-a*b)^(1/2)
/a)/(-a*b)^(1/2)*a)^(1/2),1/2*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x + b x^{\frac {1}{3}}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a\,x+b\,x^{1/3}}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(1/3))^(1/2)/x^3,x)

[Out]

int((a*x + b*x^(1/3))^(1/2)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x + b \sqrt [3]{x}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x + b*x**(1/3))/x**3, x)

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