\(\int \frac {\sqrt [3]{b x+a x^3} (b+a x^4)}{x^4} \, dx\) [2308]

Optimal result

Integrand size = 24, antiderivative size = 177 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\frac {\sqrt [3]{b x+a x^3} \left (-3 b-3 a x^2+4 a x^4\right )}{8 x^3}-\frac {\sqrt [3]{a} b \arctan \left (\frac {\sqrt {3} \sqrt [3]{a} x}{\sqrt [3]{a} x+2 \sqrt [3]{b x+a x^3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \sqrt [3]{a} b \log \left (-\sqrt [3]{a} x+\sqrt [3]{b x+a x^3}\right )+\frac {1}{12} \sqrt [3]{a} b \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{b x+a x^3}+\left (b x+a x^3\right )^{2/3}\right ) \]

[Out]

1/8*(a*x^3+b*x)^(1/3)*(4*a*x^4-3*a*x^2-3*b)/x^3-1/6*a^(1/3)*b*arctan(3^(1/2)*a^(1/3)*x/(a^(1/3)*x+2*(a*x^3+b*x
)^(1/3)))*3^(1/2)-1/6*a^(1/3)*b*ln(-a^(1/3)*x+(a*x^3+b*x)^(1/3))+1/12*a^(1/3)*b*ln(a^(2/3)*x^2+a^(1/3)*x*(a*x^
3+b*x)^(1/3)+(a*x^3+b*x)^(2/3))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2077, 2029, 2057, 335, 281, 337, 2039} \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=-\frac {\sqrt [3]{a} b x^{2/3} \left (a x^2+b\right )^{2/3} \arctan \left (\frac {\frac {2 \sqrt [3]{a} x^{2/3}}{\sqrt [3]{a x^2+b}}+1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (a x^3+b x\right )^{2/3}}+\frac {1}{2} a x \sqrt [3]{a x^3+b x}-\frac {3 \left (a x^3+b x\right )^{4/3}}{8 x^4}-\frac {\sqrt [3]{a} b x^{2/3} \left (a x^2+b\right )^{2/3} \log \left (\sqrt [3]{a} x^{2/3}-\sqrt [3]{a x^2+b}\right )}{4 \left (a x^3+b x\right )^{2/3}} \]

[In]

Int[((b*x + a*x^3)^(1/3)*(b + a*x^4))/x^4,x]

[Out]

(a*x*(b*x + a*x^3)^(1/3))/2 - (3*(b*x + a*x^3)^(4/3))/(8*x^4) - (a^(1/3)*b*x^(2/3)*(b + a*x^2)^(2/3)*ArcTan[(1
+ (2*a^(1/3)*x^(2/3))/(b + a*x^2)^(1/3))/Sqrt[3]])/(2*Sqrt[3]*(b*x + a*x^3)^(2/3)) - (a^(1/3)*b*x^(2/3)*(b +
a*x^2)^(2/3)*Log[a^(1/3)*x^(2/3) - (b + a*x^2)^(1/3)])/(4*(b*x + a*x^3)^(2/3))

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rule 2029

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 2039

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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 2077

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (a \sqrt [3]{b x+a x^3}+\frac {b \sqrt [3]{b x+a x^3}}{x^4}\right ) \, dx \\ & = a \int \sqrt [3]{b x+a x^3} \, dx+b \int \frac {\sqrt [3]{b x+a x^3}}{x^4} \, dx \\ & = \frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {1}{3} (a b) \int \frac {x}{\left (b x+a x^3\right )^{2/3}} \, dx \\ & = \frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (b+a x^2\right )^{2/3}} \, dx}{3 \left (b x+a x^3\right )^{2/3}} \\ & = \frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {x^3}{\left (b+a x^6\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{\left (b x+a x^3\right )^{2/3}} \\ & = \frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}+\frac {\left (a b x^{2/3} \left (b+a x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {x}{\left (b+a x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{2 \left (b x+a x^3\right )^{2/3}} \\ & = \frac {1}{2} a x \sqrt [3]{b x+a x^3}-\frac {3 \left (b x+a x^3\right )^{4/3}}{8 x^4}-\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a} x^{2/3}}{\sqrt [3]{b+a x^2}}}{\sqrt {3}}\right )}{2 \sqrt {3} \left (b x+a x^3\right )^{2/3}}-\frac {\sqrt [3]{a} b x^{2/3} \left (b+a x^2\right )^{2/3} \log \left (\sqrt [3]{a} x^{2/3}-\sqrt [3]{b+a x^2}\right )}{4 \left (b x+a x^3\right )^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.68 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=-\frac {\sqrt [3]{x \left (b+a x^2\right )} \left (9 b \sqrt [3]{b+a x^2}+9 a x^2 \sqrt [3]{b+a x^2}-12 a x^4 \sqrt [3]{b+a x^2}+4 \sqrt {3} \sqrt [3]{a} b x^{8/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a} x^{2/3}}{\sqrt [3]{a} x^{2/3}+2 \sqrt [3]{b+a x^2}}\right )+4 \sqrt [3]{a} b x^{8/3} \log \left (-\sqrt [3]{a} x^{2/3}+\sqrt [3]{b+a x^2}\right )-2 \sqrt [3]{a} b x^{8/3} \log \left (a^{2/3} x^{4/3}+\sqrt [3]{a} x^{2/3} \sqrt [3]{b+a x^2}+\left (b+a x^2\right )^{2/3}\right )\right )}{24 x^3 \sqrt [3]{b+a x^2}} \]

[In]

Integrate[((b*x + a*x^3)^(1/3)*(b + a*x^4))/x^4,x]

[Out]

-1/24*((x*(b + a*x^2))^(1/3)*(9*b*(b + a*x^2)^(1/3) + 9*a*x^2*(b + a*x^2)^(1/3) - 12*a*x^4*(b + a*x^2)^(1/3) +
4*Sqrt[3]*a^(1/3)*b*x^(8/3)*ArcTan[(Sqrt[3]*a^(1/3)*x^(2/3))/(a^(1/3)*x^(2/3) + 2*(b + a*x^2)^(1/3))] + 4*a^(
1/3)*b*x^(8/3)*Log[-(a^(1/3)*x^(2/3)) + (b + a*x^2)^(1/3)] - 2*a^(1/3)*b*x^(8/3)*Log[a^(2/3)*x^(4/3) + a^(1/3)
*x^(2/3)*(b + a*x^2)^(1/3) + (b + a*x^2)^(2/3)]))/(x^3*(b + a*x^2)^(1/3))

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {\left (12 a \,x^{4}-9 a \,x^{2}-9 b \right ) {\left (\left (a \,x^{2}+b \right ) x \right )}^{\frac {1}{3}}+2 b \,x^{3} a^{\frac {1}{3}} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (a^{\frac {1}{3}} x +2 {\left (\left (a \,x^{2}+b \right ) x \right )}^{\frac {1}{3}}\right )}{3 a^{\frac {1}{3}} x}\right ) \sqrt {3}-2 \ln \left (\frac {-a^{\frac {1}{3}} x +{\left (\left (a \,x^{2}+b \right ) x \right )}^{\frac {1}{3}}}{x}\right )+\ln \left (\frac {a^{\frac {2}{3}} x^{2}+a^{\frac {1}{3}} {\left (\left (a \,x^{2}+b \right ) x \right )}^{\frac {1}{3}} x +{\left (\left (a \,x^{2}+b \right ) x \right )}^{\frac {2}{3}}}{x^{2}}\right )\right )}{24 x^{3}}\)	\(146\)

[In]

int((a*x^3+b*x)^(1/3)*(a*x^4+b)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/24*((12*a*x^4-9*a*x^2-9*b)*((a*x^2+b)*x)^(1/3)+2*b*x^3*a^(1/3)*(2*arctan(1/3*3^(1/2)*(a^(1/3)*x+2*((a*x^2+b)
*x)^(1/3))/a^(1/3)/x)*3^(1/2)-2*ln((-a^(1/3)*x+((a*x^2+b)*x)^(1/3))/x)+ln((a^(2/3)*x^2+a^(1/3)*((a*x^2+b)*x)^(
1/3)*x+((a*x^2+b)*x)^(2/3))/x^2)))/x^3

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\int \frac {\sqrt [3]{x \left (a x^{2} + b\right )} \left (a x^{4} + b\right )}{x^{4}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\int { \frac {{\left (a x^{4} + b\right )} {\left (a x^{3} + b x\right )}^{\frac {1}{3}}}{x^{4}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 8.23 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\frac {12 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} a b x^{2} + 4 \, \sqrt {3} a^{\frac {1}{3}} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) + 2 \, a^{\frac {1}{3}} b^{2} \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {2}{3}} + {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, a^{\frac {1}{3}} b^{2} \log \left ({\left | {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) - 9 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {4}{3}} b}{24 \, b} \]

[In]

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

[Out]

1/24*(12*(a + b/x^2)^(1/3)*a*b*x^2 + 4*sqrt(3)*a^(1/3)*b^2*arctan(1/3*sqrt(3)*(2*(a + b/x^2)^(1/3) + a^(1/3))/
a^(1/3)) + 2*a^(1/3)*b^2*log((a + b/x^2)^(2/3) + (a + b/x^2)^(1/3)*a^(1/3) + a^(2/3)) - 4*a^(1/3)*b^2*log(abs(
(a + b/x^2)^(1/3) - a^(1/3))) - 9*(a + b/x^2)^(4/3)*b)/b

Mupad [B] (verification not implemented)

Time = 7.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.37 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\frac {3\,a\,x\,{\left (a\,x^3+b\,x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {2}{3};\ \frac {5}{3};\ -\frac {a\,x^2}{b}\right )}{4\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/3}}-\frac {3\,{\left (a\,x^3+b\,x\right )}^{1/3}\,\left (a\,x^2+b\right )}{8\,x^3} \]

[In]

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

[Out]

(3*a*x*(b*x + a*x^3)^(1/3)*hypergeom([-1/3, 2/3], 5/3, -(a*x^2)/b))/(4*((a*x^2)/b + 1)^(1/3)) - (3*(b*x + a*x^
3)^(1/3)*(b + a*x^2))/(8*x^3)