\(\int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx\) [3088]

Optimal result

Integrand size = 27, antiderivative size = 526 \[ \int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx=\frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{a}+\frac {\left (3 a^2 b-b^2\right ) \arctan \left (\frac {\sqrt {3} a x}{a x+2 \sqrt [3]{b^2 x^2+a^3 x^3}}\right )}{\sqrt {3} a^3}+\frac {\sqrt {-3-3 i \sqrt {3}} \sqrt [3]{a^2-b} b \arctan \left (\frac {\sqrt {3} \sqrt [3]{a} \sqrt [3]{a^2-b} x}{\sqrt [3]{a} \sqrt [3]{a^2-b} x-2 \sqrt [3]{-1} \sqrt [3]{b^2 x^2+a^3 x^3}}\right )}{\sqrt {2} a^{5/3}}+\frac {\left (3 a^2 b-b^2\right ) \log \left (-a x+\sqrt [3]{b^2 x^2+a^3 x^3}\right )}{3 a^3}+\frac {\left (\sqrt [3]{a^2-b} b-i \sqrt {3} \sqrt [3]{a^2-b} b\right ) \log \left (\sqrt [3]{a} \sqrt [3]{a^2-b} x+\sqrt [3]{-1} \sqrt [3]{b^2 x^2+a^3 x^3}\right )}{2 a^{5/3}}+\frac {\left (-3 a^2 b+b^2\right ) \log \left (a^2 x^2+a x \sqrt [3]{b^2 x^2+a^3 x^3}+\left (b^2 x^2+a^3 x^3\right )^{2/3}\right )}{6 a^3}+\frac {i \left (i \sqrt [3]{a^2-b} b+\sqrt {3} \sqrt [3]{a^2-b} b\right ) \log \left (a^{2/3} \left (a^2-b\right )^{2/3} x^2-\sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{a^2-b} x \sqrt [3]{b^2 x^2+a^3 x^3}+(-1)^{2/3} \left (b^2 x^2+a^3 x^3\right )^{2/3}\right )}{4 a^{5/3}} \]

[Out]

(a^3*x^3+b^2*x^2)^(1/3)/a+1/3*(3*a^2*b-b^2)*arctan(3^(1/2)*a*x/(a*x+2*(a^3*x^3+b^2*x^2)^(1/3)))*3^(1/2)/a^3+1/
2*(-3-3*I*3^(1/2))^(1/2)*(a^2-b)^(1/3)*b*arctan(3^(1/2)*a^(1/3)*(a^2-b)^(1/3)*x/(a^(1/3)*(a^2-b)^(1/3)*x-2*(-1
)^(1/3)*(a^3*x^3+b^2*x^2)^(1/3)))*2^(1/2)/a^(5/3)+1/3*(3*a^2*b-b^2)*ln(-a*x+(a^3*x^3+b^2*x^2)^(1/3))/a^3+1/2*(
(a^2-b)^(1/3)*b-I*3^(1/2)*(a^2-b)^(1/3)*b)*ln(a^(1/3)*(a^2-b)^(1/3)*x+(-1)^(1/3)*(a^3*x^3+b^2*x^2)^(1/3))/a^(5
/3)+1/6*(-3*a^2*b+b^2)*ln(a^2*x^2+a*x*(a^3*x^3+b^2*x^2)^(1/3)+(a^3*x^3+b^2*x^2)^(2/3))/a^3+1/4*I*(I*(a^2-b)^(1
/3)*b+3^(1/2)*(a^2-b)^(1/3)*b)*ln(a^(2/3)*(a^2-b)^(2/3)*x^2-(-1)^(1/3)*a^(1/3)*(a^2-b)^(1/3)*x*(a^3*x^3+b^2*x^
2)^(1/3)+(-1)^(2/3)*(a^3*x^3+b^2*x^2)^(2/3))/a^(5/3)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2081, 103, 163, 61, 93} \[ \int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx=\frac {\sqrt [3]{a^3 x^3+b^2 x^2}}{a}+\frac {b \left (3 a^2-b\right ) \sqrt [3]{a^3 x^3+b^2 x^2} \arctan \left (\frac {2 a \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a^3 x+b^2}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} a^3 x^{2/3} \sqrt [3]{a^3 x+b^2}}+\frac {b \left (3 a^2-b\right ) \sqrt [3]{a^3 x^3+b^2 x^2} \log \left (a^3 x+b^2\right )}{6 a^3 x^{2/3} \sqrt [3]{a^3 x+b^2}}+\frac {b \left (3 a^2-b\right ) \sqrt [3]{a^3 x^3+b^2 x^2} \log \left (\frac {a \sqrt [3]{x}}{\sqrt [3]{a^3 x+b^2}}-1\right )}{2 a^3 x^{2/3} \sqrt [3]{a^3 x+b^2}}-\frac {\sqrt {3} b \sqrt [3]{a^2-b} \sqrt [3]{a^3 x^3+b^2 x^2} \arctan \left (\frac {2 \sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{a^2-b}}{\sqrt {3} \sqrt [3]{a^3 x+b^2}}+\frac {1}{\sqrt {3}}\right )}{a^{5/3} x^{2/3} \sqrt [3]{a^3 x+b^2}}+\frac {b \sqrt [3]{a^2-b} \sqrt [3]{a^3 x^3+b^2 x^2} \log (a x+b)}{2 a^{5/3} x^{2/3} \sqrt [3]{a^3 x+b^2}}-\frac {3 b \sqrt [3]{a^2-b} \sqrt [3]{a^3 x^3+b^2 x^2} \log \left (\sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{a^2-b}-\sqrt [3]{a^3 x+b^2}\right )}{2 a^{5/3} x^{2/3} \sqrt [3]{a^3 x+b^2}} \]

[In]

Int[(b^2*x^2 + a^3*x^3)^(1/3)/(b + a*x),x]

[Out]

(b^2*x^2 + a^3*x^3)^(1/3)/a + ((3*a^2 - b)*b*(b^2*x^2 + a^3*x^3)^(1/3)*ArcTan[1/Sqrt[3] + (2*a*x^(1/3))/(Sqrt[
3]*(b^2 + a^3*x)^(1/3))])/(Sqrt[3]*a^3*x^(2/3)*(b^2 + a^3*x)^(1/3)) - (Sqrt[3]*(a^2 - b)^(1/3)*b*(b^2*x^2 + a^
3*x^3)^(1/3)*ArcTan[1/Sqrt[3] + (2*a^(1/3)*(a^2 - b)^(1/3)*x^(1/3))/(Sqrt[3]*(b^2 + a^3*x)^(1/3))])/(a^(5/3)*x
^(2/3)*(b^2 + a^3*x)^(1/3)) + ((a^2 - b)^(1/3)*b*(b^2*x^2 + a^3*x^3)^(1/3)*Log[b + a*x])/(2*a^(5/3)*x^(2/3)*(b
^2 + a^3*x)^(1/3)) + ((3*a^2 - b)*b*(b^2*x^2 + a^3*x^3)^(1/3)*Log[b^2 + a^3*x])/(6*a^3*x^(2/3)*(b^2 + a^3*x)^(
1/3)) + ((3*a^2 - b)*b*(b^2*x^2 + a^3*x^3)^(1/3)*Log[-1 + (a*x^(1/3))/(b^2 + a^3*x)^(1/3)])/(2*a^3*x^(2/3)*(b^
2 + a^3*x)^(1/3)) - (3*(a^2 - b)^(1/3)*b*(b^2*x^2 + a^3*x^3)^(1/3)*Log[a^(1/3)*(a^2 - b)^(1/3)*x^(1/3) - (b^2
+ a^3*x)^(1/3)])/(2*a^(5/3)*x^(2/3)*(b^2 + a^3*x)^(1/3))

Rule 61

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt
[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*
((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && Ne
Q[b*c - a*d, 0] && PosQ[d/b]

Rule 93

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b
*x)^m*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + p + 1))), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{b^2 x^2+a^3 x^3} \int \frac {x^{2/3} \sqrt [3]{b^2+a^3 x}}{b+a x} \, dx}{x^{2/3} \sqrt [3]{b^2+a^3 x}} \\ & = \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{a}-\frac {\sqrt [3]{b^2 x^2+a^3 x^3} \int \frac {\frac {2 b^3}{3}+\frac {1}{3} a \left (3 a^2-b\right ) b x}{\sqrt [3]{x} (b+a x) \left (b^2+a^3 x\right )^{2/3}} \, dx}{a x^{2/3} \sqrt [3]{b^2+a^3 x}} \\ & = \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{a}-\frac {\left (\left (3 a^2-b\right ) b \sqrt [3]{b^2 x^2+a^3 x^3}\right ) \int \frac {1}{\sqrt [3]{x} \left (b^2+a^3 x\right )^{2/3}} \, dx}{3 a x^{2/3} \sqrt [3]{b^2+a^3 x}}+\frac {\left (\left (a^2-b\right ) b^2 \sqrt [3]{b^2 x^2+a^3 x^3}\right ) \int \frac {1}{\sqrt [3]{x} (b+a x) \left (b^2+a^3 x\right )^{2/3}} \, dx}{a x^{2/3} \sqrt [3]{b^2+a^3 x}} \\ & = \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{a}+\frac {\left (3 a^2-b\right ) b \sqrt [3]{b^2 x^2+a^3 x^3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 a \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{b^2+a^3 x}}\right )}{\sqrt {3} a^3 x^{2/3} \sqrt [3]{b^2+a^3 x}}-\frac {\sqrt {3} \sqrt [3]{a^2-b} b \sqrt [3]{b^2 x^2+a^3 x^3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{b^2+a^3 x}}\right )}{a^{5/3} x^{2/3} \sqrt [3]{b^2+a^3 x}}+\frac {\sqrt [3]{a^2-b} b \sqrt [3]{b^2 x^2+a^3 x^3} \log (b+a x)}{2 a^{5/3} x^{2/3} \sqrt [3]{b^2+a^3 x}}+\frac {\left (3 a^2-b\right ) b \sqrt [3]{b^2 x^2+a^3 x^3} \log \left (b^2+a^3 x\right )}{6 a^3 x^{2/3} \sqrt [3]{b^2+a^3 x}}+\frac {\left (3 a^2-b\right ) b \sqrt [3]{b^2 x^2+a^3 x^3} \log \left (-1+\frac {a \sqrt [3]{x}}{\sqrt [3]{b^2+a^3 x}}\right )}{2 a^3 x^{2/3} \sqrt [3]{b^2+a^3 x}}-\frac {3 \sqrt [3]{a^2-b} b \sqrt [3]{b^2 x^2+a^3 x^3} \log \left (\sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}-\sqrt [3]{b^2+a^3 x}\right )}{2 a^{5/3} x^{2/3} \sqrt [3]{b^2+a^3 x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 495, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx=\frac {x^{4/3} \left (b^2+a^3 x\right )^{2/3} \left (12 a^2 x^{2/3} \sqrt [3]{b^2+a^3 x}+4 \sqrt {3} \left (3 a^2-b\right ) b \arctan \left (\frac {\sqrt {3} a \sqrt [3]{x}}{a \sqrt [3]{x}+2 \sqrt [3]{b^2+a^3 x}}\right )+6 \left (3+i \sqrt {3}\right ) a^{4/3} \sqrt [3]{a^2-b} b \text {arctanh}\left (\frac {i+\frac {\left (-i+\sqrt {3}\right ) \sqrt [3]{b^2+a^3 x}}{\sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}}}{\sqrt {3}}\right )+4 \left (3 a^2-b\right ) b \log \left (-a \sqrt [3]{x}+\sqrt [3]{b^2+a^3 x}\right )+6 \left (1-i \sqrt {3}\right ) a^{4/3} \sqrt [3]{a^2-b} b \log \left (2 \sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}+\left (1+i \sqrt {3}\right ) \sqrt [3]{b^2+a^3 x}\right )+3 i \left (i+\sqrt {3}\right ) a^{4/3} \sqrt [3]{a^2-b} b \log \left (\left (-\sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}+\sqrt [3]{b^2+a^3 x}\right ) \left (2 i \sqrt [3]{a} \sqrt [3]{a^2-b} \sqrt [3]{x}+\left (i+\sqrt {3}\right ) \sqrt [3]{b^2+a^3 x}\right )\right )-2 \left (3 a^2-b\right ) b \log \left (a^2 x^{2/3}+a \sqrt [3]{x} \sqrt [3]{b^2+a^3 x}+\left (b^2+a^3 x\right )^{2/3}\right )\right )}{12 a^3 \left (x^2 \left (b^2+a^3 x\right )\right )^{2/3}} \]

[In]

Integrate[(b^2*x^2 + a^3*x^3)^(1/3)/(b + a*x),x]

[Out]

(x^(4/3)*(b^2 + a^3*x)^(2/3)*(12*a^2*x^(2/3)*(b^2 + a^3*x)^(1/3) + 4*Sqrt[3]*(3*a^2 - b)*b*ArcTan[(Sqrt[3]*a*x
^(1/3))/(a*x^(1/3) + 2*(b^2 + a^3*x)^(1/3))] + 6*(3 + I*Sqrt[3])*a^(4/3)*(a^2 - b)^(1/3)*b*ArcTanh[(I + ((-I +
Sqrt[3])*(b^2 + a^3*x)^(1/3))/(a^(1/3)*(a^2 - b)^(1/3)*x^(1/3)))/Sqrt[3]] + 4*(3*a^2 - b)*b*Log[-(a*x^(1/3))
+ (b^2 + a^3*x)^(1/3)] + 6*(1 - I*Sqrt[3])*a^(4/3)*(a^2 - b)^(1/3)*b*Log[2*a^(1/3)*(a^2 - b)^(1/3)*x^(1/3) + (
1 + I*Sqrt[3])*(b^2 + a^3*x)^(1/3)] + (3*I)*(I + Sqrt[3])*a^(4/3)*(a^2 - b)^(1/3)*b*Log[(-(a^(1/3)*(a^2 - b)^(
1/3)*x^(1/3)) + (b^2 + a^3*x)^(1/3))*((2*I)*a^(1/3)*(a^2 - b)^(1/3)*x^(1/3) + (I + Sqrt[3])*(b^2 + a^3*x)^(1/3
))] - 2*(3*a^2 - b)*b*Log[a^2*x^(2/3) + a*x^(1/3)*(b^2 + a^3*x)^(1/3) + (b^2 + a^3*x)^(2/3)]))/(12*a^3*(x^2*(b
^2 + a^3*x))^(2/3))

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 429, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(-\frac {b^{2} x^{2} \left (2 \left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}} {\left (a \left (a^{2}-b \right )\right )}^{\frac {2}{3}} a^{2}+\frac {b \left (-\left (3 a^{2}-b \right ) \left (2 \sqrt {3}\, \arctan \left (\frac {\left (a x +2 \left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a x}\right )+\ln \left (\frac {a^{2} x^{2}+\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}} a x +\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-a x +\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}}}{x}\right )\right ) {\left (a \left (a^{2}-b \right )\right )}^{\frac {2}{3}}+3 a^{2} \left (a^{2}-b \right ) \left (2 \arctan \left (\frac {\sqrt {3}\, \left ({\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} x +2 \left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}}\right )}{3 {\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {{\left (a \left (a^{2}-b \right )\right )}^{\frac {2}{3}} x^{2}+{\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} \left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-{\left (a \left (a^{2}-b \right )\right )}^{\frac {1}{3}} x +\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}}}{x}\right )\right )\right )}{3}\right )}{2 {\left (a \left (a^{2}-b \right )\right )}^{\frac {2}{3}} \left (a x -\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}}\right ) a^{3} \left (a^{2} x^{2}+\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {1}{3}} a x +\left (x^{2} \left (a^{3} x +b^{2}\right )\right )^{\frac {2}{3}}\right )}\)	\(429\)

[In]

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

[Out]

-1/2*b^2/(a*(a^2-b))^(2/3)*x^2*(2*(x^2*(a^3*x+b^2))^(1/3)*(a*(a^2-b))^(2/3)*a^2+1/3*b*(-(3*a^2-b)*(2*3^(1/2)*a
rctan(1/3*(a*x+2*(x^2*(a^3*x+b^2))^(1/3))*3^(1/2)/a/x)+ln((a^2*x^2+(x^2*(a^3*x+b^2))^(1/3)*a*x+(x^2*(a^3*x+b^2
))^(2/3))/x^2)-2*ln((-a*x+(x^2*(a^3*x+b^2))^(1/3))/x))*(a*(a^2-b))^(2/3)+3*a^2*(a^2-b)*(2*arctan(1/3*3^(1/2)*(
(a*(a^2-b))^(1/3)*x+2*(x^2*(a^3*x+b^2))^(1/3))/(a*(a^2-b))^(1/3)/x)*3^(1/2)+ln(((a*(a^2-b))^(2/3)*x^2+(a*(a^2-
b))^(1/3)*(x^2*(a^3*x+b^2))^(1/3)*x+(x^2*(a^3*x+b^2))^(2/3))/x^2)-2*ln((-(a*(a^2-b))^(1/3)*x+(x^2*(a^3*x+b^2))
^(1/3))/x))))/(a*x-(x^2*(a^3*x+b^2))^(1/3))/a^3/(a^2*x^2+(x^2*(a^3*x+b^2))^(1/3)*a*x+(x^2*(a^3*x+b^2))^(2/3))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 437, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx=\frac {6 \, \sqrt {3} a^{2} b \left (-\frac {a^{2} - b}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (a^{2} - b\right )} x + 2 \, \sqrt {3} {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}} a \left (-\frac {a^{2} - b}{a^{2}}\right )^{\frac {2}{3}}}{3 \, {\left (a^{2} - b\right )} x}\right ) + 6 \, a^{2} b \left (-\frac {a^{2} - b}{a^{2}}\right )^{\frac {1}{3}} \log \left (\frac {a x \left (-\frac {a^{2} - b}{a^{2}}\right )^{\frac {1}{3}} + {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - 3 \, a^{2} b \left (-\frac {a^{2} - b}{a^{2}}\right )^{\frac {1}{3}} \log \left (\frac {a^{2} x^{2} \left (-\frac {a^{2} - b}{a^{2}}\right )^{\frac {2}{3}} - {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}} a x \left (-\frac {a^{2} - b}{a^{2}}\right )^{\frac {1}{3}} + {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 2 \, \sqrt {3} {\left (3 \, a^{2} b - b^{2}\right )} \arctan \left (\frac {\sqrt {3} a x + 2 \, \sqrt {3} {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}}}{3 \, a x}\right ) + 6 \, {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}} a^{2} + 2 \, {\left (3 \, a^{2} b - b^{2}\right )} \log \left (-\frac {a x - {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - {\left (3 \, a^{2} b - b^{2}\right )} \log \left (\frac {a^{2} x^{2} + {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}} a x + {\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{6 \, a^{3}} \]

[In]

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

[Out]

1/6*(6*sqrt(3)*a^2*b*(-(a^2 - b)/a^2)^(1/3)*arctan(-1/3*(sqrt(3)*(a^2 - b)*x + 2*sqrt(3)*(a^3*x^3 + b^2*x^2)^(
1/3)*a*(-(a^2 - b)/a^2)^(2/3))/((a^2 - b)*x)) + 6*a^2*b*(-(a^2 - b)/a^2)^(1/3)*log((a*x*(-(a^2 - b)/a^2)^(1/3)
+ (a^3*x^3 + b^2*x^2)^(1/3))/x) - 3*a^2*b*(-(a^2 - b)/a^2)^(1/3)*log((a^2*x^2*(-(a^2 - b)/a^2)^(2/3) - (a^3*x
^3 + b^2*x^2)^(1/3)*a*x*(-(a^2 - b)/a^2)^(1/3) + (a^3*x^3 + b^2*x^2)^(2/3))/x^2) - 2*sqrt(3)*(3*a^2*b - b^2)*a
rctan(1/3*(sqrt(3)*a*x + 2*sqrt(3)*(a^3*x^3 + b^2*x^2)^(1/3))/(a*x)) + 6*(a^3*x^3 + b^2*x^2)^(1/3)*a^2 + 2*(3*
a^2*b - b^2)*log(-(a*x - (a^3*x^3 + b^2*x^2)^(1/3))/x) - (3*a^2*b - b^2)*log((a^2*x^2 + (a^3*x^3 + b^2*x^2)^(1
/3)*a*x + (a^3*x^3 + b^2*x^2)^(2/3))/x^2))/a^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx=\int { \frac {{\left (a^{3} x^{3} + b^{2} x^{2}\right )}^{\frac {1}{3}}}{a x + b} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 19.27 (sec) , antiderivative size = 339, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx=-\frac {{\left (a^{3} - a b\right )}^{\frac {1}{3}} {\left (a^{2} b - b^{2}\right )} \log \left ({\left | -{\left (a^{3} - a b\right )}^{\frac {1}{3}} + {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {1}{3}} \right |}\right )}{a^{4} - a^{2} b} + \frac {\sqrt {3} {\left (a^{3} - a b\right )}^{\frac {1}{3}} b \arctan \left (\frac {\sqrt {3} {\left ({\left (a^{3} - a b\right )}^{\frac {1}{3}} + 2 \, {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a^{3} - a b\right )}^{\frac {1}{3}}}\right )}{a^{2}} + \frac {{\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {1}{3}} x}{a} + \frac {{\left (a^{3} - a b\right )}^{\frac {1}{3}} b \log \left ({\left (a^{3} - a b\right )}^{\frac {2}{3}} + {\left (a^{3} - a b\right )}^{\frac {1}{3}} {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {1}{3}} + {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {2}{3}}\right )}{2 \, a^{2}} - \frac {\sqrt {3} {\left (3 \, a^{2} b - b^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (a + 2 \, {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {1}{3}}\right )}}{3 \, a}\right )}{3 \, a^{3}} - \frac {{\left (3 \, a^{2} b - b^{2}\right )} \log \left (a^{2} + {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {1}{3}} a + {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {2}{3}}\right )}{6 \, a^{3}} + \frac {{\left (3 \, a^{2} b - b^{2}\right )} \log \left ({\left | -a + {\left (a^{3} + \frac {b^{2}}{x}\right )}^{\frac {1}{3}} \right |}\right )}{3 \, a^{3}} \]

[In]

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

[Out]

-(a^3 - a*b)^(1/3)*(a^2*b - b^2)*log(abs(-(a^3 - a*b)^(1/3) + (a^3 + b^2/x)^(1/3)))/(a^4 - a^2*b) + sqrt(3)*(a
^3 - a*b)^(1/3)*b*arctan(1/3*sqrt(3)*((a^3 - a*b)^(1/3) + 2*(a^3 + b^2/x)^(1/3))/(a^3 - a*b)^(1/3))/a^2 + (a^3
+ b^2/x)^(1/3)*x/a + 1/2*(a^3 - a*b)^(1/3)*b*log((a^3 - a*b)^(2/3) + (a^3 - a*b)^(1/3)*(a^3 + b^2/x)^(1/3) +
(a^3 + b^2/x)^(2/3))/a^2 - 1/3*sqrt(3)*(3*a^2*b - b^2)*arctan(1/3*sqrt(3)*(a + 2*(a^3 + b^2/x)^(1/3))/a)/a^3 -
1/6*(3*a^2*b - b^2)*log(a^2 + (a^3 + b^2/x)^(1/3)*a + (a^3 + b^2/x)^(2/3))/a^3 + 1/3*(3*a^2*b - b^2)*log(abs(
-a + (a^3 + b^2/x)^(1/3)))/a^3

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{b^2 x^2+a^3 x^3}}{b+a x} \, dx=\int \frac {{\left (a^3\,x^3+b^2\,x^2\right )}^{1/3}}{b+a\,x} \,d x \]

[In]

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

[Out]

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