\(\int \frac {b+a x^2}{(d+c x^2) \sqrt [3]{x+x^3}} \, dx\) [2889]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F(-1)]
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 26, antiderivative size = 315 \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\frac {\sqrt {3} a \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 c}-\frac {\sqrt {3} (b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{c-d} x}{\sqrt [3]{c-d} x-2 \sqrt [3]{d} \sqrt [3]{x+x^3}}\right )}{2 c \sqrt [3]{c-d} d^{2/3}}-\frac {a \log \left (-x+\sqrt [3]{x+x^3}\right )}{2 c}+\frac {(b c-a d) \log \left (\sqrt [3]{c-d} x+\sqrt [3]{d} \sqrt [3]{x+x^3}\right )}{2 c \sqrt [3]{c-d} d^{2/3}}+\frac {a \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right )}{4 c}+\frac {(-b c+a d) \log \left ((c-d)^{2/3} x^2-\sqrt [3]{c-d} \sqrt [3]{d} x \sqrt [3]{x+x^3}+d^{2/3} \left (x+x^3\right )^{2/3}\right )}{4 c \sqrt [3]{c-d} d^{2/3}} \]

[Out]

1/2*3^(1/2)*a*arctan(3^(1/2)*x/(x+2*(x^3+x)^(1/3)))/c-1/2*3^(1/2)*(-a*d+b*c)*arctan(3^(1/2)*(c-d)^(1/3)*x/((c-
d)^(1/3)*x-2*d^(1/3)*(x^3+x)^(1/3)))/c/(c-d)^(1/3)/d^(2/3)-1/2*a*ln(-x+(x^3+x)^(1/3))/c+1/2*(-a*d+b*c)*ln((c-d
)^(1/3)*x+d^(1/3)*(x^3+x)^(1/3))/c/(c-d)^(1/3)/d^(2/3)+1/4*a*ln(x^2+x*(x^3+x)^(1/3)+(x^3+x)^(2/3))/c+1/4*(a*d-
b*c)*ln((c-d)^(2/3)*x^2-(c-d)^(1/3)*d^(1/3)*x*(x^3+x)^(1/3)+d^(2/3)*(x^3+x)^(2/3))/c/(c-d)^(1/3)/d^(2/3)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2081, 598, 335, 281, 245, 477, 476, 384} \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=-\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{x^2+1} (b c-a d) \arctan \left (\frac {1-\frac {2 x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{d} \sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{2 c d^{2/3} \sqrt [3]{x^3+x} \sqrt [3]{c-d}}+\frac {\sqrt {3} a \sqrt [3]{x} \sqrt [3]{x^2+1} \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 c \sqrt [3]{x^3+x}}-\frac {\sqrt [3]{x} \sqrt [3]{x^2+1} (b c-a d) \log \left (c x^2+d\right )}{4 c d^{2/3} \sqrt [3]{x^3+x} \sqrt [3]{c-d}}+\frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} (b c-a d) \log \left (x^{2/3} \sqrt [3]{c-d}+\sqrt [3]{d} \sqrt [3]{x^2+1}\right )}{4 c d^{2/3} \sqrt [3]{x^3+x} \sqrt [3]{c-d}}-\frac {3 a \sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{4 c \sqrt [3]{x^3+x}} \]

[In]

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

[Out]

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

Rule 245

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

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 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 476

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

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 10.37 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.04 \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \left (2 \sqrt {3} a \arctan \left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )-\frac {2 \sqrt {3} (b c-a d) \arctan \left (\frac {1-\frac {2 \sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{d} \sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{\sqrt [3]{c-d} d^{2/3}}-2 a \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )+a \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )+\frac {2 (b c-a d) \log \left (\sqrt [3]{d}+\frac {\sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{c-d} d^{2/3}}-\frac {(b c-a d) \log \left (d^{2/3}+\frac {(c-d)^{2/3} x^{4/3}}{\left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{c-d} \sqrt [3]{d} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{c-d} d^{2/3}}\right )}{4 c \sqrt [3]{x+x^3}} \]

[In]

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

[Out]

(x^(1/3)*(1 + x^2)^(1/3)*(2*Sqrt[3]*a*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]] - (2*Sqrt[3]*(b*c - a*
d)*ArcTan[(1 - (2*(c - d)^(1/3)*x^(2/3))/(d^(1/3)*(1 + x^2)^(1/3)))/Sqrt[3]])/((c - d)^(1/3)*d^(2/3)) - 2*a*Lo
g[1 - x^(2/3)/(1 + x^2)^(1/3)] + a*Log[1 + x^(4/3)/(1 + x^2)^(2/3) + x^(2/3)/(1 + x^2)^(1/3)] + (2*(b*c - a*d)
*Log[d^(1/3) + ((c - d)^(1/3)*x^(2/3))/(1 + x^2)^(1/3)])/((c - d)^(1/3)*d^(2/3)) - ((b*c - a*d)*Log[d^(2/3) +
((c - d)^(2/3)*x^(4/3))/(1 + x^2)^(2/3) - ((c - d)^(1/3)*d^(1/3)*x^(2/3))/(1 + x^2)^(1/3)])/((c - d)^(1/3)*d^(
2/3))))/(4*c*(x + x^3)^(1/3))

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.91

method result size
pseudoelliptic \(\frac {\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right ) a d \left (\frac {c -d}{d}\right )^{\frac {1}{3}}+\left (-2 a d +2 b c \right ) \ln \left (\frac {\left (\frac {c -d}{d}\right )^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}}{x}\right )-2 \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right ) a d \left (\frac {c -d}{d}\right )^{\frac {1}{3}}-2 a \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right ) d \left (\frac {c -d}{d}\right )^{\frac {1}{3}}+\left (a d -b c \right ) \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {c -d}{d}\right )^{\frac {1}{3}} x -2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}\right )}{3 \left (\frac {c -d}{d}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {c -d}{d}\right )^{\frac {2}{3}} x^{2}-\left (\frac {c -d}{d}\right )^{\frac {1}{3}} {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}}{x^{2}}\right )\right )}{4 \left (\frac {c -d}{d}\right )^{\frac {1}{3}} d c}\) \(288\)

[In]

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

[Out]

1/4/((c-d)/d)^(1/3)*(ln(((x*(x^2+1))^(2/3)+(x*(x^2+1))^(1/3)*x+x^2)/x^2)*a*d*((c-d)/d)^(1/3)+(-2*a*d+2*b*c)*ln
((((c-d)/d)^(1/3)*x+(x*(x^2+1))^(1/3))/x)-2*3^(1/2)*arctan(1/3*(2*(x*(x^2+1))^(1/3)+x)*3^(1/2)/x)*a*d*((c-d)/d
)^(1/3)-2*a*ln(((x*(x^2+1))^(1/3)-x)/x)*d*((c-d)/d)^(1/3)+(a*d-b*c)*(-2*arctan(1/3*3^(1/2)*(((c-d)/d)^(1/3)*x-
2*(x*(x^2+1))^(1/3))/((c-d)/d)^(1/3)/x)*3^(1/2)+ln((((c-d)/d)^(2/3)*x^2-((c-d)/d)^(1/3)*(x*(x^2+1))^(1/3)*x+(x
*(x^2+1))^(2/3))/x^2)))/d/c

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.90 \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=-\frac {\sqrt {3} a \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right )}{2 \, c} + \frac {{\left (b c \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} - a d \left (-\frac {c - d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (c^{2} - c d\right )}} + \frac {a \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}{4 \, c} - \frac {a \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right )}{2 \, c} - \frac {{\left (\sqrt {3} b c - \sqrt {3} a d\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c - d}{d}\right )^{\frac {1}{3}}}\right )}{2 \, {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} c} + \frac {{\left (b c - a d\right )} \log \left (\left (-\frac {c - d}{d}\right )^{\frac {2}{3}} + \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right )}{4 \, {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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