3.30.66 \(\int \frac {(b+a x^2) \sqrt [3]{x+x^3}}{d+c x^2} \, dx\) [2966]

Optimal. Leaf size=367 \[ \frac {a x \sqrt [3]{x+x^3}}{2 c}-\frac {(a c+3 b c-3 a d) \text {ArcTan}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 \sqrt {3} c^2}-\frac {\sqrt {3} \sqrt [3]{c-d} (b c-a d) \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{c-d} x}{\sqrt [3]{c-d} x-2 \sqrt [3]{d} \sqrt [3]{x+x^3}}\right )}{2 c^2 \sqrt [3]{d}}+\frac {(-a c-3 b c+3 a d) \log \left (-x+\sqrt [3]{x+x^3}\right )}{6 c^2}-\frac {\sqrt [3]{c-d} (b c-a d) \log \left (\sqrt [3]{c-d} x+\sqrt [3]{d} \sqrt [3]{x+x^3}\right )}{2 c^2 \sqrt [3]{d}}+\frac {(a c+3 b c-3 a d) \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right )}{12 c^2}+\frac {\sqrt [3]{c-d} (b c-a d) \log \left ((c-d)^{2/3} x^2-\sqrt [3]{c-d} \sqrt [3]{d} x \sqrt [3]{x+x^3}+d^{2/3} \left (x+x^3\right )^{2/3}\right )}{4 c^2 \sqrt [3]{d}} \]

[Out]

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

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Rubi [A]
time = 0.46, antiderivative size = 391, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {2081, 595, 598, 335, 281, 337, 477, 476, 503} \begin {gather*} -\frac {\sqrt [3]{x^3+x} \text {ArcTan}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right ) (a (c-3 d)+3 b c)}{2 \sqrt {3} c^2 \sqrt [3]{x^2+1} \sqrt [3]{x}}-\frac {\sqrt {3} \sqrt [3]{x^3+x} \sqrt [3]{c-d} (b c-a d) \text {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^2 \sqrt [3]{d} \sqrt [3]{x^2+1} \sqrt [3]{x}}+\frac {\sqrt [3]{x^3+x} \sqrt [3]{c-d} (b c-a d) \log \left (c x^2+d\right )}{4 c^2 \sqrt [3]{d} \sqrt [3]{x^2+1} \sqrt [3]{x}}-\frac {\sqrt [3]{x^3+x} \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right ) (a (c-3 d)+3 b c)}{4 c^2 \sqrt [3]{x^2+1} \sqrt [3]{x}}-\frac {3 \sqrt [3]{x^3+x} \sqrt [3]{c-d} (b c-a d) \log \left (x^{2/3} \sqrt [3]{c-d}+\sqrt [3]{d} \sqrt [3]{x^2+1}\right )}{4 c^2 \sqrt [3]{d} \sqrt [3]{x^2+1} \sqrt [3]{x}}+\frac {a \sqrt [3]{x^3+x} x}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 503

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

Rule 595

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[f*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*g*(m + n*(p + q + 1) + 1))), x] + Dis
t[1/(b*(m + n*(p + q + 1) + 1)), Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*((b*e - a*f)*(m + 1) + b
*e*n*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*n*q*(b*c - a*d) + b*e*d*n*(p + q + 1))*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

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*} \int \frac {\left (b+a x^2\right ) \sqrt [3]{x+x^3}}{d+c x^2} \, dx &=\frac {\sqrt [3]{x+x^3} \int \frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \left (b+a x^2\right )}{d+c x^2} \, dx}{\sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}+\frac {\sqrt [3]{x+x^3} \int \frac {\sqrt [3]{x} \left (\frac {2}{3} (3 b c-2 a d)+\frac {2}{3} (3 b c+a (c-3 d)) x^2\right )}{\left (1+x^2\right )^{2/3} \left (d+c x^2\right )} \, dx}{2 c \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}+\frac {\sqrt [3]{x+x^3} \int \left (\frac {2 (3 b c+a (c-3 d)) \sqrt [3]{x}}{3 c \left (1+x^2\right )^{2/3}}+\frac {\left (-\frac {2}{3} (3 b c+a (c-3 d)) d+\frac {2}{3} c (3 b c-2 a d)\right ) \sqrt [3]{x}}{c \left (1+x^2\right )^{2/3} \left (d+c x^2\right )}\right ) \, dx}{2 c \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}+\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^2\right )^{2/3}} \, dx}{3 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left ((c-d) (b c-a d) \sqrt [3]{x+x^3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^2\right )^{2/3} \left (d+c x^2\right )} \, dx}{c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}+\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (3 (c-d) (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {x^3}{\left (1+x^6\right )^{2/3} \left (d+c x^6\right )} \, dx,x,\sqrt [3]{x}\right )}{c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}+\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {x}{\left (1+x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{2 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (3 (c-d) (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {x}{\left (1+x^3\right )^{2/3} \left (d+c x^3\right )} \, dx,x,x^{2/3}\right )}{2 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}+\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (3 (c-d) (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {x}{d-(-c+d) x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}+\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\left ((c-d)^{2/3} (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{d}+\sqrt [3]{c-d} x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left ((c-d)^{2/3} (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{d}+\sqrt [3]{c-d} x}{d^{2/3}-\sqrt [3]{c-d} \sqrt [3]{d} x+(c-d)^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}-\frac {(3 b c+a (c-3 d)) \sqrt [3]{x+x^3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3} \log \left (\sqrt [3]{d}+\frac {\sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (3 (c-d)^{2/3} (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1}{d^{2/3}-\sqrt [3]{c-d} \sqrt [3]{d} x+(c-d)^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (\sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {-\sqrt [3]{c-d} \sqrt [3]{d}+2 (c-d)^{2/3} x}{d^{2/3}-\sqrt [3]{c-d} \sqrt [3]{d} x+(c-d)^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}-\frac {(3 b c+a (c-3 d)) \sqrt [3]{x+x^3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {(3 b c+a (c-3 d)) \sqrt [3]{x+x^3} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3} \log \left (\sqrt [3]{d}+\frac {\sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3} \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 )}{4 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left ((3 b c+a (c-3 d)) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (3 \sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{d} \sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {a x \sqrt [3]{x+x^3}}{2 c}-\frac {(3 b c+a (c-3 d)) \sqrt [3]{x+x^3} \tan ^{-1}\left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt {3} c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\sqrt {3} \sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3} \tan ^{-1}\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^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {(3 b c+a (c-3 d)) \sqrt [3]{x+x^3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {(3 b c+a (c-3 d)) \sqrt [3]{x+x^3} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 c^2 \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3} \log \left (\sqrt [3]{d}+\frac {\sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\sqrt [3]{c-d} (b c-a d) \sqrt [3]{x+x^3} \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 )}{4 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ \end {align*}

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Mathematica [A]
time = 8.75, size = 411, normalized size = 1.12 \begin {gather*} \frac {\sqrt [3]{x+x^3} \left (6 a c \sqrt [3]{d} x^{4/3} \sqrt [3]{1+x^2}-2 \sqrt {3} (3 b c+a (c-3 d)) \sqrt [3]{d} \text {ArcTan}\left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )-6 \sqrt {3} \sqrt [3]{c-d} (b c-a d) \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{c-d} x^{2/3}-2 \sqrt [3]{d} \sqrt [3]{1+x^2}}\right )-2 (3 b c+a (c-3 d)) \sqrt [3]{d} \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )-6 \sqrt [3]{c-d} (b c-a d) \log \left (\sqrt [3]{c-d} x^{2/3}+\sqrt [3]{d} \sqrt [3]{1+x^2}\right )+(3 b c+a (c-3 d)) \sqrt [3]{d} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )+3 \sqrt [3]{c-d} (b c-a d) \log \left ((c-d)^{2/3} x^{4/3}-\sqrt [3]{c-d} \sqrt [3]{d} x^{2/3} \sqrt [3]{1+x^2}+d^{2/3} \left (1+x^2\right )^{2/3}\right )\right )}{12 c^2 \sqrt [3]{d} \sqrt [3]{x} \sqrt [3]{1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{2}+b \right ) \left (x^{3}+x \right )^{\frac {1}{3}}}{c \,x^{2}+d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{x \left (x^{2} + 1\right )} \left (a x^{2} + b\right )}{c x^{2} + d}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 354, normalized size = 0.96 \begin {gather*} \frac {a x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}}{2 \, c} + \frac {{\left (b c^{2} - a c d - b c d + a d^{2}\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^{3} - c^{2} d\right )}} + \frac {\sqrt {3} {\left (a c + 3 \, b c - 3 \, a d\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right )}{6 \, c^{2}} + \frac {{\left (a c + 3 \, b c - 3 \, a d\right )} \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}{12 \, c^{2}} - \frac {{\left (a c + 3 \, b c - 3 \, a d\right )} \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right )}{6 \, c^{2}} - \frac {{\left (\sqrt {3} {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} b c - \sqrt {3} {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{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 \, c^{2} d} - \frac {{\left ({\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} b c - {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} 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 \, c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2*(1/x^2 + 1)^(1/3)/c + 1/2*(b*c^2 - a*c*d - b*c*d + a*d^2)*(-(c - d)/d)^(1/3)*log(abs(-(-(c - d)/d)^(
1/3) + (1/x^2 + 1)^(1/3)))/(c^3 - c^2*d) + 1/6*sqrt(3)*(a*c + 3*b*c - 3*a*d)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)
^(1/3) + 1))/c^2 + 1/12*(a*c + 3*b*c - 3*a*d)*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1)/c^2 - 1/6*(a*c +
3*b*c - 3*a*d)*log(abs((1/x^2 + 1)^(1/3) - 1))/c^2 - 1/2*(sqrt(3)*(-c*d^2 + d^3)^(1/3)*b*c - sqrt(3)*(-c*d^2 +
d^3)^(1/3)*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^2*d) - 1
/4*((-c*d^2 + d^3)^(1/3)*b*c - (-c*d^2 + d^3)^(1/3)*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^2*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a\,x^2+b\right )\,{\left (x^3+x\right )}^{1/3}}{c\,x^2+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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