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

Optimal. Leaf size=367 \[ \frac {\sqrt [3]{c-d} (b c-a d) \log \left (-\sqrt [3]{d} \sqrt [3]{x^3+x} x \sqrt [3]{c-d}+x^2 (c-d)^{2/3}+d^{2/3} \left (x^3+x\right )^{2/3}\right )}{4 c^2 \sqrt [3]{d}}+\frac {\log \left (\sqrt [3]{x^3+x}-x\right ) (-a c+3 a d-3 b c)}{6 c^2}-\frac {\sqrt [3]{c-d} (b c-a d) \log \left (x \sqrt [3]{c-d}+\sqrt [3]{d} \sqrt [3]{x^3+x}\right )}{2 c^2 \sqrt [3]{d}}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x}+x}\right ) (a c-3 a d+3 b c)}{2 \sqrt {3} c^2}-\frac {\sqrt {3} \sqrt [3]{c-d} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {3} x \sqrt [3]{c-d}}{x \sqrt [3]{c-d}-2 \sqrt [3]{d} \sqrt [3]{x^3+x}}\right )}{2 c^2 \sqrt [3]{d}}+\frac {\log \left (\sqrt [3]{x^3+x} x+\left (x^3+x\right )^{2/3}+x^2\right ) (a c-3 a d+3 b c)}{12 c^2}+\frac {a \sqrt [3]{x^3+x} x}{2 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.86, antiderivative size = 523, normalized size of antiderivative = 1.43, number of steps used = 22, number of rules used = 16, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {2056, 581, 584, 329, 275, 331, 292, 31, 634, 618, 204, 628, 466, 465, 494, 617} \begin {gather*} \frac {\sqrt [3]{x^3+x} \sqrt [3]{c-d} (b c-a d) \log \left (\frac {x^{4/3} (c-d)^{2/3}}{\left (x^2+1\right )^{2/3}}-\frac {\sqrt [3]{d} x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2+1}}+d^{2/3}\right )}{4 c^2 \sqrt [3]{d} \sqrt [3]{x^2+1} \sqrt [3]{x}}-\frac {\sqrt [3]{x^3+x} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right ) (a (c-3 d)+3 b c)}{6 c^2 \sqrt [3]{x^2+1} \sqrt [3]{x}}+\frac {\sqrt [3]{x^3+x} \log \left (\frac {x^{4/3}}{\left (x^2+1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2+1}}+1\right ) (a (c-3 d)+3 b c)}{12 c^2 \sqrt [3]{x^2+1} \sqrt [3]{x}}-\frac {\sqrt [3]{x^3+x} \sqrt [3]{c-d} (b c-a d) \log \left (\frac {x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2+1}}+\sqrt [3]{d}\right )}{2 c^2 \sqrt [3]{d} \sqrt [3]{x^2+1} \sqrt [3]{x}}-\frac {\sqrt [3]{x^3+x} \tan ^{-1}\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) \tan ^{-1}\left (\frac {\sqrt [3]{d}-\frac {2 x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2+1}}}{\sqrt {3} \sqrt [3]{d}}\right )}{2 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[
(d^(1/3) - (2*(c - d)^(1/3)*x^(2/3))/(1 + x^2)^(1/3))/(Sqrt[3]*d^(1/3))])/(2*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[1 - x^(2/3)/(1 + x^2)^(1/3)])/(6*c^2*x^(1/3)*(1 + x^2)^(1/3))
+ ((3*b*c + a*(c - 3*d))*(x + x^3)^(1/3)*Log[1 + x^(4/3)/(1 + x^2)^(2/3) + x^(2/3)/(1 + x^2)^(1/3)])/(12*c^2*
x^(1/3)*(1 + x^2)^(1/3)) - ((c - d)^(1/3)*(b*c - a*d)*(x + x^3)^(1/3)*Log[d^(1/3) + ((c - d)^(1/3)*x^(2/3))/(1
+ x^2)^(1/3)])/(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^(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)])/(4*c^2*d^(1/3
)*x^(1/3)*(1 + x^2)^(1/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 275

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 292

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

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 465

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 466

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 494

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

Rule 581

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 584

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2056

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

________________________________________________________________________________________

Mathematica [C]  time = 0.26, size = 168, normalized size = 0.46 \begin {gather*} \frac {x \sqrt [3]{x^3+x} \left (2 x^2 \left (\frac {c x^2}{d}+1\right )^{2/3} (a (c-3 d)+3 b c) F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};-x^2,-\frac {c x^2}{d}\right )+5 \left ((3 b c-2 a d) \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {(c-d) x^2}{c x^2+d}\right )+2 a d \sqrt [3]{x^2+1} \left (\frac {c x^2}{d}+1\right )^{2/3}\right )\right )}{20 c d \sqrt [3]{x^2+1} \left (\frac {c x^2}{d}+1\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(x + x^3)^(1/3)*(2*(3*b*c + a*(c - 3*d))*x^2*(1 + (c*x^2)/d)^(2/3)*AppellF1[5/3, 2/3, 1, 8/3, -x^2, -((c*x^
2)/d)] + 5*(2*a*d*(1 + x^2)^(1/3)*(1 + (c*x^2)/d)^(2/3) + (3*b*c - 2*a*d)*Hypergeometric2F1[2/3, 2/3, 5/3, ((c
- d)*x^2)/(d + c*x^2)])))/(20*c*d*(1 + x^2)^(1/3)*(1 + (c*x^2)/d)^(2/3))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.20, size = 367, normalized size = 1.00 \begin {gather*} \frac {a x \sqrt [3]{x+x^3}}{2 c}-\frac {(a c+3 b c-3 a d) \tan ^{-1}\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) \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F(-1)]  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

________________________________________________________________________________________

giac [A]  time = 0.89, 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)

________________________________________________________________________________________

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)

________________________________________________________________________________________

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

________________________________________________________________________________________

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)

________________________________________________________________________________________

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)

________________________________________________________________________________________