3.1.28 \(\int \frac {\sqrt [3]{a+b x^3}}{(c+d x)^2} \, dx\) [28]

Optimal. Leaf size=818 \[ -\frac {c^2 \sqrt [3]{a+b x^3}}{d \left (c^3+d^3 x^3\right )}-\frac {d x^2 \sqrt [3]{a+b x^3}}{c^3+d^3 x^3}+\frac {x \sqrt [3]{a+b x^3} F_1\left (\frac {1}{3};-\frac {1}{3},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c^2 \sqrt [3]{1+\frac {b x^3}{a}}}-\frac {d^3 x^4 \sqrt [3]{a+b x^3} F_1\left (\frac {4}{3};-\frac {1}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{2 c^5 \sqrt [3]{1+\frac {b x^3}{a}}}-\frac {\sqrt [3]{b} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {2 a d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c \left (b c^3-a d^3\right )^{2/3}}+\frac {\left (3 b c^3-2 a d^3\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {b c^2 \tan ^{-1}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {b c^2 \log \left (c^3+d^3 x^3\right )}{6 d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {a d \log \left (c^3+d^3 x^3\right )}{9 c \left (b c^3-a d^3\right )^{2/3}}-\frac {\left (3 b c^3-2 a d^3\right ) \log \left (c^3+d^3 x^3\right )}{18 c d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 d^2}+\frac {a d \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{3 c \left (b c^3-a d^3\right )^{2/3}}+\frac {\left (3 b c^3-2 a d^3\right ) \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{6 c d^2 \left (b c^3-a d^3\right )^{2/3}}+\frac {b c^2 \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{2 d^2 \left (b c^3-a d^3\right )^{2/3}} \]

[Out]

-c^2*(b*x^3+a)^(1/3)/d/(d^3*x^3+c^3)-d*x^2*(b*x^3+a)^(1/3)/(d^3*x^3+c^3)+x*(b*x^3+a)^(1/3)*AppellF1(1/3,-1/3,2
,4/3,-b*x^3/a,-d^3*x^3/c^3)/c^2/(1+b*x^3/a)^(1/3)-1/2*d^3*x^4*(b*x^3+a)^(1/3)*AppellF1(4/3,-1/3,2,7/3,-b*x^3/a
,-d^3*x^3/c^3)/c^5/(1+b*x^3/a)^(1/3)-1/6*b*c^2*ln(d^3*x^3+c^3)/d^2/(-a*d^3+b*c^3)^(2/3)-1/9*a*d*ln(d^3*x^3+c^3
)/c/(-a*d^3+b*c^3)^(2/3)-1/18*(-2*a*d^3+3*b*c^3)*ln(d^3*x^3+c^3)/c/d^2/(-a*d^3+b*c^3)^(2/3)-1/2*b^(1/3)*ln(b^(
1/3)*x-(b*x^3+a)^(1/3))/d^2+1/3*a*d*ln((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/c/(-a*d^3+b*c^3)^(2/3)+1/6*(-
2*a*d^3+3*b*c^3)*ln((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/c/d^2/(-a*d^3+b*c^3)^(2/3)+1/2*b*c^2*ln((-a*d^3+
b*c^3)^(1/3)+d*(b*x^3+a)^(1/3))/d^2/(-a*d^3+b*c^3)^(2/3)-1/3*b^(1/3)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3)
)*3^(1/2))/d^2*3^(1/2)+2/9*a*d*arctan(1/3*(1+2*(-a*d^3+b*c^3)^(1/3)*x/c/(b*x^3+a)^(1/3))*3^(1/2))/c/(-a*d^3+b*
c^3)^(2/3)*3^(1/2)+1/9*(-2*a*d^3+3*b*c^3)*arctan(1/3*(1+2*(-a*d^3+b*c^3)^(1/3)*x/c/(b*x^3+a)^(1/3))*3^(1/2))/c
/d^2/(-a*d^3+b*c^3)^(2/3)*3^(1/2)-1/3*b*c^2*arctan(1/3*(1-2*d*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)^(1/3))*3^(1/2))/d
^2/(-a*d^3+b*c^3)^(2/3)*3^(1/2)

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Rubi [A]
time = 0.53, antiderivative size = 818, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 17, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.895, Rules used = {2181, 441, 440, 480, 12, 503, 455, 43, 60, 631, 210, 31, 525, 524, 478, 598, 337} \begin {gather*} -\frac {d^3 \sqrt [3]{b x^3+a} F_1\left (\frac {4}{3};-\frac {1}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right ) x^4}{2 c^5 \sqrt [3]{\frac {b x^3}{a}+1}}-\frac {d \sqrt [3]{b x^3+a} x^2}{c^3+d^3 x^3}+\frac {\sqrt [3]{b x^3+a} F_1\left (\frac {1}{3};-\frac {1}{3},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right ) x}{c^2 \sqrt [3]{\frac {b x^3}{a}+1}}-\frac {\sqrt [3]{b} \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{b x^3+a}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {2 a d \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{b x^3+a}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c \left (b c^3-a d^3\right )^{2/3}}+\frac {\left (3 b c^3-2 a d^3\right ) \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{b x^3+a}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {b c^2 \text {ArcTan}\left (\frac {1-\frac {2 d \sqrt [3]{b x^3+a}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {a d \log \left (c^3+d^3 x^3\right )}{9 c \left (b c^3-a d^3\right )^{2/3}}-\frac {\left (3 b c^3-2 a d^3\right ) \log \left (c^3+d^3 x^3\right )}{18 c d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {b c^2 \log \left (c^3+d^3 x^3\right )}{6 d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{b} x-\sqrt [3]{b x^3+a}\right )}{2 d^2}+\frac {a d \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{b x^3+a}\right )}{3 c \left (b c^3-a d^3\right )^{2/3}}+\frac {\left (3 b c^3-2 a d^3\right ) \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{b x^3+a}\right )}{6 c d^2 \left (b c^3-a d^3\right )^{2/3}}+\frac {b c^2 \log \left (\sqrt [3]{b x^3+a} d+\sqrt [3]{b c^3-a d^3}\right )}{2 d^2 \left (b c^3-a d^3\right )^{2/3}}-\frac {c^2 \sqrt [3]{b x^3+a}}{d \left (c^3+d^3 x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/(c + d*x)^2,x]

[Out]

-((c^2*(a + b*x^3)^(1/3))/(d*(c^3 + d^3*x^3))) - (d*x^2*(a + b*x^3)^(1/3))/(c^3 + d^3*x^3) + (x*(a + b*x^3)^(1
/3)*AppellF1[1/3, -1/3, 2, 4/3, -((b*x^3)/a), -((d^3*x^3)/c^3)])/(c^2*(1 + (b*x^3)/a)^(1/3)) - (d^3*x^4*(a + b
*x^3)^(1/3)*AppellF1[4/3, -1/3, 2, 7/3, -((b*x^3)/a), -((d^3*x^3)/c^3)])/(2*c^5*(1 + (b*x^3)/a)^(1/3)) - (b^(1
/3)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^2) + (2*a*d*ArcTan[(1 + (2*(b*c^3 - a*d^
3)^(1/3)*x)/(c*(a + b*x^3)^(1/3)))/Sqrt[3]])/(3*Sqrt[3]*c*(b*c^3 - a*d^3)^(2/3)) + ((3*b*c^3 - 2*a*d^3)*ArcTan
[(1 + (2*(b*c^3 - a*d^3)^(1/3)*x)/(c*(a + b*x^3)^(1/3)))/Sqrt[3]])/(3*Sqrt[3]*c*d^2*(b*c^3 - a*d^3)^(2/3)) - (
b*c^2*ArcTan[(1 - (2*d*(a + b*x^3)^(1/3))/(b*c^3 - a*d^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^2*(b*c^3 - a*d^3)^(2/3))
- (b*c^2*Log[c^3 + d^3*x^3])/(6*d^2*(b*c^3 - a*d^3)^(2/3)) - (a*d*Log[c^3 + d^3*x^3])/(9*c*(b*c^3 - a*d^3)^(2
/3)) - ((3*b*c^3 - 2*a*d^3)*Log[c^3 + d^3*x^3])/(18*c*d^2*(b*c^3 - a*d^3)^(2/3)) - (b^(1/3)*Log[b^(1/3)*x - (a
+ b*x^3)^(1/3)])/(2*d^2) + (a*d*Log[((b*c^3 - a*d^3)^(1/3)*x)/c - (a + b*x^3)^(1/3)])/(3*c*(b*c^3 - a*d^3)^(2
/3)) + ((3*b*c^3 - 2*a*d^3)*Log[((b*c^3 - a*d^3)^(1/3)*x)/c - (a + b*x^3)^(1/3)])/(6*c*d^2*(b*c^3 - a*d^3)^(2/
3)) + (b*c^2*Log[(b*c^3 - a*d^3)^(1/3) + d*(a + b*x^3)^(1/3)])/(2*d^2*(b*c^3 - a*d^3)^(2/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 60

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

Rule 210

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

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 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 455

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

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 480

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-(e*x)^
(m + 1))*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*e*n*(p + 1))), x] + Dist[1/(a*n*(p + 1)), Int[(e*x)^m*(a + b*x^
n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m + n*(p + 1) + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b,
c, d, e, m, n, p, q, x]

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 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

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 631

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 2181

Int[(Px_.)*((c_) + (d_.)*(x_))^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c^3 + d^3*x
^3)^q*(a + b*x^3)^p, Px/(c^2 - c*d*x + d^2*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p}, x] && PolyQ[Px, x] && ILtQ
[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{a+b x^3}}{(c+d x)^2} \, dx &=\int \frac {\sqrt [3]{a+b x^3}}{(c+d x)^2} \, dx\\ \end {align*}

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Mathematica [F]
time = 17.82, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{a+b x^3}}{(c+d x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^3 + a)^(1/3)/(d*x + c)^2, 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((b*x^3+a)^(1/3)/(d*x+c)^2,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]{a + b x^{3}}}{\left (c + d x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)/(c + d*x)^2,x)

[Out]

int((a + b*x^3)^(1/3)/(c + d*x)^2, x)

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