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

Optimal. Leaf size=217 \[ \frac {(3 b c-2 a d) \log \left (c+d x^3\right )}{18 c^{5/3} (b c-a d)^{4/3}}-\frac {(3 b c-2 a d) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{6 c^{5/3} (b c-a d)^{4/3}}+\frac {(3 b c-2 a d) \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{4/3}}-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c \left (c+d x^3\right ) (b c-a d)} \]

[Out]

-1/3*d*x*(b*x^3+a)^(2/3)/c/(-a*d+b*c)/(d*x^3+c)+1/18*(-2*a*d+3*b*c)*ln(d*x^3+c)/c^(5/3)/(-a*d+b*c)^(4/3)-1/6*(
-2*a*d+3*b*c)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(5/3)/(-a*d+b*c)^(4/3)+1/9*(-2*a*d+3*b*c)*arcta
n(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))/c^(5/3)/(-a*d+b*c)^(4/3)*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 276, normalized size of antiderivative = 1.27, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {382, 377, 200, 31, 634, 617, 204, 628} \[ -\frac {(3 b c-2 a d) \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{9 c^{5/3} (b c-a d)^{4/3}}+\frac {(3 b c-2 a d) \log \left (\frac {x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{18 c^{5/3} (b c-a d)^{4/3}}+\frac {(3 b c-2 a d) \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt {3} \sqrt [3]{c}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{4/3}}-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c \left (c+d x^3\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 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 377

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

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )^2} \, dx &=-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c (b c-a d) \left (c+d x^3\right )}+\frac {(3 b c-2 a d) \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{3 c (b c-a d)}\\ &=-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c (b c-a d) \left (c+d x^3\right )}+\frac {(3 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{c-(b c-a d) x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 c (b c-a d)}\\ &=-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c (b c-a d) \left (c+d x^3\right )}+\frac {(3 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{c}-\sqrt [3]{b c-a d} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 c^{5/3} (b c-a d)}+\frac {(3 b c-2 a d) \operatorname {Subst}\left (\int \frac {2 \sqrt [3]{c}+\sqrt [3]{b c-a d} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 c^{5/3} (b c-a d)}\\ &=-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c (b c-a d) \left (c+d x^3\right )}-\frac {(3 b c-2 a d) \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{9 c^{5/3} (b c-a d)^{4/3}}+\frac {(3 b c-2 a d) \operatorname {Subst}\left (\int \frac {\sqrt [3]{c} \sqrt [3]{b c-a d}+2 (b c-a d)^{2/3} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{18 c^{5/3} (b c-a d)^{4/3}}+\frac {(3 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{6 c^{4/3} (b c-a d)}\\ &=-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c (b c-a d) \left (c+d x^3\right )}-\frac {(3 b c-2 a d) \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{9 c^{5/3} (b c-a d)^{4/3}}+\frac {(3 b c-2 a d) \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{18 c^{5/3} (b c-a d)^{4/3}}-\frac {(3 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{3 c^{5/3} (b c-a d)^{4/3}}\\ &=-\frac {d x \left (a+b x^3\right )^{2/3}}{3 c (b c-a d) \left (c+d x^3\right )}+\frac {(3 b c-2 a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{4/3}}-\frac {(3 b c-2 a d) \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{9 c^{5/3} (b c-a d)^{4/3}}+\frac {(3 b c-2 a d) \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{18 c^{5/3} (b c-a d)^{4/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.14, size = 99, normalized size = 0.46 \[ \frac {x \left (\left (c+d x^3\right ) (3 b c-2 a d) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-c d \left (a+b x^3\right )\right )}{3 c^2 \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-(c*d*(a + b*x^3)) + (3*b*c - 2*a*d)*(c + d*x^3)*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a +
b*x^3))]))/(3*c^2*(b*c - a*d)*(a + b*x^3)^(1/3)*(c + d*x^3))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.56, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (d \,x^{3}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________