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

Optimal. Leaf size=148 \[ \frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}+\frac {\tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 207, normalized size of antiderivative = 1.40, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {377, 200, 31, 634, 617, 204, 628} \begin {gather*} -\frac {\log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 c^{2/3} \sqrt [3]{b c-a d}}+\frac {\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 )}{6 c^{2/3} \sqrt [3]{b c-a d}}+\frac {\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 )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 168, normalized size = 1.14 \begin {gather*} \frac {\log \left (\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\frac {x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+c^{2/3}\right )-2 \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt {3} \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 )}{6 c^{2/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 0.00, size = 320, normalized size = 2.16 \begin {gather*} \frac {\left (1+i \sqrt {3}\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\sqrt {-1+i \sqrt {3}} \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt {6} c^{2/3} \sqrt [3]{b c-a d}}-\frac {i \left (\sqrt {3}-i\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 c^{2/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[-1 + I*Sqrt[3]]*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3
)^(1/3) - Sqrt[3]*c^(1/3)*(a + b*x^3)^(1/3))])/(Sqrt[6]*c^(2/3)*(b*c - a*d)^(1/3))) + ((1 + I*Sqrt[3])*Log[2*(
b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(6*c^(2/3)*(b*c - a*d)^(1/3)) - ((I/12)*(-I +
Sqrt[3])*Log[(-2*I)*(b*c - a*d)^(2/3)*x^2 + c^(1/3)*(b*c - a*d)^(1/3)*(I*x - Sqrt[3]*x)*(a + b*x^3)^(1/3) + (
I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(c^(2/3)*(b*c - a*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(1/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,x^3+a\right )}^{1/3}\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________