3.76 \(\int \frac {x}{(8-d x^3) \sqrt {1+d x^3}} \, dx\)

Optimal. Leaf size=103 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{d} x+1\right )}{\sqrt {d x^3+1}}\right )}{6 \sqrt {3} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [3]{d} x+1\right )^2}{3 \sqrt {d x^3+1}}\right )}{18 d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {1}{3} \sqrt {d x^3+1}\right )}{18 d^{2/3}} \]

[Out]

1/18*arctanh(1/3*(1+d^(1/3)*x)^2/(d*x^3+1)^(1/2))/d^(2/3)-1/18*arctanh(1/3*(d*x^3+1)^(1/2))/d^(2/3)-1/18*arcta
n((1+d^(1/3)*x)*3^(1/2)/(d*x^3+1)^(1/2))/d^(2/3)*3^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {486, 444, 63, 206, 2138, 2145, 205} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \left (\sqrt [3]{d} x+1\right )}{\sqrt {d x^3+1}}\right )}{6 \sqrt {3} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [3]{d} x+1\right )^2}{3 \sqrt {d x^3+1}}\right )}{18 d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {1}{3} \sqrt {d x^3+1}\right )}{18 d^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x/((8 - d*x^3)*Sqrt[1 + d*x^3]),x]

[Out]

-ArcTan[(Sqrt[3]*(1 + d^(1/3)*x))/Sqrt[1 + d*x^3]]/(6*Sqrt[3]*d^(2/3)) + ArcTanh[(1 + d^(1/3)*x)^2/(3*Sqrt[1 +
d*x^3])]/(18*d^(2/3)) - ArcTanh[Sqrt[1 + d*x^3]/3]/(18*d^(2/3))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

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

Rule 206

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

Rule 444

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 486

Int[(x_)/(((a_) + (b_.)*(x_)^3)*Sqrt[(c_) + (d_.)*(x_)^3]), x_Symbol] :> With[{q = Rt[d/c, 3]}, Dist[(d*q)/(4*
b), Int[x^2/((8*c - d*x^3)*Sqrt[c + d*x^3]), x], x] + (-Dist[q^2/(12*b), Int[(1 + q*x)/((2 - q*x)*Sqrt[c + d*x
^3]), x], x] + Dist[1/(12*b*c), Int[(2*c*q^2 - 2*d*x - d*q*x^2)/((4 + 2*q*x + q^2*x^2)*Sqrt[c + d*x^3]), x], x
])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[8*b*c + a*d, 0]

Rule 2138

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(-2*e)/d, Subst[Int
[1/(9 - a*x^2), x], x, (1 + (f*x)/e)^2/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 + 8*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rule 2145

Int[((f_) + (g_.)*(x_) + (h_.)*(x_)^2)/(((c_) + (d_.)*(x_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbo
l] :> Dist[-2*g*h, Subst[Int[1/(2*e*h - (b*d*f - 2*a*e*h)*x^2), x], x, (1 + (2*h*x)/g)/Sqrt[a + b*x^3]], x] /;
FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b*d*f - 2*a*e*h, 0] && EqQ[b*g^3 - 8*a*h^3, 0] && EqQ[g^2 + 2*f*h,
0] && EqQ[b*d*f + b*c*g - 4*a*e*h, 0]

Rubi steps

\begin {align*} \int \frac {x}{\left (8-d x^3\right ) \sqrt {1+d x^3}} \, dx &=-\frac {\int \frac {2 d^{2/3}-2 d x-d^{4/3} x^2}{\left (4+2 \sqrt [3]{d} x+d^{2/3} x^2\right ) \sqrt {1+d x^3}} \, dx}{12 d}+\frac {\int \frac {1+\sqrt [3]{d} x}{\left (2-\sqrt [3]{d} x\right ) \sqrt {1+d x^3}} \, dx}{12 \sqrt [3]{d}}-\frac {1}{4} \sqrt [3]{d} \int \frac {x^2}{\left (8-d x^3\right ) \sqrt {1+d x^3}} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{9-x^2} \, dx,x,\frac {\left (1+\sqrt [3]{d} x\right )^2}{\sqrt {1+d x^3}}\right )}{6 d^{2/3}}-\frac {1}{12} \sqrt [3]{d} \operatorname {Subst}\left (\int \frac {1}{(8-d x) \sqrt {1+d x}} \, dx,x,x^3\right )+\frac {1}{3} d^{4/3} \operatorname {Subst}\left (\int \frac {1}{-2 d^2-6 d^2 x^2} \, dx,x,\frac {1+\sqrt [3]{d} x}{\sqrt {1+d x^3}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \left (1+\sqrt [3]{d} x\right )}{\sqrt {1+d x^3}}\right )}{6 \sqrt {3} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (1+\sqrt [3]{d} x\right )^2}{3 \sqrt {1+d x^3}}\right )}{18 d^{2/3}}-\frac {\operatorname {Subst}\left (\int \frac {1}{9-x^2} \, dx,x,\sqrt {1+d x^3}\right )}{6 d^{2/3}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \left (1+\sqrt [3]{d} x\right )}{\sqrt {1+d x^3}}\right )}{6 \sqrt {3} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (1+\sqrt [3]{d} x\right )^2}{3 \sqrt {1+d x^3}}\right )}{18 d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {1}{3} \sqrt {1+d x^3}\right )}{18 d^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 32, normalized size = 0.31 \[ \frac {1}{16} x^2 F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-d x^3,\frac {d x^3}{8}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x/((8 - d*x^3)*Sqrt[1 + d*x^3]),x]

[Out]

(x^2*AppellF1[2/3, 1/2, 1, 5/3, -(d*x^3), (d*x^3)/8])/16

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fricas [B]  time = 1.43, size = 497, normalized size = 4.83 \[ \frac {2 \, \sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} d \arctan \left (-\frac {{\left (9 \, \sqrt {3} d^{3} x^{5} - \sqrt {3} {\left (d^{2} x^{6} - 40 \, d x^{3} - 32\right )} {\left (d^{2}\right )}^{\frac {2}{3}} + 3 \, \sqrt {3} {\left (5 \, d^{2} x^{4} + 8 \, d x\right )} {\left (d^{2}\right )}^{\frac {1}{3}}\right )} \sqrt {d x^{3} + 1} {\left (d^{2}\right )}^{\frac {1}{6}}}{9 \, {\left (d^{4} x^{7} - 7 \, d^{3} x^{4} - 8 \, d^{2} x\right )}}\right ) + 2 \, {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (\frac {d^{4} x^{9} + 318 \, d^{3} x^{6} + 1200 \, d^{2} x^{3} + 18 \, {\left (5 \, d^{2} x^{7} + 64 \, d x^{4} + 32 \, x\right )} {\left (d^{2}\right )}^{\frac {2}{3}} + 6 \, {\left (7 \, d^{3} x^{6} + 152 \, d^{2} x^{3} + {\left (d^{2} x^{7} + 80 \, d x^{4} + 160 \, x\right )} {\left (d^{2}\right )}^{\frac {2}{3}} + 6 \, {\left (5 \, d^{2} x^{5} + 32 \, d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} + 64 \, d\right )} \sqrt {d x^{3} + 1} + 18 \, {\left (d^{3} x^{8} + 38 \, d^{2} x^{5} + 64 \, d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} + 640 \, d}{d^{3} x^{9} - 24 \, d^{2} x^{6} + 192 \, d x^{3} - 512}\right ) - {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (\frac {d^{4} x^{9} - 276 \, d^{3} x^{6} - 1608 \, d^{2} x^{3} - 18 \, {\left (d^{2} x^{7} - 52 \, d x^{4} - 80 \, x\right )} {\left (d^{2}\right )}^{\frac {2}{3}} - 6 \, {\left (4 \, d^{3} x^{6} + 164 \, d^{2} x^{3} + {\left (d^{2} x^{7} - 28 \, d x^{4} - 272 \, x\right )} {\left (d^{2}\right )}^{\frac {2}{3}} - 24 \, {\left (d^{2} x^{5} + d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} + 160 \, d\right )} \sqrt {d x^{3} + 1} + 18 \, {\left (d^{3} x^{8} + 20 \, d^{2} x^{5} - 8 \, d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} - 1088 \, d}{d^{3} x^{9} - 24 \, d^{2} x^{6} + 192 \, d x^{3} - 512}\right )}{108 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x^3+8)/(d*x^3+1)^(1/2),x, algorithm="fricas")

[Out]

1/108*(2*sqrt(3)*(d^2)^(1/6)*d*arctan(-1/9*(9*sqrt(3)*d^3*x^5 - sqrt(3)*(d^2*x^6 - 40*d*x^3 - 32)*(d^2)^(2/3)
+ 3*sqrt(3)*(5*d^2*x^4 + 8*d*x)*(d^2)^(1/3))*sqrt(d*x^3 + 1)*(d^2)^(1/6)/(d^4*x^7 - 7*d^3*x^4 - 8*d^2*x)) + 2*
(d^2)^(2/3)*log((d^4*x^9 + 318*d^3*x^6 + 1200*d^2*x^3 + 18*(5*d^2*x^7 + 64*d*x^4 + 32*x)*(d^2)^(2/3) + 6*(7*d^
3*x^6 + 152*d^2*x^3 + (d^2*x^7 + 80*d*x^4 + 160*x)*(d^2)^(2/3) + 6*(5*d^2*x^5 + 32*d*x^2)*(d^2)^(1/3) + 64*d)*
sqrt(d*x^3 + 1) + 18*(d^3*x^8 + 38*d^2*x^5 + 64*d*x^2)*(d^2)^(1/3) + 640*d)/(d^3*x^9 - 24*d^2*x^6 + 192*d*x^3
- 512)) - (d^2)^(2/3)*log((d^4*x^9 - 276*d^3*x^6 - 1608*d^2*x^3 - 18*(d^2*x^7 - 52*d*x^4 - 80*x)*(d^2)^(2/3) -
6*(4*d^3*x^6 + 164*d^2*x^3 + (d^2*x^7 - 28*d*x^4 - 272*x)*(d^2)^(2/3) - 24*(d^2*x^5 + d*x^2)*(d^2)^(1/3) + 16
0*d)*sqrt(d*x^3 + 1) + 18*(d^3*x^8 + 20*d^2*x^5 - 8*d*x^2)*(d^2)^(1/3) - 1088*d)/(d^3*x^9 - 24*d^2*x^6 + 192*d
*x^3 - 512)))/d^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x}{\sqrt {d x^{3} + 1} {\left (d x^{3} - 8\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x^3+8)/(d*x^3+1)^(1/2),x, algorithm="giac")

[Out]

integrate(-x/(sqrt(d*x^3 + 1)*(d*x^3 - 8)), x)

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maple [C]  time = 0.35, size = 383, normalized size = 3.72 \[ -\frac {i \left (-d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-d^{2}\right )^{\frac {1}{3}}+\left (-d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-d^{2}\right )^{\frac {1}{3}}+\left (-d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}-8\right )^{2} d^{2}+i \left (-d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8\right ) d -\left (-d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8\right ) d -i \sqrt {3}\, \left (-d^{2}\right )^{\frac {2}{3}}-\left (-d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \sqrt {3}\, \left (-d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8\right )^{2} d -i \sqrt {3}\, \left (-d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8\right )-3 \left (-d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8\right )+i \sqrt {3}\, d -3 d}{18 d}, \sqrt {\frac {i \sqrt {3}\, \left (-d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{27 d^{3} \sqrt {d \,x^{3}+1}\, \RootOf \left (d \,\textit {\_Z}^{3}-8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(-d*x^3+8)/(d*x^3+1)^(1/2),x)

[Out]

-1/27*I/d^3*2^(1/2)*sum(1/_alpha*(-d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2)^(1/3)+(-d^2)^(1/3)))/(-d^2)
^(1/3))^(1/2)*(d*(x-1/d*(-d^2)^(1/3))/(-3*(-d^2)^(1/3)+I*3^(1/2)*(-d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^
(1/2)*(-d^2)^(1/3)+(-d^2)^(1/3)))/(-d^2)^(1/3))^(1/2)/(d*x^3+1)^(1/2)*(I*(-d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/
2)*(-d^2)^(2/3)+2*_alpha^2*d^2-(-d^2)^(1/3)*_alpha*d-(-d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2)^(
1/3)-1/2*I*3^(1/2)/d*(-d^2)^(1/3))*3^(1/2)*d/(-d^2)^(1/3))^(1/2),-1/18/d*(2*I*3^(1/2)*(-d^2)^(1/3)*_alpha^2*d-
I*3^(1/2)*(-d^2)^(2/3)*_alpha+I*3^(1/2)*d-3*(-d^2)^(2/3)*_alpha-3*d),(I*3^(1/2)/d*(-d^2)^(1/3)/(-3/2/d*(-d^2)^
(1/3)+1/2*I*3^(1/2)/d*(-d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {x}{\sqrt {d x^{3} + 1} {\left (d x^{3} - 8\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x^3+8)/(d*x^3+1)^(1/2),x, algorithm="maxima")

[Out]

-integrate(x/(sqrt(d*x^3 + 1)*(d*x^3 - 8)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {x}{\sqrt {d\,x^3+1}\,\left (d\,x^3-8\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x/((d*x^3 + 1)^(1/2)*(d*x^3 - 8)),x)

[Out]

-int(x/((d*x^3 + 1)^(1/2)*(d*x^3 - 8)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x}{d x^{3} \sqrt {d x^{3} + 1} - 8 \sqrt {d x^{3} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x**3+8)/(d*x**3+1)**(1/2),x)

[Out]

-Integral(x/(d*x**3*sqrt(d*x**3 + 1) - 8*sqrt(d*x**3 + 1)), x)

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