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

Optimal. Leaf size=141 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{6 \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{18 c^{5/6} d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{18 c^{5/6} d^{2/3}} \]

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Rubi [A]  time = 0.41, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {486, 444, 63, 206, 2138, 2145, 205} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{6 \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{18 c^{5/6} d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{18 c^{5/6} d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + d^(1/3)*x))/Sqrt[c + d*x^3]]/(6*Sqrt[3]*c^(5/6)*d^(2/3)) + ArcTanh[(c^(1/3
) + d^(1/3)*x)^2/(3*c^(1/6)*Sqrt[c + d*x^3])]/(18*c^(5/6)*d^(2/3)) - ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/(18*
c^(5/6)*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 c-d x^3\right ) \sqrt {c+d x^3}} \, dx &=-\frac {\int \frac {2 \sqrt [3]{c} d^{2/3}-2 d x-\frac {d^{4/3} x^2}{\sqrt [3]{c}}}{\left (4+\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}+\frac {d^{2/3} x^2}{c^{2/3}}\right ) \sqrt {c+d x^3}} \, dx}{12 c d}+\frac {\int \frac {1+\frac {\sqrt [3]{d} x}{\sqrt [3]{c}}}{\left (2-\frac {\sqrt [3]{d} x}{\sqrt [3]{c}}\right ) \sqrt {c+d x^3}} \, dx}{12 c^{2/3} \sqrt [3]{d}}-\frac {\sqrt [3]{d} \int \frac {x^2}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx}{4 \sqrt [3]{c}}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{9-c x^2} \, dx,x,\frac {\left (1+\frac {\sqrt [3]{d} x}{\sqrt [3]{c}}\right )^2}{\sqrt {c+d x^3}}\right )}{6 \sqrt [3]{c} d^{2/3}}-\frac {\sqrt [3]{d} \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{12 \sqrt [3]{c}}+\frac {d^{4/3} \operatorname {Subst}\left (\int \frac {1}{-\frac {2 d^2}{c}-6 d^2 x^2} \, dx,x,\frac {1+\frac {\sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {c+d x^3}}\right )}{3 c^{4/3}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{6 \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{18 c^{5/6} d^{2/3}}-\frac {\operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{6 \sqrt [3]{c} d^{2/3}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{6 \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{18 c^{5/6} d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{18 c^{5/6} d^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 67, normalized size = 0.48 \begin {gather*} \frac {x^2 \sqrt {\frac {c+d x^3}{c}} F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{16 c \sqrt {c+d x^3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 15.45, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x/((8*c - d*x^3)*Sqrt[c + d*x^3]),x]

[Out]

Defer[IntegrateAlgebraic][x/((8*c - d*x^3)*Sqrt[c + d*x^3]), x]

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fricas [B]  time = 2.47, size = 2459, normalized size = 17.44

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54*sqrt(3)*(1/(c^5*d^4))^(1/6)*arctan(1/9*((9*sqrt(3)*c*d^2*x^5*(1/(c^5*d^4))^(1/6) - sqrt(3)*(c^4*d^5*x^6
- 40*c^5*d^4*x^3 - 32*c^6*d^3)*(1/(c^5*d^4))^(5/6) + 3*sqrt(3)*(5*c^3*d^3*x^4 + 8*c^4*d^2*x)*sqrt(1/(c^5*d^4))
)*sqrt(d*x^3 + c) + (18*sqrt(3)*(c^4*d^4*x^5 + c^5*d^3*x^2)*(1/(c^5*d^4))^(2/3) + 12*sqrt(3)*(c^2*d^3*x^6 - c^
3*d^2*x^3 - 2*c^4*d)*(1/(c^5*d^4))^(1/3) + 3*sqrt(3)*(d^2*x^7 + 5*c*d*x^4 + 4*c^2*x) + sqrt(d*x^3 + c)*(sqrt(3
)*(c^4*d^5*x^6 + 32*c^5*d^4*x^3 + 40*c^6*d^3)*(1/(c^5*d^4))^(5/6) + 3*sqrt(3)*(7*c^3*d^3*x^4 + 4*c^4*d^2*x)*sq
rt(1/(c^5*d^4)) + 9*sqrt(3)*(c*d^2*x^5 + 2*c^2*d*x^2)*(1/(c^5*d^4))^(1/6)))*sqrt((d^3*x^9 - 276*c*d^2*x^6 - 16
08*c^2*d*x^3 - 1088*c^3 - 18*(c^4*d^5*x^7 - 52*c^5*d^4*x^4 - 80*c^6*d^3*x)*(1/(c^5*d^4))^(2/3) + 6*sqrt(d*x^3
+ c)*(24*(c^5*d^5*x^5 + c^6*d^4*x^2)*(1/(c^5*d^4))^(5/6) - 4*(c^3*d^4*x^6 + 41*c^4*d^3*x^3 + 40*c^5*d^2)*sqrt(
1/(c^5*d^4)) - (c*d^3*x^7 - 28*c^2*d^2*x^4 - 272*c^3*d*x)*(1/(c^5*d^4))^(1/6)) + 18*(c^2*d^4*x^8 + 20*c^3*d^3*
x^5 - 8*c^4*d^2*x^2)*(1/(c^5*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)))/(d^2*x^7 - 7*c*
d*x^4 - 8*c^2*x)) - 1/54*sqrt(3)*(1/(c^5*d^4))^(1/6)*arctan(1/9*((9*sqrt(3)*c*d^2*x^5*(1/(c^5*d^4))^(1/6) - sq
rt(3)*(c^4*d^5*x^6 - 40*c^5*d^4*x^3 - 32*c^6*d^3)*(1/(c^5*d^4))^(5/6) + 3*sqrt(3)*(5*c^3*d^3*x^4 + 8*c^4*d^2*x
)*sqrt(1/(c^5*d^4)))*sqrt(d*x^3 + c) - (18*sqrt(3)*(c^4*d^4*x^5 + c^5*d^3*x^2)*(1/(c^5*d^4))^(2/3) + 12*sqrt(3
)*(c^2*d^3*x^6 - c^3*d^2*x^3 - 2*c^4*d)*(1/(c^5*d^4))^(1/3) + 3*sqrt(3)*(d^2*x^7 + 5*c*d*x^4 + 4*c^2*x) - sqrt
(d*x^3 + c)*(sqrt(3)*(c^4*d^5*x^6 + 32*c^5*d^4*x^3 + 40*c^6*d^3)*(1/(c^5*d^4))^(5/6) + 3*sqrt(3)*(7*c^3*d^3*x^
4 + 4*c^4*d^2*x)*sqrt(1/(c^5*d^4)) + 9*sqrt(3)*(c*d^2*x^5 + 2*c^2*d*x^2)*(1/(c^5*d^4))^(1/6)))*sqrt((d^3*x^9 -
276*c*d^2*x^6 - 1608*c^2*d*x^3 - 1088*c^3 - 18*(c^4*d^5*x^7 - 52*c^5*d^4*x^4 - 80*c^6*d^3*x)*(1/(c^5*d^4))^(2
/3) - 6*sqrt(d*x^3 + c)*(24*(c^5*d^5*x^5 + c^6*d^4*x^2)*(1/(c^5*d^4))^(5/6) - 4*(c^3*d^4*x^6 + 41*c^4*d^3*x^3
+ 40*c^5*d^2)*sqrt(1/(c^5*d^4)) - (c*d^3*x^7 - 28*c^2*d^2*x^4 - 272*c^3*d*x)*(1/(c^5*d^4))^(1/6)) + 18*(c^2*d^
4*x^8 + 20*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(1/(c^5*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3
)))/(d^2*x^7 - 7*c*d*x^4 - 8*c^2*x)) + 1/108*(1/(c^5*d^4))^(1/6)*log((d^3*x^9 + 318*c*d^2*x^6 + 1200*c^2*d*x^3
+ 640*c^3 + 18*(5*c^4*d^5*x^7 + 64*c^5*d^4*x^4 + 32*c^6*d^3*x)*(1/(c^5*d^4))^(2/3) + 6*sqrt(d*x^3 + c)*(6*(5*
c^5*d^5*x^5 + 32*c^6*d^4*x^2)*(1/(c^5*d^4))^(5/6) + (7*c^3*d^4*x^6 + 152*c^4*d^3*x^3 + 64*c^5*d^2)*sqrt(1/(c^5
*d^4)) + (c*d^3*x^7 + 80*c^2*d^2*x^4 + 160*c^3*d*x)*(1/(c^5*d^4))^(1/6)) + 18*(c^2*d^4*x^8 + 38*c^3*d^3*x^5 +
64*c^4*d^2*x^2)*(1/(c^5*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 1/108*(1/(c^5*d^4))
^(1/6)*log((d^3*x^9 + 318*c*d^2*x^6 + 1200*c^2*d*x^3 + 640*c^3 + 18*(5*c^4*d^5*x^7 + 64*c^5*d^4*x^4 + 32*c^6*d
^3*x)*(1/(c^5*d^4))^(2/3) - 6*sqrt(d*x^3 + c)*(6*(5*c^5*d^5*x^5 + 32*c^6*d^4*x^2)*(1/(c^5*d^4))^(5/6) + (7*c^3
*d^4*x^6 + 152*c^4*d^3*x^3 + 64*c^5*d^2)*sqrt(1/(c^5*d^4)) + (c*d^3*x^7 + 80*c^2*d^2*x^4 + 160*c^3*d*x)*(1/(c^
5*d^4))^(1/6)) + 18*(c^2*d^4*x^8 + 38*c^3*d^3*x^5 + 64*c^4*d^2*x^2)*(1/(c^5*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x
^6 + 192*c^2*d*x^3 - 512*c^3)) - 1/216*(1/(c^5*d^4))^(1/6)*log((d^3*x^9 - 276*c*d^2*x^6 - 1608*c^2*d*x^3 - 108
8*c^3 - 18*(c^4*d^5*x^7 - 52*c^5*d^4*x^4 - 80*c^6*d^3*x)*(1/(c^5*d^4))^(2/3) + 6*sqrt(d*x^3 + c)*(24*(c^5*d^5*
x^5 + c^6*d^4*x^2)*(1/(c^5*d^4))^(5/6) - 4*(c^3*d^4*x^6 + 41*c^4*d^3*x^3 + 40*c^5*d^2)*sqrt(1/(c^5*d^4)) - (c*
d^3*x^7 - 28*c^2*d^2*x^4 - 272*c^3*d*x)*(1/(c^5*d^4))^(1/6)) + 18*(c^2*d^4*x^8 + 20*c^3*d^3*x^5 - 8*c^4*d^2*x^
2)*(1/(c^5*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) + 1/216*(1/(c^5*d^4))^(1/6)*log((d
^3*x^9 - 276*c*d^2*x^6 - 1608*c^2*d*x^3 - 1088*c^3 - 18*(c^4*d^5*x^7 - 52*c^5*d^4*x^4 - 80*c^6*d^3*x)*(1/(c^5*
d^4))^(2/3) - 6*sqrt(d*x^3 + c)*(24*(c^5*d^5*x^5 + c^6*d^4*x^2)*(1/(c^5*d^4))^(5/6) - 4*(c^3*d^4*x^6 + 41*c^4*
d^3*x^3 + 40*c^5*d^2)*sqrt(1/(c^5*d^4)) - (c*d^3*x^7 - 28*c^2*d^2*x^4 - 272*c^3*d*x)*(1/(c^5*d^4))^(1/6)) + 18
*(c^2*d^4*x^8 + 20*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(1/(c^5*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 -
512*c^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.16, size = 416, normalized size = 2.95 \begin {gather*} -\frac {i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{18 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{27 c \,d^{3} \sqrt {d \,x^{3}+c}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 40.22, size = 272, normalized size = 1.93 \begin {gather*} \frac {\ln \left (\frac {\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}-\sqrt {c}+2\,c^{1/6}\,d^{1/3}\,x\right )}^3}{x^3\,{\left (d^{1/3}\,x-2\,c^{1/3}\right )}^3}\right )}{54\,c^{5/6}\,d^{2/3}}+\frac {\sqrt {2}\,\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (-\sqrt {3}\,c^{1/6}\,d^{1/3}\,x+\sqrt {d\,x^3+c}\,1{}\mathrm {i}+\sqrt {c}\,1{}\mathrm {i}+c^{1/6}\,d^{1/3}\,x\,1{}\mathrm {i}\right )}^3}{x^3\,{\left (d^{1/3}\,x+c^{1/3}-\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )}^3}\right )\,\sqrt {-1+\sqrt {3}\,1{}\mathrm {i}}}{108\,c^{5/6}\,d^{2/3}}+\frac {\sqrt {2}\,\ln \left (\frac {\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )\,{\left (\sqrt {3}\,c^{1/6}\,d^{1/3}\,x-\sqrt {d\,x^3+c}\,1{}\mathrm {i}+\sqrt {c}\,1{}\mathrm {i}+c^{1/6}\,d^{1/3}\,x\,1{}\mathrm {i}\right )}^3}{x^3\,{\left (d^{1/3}\,x+c^{1/3}+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )}^3}\right )\,\sqrt {1+\sqrt {3}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{108\,c^{5/6}\,d^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((((c + d*x^3)^(1/2) + c^(1/2))*((c + d*x^3)^(1/2) - c^(1/2) + 2*c^(1/6)*d^(1/3)*x)^3)/(x^3*(d^(1/3)*x - 2*
c^(1/3))^3))/(54*c^(5/6)*d^(2/3)) + (2^(1/2)*log((((c + d*x^3)^(1/2) - c^(1/2))*((c + d*x^3)^(1/2)*1i + c^(1/2
)*1i + c^(1/6)*d^(1/3)*x*1i - 3^(1/2)*c^(1/6)*d^(1/3)*x)^3)/(x^3*(d^(1/3)*x - 3^(1/2)*c^(1/3)*1i + c^(1/3))^3)
)*(3^(1/2)*1i - 1)^(1/2))/(108*c^(5/6)*d^(2/3)) + (2^(1/2)*log((((c + d*x^3)^(1/2) + c^(1/2))*(c^(1/2)*1i - (c
+ d*x^3)^(1/2)*1i + c^(1/6)*d^(1/3)*x*1i + 3^(1/2)*c^(1/6)*d^(1/3)*x)^3)/(x^3*(3^(1/2)*c^(1/3)*1i + d^(1/3)*x
+ c^(1/3))^3))*(3^(1/2)*1i + 1)^(1/2)*1i)/(108*c^(5/6)*d^(2/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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