3.3.71 \(\int \frac {x^5}{\sqrt {c+d x^3} (4 c+d x^3)} \, dx\) [271]

Optimal. Leaf size=59 \[ \frac {2 \sqrt {c+d x^3}}{3 d^2}-\frac {8 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^2} \]

[Out]

-8/9*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))*c^(1/2)/d^2*3^(1/2)+2/3*(d*x^3+c)^(1/2)/d^2

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {457, 81, 65, 209} \begin {gather*} \frac {2 \sqrt {c+d x^3}}{3 d^2}-\frac {8 \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

(2*Sqrt[c + d*x^3])/(3*d^2) - (8*Sqrt[c]*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(3*Sqrt[3]*d^2)

Rule 65

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 81

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

Rule 209

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )\\ &=\frac {2 \sqrt {c+d x^3}}{3 d^2}-\frac {(4 c) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )}{3 d}\\ &=\frac {2 \sqrt {c+d x^3}}{3 d^2}-\frac {(8 c) \text {Subst}\left (\int \frac {1}{3 c+x^2} \, dx,x,\sqrt {c+d x^3}\right )}{3 d^2}\\ &=\frac {2 \sqrt {c+d x^3}}{3 d^2}-\frac {8 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 56, normalized size = 0.95 \begin {gather*} \frac {6 \sqrt {c+d x^3}-8 \sqrt {3} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

(6*Sqrt[c + d*x^3] - 8*Sqrt[3]*Sqrt[c]*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(9*d^2)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.39, size = 425, normalized size = 7.20 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(d*x^3+4*c)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*(d*x^3+c)^(1/2)/d^2+4/9*I/d^4*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*
d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/
2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*
d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*El
lipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2
),1/6/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*
_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_al
pha=RootOf(_Z^3*d+4*c))

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 43, normalized size = 0.73 \begin {gather*} -\frac {2 \, {\left (4 \, \sqrt {3} \sqrt {c} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) - 3 \, \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

-2/9*(4*sqrt(3)*sqrt(c)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) - 3*sqrt(d*x^3 + c))/d^2

________________________________________________________________________________________

Fricas [A]
time = 2.55, size = 112, normalized size = 1.90 \begin {gather*} \left [\frac {2 \, {\left (2 \, \sqrt {3} \sqrt {-c} \log \left (\frac {d x^{3} - 2 \, \sqrt {3} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) + 3 \, \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}}, -\frac {2 \, {\left (4 \, \sqrt {3} \sqrt {c} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) - 3 \, \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

[2/9*(2*sqrt(3)*sqrt(-c)*log((d*x^3 - 2*sqrt(3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d*x^3 + 4*c)) + 3*sqrt(d*x^3
+ c))/d^2, -2/9*(4*sqrt(3)*sqrt(c)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) - 3*sqrt(d*x^3 + c))/d^2]

________________________________________________________________________________________

Sympy [A]
time = 7.00, size = 65, normalized size = 1.10 \begin {gather*} \begin {cases} \frac {2 \left (- \frac {4 \sqrt {3} \sqrt {c} \operatorname {atan}{\left (\frac {\sqrt {3} \sqrt {c + d x^{3}}}{3 \sqrt {c}} \right )}}{9 d} + \frac {\sqrt {c + d x^{3}}}{3 d}\right )}{d} & \text {for}\: d \neq 0 \\\frac {x^{6}}{24 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(d*x**3+4*c)/(d*x**3+c)**(1/2),x)

[Out]

Piecewise((2*(-4*sqrt(3)*sqrt(c)*atan(sqrt(3)*sqrt(c + d*x**3)/(3*sqrt(c)))/(9*d) + sqrt(c + d*x**3)/(3*d))/d,
Ne(d, 0)), (x**6/(24*c**(3/2)), True))

________________________________________________________________________________________

Giac [A]
time = 1.74, size = 49, normalized size = 0.83 \begin {gather*} -\frac {2 \, {\left (\frac {4 \, \sqrt {3} \sqrt {c} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{d} - \frac {3 \, \sqrt {d x^{3} + c}}{d}\right )}}{9 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

-2/9*(4*sqrt(3)*sqrt(c)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/d - 3*sqrt(d*x^3 + c)/d)/d

________________________________________________________________________________________

Mupad [B]
time = 4.86, size = 71, normalized size = 1.20 \begin {gather*} \frac {2\,\sqrt {d\,x^3+c}}{3\,d^2}+\frac {\sqrt {3}\,\sqrt {c}\,\ln \left (\frac {2\,\sqrt {3}\,c-\sqrt {3}\,d\,x^3+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,4{}\mathrm {i}}{9\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((c + d*x^3)^(1/2)*(4*c + d*x^3)),x)

[Out]

(2*(c + d*x^3)^(1/2))/(3*d^2) + (3^(1/2)*c^(1/2)*log((2*3^(1/2)*c + c^(1/2)*(c + d*x^3)^(1/2)*6i - 3^(1/2)*d*x
^3)/(4*c + d*x^3))*4i)/(9*d^2)

________________________________________________________________________________________