3.3.60 \(\int \frac {x^5 \sqrt {c+d x^3}}{4 c+d x^3} \, dx\) [260]

Optimal. Leaf size=76 \[ -\frac {8 c \sqrt {c+d x^3}}{3 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {8 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {457, 81, 52, 65, 209} \begin {gather*} \frac {8 c^{3/2} \text {ArcTan}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^2}-\frac {8 c \sqrt {c+d x^3}}{3 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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}}{4 c+d x^3} \, 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 \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac {(4 c) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{4 c+d x} \, dx,x,x^3\right )}{3 d}\\ &=-\frac {8 c \sqrt {c+d x^3}}{3 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {\left (4 c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )}{d}\\ &=-\frac {8 c \sqrt {c+d x^3}}{3 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {\left (8 c^2\right ) \text {Subst}\left (\int \frac {1}{3 c+x^2} \, dx,x,\sqrt {c+d x^3}\right )}{d^2}\\ &=-\frac {8 c \sqrt {c+d x^3}}{3 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {8 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^2}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 66, normalized size = 0.87 \begin {gather*} \frac {2 \left (-11 c+d x^3\right ) \sqrt {c+d x^3}}{9 d^2}+\frac {8 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-11*c + d*x^3)*Sqrt[c + d*x^3])/(9*d^2) + (8*c^(3/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(Sqrt[3]*d
^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.40, size = 446, normalized size = 5.87 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(d*x^3+c)^(3/2)/d^2-4*c/d*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*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))*EllipticPi(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)),_alpha=RootOf(_Z^3*d+4*c)))

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Maxima [A]
time = 0.52, size = 53, normalized size = 0.70 \begin {gather*} \frac {2 \, {\left (12 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} - 12 \, \sqrt {d x^{3} + c} c\right )}}{9 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(12*sqrt(3)*c^(3/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) + (d*x^3 + c)^(3/2) - 12*sqrt(d*x^3 + c)*c
)/d^2

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Fricas [A]
time = 2.41, size = 129, normalized size = 1.70 \begin {gather*} \left [\frac {2 \, {\left (6 \, \sqrt {3} \sqrt {-c} 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 ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 11 \, c\right )}\right )}}{9 \, d^{2}}, \frac {2 \, {\left (12 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 11 \, c\right )}\right )}}{9 \, d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/9*(6*sqrt(3)*sqrt(-c)*c*log((d*x^3 + 2*sqrt(3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d*x^3 + 4*c)) + sqrt(d*x^3
+ c)*(d*x^3 - 11*c))/d^2, 2/9*(12*sqrt(3)*c^(3/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) + sqrt(d*x^3 + c
)*(d*x^3 - 11*c))/d^2]

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Sympy [A]
time = 6.45, size = 68, normalized size = 0.89 \begin {gather*} \frac {2 \cdot \left (\frac {4 \sqrt {3} c^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {3} \sqrt {c + d x^{3}}}{3 \sqrt {c}} \right )}}{3} - \frac {4 c \sqrt {c + d x^{3}}}{3} + \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{9}\right )}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(4*sqrt(3)*c**(3/2)*atan(sqrt(3)*sqrt(c + d*x**3)/(3*sqrt(c)))/3 - 4*c*sqrt(c + d*x**3)/3 + (c + d*x**3)**(3
/2)/9)/d**2

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Giac [A]
time = 1.52, size = 64, normalized size = 0.84 \begin {gather*} \frac {8 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{3 \, d^{2}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{4} - 12 \, \sqrt {d x^{3} + c} c d^{4}\right )}}{9 \, d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/3*sqrt(3)*c^(3/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/d^2 + 2/9*((d*x^3 + c)^(3/2)*d^4 - 12*sqrt(d*x
^3 + c)*c*d^4)/d^6

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Mupad [B]
time = 4.28, size = 88, normalized size = 1.16 \begin {gather*} \frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d}-\frac {22\,c\,\sqrt {d\,x^3+c}}{9\,d^2}+\frac {\sqrt {3}\,c^{3/2}\,\ln \left (\frac {\sqrt {3}\,d\,x^3-2\,\sqrt {3}\,c+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,4{}\mathrm {i}}{3\,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*x^3*(c + d*x^3)^(1/2))/(9*d) - (22*c*(c + d*x^3)^(1/2))/(9*d^2) + (3^(1/2)*c^(3/2)*log((c^(1/2)*(c + d*x^3)
^(1/2)*6i - 2*3^(1/2)*c + 3^(1/2)*d*x^3)/(4*c + d*x^3))*4i)/(3*d^2)

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