3.260 \(\int \frac {x^5 \sqrt {c+d x^3}}{4 c+d x^3} \, dx\)

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

[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 = 0.06, 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 = {446, 80, 50, 63, 203} \[ \frac {8 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^2}-\frac {8 c \sqrt {c+d x^3}}{3 d^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2} \]

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 50

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 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 80

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 203

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

Rule 446

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} \operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.04, size = 65, normalized size = 0.86 \[ \frac {24 \sqrt {3} c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )+2 \left (d x^3-11 c\right ) \sqrt {c+d x^3}}{9 d^2} \]

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] + 24*Sqrt[3]*c^(3/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(9*d^2)

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fricas [A]  time = 0.69, size = 129, normalized size = 1.70 \[ \left [\frac {2 \, {\left (6 \, \sqrt {3} \sqrt {-c} c \log \left (\frac {d x^{3} + 2 \, \sqrt {3} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 11 \, c\right )}\right )}}{9 \, d^{2}}, \frac {2 \, {\left (12 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 11 \, c\right )}\right )}}{9 \, d^{2}}\right ] \]

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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giac [A]  time = 0.19, size = 64, normalized size = 0.84 \[ \frac {8 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{3 \, d^{2}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{4} - 12 \, \sqrt {d x^{3} + c} c d^{4}\right )}}{9 \, d^{6}} \]

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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maple [C]  time = 0.17, size = 446, normalized size = 5.87 \[ -\frac {4 \left (\frac {2 \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )}{6 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{3 d^{3} \sqrt {d \,x^{3}+c}}\right ) c}{d}+\frac {2 \left (d \,x^{3}+c \right )^{\frac {3}{2}}}{9 d^{2}} \]

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)

[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*(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),1/6*(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+4*c)))

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maxima [A]  time = 1.30, size = 53, normalized size = 0.70 \[ \frac {2 \, {\left (12 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} - 12 \, \sqrt {d x^{3} + c} c\right )}}{9 \, d^{2}} \]

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

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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sympy [A]  time = 16.96, size = 68, normalized size = 0.89 \[ \frac {2 \left (\frac {4 \sqrt {3} c^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {3} \sqrt {c + d x^{3}}}{3 \sqrt {c}} \right )}}{3} - \frac {4 c \sqrt {c + d x^{3}}}{3} + \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{9}\right )}{d^{2}} \]

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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