3.3.18 \(\int \frac {x^5 \sqrt {c+d x^3}}{8 c-d x^3} \, dx\)

Optimal. Leaf size=69 \[ \frac {16 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^2}-\frac {16 c \sqrt {c+d x^3}}{3 d^2}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {446, 80, 50, 63, 206} \begin {gather*} \frac {16 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^2}-\frac {16 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])/(8*c - d*x^3),x]

[Out]

(-16*c*Sqrt[c + d*x^3])/(3*d^2) - (2*(c + d*x^3)^(3/2))/(9*d^2) + (16*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[
c])])/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 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 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}}{8 c-d x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x \sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )\\ &=-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {(8 c) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )}{3 d}\\ &=-\frac {16 c \sqrt {c+d x^3}}{3 d^2}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {\left (24 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{d}\\ &=-\frac {16 c \sqrt {c+d x^3}}{3 d^2}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {\left (48 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{d^2}\\ &=-\frac {16 c \sqrt {c+d x^3}}{3 d^2}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^2}+\frac {16 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 58, normalized size = 0.84 \begin {gather*} \frac {144 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-2 \sqrt {c+d x^3} \left (25 c+d x^3\right )}{9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*Sqrt[c + d*x^3])/(8*c - d*x^3),x]

[Out]

(-2*Sqrt[c + d*x^3]*(25*c + d*x^3) + 144*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^2)

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IntegrateAlgebraic [A]  time = 0.06, size = 59, normalized size = 0.86 \begin {gather*} \frac {16 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^2}-\frac {2 \sqrt {c+d x^3} \left (25 c+d x^3\right )}{9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^5*Sqrt[c + d*x^3])/(8*c - d*x^3),x]

[Out]

(-2*Sqrt[c + d*x^3]*(25*c + d*x^3))/(9*d^2) + (16*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^2

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fricas [A]  time = 0.99, size = 121, normalized size = 1.75 \begin {gather*} \left [\frac {2 \, {\left (36 \, c^{\frac {3}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - {\left (d x^{3} + 25 \, c\right )} \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}}, -\frac {2 \, {\left (72 \, \sqrt {-c} c \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (d x^{3} + 25 \, c\right )} \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+c)^(1/2)/(-d*x^3+8*c),x, algorithm="fricas")

[Out]

[2/9*(36*c^(3/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (d*x^3 + 25*c)*sqrt(d*x^3 + c
))/d^2, -2/9*(72*sqrt(-c)*c*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + (d*x^3 + 25*c)*sqrt(d*x^3 + c))/d^2]

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giac [A]  time = 0.21, size = 65, normalized size = 0.94 \begin {gather*} -\frac {16 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{2}} - \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{4} + 24 \, \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+8*c),x, algorithm="giac")

[Out]

-16*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^2) - 2/9*((d*x^3 + c)^(3/2)*d^4 + 24*sqrt(d*x^3 + c)*
c*d^4)/d^6

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maple [C]  time = 0.18, size = 446, normalized size = 6.46 \begin {gather*} -\frac {8 \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}-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 )}{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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x)

[Out]

-2/9*(d*x^3+c)^(3/2)/d^2-8*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),-1/18*(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 [A]  time = 1.30, size = 66, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (36 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 24 \, \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+8*c),x, algorithm="maxima")

[Out]

-2/9*(36*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + (d*x^3 + c)^(3/2) + 24*sqr
t(d*x^3 + c)*c)/d^2

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mupad [B]  time = 3.51, size = 78, normalized size = 1.13 \begin {gather*} \frac {8\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^2}-\frac {50\,c\,\sqrt {d\,x^3+c}}{9\,d^2}-\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(c + d*x^3)^(1/2))/(8*c - d*x^3),x)

[Out]

(8*c^(3/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/d^2 - (50*c*(c + d*x^3)^(1/2))/(9*
d^2) - (2*x^3*(c + d*x^3)^(1/2))/(9*d)

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sympy [A]  time = 14.91, size = 65, normalized size = 0.94 \begin {gather*} \frac {2 \left (- \frac {8 c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{\sqrt {- c}} - \frac {8 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+8*c),x)

[Out]

2*(-8*c**2*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/sqrt(-c) - 8*c*sqrt(c + d*x**3)/3 - (c + d*x**3)**(3/2)/9)/d**2

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