3.3.75 \(\int \frac {x^5 \sqrt {c+d x^3}}{(8 c-d x^3)^2} \, dx\)

Optimal. Leaf size=82 \[ \frac {8 \left (c+d x^3\right )^{3/2}}{27 d^2 \left (8 c-d x^3\right )}+\frac {26 \sqrt {c+d x^3}}{27 d^2}-\frac {26 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 82, 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, 78, 50, 63, 206} \begin {gather*} \frac {8 \left (c+d x^3\right )^{3/2}}{27 d^2 \left (8 c-d x^3\right )}+\frac {26 \sqrt {c+d x^3}}{27 d^2}-\frac {26 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(26*Sqrt[c + d*x^3])/(27*d^2) + (8*(c + d*x^3)^(3/2))/(27*d^2*(8*c - d*x^3)) - (26*Sqrt[c]*ArcTanh[Sqrt[c + d*
x^3]/(3*Sqrt[c])])/(9*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 78

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

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}}{\left (8 c-d x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x \sqrt {c+d x}}{(8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac {8 \left (c+d x^3\right )^{3/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac {13 \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )}{27 d}\\ &=\frac {26 \sqrt {c+d x^3}}{27 d^2}+\frac {8 \left (c+d x^3\right )^{3/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac {(13 c) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 d}\\ &=\frac {26 \sqrt {c+d x^3}}{27 d^2}+\frac {8 \left (c+d x^3\right )^{3/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac {(26 c) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{3 d^2}\\ &=\frac {26 \sqrt {c+d x^3}}{27 d^2}+\frac {8 \left (c+d x^3\right )^{3/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac {26 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.96 \begin {gather*} \frac {6 \sqrt {c+d x^3} \left (d x^3-12 c\right )+26 \sqrt {c} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^2 \left (d x^3-8 c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(-12*c + d*x^3)*Sqrt[c + d*x^3] + 26*Sqrt[c]*(8*c - d*x^3)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^2*(-8
*c + d*x^3))

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IntegrateAlgebraic [A]  time = 0.07, size = 73, normalized size = 0.89 \begin {gather*} -\frac {2 \sqrt {c+d x^3} \left (12 c-d x^3\right )}{3 d^2 \left (d x^3-8 c\right )}-\frac {26 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(12*c - d*x^3)*Sqrt[c + d*x^3])/(3*d^2*(-8*c + d*x^3)) - (26*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])
/(9*d^2)

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fricas [A]  time = 0.47, size = 165, normalized size = 2.01 \begin {gather*} \left [\frac {13 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 6 \, \sqrt {d x^{3} + c} {\left (d x^{3} - 12 \, c\right )}}{9 \, {\left (d^{3} x^{3} - 8 \, c d^{2}\right )}}, \frac {2 \, {\left (13 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, \sqrt {d x^{3} + c} {\left (d x^{3} - 12 \, c\right )}\right )}}{9 \, {\left (d^{3} x^{3} - 8 \, c d^{2}\right )}}\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)^2,x, algorithm="fricas")

[Out]

[1/9*(13*(d*x^3 - 8*c)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 6*sqrt(d*x^3 +
c)*(d*x^3 - 12*c))/(d^3*x^3 - 8*c*d^2), 2/9*(13*(d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c)
+ 3*sqrt(d*x^3 + c)*(d*x^3 - 12*c))/(d^3*x^3 - 8*c*d^2)]

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giac [A]  time = 0.16, size = 69, normalized size = 0.84 \begin {gather*} \frac {26 \, c \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{9 \, \sqrt {-c} d^{2}} + \frac {2 \, \sqrt {d x^{3} + c}}{3 \, d^{2}} - \frac {8 \, \sqrt {d x^{3} + c} c}{3 \, {\left (d x^{3} - 8 \, c\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)^2,x, algorithm="giac")

[Out]

26/9*c*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^2) + 2/3*sqrt(d*x^3 + c)/d^2 - 8/3*sqrt(d*x^3 + c)*c/(
(d*x^3 - 8*c)*d^2)

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maple [C]  time = 0.17, size = 874, normalized size = 10.66

result 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+8*c)^2,x)

[Out]

8*c/d*(-1/3*(d*x^3+c)^(1/2)/(d*x^3-8*c)/d+1/54*I/d^3/c*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)*_al
pha-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)))+1/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 = 1.25, size = 79, normalized size = 0.96 \begin {gather*} \frac {13 \, \sqrt {c} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 6 \, \sqrt {d x^{3} + c} - \frac {24 \, \sqrt {d x^{3} + c} c}{d x^{3} - 8 \, c}}{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)^2,x, algorithm="maxima")

[Out]

1/9*(13*sqrt(c)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 6*sqrt(d*x^3 + c) - 24*sqrt
(d*x^3 + c)*c/(d*x^3 - 8*c))/d^2

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mupad [B]  time = 3.99, size = 87, normalized size = 1.06 \begin {gather*} \frac {2\,\sqrt {d\,x^3+c}}{3\,d^2}+\frac {13\,\sqrt {c}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{9\,d^2}+\frac {8\,c\,\sqrt {d\,x^3+c}}{3\,d^2\,\left (8\,c-d\,x^3\right )} \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)^2,x)

[Out]

(2*(c + d*x^3)^(1/2))/(3*d^2) + (13*c^(1/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(
9*d^2) + (8*c*(c + d*x^3)^(1/2))/(3*d^2*(8*c - d*x^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \sqrt {c + d x^{3}}}{\left (- 8 c + d x^{3}\right )^{2}}\, dx \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)**2,x)

[Out]

Integral(x**5*sqrt(c + d*x**3)/(-8*c + d*x**3)**2, x)

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