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

Optimal. Leaf size=85 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 c^{3/2} d^2}+\frac {8}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {2}{81 c d^2 \sqrt {c+d x^3}} \]

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Rubi [A]  time = 0.06, antiderivative size = 85, 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, 51, 63, 206} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 c^{3/2} d^2}+\frac {8}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {2}{81 c d^2 \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(81*c*d^2*Sqrt[c + d*x^3]) + 8/(27*d^2*(8*c - d*x^3)*Sqrt[c + d*x^3]) + (2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[
c])])/(243*c^(3/2)*d^2)

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x}{(8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {8}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{(8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{9 d}\\ &=-\frac {2}{81 c d^2 \sqrt {c+d x^3}}+\frac {8}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{81 c d}\\ &=-\frac {2}{81 c d^2 \sqrt {c+d x^3}}+\frac {8}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{81 c d^2}\\ &=-\frac {2}{81 c d^2 \sqrt {c+d x^3}}+\frac {8}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 c^{3/2} d^2}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 68, normalized size = 0.80 \begin {gather*} -\frac {2 \left (\left (d x^3-8 c\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3+c}{9 c}\right )+12 c\right )}{81 c d^2 \left (d x^3-8 c\right ) \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(12*c + (-8*c + d*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*x^3)/(9*c)]))/(81*c*d^2*(-8*c + d*x^3)*Sqrt[
c + d*x^3])

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IntegrateAlgebraic [A]  time = 0.08, size = 117, normalized size = 1.38 \begin {gather*} \frac {\left (\frac {16}{243 \sqrt {c} d^2}-\frac {2 x^3}{243 c^{3/2} d}\right ) \sqrt {c+d x^3} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+\frac {2 x^3}{81 c d}+\frac {8}{81 d^2}}{8 c \sqrt {c+d x^3}-d x^3 \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8/(81*d^2) + (2*x^3)/(81*c*d) + (16/(243*Sqrt[c]*d^2) - (2*x^3)/(243*c^(3/2)*d))*Sqrt[c + d*x^3]*ArcTanh[Sqrt
[c + d*x^3]/(3*Sqrt[c])])/(8*c*Sqrt[c + d*x^3] - d*x^3*Sqrt[c + d*x^3])

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fricas [A]  time = 0.71, size = 223, normalized size = 2.62 \begin {gather*} \left [\frac {{\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\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 \, {\left (c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{243 \, {\left (c^{2} d^{4} x^{6} - 7 \, c^{3} d^{3} x^{3} - 8 \, c^{4} d^{2}\right )}}, -\frac {2 \, {\left ({\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, {\left (c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{243 \, {\left (c^{2} d^{4} x^{6} - 7 \, c^{3} d^{3} x^{3} - 8 \, c^{4} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[1/243*((d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) -
6*(c*d*x^3 + 4*c^2)*sqrt(d*x^3 + c))/(c^2*d^4*x^6 - 7*c^3*d^3*x^3 - 8*c^4*d^2), -2/243*((d^2*x^6 - 7*c*d*x^3 -
8*c^2)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 3*(c*d*x^3 + 4*c^2)*sqrt(d*x^3 + c))/(c^2*d^4*x^6 -
7*c^3*d^3*x^3 - 8*c^4*d^2)]

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giac [A]  time = 0.19, size = 76, normalized size = 0.89 \begin {gather*} -\frac {2 \, {\left (\frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} c d} + \frac {3 \, {\left (d x^{3} + 4 \, c\right )}}{{\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {d x^{3} + c} c\right )} c d}\right )}}{243 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

-2/243*(arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c*d) + 3*(d*x^3 + 4*c)/(((d*x^3 + c)^(3/2) - 9*sqrt(d*x
^3 + c)*c)*c*d))/d

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maple [C]  time = 0.18, size = 908, normalized size = 10.68

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*c/d*(-1/243*(d*x^3+c)^(1/2)/(d*x^3-8*c)/c^2/d-2/243/((x^3+c/d)*d)^(1/2)/c^2/d-1/1458*I/d^3/c^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)))+1/d*(2/27/((x^3+c/d)*d)^(1
/2)/c/d+1/243*I/c^2/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 = 1.23, size = 83, normalized size = 0.98 \begin {gather*} -\frac {\frac {6 \, {\left (d x^{3} + 4 \, c\right )}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} c - 9 \, \sqrt {d x^{3} + c} c^{2}} + \frac {\log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}}}{243 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

-1/243*(6*(d*x^3 + 4*c)/((d*x^3 + c)^(3/2)*c - 9*sqrt(d*x^3 + c)*c^2) + log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqr
t(d*x^3 + c) + 3*sqrt(c)))/c^(3/2))/d^2

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mupad [B]  time = 4.26, size = 96, normalized size = 1.13 \begin {gather*} \frac {\left (\frac {8}{81\,d^2}+\frac {2\,x^3}{81\,c\,d}\right )\,\sqrt {d\,x^3+c}}{8\,c^2+7\,c\,d\,x^3-d^2\,x^6}+\frac {\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{243\,c^{3/2}\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8/(81*d^2) + (2*x^3)/(81*c*d))*(c + d*x^3)^(1/2))/(8*c^2 - d^2*x^6 + 7*c*d*x^3) + log((10*c + d*x^3 + 6*c^(1
/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3))/(243*c^(3/2)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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