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

Optimal. Leaf size=119 \[ -\frac {480 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3}+\frac {160 c^2 \sqrt {c+d x^3}}{d^3}+\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}+\frac {160 c \left (c+d x^3\right )^{3/2}}{27 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3} \]

[Out]

160/27*c*(d*x^3+c)^(3/2)/d^3+2/15*(d*x^3+c)^(5/2)/d^3+64/27*c*(d*x^3+c)^(5/2)/d^3/(-d*x^3+8*c)-480*c^(5/2)*arc
tanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^3+160*c^2*(d*x^3+c)^(1/2)/d^3

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Rubi [A]  time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {446, 89, 80, 50, 63, 206} \[ \frac {160 c^2 \sqrt {c+d x^3}}{d^3}-\frac {480 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3}+\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}+\frac {160 c \left (c+d x^3\right )^{3/2}}{27 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(160*c^2*Sqrt[c + d*x^3])/d^3 + (160*c*(c + d*x^3)^(3/2))/(27*d^3) + (2*(c + d*x^3)^(5/2))/(15*d^3) + (64*c*(c
+ d*x^3)^(5/2))/(27*d^3*(8*c - d*x^3)) - (480*c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^3

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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^8 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2 (c+d x)^{3/2}}{(8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac {\operatorname {Subst}\left (\int \frac {(c+d x)^{3/2} \left (168 c^2 d+9 c d^2 x\right )}{8 c-d x} \, dx,x,x^3\right )}{27 c d^3}\\ &=\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac {(80 c) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )}{9 d^2}\\ &=\frac {160 c \left (c+d x^3\right )^{3/2}}{27 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac {\left (80 c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )}{d^2}\\ &=\frac {160 c^2 \sqrt {c+d x^3}}{d^3}+\frac {160 c \left (c+d x^3\right )^{3/2}}{27 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac {\left (720 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{d^2}\\ &=\frac {160 c^2 \sqrt {c+d x^3}}{d^3}+\frac {160 c \left (c+d x^3\right )^{3/2}}{27 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac {\left (1440 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{d^3}\\ &=\frac {160 c^2 \sqrt {c+d x^3}}{d^3}+\frac {160 c \left (c+d x^3\right )^{3/2}}{27 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac {64 c \left (c+d x^3\right )^{5/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac {480 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 102, normalized size = 0.86 \[ \frac {21600 c^{5/2} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+2 \sqrt {c+d x^3} \left (-29944 c^3+2515 c^2 d x^3+62 c d^2 x^6+3 d^3 x^9\right )}{45 d^3 \left (d x^3-8 c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c + d*x^3]*(-29944*c^3 + 2515*c^2*d*x^3 + 62*c*d^2*x^6 + 3*d^3*x^9) + 21600*c^(5/2)*(8*c - d*x^3)*ArcT
anh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(45*d^3*(-8*c + d*x^3))

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fricas [A]  time = 0.72, size = 219, normalized size = 1.84 \[ \left [\frac {2 \, {\left (5400 \, {\left (c^{2} d x^{3} - 8 \, c^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + {\left (3 \, d^{3} x^{9} + 62 \, c d^{2} x^{6} + 2515 \, c^{2} d x^{3} - 29944 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, {\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}, \frac {2 \, {\left (10800 \, {\left (c^{2} d x^{3} - 8 \, c^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (3 \, d^{3} x^{9} + 62 \, c d^{2} x^{6} + 2515 \, c^{2} d x^{3} - 29944 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, {\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/45*(5400*(c^2*d*x^3 - 8*c^3)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + (3*d^3
*x^9 + 62*c*d^2*x^6 + 2515*c^2*d*x^3 - 29944*c^3)*sqrt(d*x^3 + c))/(d^4*x^3 - 8*c*d^3), 2/45*(10800*(c^2*d*x^3
- 8*c^3)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + (3*d^3*x^9 + 62*c*d^2*x^6 + 2515*c^2*d*x^3 - 29944
*c^3)*sqrt(d*x^3 + c))/(d^4*x^3 - 8*c*d^3)]

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giac [A]  time = 0.20, size = 111, normalized size = 0.93 \[ \frac {480 \, c^{3} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{3}} - \frac {192 \, \sqrt {d x^{3} + c} c^{3}}{{\left (d x^{3} - 8 \, c\right )} d^{3}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} d^{12} + 80 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c d^{12} + 3120 \, \sqrt {d x^{3} + c} c^{2} d^{12}\right )}}{45 \, d^{15}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

480*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 192*sqrt(d*x^3 + c)*c^3/((d*x^3 - 8*c)*d^3) + 2/
45*(3*(d*x^3 + c)^(5/2)*d^12 + 80*(d*x^3 + c)^(3/2)*c*d^12 + 3120*sqrt(d*x^3 + c)*c^2*d^12)/d^15

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maple [C]  time = 0.19, size = 920, normalized size = 7.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(d*x^3+c)^(5/2)/d^3+64*c^2/d^2*(-3*(d*x^3+c)^(1/2)/(d*x^3-8*c)*c/d+2/3*(d*x^3+c)^(1/2)/d+1/2*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)))+16*c/d^2*(2/9*(d*x^
3+c)^(1/2)*x^3+56/9*(d*x^3+c)^(1/2)*c/d+3*I*c/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 = 1.39, size = 107, normalized size = 0.90 \[ \frac {2 \, {\left (5400 \, c^{\frac {5}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} + 80 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c + 3120 \, \sqrt {d x^{3} + c} c^{2} - \frac {4320 \, \sqrt {d x^{3} + c} c^{3}}{d x^{3} - 8 \, c}\right )}}{45 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5400*c^(5/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 3*(d*x^3 + c)^(5/2) + 80
*(d*x^3 + c)^(3/2)*c + 3120*sqrt(d*x^3 + c)*c^2 - 4320*sqrt(d*x^3 + c)*c^3/(d*x^3 - 8*c))/d^3

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mupad [B]  time = 4.05, size = 127, normalized size = 1.07 \[ \frac {240\,c^{5/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^3}+\frac {6406\,c^2\,\sqrt {d\,x^3+c}}{45\,d^3}+\frac {2\,x^6\,\sqrt {d\,x^3+c}}{15\,d}+\frac {172\,c\,x^3\,\sqrt {d\,x^3+c}}{45\,d^2}+\frac {192\,c^3\,\sqrt {d\,x^3+c}}{d^3\,\left (8\,c-d\,x^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(240*c^(5/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/d^3 + (6406*c^2*(c + d*x^3)^(1/2
))/(45*d^3) + (2*x^6*(c + d*x^3)^(1/2))/(15*d) + (172*c*x^3*(c + d*x^3)^(1/2))/(45*d^2) + (192*c^3*(c + d*x^3)
^(1/2))/(d^3*(8*c - d*x^3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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