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

Optimal. Leaf size=90 \[ -\frac{128 c^2 \sqrt{c+d x^3}}{3 d^3}+\frac{128 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^3}-\frac{14 c \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3} \]

[Out]

(-128*c^2*Sqrt[c + d*x^3])/(3*d^3) - (14*c*(c + d*x^3)^(3/2))/(9*d^3) - (2*(c + d*x^3)^(5/2))/(15*d^3) + (128*
c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^3

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Rubi [A]  time = 0.0850978, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {446, 88, 50, 63, 206} \[ -\frac{128 c^2 \sqrt{c+d x^3}}{3 d^3}+\frac{128 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^3}-\frac{14 c \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-128*c^2*Sqrt[c + d*x^3])/(3*d^3) - (14*c*(c + d*x^3)^(3/2))/(9*d^3) - (2*(c + d*x^3)^(5/2))/(15*d^3) + (128*
c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^3

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]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 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])

Rubi steps

\begin{align*} \int \frac{x^8 \sqrt{c+d x^3}}{8 c-d x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 \sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{7 c \sqrt{c+d x}}{d^2}+\frac{64 c^2 \sqrt{c+d x}}{d^2 (8 c-d x)}-\frac{(c+d x)^{3/2}}{d^2}\right ) \, dx,x,x^3\right )\\ &=-\frac{14 c \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac{\left (64 c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )}{3 d^2}\\ &=-\frac{128 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{14 c \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac{\left (192 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{128 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{14 c \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac{\left (384 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{d^3}\\ &=-\frac{128 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{14 c \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac{128 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 0.0579725, size = 70, normalized size = 0.78 \[ \frac{5760 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-2 \sqrt{c+d x^3} \left (998 c^2+41 c d x^3+3 d^2 x^6\right )}{45 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3]*(998*c^2 + 41*c*d*x^3 + 3*d^2*x^6) + 5760*c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(4
5*d^3)

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Maple [C]  time = 0.011, size = 507, normalized size = 5.6 \begin{align*} -{\frac{1}{{d}^{2}} \left ( d \left ({\frac{2\,{x}^{6}}{15}\sqrt{d{x}^{3}+c}}+{\frac{2\,c{x}^{3}}{45\,d}\sqrt{d{x}^{3}+c}}-{\frac{4\,{c}^{2}}{45\,{d}^{2}}\sqrt{d{x}^{3}+c}} \right ) +{\frac{16\,c}{9\,d} \left ( d{x}^{3}+c \right ) ^{{\frac{3}{2}}}} \right ) }-64\,{\frac{{c}^{2}}{{d}^{2}} \left ( 2/3\,{\frac{\sqrt{d{x}^{3}+c}}{d}}+{\frac{i/3\sqrt{2}}{{d}^{3}}\sum _{{\it \_alpha}={\it RootOf} \left ({{\it \_Z}}^{3}d-8\,c \right ) }{\frac{\sqrt [3]{-{d}^{2}c} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{2/3} \right ) }{\sqrt{d{x}^{3}+c}}\sqrt{{\frac{i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{-i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{d}{-3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c}} \left ( x-{\frac{\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{-i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}{\it EllipticPi} \left ( 1/3\,\sqrt{3}\sqrt{{\frac{i\sqrt{3}d}{\sqrt [3]{-{d}^{2}c}} \left ( x+1/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}-{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) }},-1/18\,{\frac{2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd}{cd}},\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d} \left ( -3/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}+{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) ^{-1}}} \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^2*(d*(2/15*x^6*(d*x^3+c)^(1/2)+2/45/d*c*x^3*(d*x^3+c)^(1/2)-4/45*c^2*(d*x^3+c)^(1/2)/d^2)+16/9*c/d*(d*x^3
+c)^(3/2))-64*c^2/d^2*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2
)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)
*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d
*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha
*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d
/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2
)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^
2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77396, size = 360, normalized size = 4. \begin{align*} \left [\frac{2 \,{\left (1440 \, c^{\frac{5}{2}} \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^{2} x^{6} + 41 \, c d x^{3} + 998 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{3}}, -\frac{2 \,{\left (2880 \, \sqrt{-c} c^{2} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) +{\left (3 \, d^{2} x^{6} + 41 \, c d x^{3} + 998 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x, algorithm="fricas")

[Out]

[2/45*(1440*c^(5/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (3*d^2*x^6 + 41*c*d*x^3 +
998*c^2)*sqrt(d*x^3 + c))/d^3, -2/45*(2880*sqrt(-c)*c^2*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + (3*d^2*x^6 +
41*c*d*x^3 + 998*c^2)*sqrt(d*x^3 + c))/d^3]

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Sympy [A]  time = 47.6439, size = 82, normalized size = 0.91 \begin{align*} \frac{2 \left (- \frac{64 c^{3} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{3 \sqrt{- c}} \right )}}{\sqrt{- c}} - \frac{64 c^{2} \sqrt{c + d x^{3}}}{3} - \frac{7 c \left (c + d x^{3}\right )^{\frac{3}{2}}}{9} - \frac{\left (c + d x^{3}\right )^{\frac{5}{2}}}{15}\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(d*x**3+c)**(1/2)/(-d*x**3+8*c),x)

[Out]

2*(-64*c**3*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/sqrt(-c) - 64*c**2*sqrt(c + d*x**3)/3 - 7*c*(c + d*x**3)**(3/2
)/9 - (c + d*x**3)**(5/2)/15)/d**3

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Giac [A]  time = 1.12081, size = 112, normalized size = 1.24 \begin{align*} -\frac{128 \, c^{3} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} d^{3}} - \frac{2 \,{\left (3 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} d^{12} + 35 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} c d^{12} + 960 \, \sqrt{d x^{3} + c} c^{2} d^{12}\right )}}{45 \, d^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x, algorithm="giac")

[Out]

-128*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 2/45*(3*(d*x^3 + c)^(5/2)*d^12 + 35*(d*x^3 + c)
^(3/2)*c*d^12 + 960*sqrt(d*x^3 + c)*c^2*d^12)/d^15