3.308 \(\int \frac{x^{11}}{(8 c-d x^3) \sqrt{c+d x^3}} \, dx\)

Optimal. Leaf size=90 \[ -\frac{38 c^2 \sqrt{c+d x^3}}{d^4}+\frac{1024 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^4}-\frac{4 c \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^4} \]

[Out]

(-38*c^2*Sqrt[c + d*x^3])/d^4 - (4*c*(c + d*x^3)^(3/2))/(3*d^4) - (2*(c + d*x^3)^(5/2))/(15*d^4) + (1024*c^(5/
2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^4)

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Rubi [A]  time = 0.0806504, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {446, 88, 63, 206} \[ -\frac{38 c^2 \sqrt{c+d x^3}}{d^4}+\frac{1024 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^4}-\frac{4 c \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^4} \]

Antiderivative was successfully verified.

[In]

Int[x^11/((8*c - d*x^3)*Sqrt[c + d*x^3]),x]

[Out]

(-38*c^2*Sqrt[c + d*x^3])/d^4 - (4*c*(c + d*x^3)^(3/2))/(3*d^4) - (2*(c + d*x^3)^(5/2))/(15*d^4) + (1024*c^(5/
2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^4)

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 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^{11}}{\left (8 c-d x^3\right ) \sqrt{c+d x^3}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^3}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{57 c^2}{d^3 \sqrt{c+d x}}+\frac{512 c^3}{d^3 (8 c-d x) \sqrt{c+d x}}-\frac{6 c \sqrt{c+d x}}{d^3}-\frac{(c+d x)^{3/2}}{d^3}\right ) \, dx,x,x^3\right )\\ &=-\frac{38 c^2 \sqrt{c+d x^3}}{d^4}-\frac{4 c \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^4}+\frac{\left (512 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac{38 c^2 \sqrt{c+d x^3}}{d^4}-\frac{4 c \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^4}+\frac{\left (1024 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{3 d^4}\\ &=-\frac{38 c^2 \sqrt{c+d x^3}}{d^4}-\frac{4 c \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^4}+\frac{1024 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^4}\\ \end{align*}

Mathematica [A]  time = 0.0698614, size = 69, normalized size = 0.77 \[ \frac{5120 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-6 \sqrt{c+d x^3} \left (296 c^2+12 c d x^3+d^2 x^6\right )}{45 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/((8*c - d*x^3)*Sqrt[c + d*x^3]),x]

[Out]

(-6*Sqrt[c + d*x^3]*(296*c^2 + 12*c*d*x^3 + d^2*x^6) + 5120*c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(45*
d^4)

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Maple [C]  time = 0.036, size = 528, normalized size = 5.9 \begin{align*} -{\frac{1}{d} \left ({\frac{2\,{x}^{6}}{15\,d}\sqrt{d{x}^{3}+c}}-{\frac{8\,c{x}^{3}}{45\,{d}^{2}}\sqrt{d{x}^{3}+c}}+{\frac{16\,{c}^{2}}{45\,{d}^{3}}\sqrt{d{x}^{3}+c}} \right ) }-8\,{\frac{c}{{d}^{2}} \left ( 2/9\,{\frac{{x}^{3}\sqrt{d{x}^{3}+c}}{d}}-4/9\,{\frac{c\sqrt{d{x}^{3}+c}}{{d}^{2}}} \right ) }-{\frac{128\,{c}^{2}}{3\,{d}^{4}}\sqrt{d{x}^{3}+c}}-{\frac{{\frac{512\,i}{27}}{c}^{2}\sqrt{2}}{{d}^{6}}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}-8\,c \right ) }{\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{id\sqrt{3} \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},-{\frac{1}{18\,cd} \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x)

[Out]

-1/d*(2/15/d*x^6*(d*x^3+c)^(1/2)-8/45*c/d^2*x^3*(d*x^3+c)^(1/2)+16/45*c^2*(d*x^3+c)^(1/2)/d^3)-8*c/d^2*(2/9/d*
x^3*(d*x^3+c)^(1/2)-4/9*c*(d*x^3+c)^(1/2)/d^2)-128/3*c^2*(d*x^3+c)^(1/2)/d^4-512/27*I*c^2/d^6*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^11/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32712, size = 360, normalized size = 4. \begin{align*} \left [\frac{2 \,{\left (1280 \, 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 ) - 3 \,{\left (d^{2} x^{6} + 12 \, c d x^{3} + 296 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{4}}, -\frac{2 \,{\left (2560 \, \sqrt{-c} c^{2} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) + 3 \,{\left (d^{2} x^{6} + 12 \, c d x^{3} + 296 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

[2/45*(1280*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 + 12*c*d*x^3 +
296*c^2)*sqrt(d*x^3 + c))/d^4, -2/45*(2560*sqrt(-c)*c^2*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 3*(d^2*x^6 +
12*c*d*x^3 + 296*c^2)*sqrt(d*x^3 + c))/d^4]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(-d*x**3+8*c)/(d*x**3+c)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11246, size = 111, normalized size = 1.23 \begin{align*} -\frac{1024 \, c^{3} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{9 \, \sqrt{-c} d^{4}} - \frac{2 \,{\left ({\left (d x^{3} + c\right )}^{\frac{5}{2}} d^{16} + 10 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} c d^{16} + 285 \, \sqrt{d x^{3} + c} c^{2} d^{16}\right )}}{15 \, d^{20}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

-1024/9*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 2/15*((d*x^3 + c)^(5/2)*d^16 + 10*(d*x^3 + c
)^(3/2)*c*d^16 + 285*sqrt(d*x^3 + c)*c^2*d^16)/d^20