3.325 \(\int \frac{x^{11}}{(8 c-d x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac{2 c^2}{27 d^4 \sqrt{c+d x^3}}+\frac{1024 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{81 d^4}-\frac{4 c \sqrt{c+d x^3}}{d^4}-\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^4} \]

[Out]

(2*c^2)/(27*d^4*Sqrt[c + d*x^3]) - (4*c*Sqrt[c + d*x^3])/d^4 - (2*(c + d*x^3)^(3/2))/(9*d^4) + (1024*c^(3/2)*A
rcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(81*d^4)

________________________________________________________________________________________

Rubi [A]  time = 0.10356, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {446, 87, 43, 63, 206} \[ \frac{2 c^2}{27 d^4 \sqrt{c+d x^3}}+\frac{1024 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{81 d^4}-\frac{4 c \sqrt{c+d x^3}}{d^4}-\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^4} \]

Antiderivative was successfully verified.

[In]

Int[x^11/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(2*c^2)/(27*d^4*Sqrt[c + d*x^3]) - (4*c*Sqrt[c + d*x^3])/d^4 - (2*(c + d*x^3)^(3/2))/(9*d^4) + (1024*c^(3/2)*A
rcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(81*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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^3}{(8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{c^2}{9 d^3 (c+d x)^{3/2}}-\frac{7 c}{d^3 \sqrt{c+d x}}-\frac{x}{d^2 \sqrt{c+d x}}+\frac{512 c^2}{9 d^3 (8 c-d x) \sqrt{c+d x}}\right ) \, dx,x,x^3\right )\\ &=\frac{2 c^2}{27 d^4 \sqrt{c+d x^3}}-\frac{14 c \sqrt{c+d x^3}}{3 d^4}+\frac{\left (512 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{27 d^3}-\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{c+d x}} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac{2 c^2}{27 d^4 \sqrt{c+d x^3}}-\frac{14 c \sqrt{c+d x^3}}{3 d^4}+\frac{\left (1024 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{27 d^4}-\frac{\operatorname{Subst}\left (\int \left (-\frac{c}{d \sqrt{c+d x}}+\frac{\sqrt{c+d x}}{d}\right ) \, dx,x,x^3\right )}{3 d^2}\\ &=\frac{2 c^2}{27 d^4 \sqrt{c+d x^3}}-\frac{4 c \sqrt{c+d x^3}}{d^4}-\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^4}+\frac{1024 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{81 d^4}\\ \end{align*}

Mathematica [C]  time = 0.0447, size = 66, normalized size = 0.73 \[ -\frac{2 \left (512 c^2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^3+c}{9 c}\right )-456 c^2+60 c d x^3+3 d^2 x^6\right )}{27 d^4 \sqrt{c+d x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(-2*(-456*c^2 + 60*c*d*x^3 + 3*d^2*x^6 + 512*c^2*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*x^3)/(9*c)]))/(27*d^4*
Sqrt[c + d*x^3])

________________________________________________________________________________________

Maple [C]  time = 0.053, size = 560, normalized size = 6.2 \begin{align*} -{\frac{1}{d} \left ( -{\frac{2\,{c}^{2}}{3\,{d}^{3}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}+{\frac{2\,{x}^{3}}{9\,{d}^{2}}\sqrt{d{x}^{3}+c}}-{\frac{10\,c}{9\,{d}^{3}}\sqrt{d{x}^{3}+c}} \right ) }-8\,{\frac{c}{{d}^{2}} \left ( 2/3\,{\frac{c}{{d}^{2}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}+2/3\,{\frac{\sqrt{d{x}^{3}+c}}{{d}^{2}}} \right ) }+{\frac{128\,{c}^{2}}{3\,{d}^{4}}{\frac{1}{\sqrt{d{x}^{3}+c}}}}-512\,{\frac{{c}^{3}}{{d}^{3}} \left ({\frac{2}{27\,cd}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}+{\frac{{\frac{i}{243}}\sqrt{2}}{{d}^{3}{c}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}-8\,c \right ) }{\frac{\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 ) ^{2/3}+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{id\sqrt{3}}{\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^11/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x)

[Out]

-1/d*(-2/3/d^3*c^2/((x^3+1/d*c)*d)^(1/2)+2/9/d^2*x^3*(d*x^3+c)^(1/2)-10/9*c*(d*x^3+c)^(1/2)/d^3)-8*c/d^2*(2/3/
d^2*c/((x^3+1/d*c)*d)^(1/2)+2/3*(d*x^3+c)^(1/2)/d^2)+128/3*c^2/d^4/(d*x^3+c)^(1/2)-512*c^3/d^3*(2/27/d/c/((x^3
+1/d*c)*d)^(1/2)+1/243*I/d^3/c^2*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))*Ell
ipticPi(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)),_a
lpha=RootOf(_Z^3*d-8*c)))

________________________________________________________________________________________

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.584, size = 436, normalized size = 4.84 \begin{align*} \left [\frac{2 \,{\left (256 \,{\left (c d x^{3} + 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 ) - 3 \,{\left (3 \, d^{2} x^{6} + 60 \, c d x^{3} + 56 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{81 \,{\left (d^{5} x^{3} + c d^{4}\right )}}, -\frac{2 \,{\left (512 \,{\left (c d x^{3} + c^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) + 3 \,{\left (3 \, d^{2} x^{6} + 60 \, c d x^{3} + 56 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{81 \,{\left (d^{5} x^{3} + c d^{4}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[2/81*(256*(c*d*x^3 + c^2)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - 3*(3*d^2*x^
6 + 60*c*d*x^3 + 56*c^2)*sqrt(d*x^3 + c))/(d^5*x^3 + c*d^4), -2/81*(512*(c*d*x^3 + c^2)*sqrt(-c)*arctan(1/3*sq
rt(d*x^3 + c)*sqrt(-c)/c) + 3*(3*d^2*x^6 + 60*c*d*x^3 + 56*c^2)*sqrt(d*x^3 + c))/(d^5*x^3 + c*d^4)]

________________________________________________________________________________________

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)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.53924, size = 111, normalized size = 1.23 \begin{align*} -\frac{1024 \, c^{2} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{81 \, \sqrt{-c} d^{4}} + \frac{2 \, c^{2}}{27 \, \sqrt{d x^{3} + c} d^{4}} - \frac{2 \,{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} d^{8} + 18 \, \sqrt{d x^{3} + c} c d^{8}\right )}}{9 \, d^{12}} \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)^(3/2),x, algorithm="giac")

[Out]

-1024/81*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) + 2/27*c^2/(sqrt(d*x^3 + c)*d^4) - 2/9*((d*x^
3 + c)^(3/2)*d^8 + 18*sqrt(d*x^3 + c)*c*d^8)/d^12