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

Optimal. Leaf size=130 \[ -\frac{3072 c^4 \sqrt{c+d x^3}}{d^4}-\frac{1024 c^3 \left (c+d x^3\right )^{3/2}}{9 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}+\frac{9216 c^{9/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}-\frac{4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac{2 \left (c+d x^3\right )^{9/2}}{27 d^4} \]

[Out]

(-3072*c^4*Sqrt[c + d*x^3])/d^4 - (1024*c^3*(c + d*x^3)^(3/2))/(9*d^4) - (38*c^2*(c + d*x^3)^(5/2))/(5*d^4) -
(4*c*(c + d*x^3)^(7/2))/(7*d^4) - (2*(c + d*x^3)^(9/2))/(27*d^4) + (9216*c^(9/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sq
rt[c])])/d^4

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Rubi [A]  time = 0.112546, antiderivative size = 130, 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, 88, 50, 63, 206} \[ -\frac{3072 c^4 \sqrt{c+d x^3}}{d^4}-\frac{1024 c^3 \left (c+d x^3\right )^{3/2}}{9 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}+\frac{9216 c^{9/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}-\frac{4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac{2 \left (c+d x^3\right )^{9/2}}{27 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3072*c^4*Sqrt[c + d*x^3])/d^4 - (1024*c^3*(c + d*x^3)^(3/2))/(9*d^4) - (38*c^2*(c + d*x^3)^(5/2))/(5*d^4) -
(4*c*(c + d*x^3)^(7/2))/(7*d^4) - (2*(c + d*x^3)^(9/2))/(27*d^4) + (9216*c^(9/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sq
rt[c])])/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 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^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^3 (c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{57 c^2 (c+d x)^{3/2}}{d^3}+\frac{512 c^3 (c+d x)^{3/2}}{d^3 (8 c-d x)}-\frac{6 c (c+d x)^{5/2}}{d^3}-\frac{(c+d x)^{7/2}}{d^3}\right ) \, dx,x,x^3\right )\\ &=-\frac{38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac{2 \left (c+d x^3\right )^{9/2}}{27 d^4}+\frac{\left (512 c^3\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac{1024 c^3 \left (c+d x^3\right )^{3/2}}{9 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac{2 \left (c+d x^3\right )^{9/2}}{27 d^4}+\frac{\left (1536 c^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )}{d^3}\\ &=-\frac{3072 c^4 \sqrt{c+d x^3}}{d^4}-\frac{1024 c^3 \left (c+d x^3\right )^{3/2}}{9 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac{2 \left (c+d x^3\right )^{9/2}}{27 d^4}+\frac{\left (13824 c^5\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{d^3}\\ &=-\frac{3072 c^4 \sqrt{c+d x^3}}{d^4}-\frac{1024 c^3 \left (c+d x^3\right )^{3/2}}{9 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac{2 \left (c+d x^3\right )^{9/2}}{27 d^4}+\frac{\left (27648 c^5\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{d^4}\\ &=-\frac{3072 c^4 \sqrt{c+d x^3}}{d^4}-\frac{1024 c^3 \left (c+d x^3\right )^{3/2}}{9 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac{2 \left (c+d x^3\right )^{9/2}}{27 d^4}+\frac{9216 c^{9/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}\\ \end{align*}

Mathematica [A]  time = 0.0962615, size = 93, normalized size = 0.72 \[ \frac{9216 c^{9/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}-\frac{2 \sqrt{c+d x^3} \left (4611 c^2 d^2 x^6+61892 c^3 d x^3+1509176 c^4+410 c d^3 x^9+35 d^4 x^{12}\right )}{945 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3]*(1509176*c^4 + 61892*c^3*d*x^3 + 4611*c^2*d^2*x^6 + 410*c*d^3*x^9 + 35*d^4*x^12))/(945*d^4
) + (9216*c^(9/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^4

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Maple [C]  time = 0.043, size = 634, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(2/27*d*x^12*(d*x^3+c)^(1/2)+20/189*c*x^9*(d*x^3+c)^(1/2)+2/315/d*c^2*x^6*(d*x^3+c)^(1/2)-8/945/d^2*c^3*x
^3*(d*x^3+c)^(1/2)+16/945/d^3*c^4*(d*x^3+c)^(1/2))-8*c/d^2*(2/21*d*x^9*(d*x^3+c)^(1/2)+16/105*c*x^6*(d*x^3+c)^
(1/2)+2/105/d*c^2*x^3*(d*x^3+c)^(1/2)-4/105/d^2*c^3*(d*x^3+c)^(1/2))-128/15*c^2*(d*x^3+c)^(5/2)/d^4-512*c^3/d^
3*(2/9*x^3*(d*x^3+c)^(1/2)+56/9*c*(d*x^3+c)^(1/2)/d+3*I/d^3*c*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+c)^(3/2)/(-d*x^3+8*c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.3802, size = 495, normalized size = 3.81 \begin{align*} \left [\frac{2 \,{\left (2177280 \, c^{\frac{9}{2}} \log \left (\frac{d x^{3} + 6 \, \sqrt{d x^{3} + c} \sqrt{c} + 10 \, c}{d x^{3} - 8 \, c}\right ) -{\left (35 \, d^{4} x^{12} + 410 \, c d^{3} x^{9} + 4611 \, c^{2} d^{2} x^{6} + 61892 \, c^{3} d x^{3} + 1509176 \, c^{4}\right )} \sqrt{d x^{3} + c}\right )}}{945 \, d^{4}}, -\frac{2 \,{\left (4354560 \, \sqrt{-c} c^{4} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) +{\left (35 \, d^{4} x^{12} + 410 \, c d^{3} x^{9} + 4611 \, c^{2} d^{2} x^{6} + 61892 \, c^{3} d x^{3} + 1509176 \, c^{4}\right )} \sqrt{d x^{3} + c}\right )}}{945 \, d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/945*(2177280*c^(9/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (35*d^4*x^12 + 410*c*d
^3*x^9 + 4611*c^2*d^2*x^6 + 61892*c^3*d*x^3 + 1509176*c^4)*sqrt(d*x^3 + c))/d^4, -2/945*(4354560*sqrt(-c)*c^4*
arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + (35*d^4*x^12 + 410*c*d^3*x^9 + 4611*c^2*d^2*x^6 + 61892*c^3*d*x^3 + 1
509176*c^4)*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+c)**(3/2)/(-d*x**3+8*c),x)

[Out]

Timed out

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Giac [A]  time = 1.13079, size = 158, normalized size = 1.22 \begin{align*} -\frac{9216 \, c^{5} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} d^{4}} - \frac{2 \,{\left (35 \,{\left (d x^{3} + c\right )}^{\frac{9}{2}} d^{32} + 270 \,{\left (d x^{3} + c\right )}^{\frac{7}{2}} c d^{32} + 3591 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} c^{2} d^{32} + 53760 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} c^{3} d^{32} + 1451520 \, \sqrt{d x^{3} + c} c^{4} d^{32}\right )}}{945 \, d^{36}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9216*c^5*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 2/945*(35*(d*x^3 + c)^(9/2)*d^32 + 270*(d*x^3
+ c)^(7/2)*c*d^32 + 3591*(d*x^3 + c)^(5/2)*c^2*d^32 + 53760*(d*x^3 + c)^(3/2)*c^3*d^32 + 1451520*sqrt(d*x^3 +
c)*c^4*d^32)/d^36