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

Optimal. Leaf size=117 \[ -\frac {3968 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^4}+\frac {2 c \sqrt {c+d x^3} \left (1146 c+47 d x^3\right )}{15 d^4}+\frac {7 x^6 \sqrt {c+d x^3}}{15 d^2}+\frac {x^9 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )} \]

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Rubi [A]  time = 0.09, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {446, 97, 153, 147, 63, 206} \begin {gather*} -\frac {3968 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^4}+\frac {7 x^6 \sqrt {c+d x^3}}{15 d^2}+\frac {2 c \sqrt {c+d x^3} \left (1146 c+47 d x^3\right )}{15 d^4}+\frac {x^9 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*x^6*Sqrt[c + d*x^3])/(15*d^2) + (x^9*Sqrt[c + d*x^3])/(3*d*(8*c - d*x^3)) + (2*c*Sqrt[c + d*x^3]*(1146*c +
47*d*x^3))/(15*d^4) - (3968*c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^4)

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 97

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

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^{11} \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3 \sqrt {c+d x}}{(8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac {x^9 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 c+\frac {7 d x}{2}\right )}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 d}\\ &=\frac {7 x^6 \sqrt {c+d x^3}}{15 d^2}+\frac {x^9 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {x \left (-56 c^2 d-\frac {141}{2} c d^2 x\right )}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{15 d^3}\\ &=\frac {7 x^6 \sqrt {c+d x^3}}{15 d^2}+\frac {x^9 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}+\frac {2 c \sqrt {c+d x^3} \left (1146 c+47 d x^3\right )}{15 d^4}-\frac {\left (1984 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 {7 x^6 \sqrt {c+d x^3}}{15 d^2}+\frac {x^9 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}+\frac {2 c \sqrt {c+d x^3} \left (1146 c+47 d x^3\right )}{15 d^4}-\frac {\left (3968 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 {7 x^6 \sqrt {c+d x^3}}{15 d^2}+\frac {x^9 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}+\frac {2 c \sqrt {c+d x^3} \left (1146 c+47 d x^3\right )}{15 d^4}-\frac {3968 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^4}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 101, normalized size = 0.86 \begin {gather*} \frac {19840 c^{5/2} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+6 \sqrt {c+d x^3} \left (-9168 c^3+770 c^2 d x^3+19 c d^2 x^6+d^3 x^9\right )}{45 d^4 \left (d x^3-8 c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[c + d*x^3]*(-9168*c^3 + 770*c^2*d*x^3 + 19*c*d^2*x^6 + d^3*x^9) + 19840*c^(5/2)*(8*c - d*x^3)*ArcTanh[
Sqrt[c + d*x^3]/(3*Sqrt[c])])/(45*d^4*(-8*c + d*x^3))

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IntegrateAlgebraic [A]  time = 0.12, size = 95, normalized size = 0.81 \begin {gather*} -\frac {3968 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^4}-\frac {2 \sqrt {c+d x^3} \left (9168 c^3-770 c^2 d x^3-19 c d^2 x^6-d^3 x^9\right )}{15 d^4 \left (d x^3-8 c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3]*(9168*c^3 - 770*c^2*d*x^3 - 19*c*d^2*x^6 - d^3*x^9))/(15*d^4*(-8*c + d*x^3)) - (3968*c^(5/
2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^4)

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fricas [A]  time = 0.64, size = 219, normalized size = 1.87 \begin {gather*} \left [\frac {2 \, {\left (4960 \, {\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 ) + 3 \, {\left (d^{3} x^{9} + 19 \, c d^{2} x^{6} + 770 \, c^{2} d x^{3} - 9168 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}, \frac {2 \, {\left (9920 \, {\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 ) + 3 \, {\left (d^{3} x^{9} + 19 \, c d^{2} x^{6} + 770 \, c^{2} d x^{3} - 9168 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/45*(4960*(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 + 19*c*d^2*x^6 + 770*c^2*d*x^3 - 9168*c^3)*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4), 2/45*(9920*(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 + 19*c*d^2*x^6 + 770*c^2*d*x^3 - 9168*c^3)
*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4)]

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giac [A]  time = 0.16, size = 110, normalized size = 0.94 \begin {gather*} \frac {3968 \, c^{3} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{9 \, \sqrt {-c} d^{4}} - \frac {512 \, \sqrt {d x^{3} + c} c^{3}}{3 \, {\left (d x^{3} - 8 \, c\right )} d^{4}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {5}{2}} d^{16} + 25 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c d^{16} + 960 \, \sqrt {d x^{3} + c} c^{2} d^{16}\right )}}{15 \, d^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3968/9*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 512/3*sqrt(d*x^3 + c)*c^3/((d*x^3 - 8*c)*d^4)
+ 2/15*((d*x^3 + c)^(5/2)*d^16 + 25*(d*x^3 + c)^(3/2)*c*d^16 + 960*sqrt(d*x^3 + c)*c^2*d^16)/d^20

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maple [C]  time = 0.28, size = 952, normalized size = 8.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^3*(d*(2/15*(d*x^3+c)^(1/2)*x^6+2/45*(d*x^3+c)^(1/2)*c/d*x^3-4/45*(d*x^3+c)^(1/2)*c^2/d^2)+32/9*c/d*(d*x^3+
c)^(3/2))+512*c^3/d^3*(-1/3/d*(d*x^3+c)^(1/2)/(d*x^3-8*c)+1/54*I/d^3/c*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)))+192*c^2/d^3*(2/3*(d*x^3+c)^(1/2)/d+1/3*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)))

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maxima [A]  time = 1.24, size = 107, normalized size = 0.91 \begin {gather*} \frac {2 \, {\left (4960 \, 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}} + 75 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c + 2880 \, \sqrt {d x^{3} + c} c^{2} - \frac {3840 \, \sqrt {d x^{3} + c} c^{3}}{d x^{3} - 8 \, c}\right )}}{45 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(4960*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) + 75
*(d*x^3 + c)^(3/2)*c + 2880*sqrt(d*x^3 + c)*c^2 - 3840*sqrt(d*x^3 + c)*c^3/(d*x^3 - 8*c))/d^4

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mupad [B]  time = 4.09, size = 127, normalized size = 1.09 \begin {gather*} \frac {1984\,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 )}{9\,d^4}+\frac {1972\,c^2\,\sqrt {d\,x^3+c}}{15\,d^4}+\frac {2\,x^6\,\sqrt {d\,x^3+c}}{15\,d^2}+\frac {18\,c\,x^3\,\sqrt {d\,x^3+c}}{5\,d^3}+\frac {512\,c^3\,\sqrt {d\,x^3+c}}{3\,d^4\,\left (8\,c-d\,x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1984*c^(5/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(9*d^4) + (1972*c^2*(c + d*x^3)
^(1/2))/(15*d^4) + (2*x^6*(c + d*x^3)^(1/2))/(15*d^2) + (18*c*x^3*(c + d*x^3)^(1/2))/(5*d^3) + (512*c^3*(c + d
*x^3)^(1/2))/(3*d^4*(8*c - d*x^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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