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

Optimal. Leaf size=111 \[ \frac {1024 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}-\frac {1024 c^3 \sqrt {c+d x^3}}{3 d^4}-\frac {38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac {4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^4} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 111, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {1024 c^3 \sqrt {c+d x^3}}{3 d^4}-\frac {38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}+\frac {1024 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}-\frac {4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1024*c^3*Sqrt[c + d*x^3])/(3*d^4) - (38*c^2*(c + d*x^3)^(3/2))/(3*d^4) - (4*c*(c + d*x^3)^(5/2))/(5*d^4) - (
2*(c + d*x^3)^(7/2))/(21*d^4) + (1024*c^(7/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^4

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 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 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}}{8 c-d x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3 \sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {57 c^2 \sqrt {c+d x}}{d^3}+\frac {512 c^3 \sqrt {c+d x}}{d^3 (8 c-d x)}-\frac {6 c (c+d x)^{3/2}}{d^3}-\frac {(c+d x)^{5/2}}{d^3}\right ) \, dx,x,x^3\right )\\ &=-\frac {38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac {4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac {\left (512 c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac {1024 c^3 \sqrt {c+d x^3}}{3 d^4}-\frac {38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac {4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac {\left (1536 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{d^3}\\ &=-\frac {1024 c^3 \sqrt {c+d x^3}}{3 d^4}-\frac {38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac {4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac {\left (3072 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{d^4}\\ &=-\frac {1024 c^3 \sqrt {c+d x^3}}{3 d^4}-\frac {38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac {4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac {1024 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 81, normalized size = 0.73 \begin {gather*} \frac {107520 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-2 \sqrt {c+d x^3} \left (18632 c^3+764 c^2 d x^3+57 c d^2 x^6+5 d^3 x^9\right )}{105 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3]*(18632*c^3 + 764*c^2*d*x^3 + 57*c*d^2*x^6 + 5*d^3*x^9) + 107520*c^(7/2)*ArcTanh[Sqrt[c + d
*x^3]/(3*Sqrt[c])])/(105*d^4)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.07, size = 82, normalized size = 0.74 \begin {gather*} \frac {1024 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}-\frac {2 \sqrt {c+d x^3} \left (18632 c^3+764 c^2 d x^3+57 c d^2 x^6+5 d^3 x^9\right )}{105 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3]*(18632*c^3 + 764*c^2*d*x^3 + 57*c*d^2*x^6 + 5*d^3*x^9))/(105*d^4) + (1024*c^(7/2)*ArcTanh[
Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^4

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 169, normalized size = 1.52 \begin {gather*} \left [\frac {2 \, {\left (26880 \, c^{\frac {7}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - {\left (5 \, d^{3} x^{9} + 57 \, c d^{2} x^{6} + 764 \, c^{2} d x^{3} + 18632 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{105 \, d^{4}}, -\frac {2 \, {\left (53760 \, \sqrt {-c} c^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (5 \, d^{3} x^{9} + 57 \, c d^{2} x^{6} + 764 \, c^{2} d x^{3} + 18632 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{105 \, d^{4}}\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),x, algorithm="fricas")

[Out]

[2/105*(26880*c^(7/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (5*d^3*x^9 + 57*c*d^2*x^
6 + 764*c^2*d*x^3 + 18632*c^3)*sqrt(d*x^3 + c))/d^4, -2/105*(53760*sqrt(-c)*c^3*arctan(1/3*sqrt(d*x^3 + c)*sqr
t(-c)/c) + (5*d^3*x^9 + 57*c*d^2*x^6 + 764*c^2*d*x^3 + 18632*c^3)*sqrt(d*x^3 + c))/d^4]

________________________________________________________________________________________

giac [A]  time = 0.16, size = 100, normalized size = 0.90 \begin {gather*} -\frac {1024 \, c^{4} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{4}} - \frac {2 \, {\left (5 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} d^{24} + 42 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c d^{24} + 665 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} d^{24} + 17920 \, \sqrt {d x^{3} + c} c^{3} d^{24}\right )}}{105 \, d^{28}} \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),x, algorithm="giac")

[Out]

-1024*c^4*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 2/105*(5*(d*x^3 + c)^(7/2)*d^24 + 42*(d*x^3 +
c)^(5/2)*c*d^24 + 665*(d*x^3 + c)^(3/2)*c^2*d^24 + 17920*sqrt(d*x^3 + c)*c^3*d^24)/d^28

________________________________________________________________________________________

maple [C]  time = 0.35, size = 582, normalized size = 5.24 \begin {gather*} -\frac {512 \left (\frac {2 \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{18 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{3 d^{3} \sqrt {d \,x^{3}+c}}\right ) c^{3}}{d^{3}}-\frac {8 \left (\frac {2 \sqrt {d \,x^{3}+c}\, x^{6}}{15}+\frac {2 \sqrt {d \,x^{3}+c}\, c \,x^{3}}{45 d}-\frac {4 \sqrt {d \,x^{3}+c}\, c^{2}}{45 d^{2}}\right ) c}{d^{2}}-\frac {\frac {2 \sqrt {d \,x^{3}+c}\, x^{9}}{21}+\frac {2 \sqrt {d \,x^{3}+c}\, c \,x^{6}}{105 d}-\frac {8 \sqrt {d \,x^{3}+c}\, c^{2} x^{3}}{315 d^{2}}+\frac {16 \sqrt {d \,x^{3}+c}\, c^{3}}{315 d^{3}}}{d}-\frac {128 \left (d \,x^{3}+c \right )^{\frac {3}{2}} c^{2}}{9 d^{4}} \end {gather*}

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),x)

[Out]

-1/d*(2/21*x^9*(d*x^3+c)^(1/2)+2/105*c/d*x^6*(d*x^3+c)^(1/2)-8/315*c^2/d^2*x^3*(d*x^3+c)^(1/2)+16/315*c^3*(d*x
^3+c)^(1/2)/d^3)-8*c/d^2*(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)-
128/9*c^2*(d*x^3+c)^(3/2)/d^4-512*c^3/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)))

________________________________________________________________________________________

maxima [A]  time = 1.31, size = 96, normalized size = 0.86 \begin {gather*} -\frac {2 \, {\left (26880 \, c^{\frac {7}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 5 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} + 42 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c + 665 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} + 17920 \, \sqrt {d x^{3} + c} c^{3}\right )}}{105 \, 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),x, algorithm="maxima")

[Out]

-2/105*(26880*c^(7/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 5*(d*x^3 + c)^(7/2) +
42*(d*x^3 + c)^(5/2)*c + 665*(d*x^3 + c)^(3/2)*c^2 + 17920*sqrt(d*x^3 + c)*c^3)/d^4

________________________________________________________________________________________

mupad [B]  time = 3.51, size = 118, normalized size = 1.06 \begin {gather*} \frac {512\,c^{7/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^4}-\frac {37264\,c^3\,\sqrt {d\,x^3+c}}{105\,d^4}-\frac {2\,x^9\,\sqrt {d\,x^3+c}}{21\,d}-\frac {38\,c\,x^6\,\sqrt {d\,x^3+c}}{35\,d^2}-\frac {1528\,c^2\,x^3\,\sqrt {d\,x^3+c}}{105\,d^3} \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),x)

[Out]

(512*c^(7/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/d^4 - (37264*c^3*(c + d*x^3)^(1/
2))/(105*d^4) - (2*x^9*(c + d*x^3)^(1/2))/(21*d) - (38*c*x^6*(c + d*x^3)^(1/2))/(35*d^2) - (1528*c^2*x^3*(c +
d*x^3)^(1/2))/(105*d^3)

________________________________________________________________________________________

sympy [A]  time = 60.27, size = 99, normalized size = 0.89 \begin {gather*} \frac {2 \left (- \frac {512 c^{4} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{\sqrt {- c}} - \frac {512 c^{3} \sqrt {c + d x^{3}}}{3} - \frac {19 c^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + d x^{3}\right )^{\frac {5}{2}}}{5} - \frac {\left (c + d x^{3}\right )^{\frac {7}{2}}}{21}\right )}{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),x)

[Out]

2*(-512*c**4*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/sqrt(-c) - 512*c**3*sqrt(c + d*x**3)/3 - 19*c**2*(c + d*x**3)
**(3/2)/3 - 2*c*(c + d*x**3)**(5/2)/5 - (c + d*x**3)**(7/2)/21)/d**4

________________________________________________________________________________________