\(\int \frac {x^{11}}{(8 c-d x^3) (c+d x^3)^{3/2}} \, dx\) [499]

Optimal result

Integrand size = 27, antiderivative size = 90 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 c^2}{27 d^4 \sqrt {c+d x^3}}-\frac {4 c \sqrt {c+d x^3}}{d^4}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^4}+\frac {1024 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4} \] Output:

2/27*c^2/d^4/(d*x^3+c)^(1/2)-4*c*(d*x^3+c)^(1/2)/d^4-2/9*(d*x^3+c)^(3/2)/d 
^4+1024/81*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \left (-\frac {3 \left (56 c^2+60 c d x^3+3 d^2 x^6\right )}{\sqrt {c+d x^3}}+512 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{81 d^4} \] Input:

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

Output:

(2*((-3*(56*c^2 + 60*c*d*x^3 + 3*d^2*x^6))/Sqrt[c + d*x^3] + 512*c^(3/2)*A 
rcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]))/(81*d^4)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {948, 98, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^9}{\left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3\)

\(\Big \downarrow \) 98

\(\displaystyle \frac {1}{3} \int \left (-\frac {x^3}{d^2 \sqrt {d x^3+c}}+\frac {512 c^2}{9 d^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {7 c}{d^3 \sqrt {d x^3+c}}-\frac {c^2}{9 d^3 \left (d x^3+c\right )^{3/2}}\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {1024 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{27 d^4}+\frac {2 c^2}{9 d^4 \sqrt {c+d x^3}}-\frac {12 c \sqrt {c+d x^3}}{d^4}-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d^4}\right )\)

Input:

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

Output:

((2*c^2)/(9*d^4*Sqrt[c + d*x^3]) - (12*c*Sqrt[c + d*x^3])/d^4 - (2*(c + d* 
x^3)^(3/2))/(3*d^4) + (1024*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/ 
(27*d^4))/3
 

Defintions of rubi rules used

rule 98

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x 
_)), x_] :> Int[ExpandIntegrand[(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 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {-\frac {2 d^{2} x^{6}}{9}+\frac {1024 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {d \,x^{3}+c}}{81}-\frac {40 c d \,x^{3}}{9}-\frac {112 c^{2}}{27}}{\sqrt {d \,x^{3}+c}\, d^{4}}\)	\(65\)
risch	\(-\frac {2 \left (d \,x^{3}+19 c \right ) \sqrt {d \,x^{3}+c}}{9 d^{4}}-\frac {c^{2} \left (-\frac {2}{27 d \sqrt {d \,x^{3}+c}}-\frac {1024 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{81 \sqrt {c}\, d}\right )}{d^{3}}\)	\(71\)
default	\(-\frac {-\frac {2 c^{2}}{3 d^{3} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d^{2}}-\frac {10 c \sqrt {d \,x^{3}+c}}{9 d^{3}}}{d}+\frac {128 c^{2}}{3 d^{4} \sqrt {d \,x^{3}+c}}-\frac {8 c \left (\frac {2 c}{3 d^{2} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}\right )}{d^{2}}+\frac {1024 c^{3} \left (-\frac {1}{c \sqrt {d \,x^{3}+c}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{27 d^{4}}\)	\(161\)
elliptic	\(\frac {2 c^{2}}{27 d^{4} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d^{3}}-\frac {38 c \sqrt {d \,x^{3}+c}}{9 d^{4}}-\frac {512 i c \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \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 )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \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 )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {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}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{18 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \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 )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{243 d^{6}}\)	\(465\)

Input:

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

Output:

2/81*(-9*d^2*x^6+512*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))*(d*x^3+c 
)^(1/2)-180*c*d*x^3-168*c^2)/(d*x^3+c)^(1/2)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.07 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\left [\frac {2 \, {\left (256 \, {\left (c d x^{3} + c^{2}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 3 \, {\left (3 \, d^{2} x^{6} + 60 \, c d x^{3} + 56 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} + c d^{4}\right )}}, -\frac {2 \, {\left (512 \, {\left (c d x^{3} + c^{2}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 3 \, {\left (3 \, d^{2} x^{6} + 60 \, c d x^{3} + 56 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} + c d^{4}\right )}}\right ] \] Input:

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

Output:

[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(3*sqrt(- 
c)/sqrt(d*x^3 + 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 [A] (verification not implemented)

Time = 26.75 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{2}}{27 \sqrt {c + d x^{3}}} - \frac {512 c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{81 \sqrt {- c}} - 2 c \sqrt {c + d x^{3}} - \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{9}\right )}{d^{4}} & \text {for}\: d \neq 0 \\\frac {x^{12}}{96 c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(c**2/(27*sqrt(c + d*x**3)) - 512*c**2*atan(sqrt(c + d*x**3)/ 
(3*sqrt(-c)))/(81*sqrt(-c)) - 2*c*sqrt(c + d*x**3) - (c + d*x**3)**(3/2)/9 
)/d**4, Ne(d, 0)), (x**12/(96*c**(5/2)), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (256 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 9 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 162 \, \sqrt {d x^{3} + c} c - \frac {3 \, c^{2}}{\sqrt {d x^{3} + c}}\right )}}{81 \, d^{4}} \] Input:

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

Output:

-2/81*(256*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3* 
sqrt(c))) + 9*(d*x^3 + c)^(3/2) + 162*sqrt(d*x^3 + c)*c - 3*c^2/sqrt(d*x^3 
 + c))/d^4
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {1024 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{81 \, \sqrt {-c} d^{4}} + \frac {2 \, c^{2}}{27 \, \sqrt {d x^{3} + c} d^{4}} - \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{8} + 18 \, \sqrt {d x^{3} + c} c d^{8}\right )}}{9 \, d^{12}} \] Input:

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

Output:

-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
 

Mupad [B] (verification not implemented)

Time = 2.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.06 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {512\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{81\,d^4}-\frac {38\,c\,\sqrt {d\,x^3+c}}{9\,d^4}+\frac {2\,c^2}{27\,d^4\,\sqrt {d\,x^3+c}}-\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^3} \] Input:

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

Output:

(512*c^(3/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3 
)))/(81*d^4) - (38*c*(c + d*x^3)^(1/2))/(9*d^4) + (2*c^2)/(27*d^4*(c + d*x 
^3)^(1/2)) - (2*x^3*(c + d*x^3)^(1/2))/(9*d^3)
 

Reduce [F]

\[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-\frac {80 \sqrt {d \,x^{3}+c}\, c^{2}}{9}-\frac {40 \sqrt {d \,x^{3}+c}\, c d \,x^{3}}{9}-\frac {2 \sqrt {d \,x^{3}+c}\, d^{2} x^{6}}{9}+64 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c^{3} d^{2}+64 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c^{2} d^{3} x^{3}}{d^{4} \left (d \,x^{3}+c \right )} \] Input:

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

Output:

(2*( - 40*sqrt(c + d*x**3)*c**2 - 20*sqrt(c + d*x**3)*c*d*x**3 - sqrt(c + 
d*x**3)*d**2*x**6 + 288*int((sqrt(c + d*x**3)*x**5)/(8*c**3 + 15*c**2*d*x* 
*3 + 6*c*d**2*x**6 - d**3*x**9),x)*c**3*d**2 + 288*int((sqrt(c + d*x**3)*x 
**5)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*c**2*d**3*x* 
*3))/(9*d**4*(c + d*x**3))