\(\int \frac {x^{11} (c+d x^3)^{3/2}}{8 c-d x^3} \, dx\) [470]

Optimal result

Integrand size = 27, antiderivative size = 130 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {3072 c^4 \sqrt {c+d x^3}}{d^4}-\frac {1024 c^3 \left (c+d x^3\right )^{3/2}}{9 d^4}-\frac {38 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {4 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac {2 \left (c+d x^3\right )^{9/2}}{27 d^4}+\frac {9216 c^{9/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4} \] Output:

-3072*c^4*(d*x^3+c)^(1/2)/d^4-1024/9*c^3*(d*x^3+c)^(3/2)/d^4-38/5*c^2*(d*x 
^3+c)^(5/2)/d^4-4/7*c*(d*x^3+c)^(7/2)/d^4-2/27*(d*x^3+c)^(9/2)/d^4+9216*c^ 
(9/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.72 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {2 \sqrt {c+d x^3} \left (1509176 c^4+61892 c^3 d x^3+4611 c^2 d^2 x^6+410 c d^3 x^9+35 d^4 x^{12}\right )}{945 d^4}+\frac {9216 c^{9/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4} \] Input:

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

Output:

(-2*Sqrt[c + d*x^3]*(1509176*c^4 + 61892*c^3*d*x^3 + 4611*c^2*d^2*x^6 + 41 
0*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
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {948, 99, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {1}{3} \int \left (-\frac {\left (d x^3+c\right )^{7/2}}{d^3}-\frac {6 c \left (d x^3+c\right )^{5/2}}{d^3}+\frac {512 c^3 \left (d x^3+c\right )^{3/2}}{d^3 \left (8 c-d x^3\right )}-\frac {57 c^2 \left (d x^3+c\right )^{3/2}}{d^3}\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {27648 c^{9/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}-\frac {9216 c^4 \sqrt {c+d x^3}}{d^4}-\frac {1024 c^3 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac {114 c^2 \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac {12 c \left (c+d x^3\right )^{7/2}}{7 d^4}-\frac {2 \left (c+d x^3\right )^{9/2}}{9 d^4}\right )\)

Input:

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

Output:

((-9216*c^4*Sqrt[c + d*x^3])/d^4 - (1024*c^3*(c + d*x^3)^(3/2))/(3*d^4) - 
(114*c^2*(c + d*x^3)^(5/2))/(5*d^4) - (12*c*(c + d*x^3)^(7/2))/(7*d^4) - ( 
2*(c + d*x^3)^(9/2))/(9*d^4) + (27648*c^(9/2)*ArcTanh[Sqrt[c + d*x^3]/(3*S 
qrt[c])])/d^4)/3
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(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 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 = 1.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62

method	result	size
pseudoelliptic	\(\frac {\frac {2 \left (-35 d^{4} x^{12}-410 c \,d^{3} x^{9}-4611 c^{2} x^{6} d^{2}-61892 c^{3} d \,x^{3}-1509176 c^{4}\right ) \sqrt {d \,x^{3}+c}}{945}+9216 c^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{d^{4}}\)	\(80\)
risch	\(-\frac {2 \left (35 d^{4} x^{12}+410 c \,d^{3} x^{9}+4611 c^{2} x^{6} d^{2}+61892 c^{3} d \,x^{3}+1509176 c^{4}\right ) \sqrt {d \,x^{3}+c}}{945 d^{4}}+\frac {9216 c^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{d^{4}}\)	\(82\)
default	\(-\frac {\frac {2 d \,x^{12} \sqrt {d \,x^{3}+c}}{27}+\frac {20 c \,x^{9} \sqrt {d \,x^{3}+c}}{189}+\frac {2 c^{2} x^{6} \sqrt {d \,x^{3}+c}}{315 d}-\frac {8 c^{3} x^{3} \sqrt {d \,x^{3}+c}}{945 d^{2}}+\frac {16 c^{4} \sqrt {d \,x^{3}+c}}{945 d^{3}}}{d}-\frac {128 c^{2} \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 d^{4}}-\frac {8 c \left (\frac {2 d \,x^{9} \sqrt {d \,x^{3}+c}}{21}+\frac {16 c \,x^{6} \sqrt {d \,x^{3}+c}}{105}+\frac {2 c^{2} x^{3} \sqrt {d \,x^{3}+c}}{105 d}-\frac {4 c^{3} \sqrt {d \,x^{3}+c}}{105 d^{2}}\right )}{d^{2}}+\frac {1024 c^{3} \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{9 d^{4}}\)	\(235\)
elliptic	\(-\frac {2 x^{12} \sqrt {d \,x^{3}+c}}{27}-\frac {164 c \,x^{9} \sqrt {d \,x^{3}+c}}{189 d}-\frac {3074 c^{2} x^{6} \sqrt {d \,x^{3}+c}}{315 d^{2}}-\frac {123784 c^{3} x^{3} \sqrt {d \,x^{3}+c}}{945 d^{3}}-\frac {3018352 c^{4} \sqrt {d \,x^{3}+c}}{945 d^{4}}-\frac {1536 i c^{4} \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 )}{d^{6}}\)	\(503\)

Input:

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

Output:

2/945*((-35*d^4*x^12-410*c*d^3*x^9-4611*c^2*d^2*x^6-61892*c^3*d*x^3-150917 
6*c^4)*(d*x^3+c)^(1/2)+4354560*c^(9/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2) 
))/d^4
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.45 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\left [\frac {2 \, {\left (2177280 \, c^{\frac {9}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - {\left (35 \, d^{4} x^{12} + 410 \, c d^{3} x^{9} + 4611 \, c^{2} d^{2} x^{6} + 61892 \, c^{3} d x^{3} + 1509176 \, c^{4}\right )} \sqrt {d x^{3} + c}\right )}}{945 \, d^{4}}, -\frac {2 \, {\left (4354560 \, \sqrt {-c} c^{4} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + {\left (35 \, d^{4} x^{12} + 410 \, c d^{3} x^{9} + 4611 \, c^{2} d^{2} x^{6} + 61892 \, c^{3} d x^{3} + 1509176 \, c^{4}\right )} \sqrt {d x^{3} + c}\right )}}{945 \, d^{4}}\right ] \] Input:

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

Output:

[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*ar 
ctan(3*sqrt(-c)/sqrt(d*x^3 + 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]
 

Sympy [A] (verification not implemented)

Time = 55.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.10 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\begin {cases} \frac {2 \left (- \frac {4608 c^{5} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{d^{3} \sqrt {- c}} - \frac {1536 c^{4} \sqrt {c + d x^{3}}}{d^{3}} - \frac {512 c^{3} \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 d^{3}} - \frac {19 c^{2} \left (c + d x^{3}\right )^{\frac {5}{2}}}{5 d^{3}} - \frac {2 c \left (c + d x^{3}\right )^{\frac {7}{2}}}{7 d^{3}} - \frac {\left (c + d x^{3}\right )^{\frac {9}{2}}}{27 d^{3}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{12}}{96} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(-4608*c**5*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(d**3*sqrt(-c 
)) - 1536*c**4*sqrt(c + d*x**3)/d**3 - 512*c**3*(c + d*x**3)**(3/2)/(9*d** 
3) - 19*c**2*(c + d*x**3)**(5/2)/(5*d**3) - 2*c*(c + d*x**3)**(7/2)/(7*d** 
3) - (c + d*x**3)**(9/2)/(27*d**3))/d, Ne(d, 0)), (sqrt(c)*x**12/96, True) 
)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {2 \, {\left (2177280 \, c^{\frac {9}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 35 \, {\left (d x^{3} + c\right )}^{\frac {9}{2}} + 270 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} c + 3591 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c^{2} + 53760 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{3} + 1451520 \, \sqrt {d x^{3} + c} c^{4}\right )}}{945 \, d^{4}} \] Input:

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

Output:

-2/945*(2177280*c^(9/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) 
 + 3*sqrt(c))) + 35*(d*x^3 + c)^(9/2) + 270*(d*x^3 + c)^(7/2)*c + 3591*(d* 
x^3 + c)^(5/2)*c^2 + 53760*(d*x^3 + c)^(3/2)*c^3 + 1451520*sqrt(d*x^3 + c) 
*c^4)/d^4
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.90 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {9216 \, c^{5} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{4}} - \frac {2 \, {\left (35 \, {\left (d x^{3} + c\right )}^{\frac {9}{2}} d^{32} + 270 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} c d^{32} + 3591 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c^{2} d^{32} + 53760 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{3} d^{32} + 1451520 \, \sqrt {d x^{3} + c} c^{4} d^{32}\right )}}{945 \, d^{36}} \] Input:

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

Output:

-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
 

Mupad [B] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.04 \[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\frac {4608\,c^{9/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^4}-\frac {2\,x^{12}\,\sqrt {d\,x^3+c}}{27}-\frac {3018352\,c^4\,\sqrt {d\,x^3+c}}{945\,d^4}-\frac {164\,c\,x^9\,\sqrt {d\,x^3+c}}{189\,d}-\frac {123784\,c^3\,x^3\,\sqrt {d\,x^3+c}}{945\,d^3}-\frac {3074\,c^2\,x^6\,\sqrt {d\,x^3+c}}{315\,d^2} \] Input:

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

Output:

(4608*c^(9/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^ 
3)))/d^4 - (2*x^12*(c + d*x^3)^(1/2))/27 - (3018352*c^4*(c + d*x^3)^(1/2)) 
/(945*d^4) - (164*c*x^9*(c + d*x^3)^(1/2))/(189*d) - (123784*c^3*x^3*(c + 
d*x^3)^(1/2))/(945*d^3) - (3074*c^2*x^6*(c + d*x^3)^(1/2))/(315*d^2)
 

Reduce [F]

\[ \int \frac {x^{11} \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\frac {\frac {247568 \sqrt {d \,x^{3}+c}\, c^{4}}{945}-\frac {123784 \sqrt {d \,x^{3}+c}\, c^{3} d \,x^{3}}{945}-\frac {3074 \sqrt {d \,x^{3}+c}\, c^{2} d^{2} x^{6}}{315}-\frac {164 \sqrt {d \,x^{3}+c}\, c \,d^{3} x^{9}}{189}-\frac {2 \sqrt {d \,x^{3}+c}\, d^{4} x^{12}}{27}+5184 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{2} x^{6}+7 c d \,x^{3}+8 c^{2}}d x \right ) c^{4} d^{2}}{d^{4}} \] Input:

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

Output:

(2*(123784*sqrt(c + d*x**3)*c**4 - 61892*sqrt(c + d*x**3)*c**3*d*x**3 - 46 
11*sqrt(c + d*x**3)*c**2*d**2*x**6 - 410*sqrt(c + d*x**3)*c*d**3*x**9 - 35 
*sqrt(c + d*x**3)*d**4*x**12 + 2449440*int((sqrt(c + d*x**3)*x**5)/(8*c**2 
 + 7*c*d*x**3 - d**2*x**6),x)*c**4*d**2))/(945*d**4)