\(\int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx\) [434]

Optimal result

Integrand size = 26, antiderivative size = 97 \[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {32 c^2 \sqrt {c+d x^3}}{3 d^3}-\frac {10 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {32 c^{5/2} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^3} \] Output:

32/3*c^2*(d*x^3+c)^(1/2)/d^3-10/9*c*(d*x^3+c)^(3/2)/d^3+2/15*(d*x^3+c)^(5/ 
2)/d^3-32/3*c^(5/2)*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))*3^(1/2)/d^ 
3
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {2 \sqrt {c+d x^3} \left (218 c^2-19 c d x^3+3 d^2 x^6\right )}{45 d^3}-\frac {32 c^{5/2} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^3} \] Input:

Integrate[(x^8*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
 

Output:

(2*Sqrt[c + d*x^3]*(218*c^2 - 19*c*d*x^3 + 3*d^2*x^6))/(45*d^3) - (32*c^(5 
/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(Sqrt[3]*d^3)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, 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.

\(\displaystyle \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^6 \sqrt {d x^3+c}}{d x^3+4 c}dx^3\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {1}{3} \int \left (\frac {16 \sqrt {d x^3+c} c^2}{d^2 \left (d x^3+4 c\right )}-\frac {5 \sqrt {d x^3+c} c}{d^2}+\frac {\left (d x^3+c\right )^{3/2}}{d^2}\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (-\frac {32 \sqrt {3} c^{5/2} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{d^3}+\frac {32 c^2 \sqrt {c+d x^3}}{d^3}-\frac {10 c \left (c+d x^3\right )^{3/2}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{5/2}}{5 d^3}\right )\)

Input:

Int[(x^8*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
 

Output:

((32*c^2*Sqrt[c + d*x^3])/d^3 - (10*c*(c + d*x^3)^(3/2))/(3*d^3) + (2*(c + 
 d*x^3)^(5/2))/(5*d^3) - (32*Sqrt[3]*c^(5/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[ 
3]*Sqrt[c])])/d^3)/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 = 10.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.66

method	result	size
pseudoelliptic	\(\frac {-480 c^{\frac {5}{2}} \sqrt {3}\, \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right )+\left (6 d^{2} x^{6}-38 c d \,x^{3}+436 c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d^{3}}\)	\(64\)
risch	\(\frac {2 \left (3 d^{2} x^{6}-19 c d \,x^{3}+218 c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d^{3}}-\frac {32 c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) \sqrt {3}}{3 d^{3}}\)	\(66\)
default	\(\frac {\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15}+\frac {2 c \,x^{3} \sqrt {d \,x^{3}+c}}{45 d}-\frac {4 c^{2} \sqrt {d \,x^{3}+c}}{45 d^{2}}}{d}-\frac {8 c \left (d \,x^{3}+c \right )^{\frac {3}{2}}}{9 d^{3}}+\frac {16 c^{2} \left (2 \sqrt {d \,x^{3}+c}-2 \sqrt {c}\, \sqrt {3}\, \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right )\right )}{3 d^{3}}\)	\(117\)
elliptic	\(\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15 d}-\frac {38 c \,x^{3} \sqrt {d \,x^{3}+c}}{45 d^{2}}+\frac {436 c^{2} \sqrt {d \,x^{3}+c}}{45 d^{3}}+\frac {16 i c^{2} \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+4 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}{6 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 )}{3 d^{5}}\)	\(466\)

Input:

int(x^8*(d*x^3+c)^(1/2)/(d*x^3+4*c),x,method=_RETURNVERBOSE)
 

Output:

1/45*(-480*c^(5/2)*3^(1/2)*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))+(6* 
d^2*x^6-38*c*d*x^3+436*c^2)*(d*x^3+c)^(1/2))/d^3
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.60 \[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\left [\frac {2 \, {\left (360 \, \sqrt {\frac {1}{3}} \sqrt {-c} c^{2} \log \left (\frac {d x^{3} - 6 \, \sqrt {\frac {1}{3}} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) + {\left (3 \, d^{2} x^{6} - 19 \, c d x^{3} + 218 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, d^{3}}, \frac {2 \, {\left (720 \, \sqrt {\frac {1}{3}} c^{\frac {5}{2}} \arctan \left (\frac {3 \, \sqrt {\frac {1}{3}} \sqrt {c}}{\sqrt {d x^{3} + c}}\right ) + {\left (3 \, d^{2} x^{6} - 19 \, c d x^{3} + 218 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, d^{3}}\right ] \] Input:

integrate(x^8*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")
 

Output:

[2/45*(360*sqrt(1/3)*sqrt(-c)*c^2*log((d*x^3 - 6*sqrt(1/3)*sqrt(d*x^3 + c) 
*sqrt(-c) - 2*c)/(d*x^3 + 4*c)) + (3*d^2*x^6 - 19*c*d*x^3 + 218*c^2)*sqrt( 
d*x^3 + c))/d^3, 2/45*(720*sqrt(1/3)*c^(5/2)*arctan(3*sqrt(1/3)*sqrt(c)/sq 
rt(d*x^3 + c)) + (3*d^2*x^6 - 19*c*d*x^3 + 218*c^2)*sqrt(d*x^3 + c))/d^3]
 

Sympy [A] (verification not implemented)

Time = 9.46 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\begin {cases} \frac {2 \left (- \frac {16 \sqrt {3} c^{\frac {5}{2}} \operatorname {atan}{\left (\frac {\sqrt {3} \sqrt {c + d x^{3}}}{3 \sqrt {c}} \right )}}{3} + \frac {16 c^{2} \sqrt {c + d x^{3}}}{3} - \frac {5 c \left (c + d x^{3}\right )^{\frac {3}{2}}}{9} + \frac {\left (c + d x^{3}\right )^{\frac {5}{2}}}{15}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{9}}{36 \sqrt {c}} & \text {otherwise} \end {cases} \] Input:

integrate(x**8*(d*x**3+c)**(1/2)/(d*x**3+4*c),x)
 

Output:

Piecewise((2*(-16*sqrt(3)*c**(5/2)*atan(sqrt(3)*sqrt(c + d*x**3)/(3*sqrt(c 
)))/3 + 16*c**2*sqrt(c + d*x**3)/3 - 5*c*(c + d*x**3)**(3/2)/9 + (c + d*x* 
*3)**(5/2)/15)/d**3, Ne(d, 0)), (x**9/(36*sqrt(c)), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.71 \[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=-\frac {2 \, {\left (240 \, \sqrt {3} c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) - 3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} + 25 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c - 240 \, \sqrt {d x^{3} + c} c^{2}\right )}}{45 \, d^{3}} \] Input:

integrate(x^8*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")
 

Output:

-2/45*(240*sqrt(3)*c^(5/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) - 3 
*(d*x^3 + c)^(5/2) + 25*(d*x^3 + c)^(3/2)*c - 240*sqrt(d*x^3 + c)*c^2)/d^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=-\frac {32 \, \sqrt {3} c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{3 \, d^{3}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} d^{12} - 25 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c d^{12} + 240 \, \sqrt {d x^{3} + c} c^{2} d^{12}\right )}}{45 \, d^{15}} \] Input:

integrate(x^8*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")
 

Output:

-32/3*sqrt(3)*c^(5/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/d^3 + 2/ 
45*(3*(d*x^3 + c)^(5/2)*d^12 - 25*(d*x^3 + c)^(3/2)*c*d^12 + 240*sqrt(d*x^ 
3 + c)*c^2*d^12)/d^15
 

Mupad [B] (verification not implemented)

Time = 3.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.12 \[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {436\,c^2\,\sqrt {d\,x^3+c}}{45\,d^3}+\frac {2\,x^6\,\sqrt {d\,x^3+c}}{15\,d}-\frac {38\,c\,x^3\,\sqrt {d\,x^3+c}}{45\,d^2}+\frac {\sqrt {3}\,c^{5/2}\,\ln \left (\frac {2\,\sqrt {3}\,c-\sqrt {3}\,d\,x^3+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,16{}\mathrm {i}}{3\,d^3} \] Input:

int((x^8*(c + d*x^3)^(1/2))/(4*c + d*x^3),x)
 

Output:

(436*c^2*(c + d*x^3)^(1/2))/(45*d^3) + (2*x^6*(c + d*x^3)^(1/2))/(15*d) - 
(38*c*x^3*(c + d*x^3)^(1/2))/(45*d^2) + (3^(1/2)*c^(5/2)*log((2*3^(1/2)*c 
+ c^(1/2)*(c + d*x^3)^(1/2)*6i - 3^(1/2)*d*x^3)/(4*c + d*x^3))*16i)/(3*d^3 
)
 

Reduce [F]

\[ \int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {\frac {76 \sqrt {d \,x^{3}+c}\, c^{2}}{45}-\frac {38 \sqrt {d \,x^{3}+c}\, c d \,x^{3}}{45}+\frac {2 \sqrt {d \,x^{3}+c}\, d^{2} x^{6}}{15}+12 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \right ) c^{2} d^{2}}{d^{3}} \] Input:

int(x^8*(d*x^3+c)^(1/2)/(d*x^3+4*c),x)
 

Output:

(2*(38*sqrt(c + d*x**3)*c**2 - 19*sqrt(c + d*x**3)*c*d*x**3 + 3*sqrt(c + d 
*x**3)*d**2*x**6 + 270*int((sqrt(c + d*x**3)*x**5)/(4*c**2 + 5*c*d*x**3 + 
d**2*x**6),x)*c**2*d**2))/(45*d**3)