\(\int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx\) [58]

Optimal result

Integrand size = 29, antiderivative size = 140 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=-\frac {\left (2 a c^2-d^2\right ) x^3}{3 b^2 c^3}+\frac {2 d \left (2 a c^2-d^2\right ) \sqrt {a+b x^3}}{3 b^3 c^4}-\frac {2 d \left (a+b x^3\right )^{3/2}}{9 b^3 c^2}+\frac {\left (a+b x^3\right )^2}{6 b^3 c}+\frac {2 \left (a c^2-d^2\right )^2 \log \left (d+c \sqrt {a+b x^3}\right )}{3 b^3 c^5} \] Output:

-1/3*(2*a*c^2-d^2)*x^3/b^2/c^3+2/3*d*(2*a*c^2-d^2)*(b*x^3+a)^(1/2)/b^3/c^4 
-2/9*d*(b*x^3+a)^(3/2)/b^3/c^2+1/6*(b*x^3+a)^2/b^3/c+2/3*(a*c^2-d^2)^2*ln( 
d+c*(b*x^3+a)^(1/2))/b^3/c^5
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {3 c^2 \left (a+b x^3\right ) \left (-3 a c^2+2 d^2+b c^2 x^3\right )-4 c d \sqrt {a+b x^3} \left (-5 a c^2+3 d^2+b c^2 x^3\right )+12 \left (-a c^2+d^2\right )^2 \log \left (d+c \sqrt {a+b x^3}\right )}{18 b^3 c^5} \] Input:

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

Output:

(3*c^2*(a + b*x^3)*(-3*a*c^2 + 2*d^2 + b*c^2*x^3) - 4*c*d*Sqrt[a + b*x^3]* 
(-5*a*c^2 + 3*d^2 + b*c^2*x^3) + 12*(-(a*c^2) + d^2)^2*Log[d + c*Sqrt[a + 
b*x^3]])/(18*b^3*c^5)
 

Rubi [A] (warning: unable to verify)

Time = 0.84 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2586, 7267, 476, 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.

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

\(\Big \downarrow \) 2586

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

\(\Big \downarrow \) 7267

\(\displaystyle \frac {2 \int \frac {\left (a-x^6\right )^2}{\sqrt {b x^3+a} c+d}d\sqrt {b x^3+a}}{3 b^3}\)

\(\Big \downarrow \) 476

\(\displaystyle \frac {2 \int \left (\frac {x^9}{c}-\frac {d x^6}{c^2}+\frac {2 a c^2 d-d^3}{c^4}-\frac {\left (2 a c^2-d^2\right ) \sqrt {b x^3+a}}{c^3}+\frac {\left (a c^2-d^2\right )^2}{c^4 \left (\sqrt {b x^3+a} c+d\right )}\right )d\sqrt {b x^3+a}}{3 b^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {\left (a c^2-d^2\right )^2 \log \left (c \sqrt {a+b x^3}+d\right )}{c^5}+\frac {d \sqrt {a+b x^3} \left (2 a c^2-d^2\right )}{c^4}-\frac {x^6 \left (2 a c^2-d^2\right )}{2 c^3}-\frac {d x^9}{3 c^2}+\frac {x^{12}}{4 c}\right )}{3 b^3}\)

Input:

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

Output:

(2*(-1/2*((2*a*c^2 - d^2)*x^6)/c^3 - (d*x^9)/(3*c^2) + x^12/(4*c) + (d*(2* 
a*c^2 - d^2)*Sqrt[a + b*x^3])/c^4 + ((a*c^2 - d^2)^2*Log[d + c*Sqrt[a + b* 
x^3]])/c^5))/(3*b^3)
 

Defintions of rubi rules used

rule 476

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, 
 x] && IGtQ[p, 0]
 

rule 2009

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

rule 2586

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] 
), x_Symbol] :> Simp[1/n   Subst[Int[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a 
+ b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c - a* 
d, 0] && IntegerQ[(m + 1)/n]
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(124)=248\).

Time = 0.66 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.37

method	result	size
default	\(d \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}}}{9 b^{3} c^{2}}-\frac {\left (a^{2} c^{4}-2 c^{2} a \,d^{2}+d^{4}\right ) \left (2 c \sqrt {b \,x^{3}+a}+d \ln \left (c \sqrt {b \,x^{3}+a}-d \right )-d \ln \left (d +c \sqrt {b \,x^{3}+a}\right )\right )}{3 b^{3} c^{5} d^{2}}+\frac {2 a^{2} \sqrt {b \,x^{3}+a}}{3 b^{3} d^{2}}\right )-a c \left (-\frac {x^{3}}{3 c^{2} b^{2}}+\frac {\left (-a^{2} c^{4}+2 c^{2} a \,d^{2}-d^{4}\right ) \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 b^{3} c^{4} d^{2}}+\frac {a^{2} \ln \left (b \,x^{3}+a \right )}{3 b^{3} d^{2}}\right )-b c \left (-\frac {\frac {1}{2} b \,x^{6} c^{2}-2 a \,c^{2} x^{3}+x^{3} d^{2}}{3 b^{3} c^{4}}+\frac {\left (a^{3} c^{6}-3 a^{2} c^{4} d^{2}+3 a \,c^{2} d^{4}-d^{6}\right ) \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 b^{4} c^{6} d^{2}}-\frac {a^{3} \ln \left (b \,x^{3}+a \right )}{3 b^{4} d^{2}}\right )\)	\(332\)
elliptic	\(\frac {\sqrt {b \,x^{3}+a}\, \left (d +c \sqrt {b \,x^{3}+a}\right ) \left (c \left (\frac {\frac {1}{2} b \,x^{6} c^{2}-a \,c^{2} x^{3}+x^{3} d^{2}}{3 b^{2} c^{4}}+\frac {\left (a^{2} c^{4}-2 c^{2} a \,d^{2}+d^{4}\right ) \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 b^{3} c^{6}}\right )-\frac {2 d \,x^{3} \sqrt {b \,x^{3}+a}}{9 b^{2} c^{2}}+\frac {2 \left (\frac {\left (a \,c^{2}-d^{2}\right ) d}{b^{2} c^{4}}+\frac {2 d a}{3 b^{2} c^{2}}\right ) \sqrt {b \,x^{3}+a}}{3 b}-\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right )}{\sum }\frac {\left (a^{2} c^{4}-2 c^{2} a \,d^{2}+d^{4}\right ) \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i b \left (2 x +\frac {-i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {b \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{-3 \left (-a \,b^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i b \left (2 x +\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{2 \left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha b -i \left (-a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} b^{2}-\left (-a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha b -\left (-a \,b^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {c^{2} \left (2 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} b -i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, a b -3 \left (-a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 a b \right )}{2 b \,d^{2}}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{2 \sqrt {b \,x^{3}+a}}\right )}{3 d \,b^{5} c^{4}}\right )}{a c +b c \,x^{3}+d \sqrt {b \,x^{3}+a}}\)	\(646\)

Input:

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

Output:

d*(-2/9/b^3/c^2*(b*x^3+a)^(3/2)-1/3*(a^2*c^4-2*a*c^2*d^2+d^4)/b^3/c^5/d^2* 
(2*c*(b*x^3+a)^(1/2)+d*ln(c*(b*x^3+a)^(1/2)-d)-d*ln(d+c*(b*x^3+a)^(1/2)))+ 
2/3*a^2/b^3/d^2*(b*x^3+a)^(1/2))-a*c*(-1/3/c^2/b^2*x^3+1/3*(-a^2*c^4+2*a*c 
^2*d^2-d^4)/b^3/c^4/d^2*ln(b*c^2*x^3+a*c^2-d^2)+1/3*a^2/b^3/d^2*ln(b*x^3+a 
))-b*c*(-1/3/b^3/c^4*(1/2*b*x^6*c^2-2*a*c^2*x^3+x^3*d^2)+1/3/b^4/c^6*(a^3* 
c^6-3*a^2*c^4*d^2+3*a*c^2*d^4-d^6)/d^2*ln(b*c^2*x^3+a*c^2-d^2)-1/3/b^4*a^3 
/d^2*ln(b*x^3+a))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.36 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {3 \, b^{2} c^{4} x^{6} - 6 \, {\left (a b c^{4} - b c^{2} d^{2}\right )} x^{3} + 6 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + 6 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt {b x^{3} + a} c + d\right ) - 6 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt {b x^{3} + a} c - d\right ) - 4 \, {\left (b c^{3} d x^{3} - 5 \, a c^{3} d + 3 \, c d^{3}\right )} \sqrt {b x^{3} + a}}{18 \, b^{3} c^{5}} \] Input:

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

Output:

1/18*(3*b^2*c^4*x^6 - 6*(a*b*c^4 - b*c^2*d^2)*x^3 + 6*(a^2*c^4 - 2*a*c^2*d 
^2 + d^4)*log(b*c^2*x^3 + a*c^2 - d^2) + 6*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*l 
og(sqrt(b*x^3 + a)*c + d) - 6*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(sqrt(b*x^3 
 + a)*c - d) - 4*(b*c^3*d*x^3 - 5*a*c^3*d + 3*c*d^3)*sqrt(b*x^3 + a))/(b^3 
*c^5)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {\frac {3 \, {\left (b x^{3} + a\right )}^{2} c^{3} - 4 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} c^{2} d - 6 \, {\left (2 \, a c^{3} - c d^{2}\right )} {\left (b x^{3} + a\right )} + 12 \, {\left (2 \, a c^{2} d - d^{3}\right )} \sqrt {b x^{3} + a}}{c^{4}} + \frac {12 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt {b x^{3} + a} c + d\right )}{c^{5}}}{18 \, b^{3}} \] Input:

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

Output:

1/18*((3*(b*x^3 + a)^2*c^3 - 4*(b*x^3 + a)^(3/2)*c^2*d - 6*(2*a*c^3 - c*d^ 
2)*(b*x^3 + a) + 12*(2*a*c^2*d - d^3)*sqrt(b*x^3 + a))/c^4 + 12*(a^2*c^4 - 
 2*a*c^2*d^2 + d^4)*log(sqrt(b*x^3 + a)*c + d)/c^5)/b^3
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.14 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {\frac {12 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left ({\left | \sqrt {b x^{3} + a} c + d \right |}\right )}{b c^{5}} + \frac {3 \, {\left (b x^{3} + a\right )}^{2} b^{3} c^{3} - 12 \, {\left (b x^{3} + a\right )} a b^{3} c^{3} - 4 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3} c^{2} d + 24 \, \sqrt {b x^{3} + a} a b^{3} c^{2} d + 6 \, {\left (b x^{3} + a\right )} b^{3} c d^{2} - 12 \, \sqrt {b x^{3} + a} b^{3} d^{3}}{b^{4} c^{4}}}{18 \, b^{2}} \] Input:

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

Output:

1/18*(12*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(abs(sqrt(b*x^3 + a)*c + d))/(b* 
c^5) + (3*(b*x^3 + a)^2*b^3*c^3 - 12*(b*x^3 + a)*a*b^3*c^3 - 4*(b*x^3 + a) 
^(3/2)*b^3*c^2*d + 24*sqrt(b*x^3 + a)*a*b^3*c^2*d + 6*(b*x^3 + a)*b^3*c*d^ 
2 - 12*sqrt(b*x^3 + a)*b^3*d^3)/(b^4*c^4))/b^2
 

Mupad [B] (verification not implemented)

Time = 23.69 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.43 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {\left (\frac {2\,d\,\left (a\,c^2-d^2\right )}{b^2\,c^4}+\frac {4\,a\,d}{3\,b^2\,c^2}\right )\,\sqrt {b\,x^3+a}}{3\,b}+\frac {x^6}{6\,b\,c}-\frac {x^3\,\left (a\,c^2-d^2\right )}{3\,b^2\,c^3}+\frac {\ln \left (\frac {d+c\,\sqrt {b\,x^3+a}}{d-c\,\sqrt {b\,x^3+a}}\right )\,{\left (a\,c^2-d^2\right )}^2}{3\,b^3\,c^5}+\frac {\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )\,\left (a^2\,c^4-2\,a\,c^2\,d^2+d^4\right )}{3\,b^3\,c^5}-\frac {2\,d\,x^3\,\sqrt {b\,x^3+a}}{9\,b^2\,c^2} \] Input:

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

Output:

(((2*d*(a*c^2 - d^2))/(b^2*c^4) + (4*a*d)/(3*b^2*c^2))*(a + b*x^3)^(1/2))/ 
(3*b) + x^6/(6*b*c) - (x^3*(a*c^2 - d^2))/(3*b^2*c^3) + (log((d + c*(a + b 
*x^3)^(1/2))/(d - c*(a + b*x^3)^(1/2)))*(a*c^2 - d^2)^2)/(3*b^3*c^5) + (lo 
g(a*c^2 - d^2 + b*c^2*x^3)*(d^4 + a^2*c^4 - 2*a*c^2*d^2))/(3*b^3*c^5) - (2 
*d*x^3*(a + b*x^3)^(1/2))/(9*b^2*c^2)
 

Reduce [F]

\[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\int \frac {x^{8}}{a c +b c \,x^{3}+d \sqrt {b \,x^{3}+a}}d x \] Input:

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

Output:

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