\(\int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\) [788]

Optimal result

Integrand size = 28, antiderivative size = 268 \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {7 a x^2 \sqrt [3]{a+b x^3}}{18 b^2 d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}+\frac {11 a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3} d}-\frac {\sqrt [3]{2} a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{8/3} d}+\frac {a^2 \log \left (a d-b d x^3\right )}{3\ 2^{2/3} b^{8/3} d}+\frac {11 a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{8/3} d}-\frac {a^2 \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{8/3} d} \] Output:

-7/18*a*x^2*(b*x^3+a)^(1/3)/b^2/d-1/6*x^5*(b*x^3+a)^(1/3)/b/d+11/27*a^2*ar 
ctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(8/3)/d-1/3*2^ 
(1/3)*a^2*arctan(1/3*(1+2*2^(1/3)*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1 
/2)/b^(8/3)/d+1/6*a^2*ln(-b*d*x^3+a*d)*2^(1/3)/b^(8/3)/d+11/18*a^2*ln(b^(1 
/3)*x-(b*x^3+a)^(1/3))/b^(8/3)/d-1/2*a^2*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1 
/3))*2^(1/3)/b^(8/3)/d
 

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.22 \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {21 a b^{2/3} x^2 \sqrt [3]{a+b x^3}+9 b^{5/3} x^5 \sqrt [3]{a+b x^3}-22 \sqrt {3} a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+18 \sqrt [3]{2} \sqrt {3} a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )-22 a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+18 \sqrt [3]{2} a^2 \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+11 a^2 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-9 \sqrt [3]{2} a^2 \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{54 b^{8/3} d} \] Input:

Integrate[(x^7*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
 

Output:

-1/54*(21*a*b^(2/3)*x^2*(a + b*x^3)^(1/3) + 9*b^(5/3)*x^5*(a + b*x^3)^(1/3 
) - 22*Sqrt[3]*a^2*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^( 
1/3))] + 18*2^(1/3)*Sqrt[3]*a^2*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2^ 
(2/3)*(a + b*x^3)^(1/3))] - 22*a^2*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + 
 18*2^(1/3)*a^2*Log[-2*b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3)] + 11*a^2*Log 
[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)] - 9*2^(1/3 
)*a^2*Log[2*b^(2/3)*x^2 + 2^(2/3)*b^(1/3)*x*(a + b*x^3)^(1/3) + 2^(1/3)*(a 
 + b*x^3)^(2/3)])/(b^(8/3)*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {978, 27, 1052, 27, 1054, 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.

\(\displaystyle \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\)

\(\Big \downarrow \) 978

\(\displaystyle \frac {\int \frac {a x^4 \left (7 b x^3+5 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \int \frac {x^4 \left (7 b x^3+5 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\)

\(\Big \downarrow \) 1052

\(\displaystyle \frac {a \left (\frac {\int \frac {2 a b x \left (11 b x^3+7 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{3 b^2}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \left (\frac {2 a \int \frac {x \left (11 b x^3+7 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{3 b}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\)

\(\Big \downarrow \) 1054

\(\displaystyle \frac {a \left (\frac {2 a \int \left (\frac {18 a x}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}-\frac {11 x}{\left (b x^3+a\right )^{2/3}}\right )dx}{3 b}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \left (\frac {2 a \left (\frac {11 \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {3 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{b^{2/3}}+\frac {3 \log \left (a-b x^3\right )}{2^{2/3} b^{2/3}}+\frac {11 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}-\frac {9 \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{2/3}}\right )}{3 b}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\)

Input:

Int[(x^7*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
 

Output:

-1/6*(x^5*(a + b*x^3)^(1/3))/(b*d) + (a*((-7*x^2*(a + b*x^3)^(1/3))/(3*b) 
+ (2*a*((11*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3 
]*b^(2/3)) - (3*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x)/(a + b*x 
^3)^(1/3))/Sqrt[3]])/b^(2/3) + (3*Log[a - b*x^3])/(2^(2/3)*b^(2/3)) + (11* 
Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*b^(2/3)) - (9*Log[2^(1/3)*b^(1/3)*x 
 - (a + b*x^3)^(1/3)])/(2^(2/3)*b^(2/3))))/(3*b)))/(6*b*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 978

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* 
((c + d*x^n)^q/(b*(m + n*(p + q) + 1))), x] - Simp[e^n/(b*(m + n*(p + q) + 
1))   Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 
 1) + (a*d*(m - n + 1) - n*q*(b*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c 
, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] && GtQ[m - n 
 + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
 

rule 1052

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m 
- n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q + 1) + 
 1))), x] - Simp[g^n/(b*d*(m + n*(p + q + 1) + 1))   Int[(g*x)^(m - n)*(a + 
 b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*( 
f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x] /; FreeQ[ 
{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]
 

rule 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

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

Maple [A] (verified)

Time = 7.09 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\frac {-3 \left (3 b \,x^{3}+7 a \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2} b^{\frac {8}{3}}+a^{2} b^{2} \left (9 \left (2 \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\ln \left (\frac {b^{\frac {2}{3}} 2^{\frac {2}{3}} x^{2}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {1}{3}} 2^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right ) 2^{\frac {1}{3}}-22 \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}-11 \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )+22 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{54 b^{\frac {14}{3}} d}\)	\(242\)

Input:

int(x^7*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x,method=_RETURNVERBOSE)
 

Output:

1/54*(-3*(3*b*x^3+7*a)*(b*x^3+a)^(1/3)*x^2*b^(8/3)+a^2*b^2*(9*(2*arctan(1/ 
3*3^(1/2)*(2^(2/3)/b^(1/3)*(b*x^3+a)^(1/3)+x)/x)*3^(1/2)+ln((b^(2/3)*2^(2/ 
3)*x^2+(b*x^3+a)^(1/3)*b^(1/3)*2^(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2*ln((-2^(1 
/3)*b^(1/3)*x+(b*x^3+a)^(1/3))/x))*2^(1/3)-22*arctan(1/3*3^(1/2)*(2*(b*x^3 
+a)^(1/3)/b^(1/3)+x)/x)*3^(1/2)-11*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3) 
*x+(b*x^3+a)^(2/3))/x^2)+22*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)))/b^(14/3)/ 
d
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.34 \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {18 \, \sqrt {3} 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} + \sqrt {3} x}{3 \, x}\right ) - 18 \cdot 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 9 \cdot 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {2}{3}} b^{2} x^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 66 \, \sqrt {\frac {1}{3}} a^{2} {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left ({\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{b^{2} x}\right ) - 22 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + 11 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + 3 \, {\left (3 \, b^{3} x^{5} + 7 \, a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{4} d} \] Input:

integrate(x^7*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="fricas")
 

Output:

-1/54*(18*sqrt(3)*2^(1/3)*a^2*b^2*(-1/b^2)^(1/3)*arctan(1/3*(sqrt(3)*2^(2/ 
3)*(b*x^3 + a)^(1/3)*b*(-1/b^2)^(2/3) + sqrt(3)*x)/x) - 18*2^(1/3)*a^2*b^2 
*(-1/b^2)^(1/3)*log((2^(1/3)*b*x*(-1/b^2)^(1/3) + (b*x^3 + a)^(1/3))/x) + 
9*2^(1/3)*a^2*b^2*(-1/b^2)^(1/3)*log((2^(2/3)*b^2*x^2*(-1/b^2)^(2/3) - 2^( 
1/3)*(b*x^3 + a)^(1/3)*b*x*(-1/b^2)^(1/3) + (b*x^3 + a)^(2/3))/x^2) + 66*s 
qrt(1/3)*a^2*(b^2)^(1/6)*b*arctan(sqrt(1/3)*((b^2)^(1/3)*b*x + 2*(b*x^3 + 
a)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6)/(b^2*x)) - 22*a^2*(b^2)^(2/3)*log(-((b^2 
)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + 11*a^2*(b^2)^(2/3)*log(((b^2)^(1/3)* 
b*x^2 + (b*x^3 + a)^(1/3)*(b^2)^(2/3)*x + (b*x^3 + a)^(2/3)*b)/x^2) + 3*(3 
*b^3*x^5 + 7*a*b^2*x^2)*(b*x^3 + a)^(1/3))/(b^4*d)
 

Sympy [F]

\[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=- \frac {\int \frac {x^{7} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx}{d} \] Input:

integrate(x**7*(b*x**3+a)**(1/3)/(-b*d*x**3+a*d),x)
 

Output:

-Integral(x**7*(a + b*x**3)**(1/3)/(-a + b*x**3), x)/d
 

Maxima [F]

\[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{7}}{b d x^{3} - a d} \,d x } \] Input:

integrate(x^7*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="maxima")
 

Output:

-integrate((b*x^3 + a)^(1/3)*x^7/(b*d*x^3 - a*d), x)
 

Giac [F]

\[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{7}}{b d x^{3} - a d} \,d x } \] Input:

integrate(x^7*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="giac")
 

Output:

integrate(-(b*x^3 + a)^(1/3)*x^7/(b*d*x^3 - a*d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int \frac {x^7\,{\left (b\,x^3+a\right )}^{1/3}}{a\,d-b\,d\,x^3} \,d x \] Input:

int((x^7*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x)
 

Output:

int((x^7*(a + b*x^3)^(1/3))/(a*d - b*d*x^3), x)
 

Reduce [F]

\[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {-7 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,x^{2}-3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{5}+22 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{4}}{-b^{2} x^{6}+a^{2}}d x \right ) a^{2} b +14 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x}{-b^{2} x^{6}+a^{2}}d x \right ) a^{3}}{18 b^{2} d} \] Input:

int(x^7*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x)
 

Output:

( - 7*(a + b*x**3)**(1/3)*a*x**2 - 3*(a + b*x**3)**(1/3)*b*x**5 + 22*int(( 
(a + b*x**3)**(1/3)*x**4)/(a**2 - b**2*x**6),x)*a**2*b + 14*int(((a + b*x* 
*3)**(1/3)*x)/(a**2 - b**2*x**6),x)*a**3)/(18*b**2*d)