3.3.3 \(\int x^8 (a+b x^3+c x^6)^{3/2} \, dx\) [203]

3.3.3.1 Optimal result

Integrand size = 20, antiderivative size = 204 \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=-\frac {\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{1536 c^4}+\frac {\left (7 b^2-4 a c\right ) \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{576 c^3}-\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{180 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{18 c}+\frac {\left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3072 c^{9/2}} \]

output

1/576*(-4*a*c+7*b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(3/2)/c^3-7/180*b*(c*x^6+ 
b*x^3+a)^(5/2)/c^2+1/18*x^3*(c*x^6+b*x^3+a)^(5/2)/c+1/3072*(-4*a*c+b^2)^2* 
(-4*a*c+7*b^2)*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^(9 
/2)-1/1536*(-4*a*c+b^2)*(-4*a*c+7*b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c 
^4
 

3.3.3.2 Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.95 \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {2 \sqrt {c} \sqrt {a+b x^3+c x^6} \left (-105 b^5+70 b^4 c x^3+8 b^3 c \left (95 a-7 c x^6\right )+48 b^2 c^2 x^3 \left (-9 a+c x^6\right )+160 c^3 x^3 \left (3 a^2+14 a c x^6+8 c^2 x^{12}\right )+16 b c^2 \left (-81 a^2+18 a c x^6+104 c^2 x^{12}\right )\right )-15 \left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \log \left (b+2 c x^3-2 \sqrt {c} \sqrt {a+b x^3+c x^6}\right )}{46080 c^{9/2}} \]

input

Integrate[x^8*(a + b*x^3 + c*x^6)^(3/2),x]
 

output

(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6]*(-105*b^5 + 70*b^4*c*x^3 + 8*b^3*c*(95* 
a - 7*c*x^6) + 48*b^2*c^2*x^3*(-9*a + c*x^6) + 160*c^3*x^3*(3*a^2 + 14*a*c 
*x^6 + 8*c^2*x^12) + 16*b*c^2*(-81*a^2 + 18*a*c*x^6 + 104*c^2*x^12)) - 15* 
(b^2 - 4*a*c)^2*(7*b^2 - 4*a*c)*Log[b + 2*c*x^3 - 2*Sqrt[c]*Sqrt[a + b*x^3 
 + c*x^6]])/(46080*c^(9/2))
 

3.3.3.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1693, 1166, 27, 1160, 1087, 1087, 1092, 219}

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.

\(\displaystyle \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx\)

\(\Big \downarrow \) 1693

\(\displaystyle \frac {1}{3} \int x^6 \left (c x^6+b x^3+a\right )^{3/2}dx^3\)

\(\Big \downarrow \) 1166

\(\displaystyle \frac {1}{3} \left (\frac {\int -\frac {1}{2} \left (7 b x^3+2 a\right ) \left (c x^6+b x^3+a\right )^{3/2}dx^3}{6 c}+\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\int \left (7 b x^3+2 a\right ) \left (c x^6+b x^3+a\right )^{3/2}dx^3}{12 c}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \int \left (c x^6+b x^3+a\right )^{3/2}dx^3}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^6+b x^3+a}dx^3}{16 c}\right )}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^6+b x^3+a}}dx^3}{8 c}\right )}{16 c}\right )}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^6}d\frac {2 c x^3+b}{\sqrt {c x^6+b x^3+a}}}{4 c}\right )}{16 c}\right )}{2 c}}{12 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{2 c}}{12 c}\right )\)

input

Int[x^8*(a + b*x^3 + c*x^6)^(3/2),x]
 

output

((x^3*(a + b*x^3 + c*x^6)^(5/2))/(6*c) - ((7*b*(a + b*x^3 + c*x^6)^(5/2))/ 
(5*c) - ((7*b^2 - 4*a*c)*(((b + 2*c*x^3)*(a + b*x^3 + c*x^6)^(3/2))/(8*c) 
- (3*(b^2 - 4*a*c)*(((b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/(4*c) - ((b^2 
- 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(8*c^ 
(3/2))))/(16*c)))/(2*c))/(12*c))/3
 

3.3.3.3.1 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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1166

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[1/(c*(m + 2*p + 1))   Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m 
+ 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* 
(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

rule 1693

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, 
x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ 
[Simplify[(m + 1)/n]]
 

3.3.3.4 Maple [F]

\[\int x^{8} \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}d x\]

input

int(x^8*(c*x^6+b*x^3+a)^(3/2),x)
 

output

int(x^8*(c*x^6+b*x^3+a)^(3/2),x)
 

3.3.3.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.21 \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (7 \, b^{6} - 60 \, a b^{4} c + 144 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (1280 \, c^{6} x^{15} + 1664 \, b c^{5} x^{12} + 16 \, {\left (3 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{9} - 8 \, {\left (7 \, b^{3} c^{3} - 36 \, a b c^{4}\right )} x^{6} - 105 \, b^{5} c + 760 \, a b^{3} c^{2} - 1296 \, a^{2} b c^{3} + 2 \, {\left (35 \, b^{4} c^{2} - 216 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{92160 \, c^{5}}, -\frac {15 \, {\left (7 \, b^{6} - 60 \, a b^{4} c + 144 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \, {\left (1280 \, c^{6} x^{15} + 1664 \, b c^{5} x^{12} + 16 \, {\left (3 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{9} - 8 \, {\left (7 \, b^{3} c^{3} - 36 \, a b c^{4}\right )} x^{6} - 105 \, b^{5} c + 760 \, a b^{3} c^{2} - 1296 \, a^{2} b c^{3} + 2 \, {\left (35 \, b^{4} c^{2} - 216 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{46080 \, c^{5}}\right ] \]

input

integrate(x^8*(c*x^6+b*x^3+a)^(3/2),x, algorithm="fricas")
 

output

[-1/92160*(15*(7*b^6 - 60*a*b^4*c + 144*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)* 
log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 + 4*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b) 
*sqrt(c) - 4*a*c) - 4*(1280*c^6*x^15 + 1664*b*c^5*x^12 + 16*(3*b^2*c^4 + 1 
40*a*c^5)*x^9 - 8*(7*b^3*c^3 - 36*a*b*c^4)*x^6 - 105*b^5*c + 760*a*b^3*c^2 
 - 1296*a^2*b*c^3 + 2*(35*b^4*c^2 - 216*a*b^2*c^3 + 240*a^2*c^4)*x^3)*sqrt 
(c*x^6 + b*x^3 + a))/c^5, -1/46080*(15*(7*b^6 - 60*a*b^4*c + 144*a^2*b^2*c 
^2 - 64*a^3*c^3)*sqrt(-c)*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b) 
*sqrt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) - 2*(1280*c^6*x^15 + 1664*b*c^5*x^12 
+ 16*(3*b^2*c^4 + 140*a*c^5)*x^9 - 8*(7*b^3*c^3 - 36*a*b*c^4)*x^6 - 105*b^ 
5*c + 760*a*b^3*c^2 - 1296*a^2*b*c^3 + 2*(35*b^4*c^2 - 216*a*b^2*c^3 + 240 
*a^2*c^4)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^5]
 

3.3.3.6 Sympy [F]

\[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x^{8} \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}\, dx \]

input

integrate(x**8*(c*x**6+b*x**3+a)**(3/2),x)
 

output

Integral(x**8*(a + b*x**3 + c*x**6)**(3/2), x)
 

3.3.3.7 Maxima [F(-2)]

Exception generated. \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]

input

integrate(x^8*(c*x^6+b*x^3+a)^(3/2),x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

3.3.3.8 Giac [F]

\[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x^{8} \,d x } \]

input

integrate(x^8*(c*x^6+b*x^3+a)^(3/2),x, algorithm="giac")
 

output

integrate((c*x^6 + b*x^3 + a)^(3/2)*x^8, x)
 

3.3.3.9 Mupad [F(-1)]

Timed out. \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x^8\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2} \,d x \]

input

int(x^8*(a + b*x^3 + c*x^6)^(3/2),x)
 

output

int(x^8*(a + b*x^3 + c*x^6)^(3/2), x)