\(\int \frac {(a+b x^3+c x^6)^{3/2}}{x^{22}} \, dx\) [198]

Optimal result

Integrand size = 20, antiderivative size = 255 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=-\frac {b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{1024 a^5 x^6}+\frac {b \left (3 b^2-4 a c\right ) \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{384 a^4 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{21 a x^{21}}+\frac {b \left (a+b x^3+c x^6\right )^{5/2}}{28 a^2 x^{18}}-\frac {\left (21 b^2-16 a c\right ) \left (a+b x^3+c x^6\right )^{5/2}}{840 a^3 x^{15}}+\frac {b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{2048 a^{11/2}} \] Output:

-1/1024*b*(-4*a*c+b^2)*(-4*a*c+3*b^2)*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(1/2)/a^ 
5/x^6+1/384*b*(-4*a*c+3*b^2)*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(3/2)/a^4/x^12-1/ 
21*(c*x^6+b*x^3+a)^(5/2)/a/x^21+1/28*b*(c*x^6+b*x^3+a)^(5/2)/a^2/x^18-1/84 
0*(-16*a*c+21*b^2)*(c*x^6+b*x^3+a)^(5/2)/a^3/x^15+1/2048*b*(-4*a*c+b^2)^2* 
(-4*a*c+3*b^2)*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))/a^(1 
1/2)
 

Mathematica [A] (verified)

Time = 2.32 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=\frac {-\frac {\sqrt {a} \sqrt {a+b x^3+c x^6} \left (5120 a^6+315 b^6 x^{18}-210 a b^4 x^{15} \left (b+12 c x^3\right )+256 a^5 \left (25 b x^3+32 c x^6\right )+64 a^4 x^6 \left (2 b^2+11 b c x^3+16 c^2 x^6\right )+56 a^2 b^2 x^{12} \left (3 b^2+26 b c x^3+98 c^2 x^6\right )-16 a^3 x^9 \left (9 b^3+62 b^2 c x^3+146 b c^2 x^6+128 c^3 x^9\right )\right )}{x^{21}}-105 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{107520 a^{11/2}} \] Input:

Integrate[(a + b*x^3 + c*x^6)^(3/2)/x^22,x]
 

Output:

(-((Sqrt[a]*Sqrt[a + b*x^3 + c*x^6]*(5120*a^6 + 315*b^6*x^18 - 210*a*b^4*x 
^15*(b + 12*c*x^3) + 256*a^5*(25*b*x^3 + 32*c*x^6) + 64*a^4*x^6*(2*b^2 + 1 
1*b*c*x^3 + 16*c^2*x^6) + 56*a^2*b^2*x^12*(3*b^2 + 26*b*c*x^3 + 98*c^2*x^6 
) - 16*a^3*x^9*(9*b^3 + 62*b^2*c*x^3 + 146*b*c^2*x^6 + 128*c^3*x^9)))/x^21 
) - 105*b*(b^2 - 4*a*c)^2*(3*b^2 - 4*a*c)*ArcTanh[(Sqrt[c]*x^3 - Sqrt[a + 
b*x^3 + c*x^6])/Sqrt[a]])/(107520*a^(11/2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1693, 1167, 27, 1237, 27, 1228, 1152, 1152, 1154, 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.

\(\displaystyle \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx\)

\(\Big \downarrow \) 1693

\(\displaystyle \frac {1}{3} \int \frac {\left (c x^6+b x^3+a\right )^{3/2}}{x^{24}}dx^3\)

\(\Big \downarrow \) 1167

\(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {\left (4 c x^3+9 b\right ) \left (c x^6+b x^3+a\right )^{3/2}}{2 x^{21}}dx^3}{7 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {\left (4 c x^3+9 b\right ) \left (c x^6+b x^3+a\right )^{3/2}}{x^{21}}dx^3}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 1237

\(\displaystyle \frac {1}{3} \left (-\frac {-\frac {\int \frac {3 \left (6 b c x^3+21 b^2-16 a c\right ) \left (c x^6+b x^3+a\right )^{3/2}}{2 x^{18}}dx^3}{6 a}-\frac {3 b \left (a+b x^3+c x^6\right )^{5/2}}{2 a x^{18}}}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-\frac {-\frac {\int \frac {\left (6 b c x^3+21 b^2-16 a c\right ) \left (c x^6+b x^3+a\right )^{3/2}}{x^{18}}dx^3}{4 a}-\frac {3 b \left (a+b x^3+c x^6\right )^{5/2}}{2 a x^{18}}}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 1228

\(\displaystyle \frac {1}{3} \left (-\frac {-\frac {-\frac {7 b \left (3 b^2-4 a c\right ) \int \frac {\left (c x^6+b x^3+a\right )^{3/2}}{x^{15}}dx^3}{2 a}-\frac {\left (21 b^2-16 a c\right ) \left (a+b x^3+c x^6\right )^{5/2}}{5 a x^{15}}}{4 a}-\frac {3 b \left (a+b x^3+c x^6\right )^{5/2}}{2 a x^{18}}}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 1152

\(\displaystyle \frac {1}{3} \left (-\frac {-\frac {-\frac {7 b \left (3 b^2-4 a c\right ) \left (-\frac {3 \left (b^2-4 a c\right ) \int \frac {\sqrt {c x^6+b x^3+a}}{x^9}dx^3}{16 a}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 a x^{12}}\right )}{2 a}-\frac {\left (21 b^2-16 a c\right ) \left (a+b x^3+c x^6\right )^{5/2}}{5 a x^{15}}}{4 a}-\frac {3 b \left (a+b x^3+c x^6\right )^{5/2}}{2 a x^{18}}}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 1152

\(\displaystyle \frac {1}{3} \left (-\frac {-\frac {-\frac {7 b \left (3 b^2-4 a c\right ) \left (-\frac {3 \left (b^2-4 a c\right ) \left (-\frac {\left (b^2-4 a c\right ) \int \frac {1}{x^3 \sqrt {c x^6+b x^3+a}}dx^3}{8 a}-\frac {\left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{4 a x^6}\right )}{16 a}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 a x^{12}}\right )}{2 a}-\frac {\left (21 b^2-16 a c\right ) \left (a+b x^3+c x^6\right )^{5/2}}{5 a x^{15}}}{4 a}-\frac {3 b \left (a+b x^3+c x^6\right )^{5/2}}{2 a x^{18}}}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{3} \left (-\frac {-\frac {-\frac {7 b \left (3 b^2-4 a c\right ) \left (-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 a-x^6}d\frac {b x^3+2 a}{\sqrt {c x^6+b x^3+a}}}{4 a}-\frac {\left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{4 a x^6}\right )}{16 a}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 a x^{12}}\right )}{2 a}-\frac {\left (21 b^2-16 a c\right ) \left (a+b x^3+c x^6\right )^{5/2}}{5 a x^{15}}}{4 a}-\frac {3 b \left (a+b x^3+c x^6\right )^{5/2}}{2 a x^{18}}}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (-\frac {-\frac {-\frac {7 b \left (3 b^2-4 a c\right ) \left (-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{8 a^{3/2}}-\frac {\left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{4 a x^6}\right )}{16 a}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 a x^{12}}\right )}{2 a}-\frac {\left (21 b^2-16 a c\right ) \left (a+b x^3+c x^6\right )^{5/2}}{5 a x^{15}}}{4 a}-\frac {3 b \left (a+b x^3+c x^6\right )^{5/2}}{2 a x^{18}}}{14 a}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{7 a x^{21}}\right )\)

Input:

Int[(a + b*x^3 + c*x^6)^(3/2)/x^22,x]
 

Output:

(-1/7*(a + b*x^3 + c*x^6)^(5/2)/(a*x^21) - ((-3*b*(a + b*x^3 + c*x^6)^(5/2 
))/(2*a*x^18) - (-1/5*((21*b^2 - 16*a*c)*(a + b*x^3 + c*x^6)^(5/2))/(a*x^1 
5) - (7*b*(3*b^2 - 4*a*c)*(-1/8*((2*a + b*x^3)*(a + b*x^3 + c*x^6)^(3/2))/ 
(a*x^12) - (3*(b^2 - 4*a*c)*(-1/4*((2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/ 
(a*x^6) + ((b^2 - 4*a*c)*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + 
 c*x^6])])/(8*a^(3/2))))/(16*a)))/(2*a))/(4*a))/(14*a))/3
 

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 1152

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-(d + e*x)^(m + 1))*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b 
*x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a 
*c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)))   Int[(d + e*x)^(m + 2)*(a + b*x + 
 c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] 
 && GtQ[p, 0]
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1167

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)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2))   Int[ 
(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, 
 x]*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[m 
, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp 
lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 0])
 

rule 1228

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + 
 b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e 
*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^ 
(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 1237

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* 
x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) 
*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ 
(c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m 
+ 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 
] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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]]
 

Maple [F]

\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}{x^{22}}d x\]

Input:

int((c*x^6+b*x^3+a)^(3/2)/x^22,x)
 

Output:

int((c*x^6+b*x^3+a)^(3/2)/x^22,x)
 

Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 557, normalized size of antiderivative = 2.18 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=\left [-\frac {105 \, {\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt {a} x^{21} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) + 4 \, {\left ({\left (315 \, a b^{6} - 2520 \, a^{2} b^{4} c + 5488 \, a^{3} b^{2} c^{2} - 2048 \, a^{4} c^{3}\right )} x^{18} - 2 \, {\left (105 \, a^{2} b^{5} - 728 \, a^{3} b^{3} c + 1168 \, a^{4} b c^{2}\right )} x^{15} + 8 \, {\left (21 \, a^{3} b^{4} - 124 \, a^{4} b^{2} c + 128 \, a^{5} c^{2}\right )} x^{12} + 6400 \, a^{6} b x^{3} - 16 \, {\left (9 \, a^{4} b^{3} - 44 \, a^{5} b c\right )} x^{9} + 5120 \, a^{7} + 128 \, {\left (a^{5} b^{2} + 64 \, a^{6} c\right )} x^{6}\right )} \sqrt {c x^{6} + b x^{3} + a}}{430080 \, a^{6} x^{21}}, -\frac {105 \, {\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt {-a} x^{21} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (315 \, a b^{6} - 2520 \, a^{2} b^{4} c + 5488 \, a^{3} b^{2} c^{2} - 2048 \, a^{4} c^{3}\right )} x^{18} - 2 \, {\left (105 \, a^{2} b^{5} - 728 \, a^{3} b^{3} c + 1168 \, a^{4} b c^{2}\right )} x^{15} + 8 \, {\left (21 \, a^{3} b^{4} - 124 \, a^{4} b^{2} c + 128 \, a^{5} c^{2}\right )} x^{12} + 6400 \, a^{6} b x^{3} - 16 \, {\left (9 \, a^{4} b^{3} - 44 \, a^{5} b c\right )} x^{9} + 5120 \, a^{7} + 128 \, {\left (a^{5} b^{2} + 64 \, a^{6} c\right )} x^{6}\right )} \sqrt {c x^{6} + b x^{3} + a}}{215040 \, a^{6} x^{21}}\right ] \] Input:

integrate((c*x^6+b*x^3+a)^(3/2)/x^22,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

[-1/430080*(105*(3*b^7 - 28*a*b^5*c + 80*a^2*b^3*c^2 - 64*a^3*b*c^3)*sqrt( 
a)*x^21*log(-((b^2 + 4*a*c)*x^6 + 8*a*b*x^3 - 4*sqrt(c*x^6 + b*x^3 + a)*(b 
*x^3 + 2*a)*sqrt(a) + 8*a^2)/x^6) + 4*((315*a*b^6 - 2520*a^2*b^4*c + 5488* 
a^3*b^2*c^2 - 2048*a^4*c^3)*x^18 - 2*(105*a^2*b^5 - 728*a^3*b^3*c + 1168*a 
^4*b*c^2)*x^15 + 8*(21*a^3*b^4 - 124*a^4*b^2*c + 128*a^5*c^2)*x^12 + 6400* 
a^6*b*x^3 - 16*(9*a^4*b^3 - 44*a^5*b*c)*x^9 + 5120*a^7 + 128*(a^5*b^2 + 64 
*a^6*c)*x^6)*sqrt(c*x^6 + b*x^3 + a))/(a^6*x^21), -1/215040*(105*(3*b^7 - 
28*a*b^5*c + 80*a^2*b^3*c^2 - 64*a^3*b*c^3)*sqrt(-a)*x^21*arctan(1/2*sqrt( 
c*x^6 + b*x^3 + a)*(b*x^3 + 2*a)*sqrt(-a)/(a*c*x^6 + a*b*x^3 + a^2)) + 2*( 
(315*a*b^6 - 2520*a^2*b^4*c + 5488*a^3*b^2*c^2 - 2048*a^4*c^3)*x^18 - 2*(1 
05*a^2*b^5 - 728*a^3*b^3*c + 1168*a^4*b*c^2)*x^15 + 8*(21*a^3*b^4 - 124*a^ 
4*b^2*c + 128*a^5*c^2)*x^12 + 6400*a^6*b*x^3 - 16*(9*a^4*b^3 - 44*a^5*b*c) 
*x^9 + 5120*a^7 + 128*(a^5*b^2 + 64*a^6*c)*x^6)*sqrt(c*x^6 + b*x^3 + a))/( 
a^6*x^21)]
 

Sympy [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=\int \frac {\left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}{x^{22}}\, dx \] Input:

integrate((c*x**6+b*x**3+a)**(3/2)/x**22,x)
 

Output:

Integral((a + b*x**3 + c*x**6)**(3/2)/x**22, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*x^6+b*x^3+a)^(3/2)/x^22,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
 

Giac [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}{x^{22}} \,d x } \] Input:

integrate((c*x^6+b*x^3+a)^(3/2)/x^22,x, algorithm="giac")
 

Output:

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^22, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{x^{22}} \,d x \] Input:

int((a + b*x^3 + c*x^6)^(3/2)/x^22,x)
 

Output:

int((a + b*x^3 + c*x^6)^(3/2)/x^22, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{22}} \, dx=\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}{x^{22}}d x \] Input:

int((c*x^6+b*x^3+a)^(3/2)/x^22,x)
 

Output:

int((c*x^6+b*x^3+a)^(3/2)/x^22,x)