\(\int \frac {(a+b x^3+c x^6)^{3/2}}{x} \, dx\) [191]

Optimal result

Integrand size = 20, antiderivative size = 155 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x} \, dx=\frac {\left (b^2+8 a c+2 b c x^3\right ) \sqrt {a+b x^3+c x^6}}{24 c}+\frac {1}{9} \left (a+b x^3+c x^6\right )^{3/2}-\frac {1}{3} a^{3/2} \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )-\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{48 c^{3/2}} \] Output:

1/24*(2*b*c*x^3+8*a*c+b^2)*(c*x^6+b*x^3+a)^(1/2)/c+1/9*(c*x^6+b*x^3+a)^(3/ 
2)-1/3*a^(3/2)*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))-1/48 
*b*(-12*a*c+b^2)*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^ 
(3/2)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x} \, dx=\frac {1}{144} \left (\frac {2 \sqrt {a+b x^3+c x^6} \left (3 b^2+14 b c x^3+8 c \left (4 a+c x^6\right )\right )}{c}-\frac {3 \left (b^3-12 a b c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{c^{3/2}}+96 a^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )\right ) \] Input:

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

Output:

((2*Sqrt[a + b*x^3 + c*x^6]*(3*b^2 + 14*b*c*x^3 + 8*c*(4*a + c*x^6)))/c - 
(3*(b^3 - 12*a*b*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^ 
6])])/c^(3/2) + 96*a^(3/2)*ArcTanh[(Sqrt[c]*x^3 - Sqrt[a + b*x^3 + c*x^6]) 
/Sqrt[a]])/144
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1693, 1162, 25, 1231, 27, 1269, 1092, 219, 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.

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

\(\Big \downarrow \) 1693

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

\(\Big \downarrow \) 1162

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1231

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {-32 a^2 c \int \frac {1}{4 a-x^6}d\frac {b x^3+2 a}{\sqrt {c x^6+b x^3+a}}-\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{\sqrt {c}}}{8 c}+\frac {\sqrt {a+b x^3+c x^6} \left (8 a c+b^2+2 b c x^3\right )}{4 c}\right )+\frac {1}{3} \left (a+b x^3+c x^6\right )^{3/2}\right )\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 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 1162

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

rule 1231

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) 
 - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ 
(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 
 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* 
a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* 
c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c 
^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  !R 
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Integer 
Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1269

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

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 727, normalized size of antiderivative = 4.69 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/288*(48*a^(3/2)*c^2*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) - 3*(b^3 - 12*a*b*c)*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*(8*c^3*x^6 + 14*b*c^2*x^3 + 3*b^2*c + 32*a*c^2)* 
sqrt(c*x^6 + b*x^3 + a))/c^2, 1/144*(24*a^(3/2)*c^2*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) + 3*(b^3 - 12*a*b*c)*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*(8*c^3*x^6 + 14*b*c^2*x^3 
 + 3*b^2*c + 32*a*c^2)*sqrt(c*x^6 + b*x^3 + a))/c^2, 1/288*(96*sqrt(-a)*a* 
c^2*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)) - 3*(b^3 - 12*a*b*c)*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*(8*c^3*x 
^6 + 14*b*c^2*x^3 + 3*b^2*c + 32*a*c^2)*sqrt(c*x^6 + b*x^3 + a))/c^2, 1/14 
4*(48*sqrt(-a)*a*c^2*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)) + 3*(b^3 - 12*a*b*c)*sqrt(-c)*arctan(1/2*s 
qrt(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*(8*c^3*x^6 + 14*b*c^2*x^3 + 3*b^2*c + 32*a*c^2)*sqrt(c*x^6 + b*x^3 + a) 
)/c^2]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(24*sqrt(a + b*x**3 + c*x**6)*a**2*c**3 + 88*sqrt(a + b*x**3 + c*x**6)*a*b 
**2*c**2 + 84*sqrt(a + b*x**3 + c*x**6)*a*b*c**3*x**3 + 48*sqrt(a + b*x**3 
 + c*x**6)*a*c**4*x**6 + 6*sqrt(a + b*x**3 + c*x**6)*b**4*c + 28*sqrt(a + 
b*x**3 + c*x**6)*b**3*c**2*x**3 + 16*sqrt(a + b*x**3 + c*x**6)*b**2*c**3*x 
**6 + 144*sqrt(a)*log(sqrt(a + b*x**3 + c*x**6) - sqrt(a))*a**2*c**3 + 48* 
sqrt(a)*log(sqrt(a + b*x**3 + c*x**6) - sqrt(a))*a*b**2*c**2 - 144*sqrt(a) 
*log(sqrt(a + b*x**3 + c*x**6) + sqrt(a))*a**2*c**3 - 48*sqrt(a)*log(sqrt( 
a + b*x**3 + c*x**6) + sqrt(a))*a*b**2*c**2 - 108*sqrt(c)*log(sqrt(a + b*x 
**3 + c*x**6) - sqrt(c)*x**3)*a**2*b*c**2 - 27*sqrt(c)*log(sqrt(a + b*x**3 
 + c*x**6) - sqrt(c)*x**3)*a*b**3*c + 3*sqrt(c)*log(sqrt(a + b*x**3 + c*x* 
*6) - sqrt(c)*x**3)*b**5 + 108*sqrt(c)*log(sqrt(a + b*x**3 + c*x**6) + sqr 
t(c)*x**3)*a**2*b*c**2 + 27*sqrt(c)*log(sqrt(a + b*x**3 + c*x**6) + sqrt(c 
)*x**3)*a*b**3*c - 3*sqrt(c)*log(sqrt(a + b*x**3 + c*x**6) + sqrt(c)*x**3) 
*b**5 + 1512*int((sqrt(a + b*x**3 + c*x**6)*x**11)/(3*a**3*b*c + 3*a**3*c* 
*2*x**3 + a**2*b**3 + 7*a**2*b**2*c*x**3 + 9*a**2*b*c**2*x**6 + 3*a**2*c** 
3*x**9 + 2*a*b**4*x**3 + 6*a*b**3*c*x**6 + 7*a*b**2*c**2*x**9 + 3*a*b*c**3 
*x**12 + b**5*x**6 + 2*b**4*c*x**9 + b**3*c**2*x**12),x)*a**3*b*c**6 + 450 
*int((sqrt(a + b*x**3 + c*x**6)*x**11)/(3*a**3*b*c + 3*a**3*c**2*x**3 + a* 
*2*b**3 + 7*a**2*b**2*c*x**3 + 9*a**2*b*c**2*x**6 + 3*a**2*c**3*x**9 + 2*a 
*b**4*x**3 + 6*a*b**3*c*x**6 + 7*a*b**2*c**2*x**9 + 3*a*b*c**3*x**12 + ...