\(\int \frac {1}{x^{3/2} (a x+b x^3+c x^5)^{3/2}} \, dx\) [67]

Optimal result

Integrand size = 24, antiderivative size = 154 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^{3/2} \sqrt {a x+b x^3+c x^5}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a x+b x^3+c x^5}}{2 a^2 \left (b^2-4 a c\right ) x^{5/2}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{4 a^{5/2}} \] Output:

(b*c*x^2-2*a*c+b^2)/a/(-4*a*c+b^2)/x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2)-1/2*(-8 
*a*c+3*b^2)*(c*x^5+b*x^3+a*x)^(1/2)/a^2/(-4*a*c+b^2)/x^(5/2)+3/4*b*arctanh 
(1/2*x^(1/2)*(b*x^2+2*a)/a^(1/2)/(c*x^5+b*x^3+a*x)^(1/2))/a^(5/2)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-4 a^2 c+3 b^2 x^2 \left (b+c x^2\right )+a \left (b^2-10 b c x^2-8 c^2 x^4\right )\right )+3 b \left (b^2-4 a c\right ) x^2 \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{2 a^{5/2} \left (-b^2+4 a c\right ) x^{3/2} \sqrt {x \left (a+b x^2+c x^4\right )}} \] Input:

Integrate[1/(x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2)),x]
 

Output:

(Sqrt[a]*(-4*a^2*c + 3*b^2*x^2*(b + c*x^2) + a*(b^2 - 10*b*c*x^2 - 8*c^2*x 
^4)) + 3*b*(b^2 - 4*a*c)*x^2*Sqrt[a + b*x^2 + c*x^4]*ArcTanh[(Sqrt[c]*x^2 
- Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/(2*a^(5/2)*(-b^2 + 4*a*c)*x^(3/2)*Sqr 
t[x*(a + b*x^2 + c*x^4)])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1971, 25, 1998, 27, 1960, 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.

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

\(\Big \downarrow \) 1971

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1998

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1960

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

\(\Big \downarrow \) 219

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

Input:

Int[1/(x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2)),x]
 

Output:

(b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*x^(3/2)*Sqrt[a*x + b*x^3 + c*x^5] 
) + (-1/2*((3*b^2 - 8*a*c)*Sqrt[a*x + b*x^3 + c*x^5])/(a*x^(5/2)) + (3*b*( 
b^2 - 4*a*c)*ArcTanh[(Sqrt[x]*(2*a + b*x^2))/(2*Sqrt[a]*Sqrt[a*x + b*x^3 + 
 c*x^5])])/(4*a^(3/2)))/(a*(b^2 - 4*a*c))
 

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 1960

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)] 
, x_Symbol] :> Simp[-2/(n - q)   Subst[Int[1/(4*a - x^2), x], x, x^(m + 1)* 
((2*a + b*x^(n - q))/Sqrt[a*x^q + b*x^n + c*x^r])], x] /; FreeQ[{a, b, c, m 
, n, q, r}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && NeQ[b^2 - 4*a*c, 0] && 
EqQ[m, q/2 - 1]
 

rule 1971

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[(-x^(m - q + 1))*(b^2 - 2*a*c + b*c*x^(n - q))*((a*x^q 
 + b*x^n + c*x^(2*n - q))^(p + 1)/(a*(n - q)*(p + 1)*(b^2 - 4*a*c))), x] + 
Simp[1/(a*(n - q)*(p + 1)*(b^2 - 4*a*c))   Int[x^(m - q)*(b^2*(m + p*q + (n 
 - q)*(p + 1) + 1) - 2*a*c*(m + p*q + 2*(n - q)*(p + 1) + 1) + b*c*(m + p*q 
 + (n - q)*(2*p + 3) + 1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1 
), x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !Int 
egerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[p, -1] && RationalQ[m, 
 q] && LtQ[m + p*q + 1, n - q]
 

rule 1998

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ 
.)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[A*x^(m - q + 1)*((a*x^q + b 
*x^n + c*x^(2*n - q))^(p + 1)/(a*(m + p*q + 1))), x] + Simp[1/(a*(m + p*q + 
 1))   Int[x^(m + n - q)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p 
 + 1) + 1) - A*c*(m + p*q + 2*(n - q)*(p + 1) + 1)*x^(n - q), x]*(a*x^q + b 
*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - 
q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] 
 && RationalQ[m, p, q] && ((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - 
q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*q + 1, 0]
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.43

method	result	size
default	\(\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \left (-16 a^{\frac {3}{2}} c^{2} x^{4}+6 b^{2} c \,x^{4} \sqrt {a}+12 \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) a b c \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}-3 \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) b^{3} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}-20 a^{\frac {3}{2}} b c \,x^{2}+6 \sqrt {a}\, b^{3} x^{2}-8 a^{\frac {5}{2}} c +2 a^{\frac {3}{2}} b^{2}\right )}{4 a^{\frac {5}{2}} x^{\frac {5}{2}} \left (c \,x^{4}+b \,x^{2}+a \right ) \left (4 a c -b^{2}\right )}\)	\(220\)
risch	\(-\frac {c \,x^{4}+b \,x^{2}+a}{2 a^{2} x^{\frac {3}{2}} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}+\frac {\left (\frac {b^{2} x^{2} c}{a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {b^{3}}{4 a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {2 c^{2} x^{2}}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {3 b}{4 a^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {3 b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(240\)

Input:

int(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/4*(x*(c*x^4+b*x^2+a))^(1/2)/a^(5/2)*(-16*a^(3/2)*c^2*x^4+6*b^2*c*x^4*a^( 
1/2)+12*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)*a*b*c*x^2*(c*x 
^4+b*x^2+a)^(1/2)-3*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)*b^ 
3*x^2*(c*x^4+b*x^2+a)^(1/2)-20*a^(3/2)*b*c*x^2+6*a^(1/2)*b^3*x^2-8*a^(5/2) 
*c+2*a^(3/2)*b^2)/x^(5/2)/(c*x^4+b*x^2+a)/(4*a*c-b^2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 508, normalized size of antiderivative = 3.30 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{7} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{5} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{5}}\right ) - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {x}}{8 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{7} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{5} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}, -\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{7} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{5} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{5} + b x^{3} + a x} {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {x}}{4 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{7} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{5} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}\right ] \] Input:

integrate(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="fricas")
 

Output:

[1/8*(3*((b^3*c - 4*a*b*c^2)*x^7 + (b^4 - 4*a*b^2*c)*x^5 + (a*b^3 - 4*a^2* 
b*c)*x^3)*sqrt(a)*log(-((b^2 + 4*a*c)*x^5 + 8*a*b*x^3 + 8*a^2*x + 4*sqrt(c 
*x^5 + b*x^3 + a*x)*(b*x^2 + 2*a)*sqrt(a)*sqrt(x))/x^5) - 4*sqrt(c*x^5 + b 
*x^3 + a*x)*((3*a*b^2*c - 8*a^2*c^2)*x^4 + a^2*b^2 - 4*a^3*c + (3*a*b^3 - 
10*a^2*b*c)*x^2)*sqrt(x))/((a^3*b^2*c - 4*a^4*c^2)*x^7 + (a^3*b^3 - 4*a^4* 
b*c)*x^5 + (a^4*b^2 - 4*a^5*c)*x^3), -1/4*(3*((b^3*c - 4*a*b*c^2)*x^7 + (b 
^4 - 4*a*b^2*c)*x^5 + (a*b^3 - 4*a^2*b*c)*x^3)*sqrt(-a)*arctan(1/2*sqrt(c* 
x^5 + b*x^3 + a*x)*(b*x^2 + 2*a)*sqrt(-a)*sqrt(x)/(a*c*x^5 + a*b*x^3 + a^2 
*x)) + 2*sqrt(c*x^5 + b*x^3 + a*x)*((3*a*b^2*c - 8*a^2*c^2)*x^4 + a^2*b^2 
- 4*a^3*c + (3*a*b^3 - 10*a^2*b*c)*x^2)*sqrt(x))/((a^3*b^2*c - 4*a^4*c^2)* 
x^7 + (a^3*b^3 - 4*a^4*b*c)*x^5 + (a^4*b^2 - 4*a^5*c)*x^3)]
 

Sympy [F]

\[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \left (x \left (a + b x^{2} + c x^{4}\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/x**(3/2)/(c*x**5+b*x**3+a*x)**(3/2),x)
 

Output:

Integral(1/(x**(3/2)*(x*(a + b*x**2 + c*x**4))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac {3}{2}} x^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((c*x^5 + b*x^3 + a*x)^(3/2)*x^(3/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=-\frac {\frac {{\left (a^{2} b^{2} c - 2 \, a^{3} c^{2}\right )} x^{2}}{a^{4} b^{2} - 4 \, a^{5} c} + \frac {a^{2} b^{3} - 3 \, a^{3} b c}{a^{4} b^{2} - 4 \, a^{5} c}}{\sqrt {c x^{4} + b x^{2} + a}} - \frac {3 \, b \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{2}} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} b + 2 \, a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \] Input:

integrate(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="giac")
 

Output:

-((a^2*b^2*c - 2*a^3*c^2)*x^2/(a^4*b^2 - 4*a^5*c) + (a^2*b^3 - 3*a^3*b*c)/ 
(a^4*b^2 - 4*a^5*c))/sqrt(c*x^4 + b*x^2 + a) - 3/2*b*arctan(-(sqrt(c)*x^2 
- sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) + 1/2*((sqrt(c)*x^2 - 
sqrt(c*x^4 + b*x^2 + a))*b + 2*a*sqrt(c))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b* 
x^2 + a))^2 - a)*a^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\int \frac {1}{x^{3/2}\,{\left (c\,x^5+b\,x^3+a\,x\right )}^{3/2}} \,d x \] Input:

int(1/(x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2)),x)
 

Output:

int(1/(x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2)), x)
 

Reduce [F]

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

int(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(3/2),x)
 

Output:

int(1/(sqrt(a + b*x**2 + c*x**4)*a*x**3 + sqrt(a + b*x**2 + c*x**4)*b*x**5 
 + sqrt(a + b*x**2 + c*x**4)*c*x**7),x)