\(\int \frac {1}{x^{3/2} (a x+b x^3+c x^5)^{3/2}} \, dx\) [120]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1938, 1965, 12, 1927, 212} \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\frac {3 b \text {arctanh}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{4 a^{5/2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a x+b x^3+c x^5}}{2 a^2 x^{5/2} \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{a x^{3/2} \left (b^2-4 a c\right ) \sqrt {a x+b x^3+c x^5}} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1927

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - q), Sub
st[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 1938

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] + Dist[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] &&  !IntegerQ[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 1965

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> 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] + Dist[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] && Eq
Q[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]

Rubi steps \begin{align*} \text {integral}& = \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 {\int \frac {-3 b^2+8 a c-2 b c x^2}{x^{5/2} \sqrt {a x+b x^3+c x^5}} \, dx}{a \left (b^2-4 a c\right )} \\ & = \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 {\int -\frac {3 b \left (b^2-4 a c\right )}{\sqrt {x} \sqrt {a x+b x^3+c x^5}} \, dx}{2 a^2 \left (b^2-4 a c\right )} \\ & = \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) \int \frac {1}{\sqrt {x} \sqrt {a x+b x^3+c x^5}} \, dx}{2 a^2} \\ & = \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 {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {\sqrt {x} \left (2 a+b x^2\right )}{\sqrt {a x+b x^3+c x^5}}\right )}{2 a^2} \\ & = \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 \tanh ^{-1}\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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-4 a^2 c+3 b^2 x^2 \left (b+c x^2\right )+a \left (b^2-10 b c x^2-8 c^2 x^4\right )\right )+3 b \left (b^2-4 a c\right ) x^2 \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{2 a^{5/2} \left (-b^2+4 a c\right ) x^{3/2} \sqrt {x \left (a+b x^2+c x^4\right )}} \]

[In]

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

[Out]

(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)*S
qrt[x*(a + b*x^2 + c*x^4)])

Maple [A] (verified)

Time = 0.12 (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	\(-\frac {c \,x^{4}+b \,x^{2}+a}{2 a^{2} x^{\frac {3}{2}} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}+\frac {\left (\frac {b^{2} c \,x^{2}}{a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {b^{3}}{4 a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {2 c^{2} x^{2}}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {3 b}{4 a^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {3 b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(240\)

[In]

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

[Out]

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)

none

Time = 0.33 (sec) , antiderivative size = 508, normalized size of antiderivative = 3.30 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{7} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{5} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{5}}\right ) - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {x}}{8 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{7} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{5} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}, -\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{7} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{5} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{5} + b x^{3} + a x} {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {x}}{4 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{7} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{5} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}\right ] \]

[In]

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

[Out]

[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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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