\(\int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx\) [113]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c} \sqrt {a x+b x^3+c x^5}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1928, 1121, 635, 212} \[ \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c} \sqrt {a x+b x^3+c x^5}} \]

[In]

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

[Out]

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

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 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1121

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rule 1928

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[x^(q/2)*(Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))]/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)]), Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx=-\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{2 \sqrt {c} \sqrt {x \left (a+b x^2+c x^4\right )}} \]

[In]

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

[Out]

-1/2*(Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(Sqrt[c]*Sqrt[x*(a
+ b*x^2 + c*x^4)])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right )}{2 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}\)	\(72\)

[In]

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

[Out]

1/2/x^(1/2)*(x*(c*x^4+b*x^2+a))^(1/2)/(c*x^4+b*x^2+a)^(1/2)*ln(1/2*(2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/2)+b)
/c^(1/2))/c^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.65 \[ \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx=\left [\frac {\log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right )}{4 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right )}{2 \, c}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {x^{\frac {3}{2}}}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68 \[ \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx=-\frac {\log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, \sqrt {c}} + \frac {\log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right )}{2 \, \sqrt {c}} \]

[In]

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

[Out]

-1/2*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/sqrt(c) + 1/2*log(abs(b - 2*sqrt(a)*sqrt(
c)))/sqrt(c)

Mupad [F(-1)]

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

[In]

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

[Out]

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