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

Optimal. Leaf size=103 \[ \frac{\sqrt{x} \left (-2 a c+b^2+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x} \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x+b x^3+c x^5}}\right )}{2 a^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.072923, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1922, 1913, 206} \[ \frac{\sqrt{x} \left (-2 a c+b^2+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x} \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x+b x^3+c x^5}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1922

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[(2*a*c - b^2*(p + 2))/(a*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - 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] && I
GtQ[n, 0] && LtQ[p, -1] && RationalQ[m, p, q] && EqQ[m + p*q + 1, -((n - q)*(2*p + 3))]

Rule 1913

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{\left (a x+b x^3+c x^5\right )^{3/2}} \, dx &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}+\frac{\int \frac{1}{\sqrt{x} \sqrt{a x+b x^3+c x^5}} \, dx}{a}\\ &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}-\frac{\operatorname{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 )}{a}\\ &=\frac{\sqrt{x} \left (b^2-2 a c+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}-\frac{\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 )}{2 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0829563, size = 126, normalized size = 1.22 \[ \frac{\sqrt{x} \left (\left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )-2 \sqrt{a} \left (-2 a c+b^2+b c x^2\right )\right )}{2 a^{3/2} \left (4 a c-b^2\right ) \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.019, size = 179, normalized size = 1.7 \begin{align*} -{\frac{1}{ \left ( 2\,c{x}^{4}+2\,b{x}^{2}+2\,a \right ) \left ( 4\,ac-{b}^{2} \right ) }\sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) } \left ( 2\,{x}^{2}bc\sqrt{a}+4\,\ln \left ({\frac{2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a}}{{x}^{2}}} \right ) ac\sqrt{c{x}^{4}+b{x}^{2}+a}-\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){b}^{2}\sqrt{c{x}^{4}+b{x}^{2}+a}-4\,{a}^{3/2}c+2\,{b}^{2}\sqrt{a} \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(x*(c*x^4+b*x^2+a))^(1/2)/a^(3/2)*(2*x^2*b*c*a^(1/2)+4*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2
)*a*c*(c*x^4+b*x^2+a)^(1/2)-ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)*b^2*(c*x^4+b*x^2+a)^(1/2)-4*a^
(3/2)*c+2*b^2*a^(1/2))/x^(1/2)/(c*x^4+b*x^2+a)/(4*a*c-b^2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.7769, size = 918, normalized size = 8.91 \begin{align*} \left [\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{5} +{\left (b^{3} - 4 \, a b c\right )} x^{3} +{\left (a b^{2} - 4 \, a^{2} c\right )} x\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 (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt{x}}{4 \,{\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{5} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{3} +{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x\right )}}, \frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{5} +{\left (b^{3} - 4 \, a b c\right )} x^{3} +{\left (a b^{2} - 4 \, a^{2} c\right )} x\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 (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt{x}}{2 \,{\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{5} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{3} +{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{\left (x \left (a + b x^{2} + c x^{4}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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