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

Optimal. Leaf size=154 \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 x^{5/2} \left (b^2-4 a c\right )}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{x} \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x+b x^3+c x^5}}\right )}{4 a^{5/2}}+\frac{-2 a c+b^2+b c x^2}{a x^{3/2} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}} \]

[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))

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Rubi [A]  time = 0.174953, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1924, 1951, 12, 1913, 206} \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a x+b x^3+c x^5}}{2 a^2 x^{5/2} \left (b^2-4 a c\right )}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{x} \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x+b x^3+c x^5}}\right )}{4 a^{5/2}}+\frac{-2 a c+b^2+b c x^2}{a x^{3/2} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}} \]

Antiderivative was successfully verified.

[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 1924

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 1951

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]

Rule 12

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

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{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{\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) \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 )}{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]  time = 0.0820778, size = 160, normalized size = 1.04 \[ \frac{2 \sqrt{a} \left (-4 a^2 c+a \left (b^2-10 b c x^2-8 c^2 x^4\right )+3 b^2 x^2 \left (b+c x^2\right )\right )-3 b x^2 \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 a^{5/2} x^{3/2} \left (4 a c-b^2\right ) \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 220, normalized size = 1.4 \begin{align*}{\frac{1}{ \left ( 4\,c{x}^{4}+4\,b{x}^{2}+4\,a \right ) \left ( 4\,ac-{b}^{2} \right ) }\sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) } \left ( -16\,{x}^{4}{a}^{3/2}{c}^{2}+6\,{x}^{4}{b}^{2}c\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 ){x}^{2}abc\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 ){x}^{2}{b}^{3}\sqrt{c{x}^{4}+b{x}^{2}+a}-20\,{a}^{3/2}{x}^{2}bc+6\,{x}^{2}{b}^{3}\sqrt{a}-8\,{a}^{5/2}c+2\,{a}^{3/2}{b}^{2} \right ){a}^{-{\frac{5}{2}}}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.85444, size = 1084, normalized size = 7.04 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{3}{2}} \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(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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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