3.254 \(\int \frac{(b x^2+c x^4)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=73 \[ b^{3/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )\right )+\frac{b \sqrt{b x^2+c x^4}}{x}+\frac{\left (b x^2+c x^4\right )^{3/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.110463, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2021, 2008, 206} \[ b^{3/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )\right )+\frac{b \sqrt{b x^2+c x^4}}{x}+\frac{\left (b x^2+c x^4\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^4,x]

[Out]

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

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2008

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

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

Mathematica [A]  time = 0.0531942, size = 76, normalized size = 1.04 \[ \frac{x \left (-3 b^{3/2} \sqrt{b+c x^2} \tanh ^{-1}\left (\frac{\sqrt{b+c x^2}}{\sqrt{b}}\right )+4 b^2+5 b c x^2+c^2 x^4\right )}{3 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^4,x]

[Out]

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

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Maple [A]  time = 0.046, size = 78, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 3\,{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ) - \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}-3\,\sqrt{c{x}^{2}+b}b \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^4,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^4, x)

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Fricas [A]  time = 1.40754, size = 313, normalized size = 4.29 \begin{align*} \left [\frac{3 \, b^{\frac{3}{2}} x \log \left (-\frac{c x^{3} + 2 \, b x - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{b}}{x^{3}}\right ) + 2 \, \sqrt{c x^{4} + b x^{2}}{\left (c x^{2} + 4 \, b\right )}}{6 \, x}, \frac{3 \, \sqrt{-b} b x \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-b}}{c x^{3} + b x}\right ) + \sqrt{c x^{4} + b x^{2}}{\left (c x^{2} + 4 \, b\right )}}{3 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/6*(3*b^(3/2)*x*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*sqrt(c*x^4 + b*x^2)*(c*x^2 + 4
*b))/x, 1/3*(3*sqrt(-b)*b*x*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^4 + b*x^2)*(c*x^2 +
4*b))/x]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**4,x)

[Out]

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

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Giac [A]  time = 1.157, size = 119, normalized size = 1.63 \begin{align*} \frac{1}{3} \,{\left (\frac{3 \, b^{2} \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} +{\left (c x^{2} + b\right )}^{\frac{3}{2}} + 3 \, \sqrt{c x^{2} + b} b\right )} \mathrm{sgn}\left (x\right ) - \frac{{\left (3 \, b^{2} \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) + 4 \, \sqrt{-b} b^{\frac{3}{2}}\right )} \mathrm{sgn}\left (x\right )}{3 \, \sqrt{-b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

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