3.255 \(\int \frac{(b x^2+c x^4)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=79 \[ -\frac{\left (b x^2+c x^4\right )^{3/2}}{2 x^5}+\frac{3 c \sqrt{b x^2+c x^4}}{2 x}-\frac{3}{2} \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.118345, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2021, 2008, 206} \[ -\frac{\left (b x^2+c x^4\right )^{3/2}}{2 x^5}+\frac{3 c \sqrt{b x^2+c x^4}}{2 x}-\frac{3}{2} \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^6,x]

[Out]

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

Rule 2020

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 + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

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^6} \, dx &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{2 x^5}+\frac{1}{2} (3 c) \int \frac{\sqrt{b x^2+c x^4}}{x^2} \, dx\\ &=\frac{3 c \sqrt{b x^2+c x^4}}{2 x}-\frac{\left (b x^2+c x^4\right )^{3/2}}{2 x^5}+\frac{1}{2} (3 b c) \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx\\ &=\frac{3 c \sqrt{b x^2+c x^4}}{2 x}-\frac{\left (b x^2+c x^4\right )^{3/2}}{2 x^5}-\frac{1}{2} (3 b c) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{b x^2+c x^4}}\right )\\ &=\frac{3 c \sqrt{b x^2+c x^4}}{2 x}-\frac{\left (b x^2+c x^4\right )^{3/2}}{2 x^5}-\frac{3}{2} \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0170345, size = 44, normalized size = 0.56 \[ \frac{c \left (x^2 \left (b+c x^2\right )\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{c x^2}{b}+1\right )}{5 b^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^6,x]

[Out]

(c*(x^2*(b + c*x^2))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x^2)/b])/(5*b^2*x^5)

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Maple [A]  time = 0.047, size = 102, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,b{x}^{5}} \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 ){x}^{2}c- \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{2}c+ \left ( c{x}^{2}+b \right ) ^{{\frac{5}{2}}}-3\,\sqrt{c{x}^{2}+b}{x}^{2}bc \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^6,x)

[Out]

-1/2*(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)*x^2*c-(c*x^2+b)^(3/2)*x^2*c+(c*x^2+b)^
(5/2)-3*(c*x^2+b)^(1/2)*x^2*b*c)/x^5/(c*x^2+b)^(3/2)/b

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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^{6}}\,{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^6,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^6, x)

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Fricas [A]  time = 1.43626, size = 327, normalized size = 4.14 \begin{align*} \left [\frac{3 \, \sqrt{b} c x^{3} \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 (2 \, c x^{2} - b\right )}}{4 \, x^{3}}, \frac{3 \, \sqrt{-b} c x^{3} \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 (2 \, c x^{2} - b\right )}}{2 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

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

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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^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**6,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**6, x)

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Giac [A]  time = 1.20255, size = 80, normalized size = 1.01 \begin{align*} \frac{1}{2} \,{\left (\frac{3 \, b \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 2 \, \sqrt{c x^{2} + b} - \frac{\sqrt{c x^{2} + b} b}{c x^{2}}\right )} c \mathrm{sgn}\left (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^6,x, algorithm="giac")

[Out]

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