3.258 \(\int \frac{(b x^2+c x^4)^{3/2}}{x^{12}} \, dx\)

Optimal. Leaf size=137 \[ \frac{3 c^3 \sqrt{b x^2+c x^4}}{128 b^2 x^3}-\frac{3 c^4 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{128 b^{5/2}}-\frac{c^2 \sqrt{b x^2+c x^4}}{64 b x^5}-\frac{c \sqrt{b x^2+c x^4}}{16 x^7}-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}} \]

[Out]

-(c*Sqrt[b*x^2 + c*x^4])/(16*x^7) - (c^2*Sqrt[b*x^2 + c*x^4])/(64*b*x^5) + (3*c^3*Sqrt[b*x^2 + c*x^4])/(128*b^
2*x^3) - (b*x^2 + c*x^4)^(3/2)/(8*x^11) - (3*c^4*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(128*b^(5/2))

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Rubi [A]  time = 0.203488, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2008, 206} \[ \frac{3 c^3 \sqrt{b x^2+c x^4}}{128 b^2 x^3}-\frac{3 c^4 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{128 b^{5/2}}-\frac{c^2 \sqrt{b x^2+c x^4}}{64 b x^5}-\frac{c \sqrt{b x^2+c x^4}}{16 x^7}-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^12,x]

[Out]

-(c*Sqrt[b*x^2 + c*x^4])/(16*x^7) - (c^2*Sqrt[b*x^2 + c*x^4])/(64*b*x^5) + (3*c^3*Sqrt[b*x^2 + c*x^4])/(128*b^
2*x^3) - (b*x^2 + c*x^4)^(3/2)/(8*x^11) - (3*c^4*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(128*b^(5/2))

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 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*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^{12}} \, dx &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}}+\frac{1}{8} (3 c) \int \frac{\sqrt{b x^2+c x^4}}{x^8} \, dx\\ &=-\frac{c \sqrt{b x^2+c x^4}}{16 x^7}-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}}+\frac{1}{16} c^2 \int \frac{1}{x^4 \sqrt{b x^2+c x^4}} \, dx\\ &=-\frac{c \sqrt{b x^2+c x^4}}{16 x^7}-\frac{c^2 \sqrt{b x^2+c x^4}}{64 b x^5}-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}}-\frac{\left (3 c^3\right ) \int \frac{1}{x^2 \sqrt{b x^2+c x^4}} \, dx}{64 b}\\ &=-\frac{c \sqrt{b x^2+c x^4}}{16 x^7}-\frac{c^2 \sqrt{b x^2+c x^4}}{64 b x^5}+\frac{3 c^3 \sqrt{b x^2+c x^4}}{128 b^2 x^3}-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}}+\frac{\left (3 c^4\right ) \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx}{128 b^2}\\ &=-\frac{c \sqrt{b x^2+c x^4}}{16 x^7}-\frac{c^2 \sqrt{b x^2+c x^4}}{64 b x^5}+\frac{3 c^3 \sqrt{b x^2+c x^4}}{128 b^2 x^3}-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}}-\frac{\left (3 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{b x^2+c x^4}}\right )}{128 b^2}\\ &=-\frac{c \sqrt{b x^2+c x^4}}{16 x^7}-\frac{c^2 \sqrt{b x^2+c x^4}}{64 b x^5}+\frac{3 c^3 \sqrt{b x^2+c x^4}}{128 b^2 x^3}-\frac{\left (b x^2+c x^4\right )^{3/2}}{8 x^{11}}-\frac{3 c^4 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{128 b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0172812, size = 46, normalized size = 0.34 \[ -\frac{c^4 \left (x^2 \left (b+c x^2\right )\right )^{5/2} \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{c x^2}{b}+1\right )}{5 b^5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^12,x]

[Out]

-(c^4*(x^2*(b + c*x^2))^(5/2)*Hypergeometric2F1[5/2, 5, 7/2, 1 + (c*x^2)/b])/(5*b^5*x^5)

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Maple [A]  time = 0.055, size = 165, normalized size = 1.2 \begin{align*} -{\frac{1}{128\,{b}^{4}{x}^{11}} \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}^{8}{c}^{4}- \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{8}{c}^{4}+ \left ( c{x}^{2}+b \right ) ^{{\frac{5}{2}}}{x}^{6}{c}^{3}-3\,\sqrt{c{x}^{2}+b}{x}^{8}b{c}^{4}+2\, \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{4}b{c}^{2}-8\, \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{2}{b}^{2}c+16\, \left ( c{x}^{2}+b \right ) ^{5/2}{b}^{3} \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^12,x)

[Out]

-1/128*(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^8*c^4-(c*x^2+b)^(3/2)*x^8*c^4+(c*x
^2+b)^(5/2)*x^6*c^3-3*(c*x^2+b)^(1/2)*x^8*b*c^4+2*(c*x^2+b)^(5/2)*x^4*b*c^2-8*(c*x^2+b)^(5/2)*x^2*b^2*c+16*(c*
x^2+b)^(5/2)*b^3)/x^11/(c*x^2+b)^(3/2)/b^4

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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^{12}}\,{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^12,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^12, x)

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Fricas [A]  time = 1.44683, size = 462, normalized size = 3.37 \begin{align*} \left [\frac{3 \, \sqrt{b} c^{4} x^{9} \log \left (-\frac{c x^{3} + 2 \, b x - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{b}}{x^{3}}\right ) + 2 \,{\left (3 \, b c^{3} x^{6} - 2 \, b^{2} c^{2} x^{4} - 24 \, b^{3} c x^{2} - 16 \, b^{4}\right )} \sqrt{c x^{4} + b x^{2}}}{256 \, b^{3} x^{9}}, \frac{3 \, \sqrt{-b} c^{4} x^{9} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-b}}{c x^{3} + b x}\right ) +{\left (3 \, b c^{3} x^{6} - 2 \, b^{2} c^{2} x^{4} - 24 \, b^{3} c x^{2} - 16 \, b^{4}\right )} \sqrt{c x^{4} + b x^{2}}}{128 \, b^{3} x^{9}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^12,x, algorithm="fricas")

[Out]

[1/256*(3*sqrt(b)*c^4*x^9*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*b*c^3*x^6 - 2*b^2*c
^2*x^4 - 24*b^3*c*x^2 - 16*b^4)*sqrt(c*x^4 + b*x^2))/(b^3*x^9), 1/128*(3*sqrt(-b)*c^4*x^9*arctan(sqrt(c*x^4 +
b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (3*b*c^3*x^6 - 2*b^2*c^2*x^4 - 24*b^3*c*x^2 - 16*b^4)*sqrt(c*x^4 + b*x^2))/(b
^3*x^9)]

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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^{12}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**12,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**12, x)

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Giac [A]  time = 1.34524, size = 130, normalized size = 0.95 \begin{align*} \frac{1}{128} \, c^{4}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} + \frac{3 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} - 11 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b - 11 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{2} + 3 \, \sqrt{c x^{2} + b} b^{3}}{b^{2} c^{4} x^{8}}\right )} \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^12,x, algorithm="giac")

[Out]

1/128*c^4*(3*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x^2 + b)^(7/2) - 11*(c*x^2 + b)^(5/2)*b -
11*(c*x^2 + b)^(3/2)*b^2 + 3*sqrt(c*x^2 + b)*b^3)/(b^2*c^4*x^8))*sgn(x)