3.945 \(\int \frac{(a+b x^2+c x^4)^{3/2}}{x^{11}} \, dx\)

Optimal. Leaf size=162 \[ -\frac{3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^3 x^4}+\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{512 a^{7/2}}+\frac{b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}} \]

[Out]

(-3*b*(b^2 - 4*a*c)*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*a^3*x^4) + (b*(2*a + b*x^2)*(a + b*x^2 + c*x^4
)^(3/2))/(32*a^2*x^8) - (a + b*x^2 + c*x^4)^(5/2)/(10*a*x^10) + (3*b*(b^2 - 4*a*c)^2*ArcTanh[(2*a + b*x^2)/(2*
Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(512*a^(7/2))

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Rubi [A]  time = 0.14011, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1114, 730, 720, 724, 206} \[ -\frac{3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^3 x^4}+\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{512 a^{7/2}}+\frac{b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x^11,x]

[Out]

(-3*b*(b^2 - 4*a*c)*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*a^3*x^4) + (b*(2*a + b*x^2)*(a + b*x^2 + c*x^4
)^(3/2))/(32*a^2*x^8) - (a + b*x^2 + c*x^4)^(5/2)/(10*a*x^10) + (3*b*(b^2 - 4*a*c)^2*ArcTanh[(2*a + b*x^2)/(2*
Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(512*a^(7/2))

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0]

Rule 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

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 (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}-\frac{b \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{4 a}\\ &=\frac{b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}+\frac{\left (3 b \left (b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x+c x^2}}{x^3} \, dx,x,x^2\right )}{64 a^2}\\ &=-\frac{3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^3 x^4}+\frac{b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}-\frac{\left (3 b \left (b^2-4 a c\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{512 a^3}\\ &=-\frac{3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^3 x^4}+\frac{b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}+\frac{\left (3 b \left (b^2-4 a c\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x^2}{\sqrt{a+b x^2+c x^4}}\right )}{256 a^3}\\ &=-\frac{3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^3 x^4}+\frac{b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}+\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{512 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.143446, size = 167, normalized size = 1.03 \[ \frac{b \left (16 a^{3/2} \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}-3 x^4 \left (b^2-4 a c\right ) \left (2 \sqrt{a} \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}-x^4 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )\right )\right )}{512 a^{7/2} x^8}-\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^11,x]

[Out]

-(a + b*x^2 + c*x^4)^(5/2)/(10*a*x^10) + (b*(16*a^(3/2)*(2*a + b*x^2)*(a + b*x^2 + c*x^4)^(3/2) - 3*(b^2 - 4*a
*c)*x^4*(2*Sqrt[a]*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4] - (b^2 - 4*a*c)*x^4*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*
Sqrt[a + b*x^2 + c*x^4])])))/(512*a^(7/2)*x^8)

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Maple [B]  time = 0.175, size = 337, normalized size = 2.1 \begin{align*} -{\frac{c}{5\,{x}^{6}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{b}^{5}}{512}\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 ){a}^{-{\frac{7}{2}}}}-{\frac{a}{10\,{x}^{10}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{11\,b}{80\,{x}^{8}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{7\,bc}{160\,a{x}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,{b}^{2}c}{64\,{a}^{2}{x}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,{b}^{3}c}{64}\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 ){a}^{-{\frac{5}{2}}}}+{\frac{3\,b{c}^{2}}{32}\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 ){a}^{-{\frac{3}{2}}}}-{\frac{{c}^{2}}{10\,a{x}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{{b}^{2}}{160\,a{x}^{6}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{{b}^{3}}{128\,{a}^{2}{x}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,{b}^{4}}{256\,{x}^{2}{a}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(3/2)/x^11,x)

[Out]

-1/5*c/x^6*(c*x^4+b*x^2+a)^(1/2)+3/512*b^5/a^(7/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)-1/10*a/
x^10*(c*x^4+b*x^2+a)^(1/2)-11/80*b/x^8*(c*x^4+b*x^2+a)^(1/2)-7/160/a*c*b/x^4*(c*x^4+b*x^2+a)^(1/2)+5/64/a^2*c*
b^2/x^2*(c*x^4+b*x^2+a)^(1/2)-3/64/a^(5/2)*c*b^3*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)+3/32/a^(3
/2)*c^2*b*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)-1/10/a*c^2/x^2*(c*x^4+b*x^2+a)^(1/2)-1/160*b^2/a
/x^6*(c*x^4+b*x^2+a)^(1/2)+1/128*b^3/a^2/x^4*(c*x^4+b*x^2+a)^(1/2)-3/256*b^4/a^3/x^2*(c*x^4+b*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^11,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.0542, size = 887, normalized size = 5.48 \begin{align*} \left [\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt{a} x^{10} \log \left (-\frac{{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \,{\left ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{8} + 176 \, a^{4} b x^{2} - 2 \,{\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{6} + 128 \, a^{5} + 8 \,{\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{4}\right )} \sqrt{c x^{4} + b x^{2} + a}}{5120 \, a^{4} x^{10}}, -\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt{-a} x^{10} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \,{\left ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{8} + 176 \, a^{4} b x^{2} - 2 \,{\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{6} + 128 \, a^{5} + 8 \,{\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{4}\right )} \sqrt{c x^{4} + b x^{2} + a}}{2560 \, a^{4} x^{10}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^11,x, algorithm="fricas")

[Out]

[1/5120*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(a)*x^10*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 + 4*sqrt(c*x^4 +
b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) - 4*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*x^8 + 176*a^4*b*
x^2 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^6 + 128*a^5 + 8*(a^3*b^2 + 32*a^4*c)*x^4)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^1
0), -1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-a)*x^10*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a
)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*x^8 + 176*a^4*b*x^2 - 2*(5
*a^2*b^3 - 28*a^3*b*c)*x^6 + 128*a^5 + 8*(a^3*b^2 + 32*a^4*c)*x^4)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^10)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2} + c x^{4}\right )^{\frac{3}{2}}}{x^{11}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(3/2)/x**11,x)

[Out]

Integral((a + b*x**2 + c*x**4)**(3/2)/x**11, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}}}{x^{11}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^11,x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2 + a)^(3/2)/x^11, x)