3.569 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{13}} \, dx\)

Optimal. Leaf size=167 \[ -\frac{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*x^12*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*x
^10*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*x^8*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4])/(6*x^6*(a + b*x^2))

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Rubi [A]  time = 0.104971, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \[ -\frac{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^13,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*x^12*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*x
^10*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*x^8*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4])/(6*x^6*(a + b*x^2))

Rule 1111

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] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 646

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{13}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx,x,x^2\right )\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^3}{x^7} \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \operatorname{Subst}\left (\int \left (\frac{a^3 b^3}{x^7}+\frac{3 a^2 b^4}{x^6}+\frac{3 a b^5}{x^5}+\frac{b^6}{x^4}\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=-\frac{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0136877, size = 61, normalized size = 0.37 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (36 a^2 b x^2+10 a^3+45 a b^2 x^4+20 b^3 x^6\right )}{120 x^{12} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^13,x]

[Out]

-(Sqrt[(a + b*x^2)^2]*(10*a^3 + 36*a^2*b*x^2 + 45*a*b^2*x^4 + 20*b^3*x^6))/(120*x^12*(a + b*x^2))

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Maple [A]  time = 0.166, size = 58, normalized size = 0.4 \begin{align*} -{\frac{20\,{b}^{3}{x}^{6}+45\,a{x}^{4}{b}^{2}+36\,{a}^{2}b{x}^{2}+10\,{a}^{3}}{120\,{x}^{12} \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^13,x)

[Out]

-1/120*(20*b^3*x^6+45*a*b^2*x^4+36*a^2*b*x^2+10*a^3)*((b*x^2+a)^2)^(3/2)/x^12/(b*x^2+a)^3

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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((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^13,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.46861, size = 88, normalized size = 0.53 \begin{align*} -\frac{20 \, b^{3} x^{6} + 45 \, a b^{2} x^{4} + 36 \, a^{2} b x^{2} + 10 \, a^{3}}{120 \, x^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^13,x, algorithm="fricas")

[Out]

-1/120*(20*b^3*x^6 + 45*a*b^2*x^4 + 36*a^2*b*x^2 + 10*a^3)/x^12

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}{x^{13}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**13,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**13, x)

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Giac [A]  time = 1.11735, size = 93, normalized size = 0.56 \begin{align*} -\frac{20 \, b^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + 45 \, a b^{2} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 36 \, a^{2} b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 10 \, a^{3} \mathrm{sgn}\left (b x^{2} + a\right )}{120 \, x^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^13,x, algorithm="giac")

[Out]

-1/120*(20*b^3*x^6*sgn(b*x^2 + a) + 45*a*b^2*x^4*sgn(b*x^2 + a) + 36*a^2*b*x^2*sgn(b*x^2 + a) + 10*a^3*sgn(b*x
^2 + a))/x^12