3.598 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{13}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{\left (a+b x^2\right )^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 a x^{12}} \]

[Out]

-((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*a*x^12)

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Rubi [A]  time = 0.0392379, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1111, 646, 37} \[ -\frac{\left (a+b x^2\right )^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 a x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^13,x]

[Out]

-((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*a*x^12)

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 37

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

Rubi steps

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

Mathematica [A]  time = 0.0179195, size = 81, normalized size = 1.98 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (20 a^2 b^3 x^6+15 a^3 b^2 x^4+6 a^4 b x^2+a^5+15 a b^4 x^8+6 b^5 x^{10}\right )}{12 x^{12} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^13,x]

[Out]

-(Sqrt[(a + b*x^2)^2]*(a^5 + 6*a^4*b*x^2 + 15*a^3*b^2*x^4 + 20*a^2*b^3*x^6 + 15*a*b^4*x^8 + 6*b^5*x^10))/(12*x
^12*(a + b*x^2))

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Maple [B]  time = 0.159, size = 78, normalized size = 1.9 \begin{align*} -{\frac{6\,{b}^{5}{x}^{10}+15\,a{b}^{4}{x}^{8}+20\,{a}^{2}{b}^{3}{x}^{6}+15\,{b}^{2}{a}^{3}{x}^{4}+6\,{a}^{4}b{x}^{2}+{a}^{5}}{12\,{x}^{12} \left ( b{x}^{2}+a \right ) ^{5}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{5}{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)^(5/2)/x^13,x)

[Out]

-1/12*(6*b^5*x^10+15*a*b^4*x^8+20*a^2*b^3*x^6+15*a^3*b^2*x^4+6*a^4*b*x^2+a^5)*((b*x^2+a)^2)^(5/2)/x^12/(b*x^2+
a)^5

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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)^(5/2)/x^13,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.47602, size = 127, normalized size = 3.1 \begin{align*} -\frac{6 \, b^{5} x^{10} + 15 \, a b^{4} x^{8} + 20 \, a^{2} b^{3} x^{6} + 15 \, a^{3} b^{2} x^{4} + 6 \, a^{4} b x^{2} + a^{5}}{12 \, 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)^(5/2)/x^13,x, algorithm="fricas")

[Out]

-1/12*(6*b^5*x^10 + 15*a*b^4*x^8 + 20*a^2*b^3*x^6 + 15*a^3*b^2*x^4 + 6*a^4*b*x^2 + a^5)/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{5}{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)**(5/2)/x**13,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**13, x)

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Giac [B]  time = 1.11985, size = 143, normalized size = 3.49 \begin{align*} -\frac{6 \, b^{5} x^{10} \mathrm{sgn}\left (b x^{2} + a\right ) + 15 \, a b^{4} x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + 20 \, a^{2} b^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + 15 \, a^{3} b^{2} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 6 \, a^{4} b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{5} \mathrm{sgn}\left (b x^{2} + a\right )}{12 \, 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)^(5/2)/x^13,x, algorithm="giac")

[Out]

-1/12*(6*b^5*x^10*sgn(b*x^2 + a) + 15*a*b^4*x^8*sgn(b*x^2 + a) + 20*a^2*b^3*x^6*sgn(b*x^2 + a) + 15*a^3*b^2*x^
4*sgn(b*x^2 + a) + 6*a^4*b*x^2*sgn(b*x^2 + a) + a^5*sgn(b*x^2 + a))/x^12