3.600 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{17}} \, dx\)

Optimal. Leaf size=128 \[ -\frac{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{336 a^3 x^{12}}+\frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{56 a^2 x^{14}}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{16 a x^{16}} \]

[Out]

-((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*a*x^16) + (b*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4
])/(56*a^2*x^14) - (b^2*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(336*a^3*x^12)

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Rubi [A]  time = 0.0911089, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1111, 646, 45, 37} \[ -\frac{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{336 a^3 x^{12}}+\frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{56 a^2 x^{14}}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{16 a x^{16}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^17,x]

[Out]

-((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*a*x^16) + (b*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4
])/(56*a^2*x^14) - (b^2*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(336*a^3*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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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^{17}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, 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^9} \, 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}}{16 a x^{16}}-\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^8} \, dx,x,x^2\right )}{8 a b^3 \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}}{16 a x^{16}}+\frac{b \left (a+b x^2\right )^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{56 a^2 x^{14}}+\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 )}{56 a^2 b^2 \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}}{16 a x^{16}}+\frac{b \left (a+b x^2\right )^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{56 a^2 x^{14}}-\frac{b^2 \left (a+b x^2\right )^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{336 a^3 x^{12}}\\ \end{align*}

Mathematica [A]  time = 0.0177042, size = 83, normalized size = 0.65 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (336 a^2 b^3 x^6+280 a^3 b^2 x^4+120 a^4 b x^2+21 a^5+210 a b^4 x^8+56 b^5 x^{10}\right )}{336 x^{16} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^17,x]

[Out]

-(Sqrt[(a + b*x^2)^2]*(21*a^5 + 120*a^4*b*x^2 + 280*a^3*b^2*x^4 + 336*a^2*b^3*x^6 + 210*a*b^4*x^8 + 56*b^5*x^1
0))/(336*x^16*(a + b*x^2))

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Maple [A]  time = 0.165, size = 80, normalized size = 0.6 \begin{align*} -{\frac{56\,{b}^{5}{x}^{10}+210\,a{b}^{4}{x}^{8}+336\,{a}^{2}{b}^{3}{x}^{6}+280\,{b}^{2}{a}^{3}{x}^{4}+120\,{a}^{4}b{x}^{2}+21\,{a}^{5}}{336\,{x}^{16} \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^17,x)

[Out]

-1/336*(56*b^5*x^10+210*a*b^4*x^8+336*a^2*b^3*x^6+280*a^3*b^2*x^4+120*a^4*b*x^2+21*a^5)*((b*x^2+a)^2)^(5/2)/x^
16/(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^17,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.488, size = 140, normalized size = 1.09 \begin{align*} -\frac{56 \, b^{5} x^{10} + 210 \, a b^{4} x^{8} + 336 \, a^{2} b^{3} x^{6} + 280 \, a^{3} b^{2} x^{4} + 120 \, a^{4} b x^{2} + 21 \, a^{5}}{336 \, x^{16}} \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^17,x, algorithm="fricas")

[Out]

-1/336*(56*b^5*x^10 + 210*a*b^4*x^8 + 336*a^2*b^3*x^6 + 280*a^3*b^2*x^4 + 120*a^4*b*x^2 + 21*a^5)/x^16

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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^{17}}\, 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**17,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**17, x)

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Giac [A]  time = 1.1318, size = 144, normalized size = 1.12 \begin{align*} -\frac{56 \, b^{5} x^{10} \mathrm{sgn}\left (b x^{2} + a\right ) + 210 \, a b^{4} x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + 336 \, a^{2} b^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + 280 \, a^{3} b^{2} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 120 \, a^{4} b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 21 \, a^{5} \mathrm{sgn}\left (b x^{2} + a\right )}{336 \, x^{16}} \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^17,x, algorithm="giac")

[Out]

-1/336*(56*b^5*x^10*sgn(b*x^2 + a) + 210*a*b^4*x^8*sgn(b*x^2 + a) + 336*a^2*b^3*x^6*sgn(b*x^2 + a) + 280*a^3*b
^2*x^4*sgn(b*x^2 + a) + 120*a^4*b*x^2*sgn(b*x^2 + a) + 21*a^5*sgn(b*x^2 + a))/x^16