3.603 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{23}} \, dx\)

Optimal. Leaf size=255 \[ -\frac{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{22 x^{22} \left (a+b x^2\right )}-\frac{a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 x^{20} \left (a+b x^2\right )}-\frac{5 a^3 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )}-\frac{5 a^2 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac{5 a b^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac{b^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )} \]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(22*x^22*(a + b*x^2)) - (a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^20
*(a + b*x^2)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*x^18*(a + b*x^2)) - (5*a^2*b^3*Sqrt[a^2 + 2*a*b
*x^2 + b^2*x^4])/(8*x^16*(a + b*x^2)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(14*x^14*(a + b*x^2)) - (b^5
*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*x^12*(a + b*x^2))

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Rubi [A]  time = 0.153309, antiderivative size = 255, 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} \[ -\frac{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{22 x^{22} \left (a+b x^2\right )}-\frac{a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 x^{20} \left (a+b x^2\right )}-\frac{5 a^3 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )}-\frac{5 a^2 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac{5 a b^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac{b^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^23,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(22*x^22*(a + b*x^2)) - (a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^20
*(a + b*x^2)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*x^18*(a + b*x^2)) - (5*a^2*b^3*Sqrt[a^2 + 2*a*b
*x^2 + b^2*x^4])/(8*x^16*(a + b*x^2)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(14*x^14*(a + b*x^2)) - (b^5
*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*x^12*(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 )^{5/2}}{x^{23}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, 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^{12}} \, dx,x,x^2\right )}{2 b^4 \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^5 b^5}{x^{12}}+\frac{5 a^4 b^6}{x^{11}}+\frac{10 a^3 b^7}{x^{10}}+\frac{10 a^2 b^8}{x^9}+\frac{5 a b^9}{x^8}+\frac{b^{10}}{x^7}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{22 x^{22} \left (a+b x^2\right )}-\frac{a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 x^{20} \left (a+b x^2\right )}-\frac{5 a^3 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )}-\frac{5 a^2 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac{5 a b^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac{b^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0175634, size = 83, normalized size = 0.33 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (3465 a^2 b^3 x^6+3080 a^3 b^2 x^4+1386 a^4 b x^2+252 a^5+1980 a b^4 x^8+462 b^5 x^{10}\right )}{5544 x^{22} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^23,x]

[Out]

-(Sqrt[(a + b*x^2)^2]*(252*a^5 + 1386*a^4*b*x^2 + 3080*a^3*b^2*x^4 + 3465*a^2*b^3*x^6 + 1980*a*b^4*x^8 + 462*b
^5*x^10))/(5544*x^22*(a + b*x^2))

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Maple [A]  time = 0.163, size = 80, normalized size = 0.3 \begin{align*} -{\frac{462\,{b}^{5}{x}^{10}+1980\,a{b}^{4}{x}^{8}+3465\,{a}^{2}{b}^{3}{x}^{6}+3080\,{b}^{2}{a}^{3}{x}^{4}+1386\,{a}^{4}b{x}^{2}+252\,{a}^{5}}{5544\,{x}^{22} \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^23,x)

[Out]

-1/5544*(462*b^5*x^10+1980*a*b^4*x^8+3465*a^2*b^3*x^6+3080*a^3*b^2*x^4+1386*a^4*b*x^2+252*a^5)*((b*x^2+a)^2)^(
5/2)/x^22/(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^23,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.25197, size = 150, normalized size = 0.59 \begin{align*} -\frac{462 \, b^{5} x^{10} + 1980 \, a b^{4} x^{8} + 3465 \, a^{2} b^{3} x^{6} + 3080 \, a^{3} b^{2} x^{4} + 1386 \, a^{4} b x^{2} + 252 \, a^{5}}{5544 \, x^{22}} \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^23,x, algorithm="fricas")

[Out]

-1/5544*(462*b^5*x^10 + 1980*a*b^4*x^8 + 3465*a^2*b^3*x^6 + 3080*a^3*b^2*x^4 + 1386*a^4*b*x^2 + 252*a^5)/x^22

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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^{23}}\, 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**23,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**23, x)

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Giac [A]  time = 1.12091, size = 144, normalized size = 0.56 \begin{align*} -\frac{462 \, b^{5} x^{10} \mathrm{sgn}\left (b x^{2} + a\right ) + 1980 \, a b^{4} x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + 3465 \, a^{2} b^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + 3080 \, a^{3} b^{2} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 1386 \, a^{4} b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 252 \, a^{5} \mathrm{sgn}\left (b x^{2} + a\right )}{5544 \, x^{22}} \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^23,x, algorithm="giac")

[Out]

-1/5544*(462*b^5*x^10*sgn(b*x^2 + a) + 1980*a*b^4*x^8*sgn(b*x^2 + a) + 3465*a^2*b^3*x^6*sgn(b*x^2 + a) + 3080*
a^3*b^2*x^4*sgn(b*x^2 + a) + 1386*a^4*b*x^2*sgn(b*x^2 + a) + 252*a^5*sgn(b*x^2 + a))/x^22