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

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

[Out]

-1/22*a^5*((b*x^2+a)^2)^(1/2)/x^22/(b*x^2+a)-1/4*a^4*b*((b*x^2+a)^2)^(1/2)/x^20/(b*x^2+a)-5/9*a^3*b^2*((b*x^2+
a)^2)^(1/2)/x^18/(b*x^2+a)-5/8*a^2*b^3*((b*x^2+a)^2)^(1/2)/x^16/(b*x^2+a)-5/14*a*b^4*((b*x^2+a)^2)^(1/2)/x^14/
(b*x^2+a)-1/12*b^5*((b*x^2+a)^2)^(1/2)/x^12/(b*x^2+a)

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 255, normalized size of antiderivative = 1.00, 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 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])

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 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])

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 = 0.02, size = 83, normalized size = 0.33 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (252 a^5+1386 a^4 b x^2+3080 a^3 b^2 x^4+3465 a^2 b^3 x^6+1980 a b^4 x^8+462 b^5 x^{10}\right )}{5544 x^{22} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5544*(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))/(x^22*(a + b*x^2))

________________________________________________________________________________________

fricas [A]  time = 1.00, size = 59, normalized size = 0.23 \[ -\frac {462 \, b^{5} x^{10} + 1980 \, a b^{4} x^{8} + 3465 \, a^{2} b^{3} x^{6} + 3080 \, a^{3} b^{2} x^{4} + 1386 \, a^{4} b x^{2} + 252 \, a^{5}}{5544 \, x^{22}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.19, size = 107, normalized size = 0.42 \[ -\frac {462 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 1980 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 3465 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 3080 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 1386 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 252 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{5544 \, x^{22}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 80, normalized size = 0.31 \[ -\frac {\left (462 b^{5} x^{10}+1980 a \,b^{4} x^{8}+3465 a^{2} b^{3} x^{6}+3080 a^{3} b^{2} x^{4}+1386 a^{4} b \,x^{2}+252 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{5544 \left (b \,x^{2}+a \right )^{5} x^{22}} \]

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

________________________________________________________________________________________

maxima [A]  time = 1.34, size = 57, normalized size = 0.22 \[ -\frac {b^{5}}{12 \, x^{12}} - \frac {5 \, a b^{4}}{14 \, x^{14}} - \frac {5 \, a^{2} b^{3}}{8 \, x^{16}} - \frac {5 \, a^{3} b^{2}}{9 \, x^{18}} - \frac {a^{4} b}{4 \, x^{20}} - \frac {a^{5}}{22 \, x^{22}} \]

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]

-1/12*b^5/x^12 - 5/14*a*b^4/x^14 - 5/8*a^2*b^3/x^16 - 5/9*a^3*b^2/x^18 - 1/4*a^4*b/x^20 - 1/22*a^5/x^22

________________________________________________________________________________________

mupad [B]  time = 4.23, size = 231, normalized size = 0.91 \[ -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{22\,x^{22}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{12\,x^{12}\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^{20}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^{16}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^{18}\,\left (b\,x^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^23,x)

[Out]

- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(22*x^22*(a + b*x^2)) - (b^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(12*
x^12*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (a^4*b*(a^2 + b^2*x^4
+ 2*a*b*x^2)^(1/2))/(4*x^20*(a + b*x^2)) - (5*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(8*x^16*(a + b*x^2))
- (5*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(9*x^18*(a + b*x^2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{23}}\, dx \]

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)

________________________________________________________________________________________