3.179 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{x^{11}} \, dx\)

Optimal. Leaf size=231 \[ -\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)} \]

[Out]

-1/10*a^5*((b*x+a)^2)^(1/2)/x^10/(b*x+a)-5/9*a^4*b*((b*x+a)^2)^(1/2)/x^9/(b*x+a)-5/4*a^3*b^2*((b*x+a)^2)^(1/2)
/x^8/(b*x+a)-10/7*a^2*b^3*((b*x+a)^2)^(1/2)/x^7/(b*x+a)-5/6*a*b^4*((b*x+a)^2)^(1/2)/x^6/(b*x+a)-1/5*b^5*((b*x+
a)^2)^(1/2)/x^5/(b*x+a)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {646, 43} \[ -\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^11,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(10*x^10*(a + b*x)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*x^9*(a +
b*x)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^8*(a + b*x)) - (10*a^2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/(7*x^7*(a + b*x)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*x^6*(a + b*x)) - (b^5*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(5*x^5*(a + b*x))

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]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^{11}} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5 b^5}{x^{11}}+\frac {5 a^4 b^6}{x^{10}}+\frac {10 a^3 b^7}{x^9}+\frac {10 a^2 b^8}{x^8}+\frac {5 a b^9}{x^7}+\frac {b^{10}}{x^6}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 77, normalized size = 0.33 \[ -\frac {\sqrt {(a+b x)^2} \left (126 a^5+700 a^4 b x+1575 a^3 b^2 x^2+1800 a^2 b^3 x^3+1050 a b^4 x^4+252 b^5 x^5\right )}{1260 x^{10} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^11,x]

[Out]

-1/1260*(Sqrt[(a + b*x)^2]*(126*a^5 + 700*a^4*b*x + 1575*a^3*b^2*x^2 + 1800*a^2*b^3*x^3 + 1050*a*b^4*x^4 + 252
*b^5*x^5))/(x^10*(a + b*x))

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 57, normalized size = 0.25 \[ -\frac {252 \, b^{5} x^{5} + 1050 \, a b^{4} x^{4} + 1800 \, a^{2} b^{3} x^{3} + 1575 \, a^{3} b^{2} x^{2} + 700 \, a^{4} b x + 126 \, a^{5}}{1260 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^11,x, algorithm="fricas")

[Out]

-1/1260*(252*b^5*x^5 + 1050*a*b^4*x^4 + 1800*a^2*b^3*x^3 + 1575*a^3*b^2*x^2 + 700*a^4*b*x + 126*a^5)/x^10

________________________________________________________________________________________

giac [A]  time = 0.16, size = 108, normalized size = 0.47 \[ -\frac {b^{10} \mathrm {sgn}\left (b x + a\right )}{1260 \, a^{5}} - \frac {252 \, b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 1050 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 1800 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1575 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 700 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 126 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{1260 \, x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^11,x, algorithm="giac")

[Out]

-1/1260*b^10*sgn(b*x + a)/a^5 - 1/1260*(252*b^5*x^5*sgn(b*x + a) + 1050*a*b^4*x^4*sgn(b*x + a) + 1800*a^2*b^3*
x^3*sgn(b*x + a) + 1575*a^3*b^2*x^2*sgn(b*x + a) + 700*a^4*b*x*sgn(b*x + a) + 126*a^5*sgn(b*x + a))/x^10

________________________________________________________________________________________

maple [A]  time = 0.05, size = 74, normalized size = 0.32 \[ -\frac {\left (252 b^{5} x^{5}+1050 a \,b^{4} x^{4}+1800 a^{2} b^{3} x^{3}+1575 a^{3} b^{2} x^{2}+700 a^{4} b x +126 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{1260 \left (b x +a \right )^{5} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^11,x)

[Out]

-1/1260*(252*b^5*x^5+1050*a*b^4*x^4+1800*a^2*b^3*x^3+1575*a^3*b^2*x^2+700*a^4*b*x+126*a^5)*((b*x+a)^2)^(5/2)/x
^10/(b*x+a)^5

________________________________________________________________________________________

maxima [B]  time = 1.67, size = 312, normalized size = 1.35 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{10}}{6 \, a^{10}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{9}}{6 \, a^{9} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{8}}{6 \, a^{10} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{7}}{6 \, a^{9} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{6}}{6 \, a^{8} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{5}}{6 \, a^{7} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{4}}{6 \, a^{6} x^{6}} + \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{1260 \, a^{5} x^{7}} - \frac {29 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{180 \, a^{4} x^{8}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{90 \, a^{3} x^{9}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{10 \, a^{2} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^11,x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^10/a^10 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^9/(a^9*x) - 1/6*(b^2*x^2
+ 2*a*b*x + a^2)^(7/2)*b^8/(a^10*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^7/(a^9*x^3) - 1/6*(b^2*x^2 + 2*
a*b*x + a^2)^(7/2)*b^6/(a^8*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^5/(a^7*x^5) - 1/6*(b^2*x^2 + 2*a*b*x
+ a^2)^(7/2)*b^4/(a^6*x^6) + 209/1260*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^3/(a^5*x^7) - 29/180*(b^2*x^2 + 2*a*b*
x + a^2)^(7/2)*b^2/(a^4*x^8) + 13/90*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^9) - 1/10*(b^2*x^2 + 2*a*b*x + a
^2)^(7/2)/(a^2*x^10)

________________________________________________________________________________________

mupad [B]  time = 0.20, size = 207, normalized size = 0.90 \[ -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{10\,x^{10}\,\left (a+b\,x\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^8\,\left (a+b\,x\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^11,x)

[Out]

- (a^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(10*x^10*(a + b*x)) - (b^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a
+ b*x)) - (10*a^2*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (5*a^3*b^2*(a^2 + b^2*x^2 + 2*a*b*x
)^(1/2))/(4*x^8*(a + b*x)) - (5*a*b^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (5*a^4*b*(a^2 + b^2
*x^2 + 2*a*b*x)^(1/2))/(9*x^9*(a + b*x))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{11}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/x**11,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x**11, x)

________________________________________________________________________________________