\(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{x^9} \, dx\) [177]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=-\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{8 a x^8}+\frac {b (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{28 a^2 x^7}-\frac {b^2 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{168 a^3 x^6} \]

[Out]

-1/8*(b*x+a)^5*((b*x+a)^2)^(1/2)/a/x^8+1/28*b*(b*x+a)^5*((b*x+a)^2)^(1/2)/a^2/x^7-1/168*b^2*(b*x+a)^5*((b*x+a)
^2)^(1/2)/a^3/x^6

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {660, 47, 37} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{8 a x^8}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{28 a^2 x^7}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{168 a^3 x^6} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^9,x]

[Out]

-1/8*((a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a*x^8) + (b*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(28*a
^2*x^7) - (b^2*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(168*a^3*x^6)

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]

Rule 47

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 660

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*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^9} \, dx}{b^4 \left (a b+b^2 x\right )} \\ & = -\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{8 a x^8}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^8} \, dx}{4 a b^3 \left (a b+b^2 x\right )} \\ & = -\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{8 a x^8}+\frac {b (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{28 a^2 x^7}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^7} \, dx}{28 a^2 b^2 \left (a b+b^2 x\right )} \\ & = -\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{8 a x^8}+\frac {b (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{28 a^2 x^7}-\frac {b^2 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{168 a^3 x^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.66 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=-\frac {\sqrt {(a+b x)^2} \left (21 a^5+120 a^4 b x+280 a^3 b^2 x^2+336 a^2 b^3 x^3+210 a b^4 x^4+56 b^5 x^5\right )}{168 x^8 (a+b x)} \]

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^9,x]

[Out]

-1/168*(Sqrt[(a + b*x)^2]*(21*a^5 + 120*a^4*b*x + 280*a^3*b^2*x^2 + 336*a^2*b^3*x^3 + 210*a*b^4*x^4 + 56*b^5*x
^5))/(x^8*(a + b*x))

Maple [A] (verified)

Time = 2.73 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.63

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {1}{3} b^{5} x^{5}-\frac {5}{4} a \,b^{4} x^{4}-2 a^{2} b^{3} x^{3}-\frac {5}{3} a^{3} b^{2} x^{2}-\frac {5}{7} a^{4} b x -\frac {1}{8} a^{5}\right )}{\left (b x +a \right ) x^{8}}\)	\(73\)
gosper	\(-\frac {\left (56 b^{5} x^{5}+210 a \,b^{4} x^{4}+336 a^{2} b^{3} x^{3}+280 a^{3} b^{2} x^{2}+120 a^{4} b x +21 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{168 x^{8} \left (b x +a \right )^{5}}\)	\(74\)
default	\(-\frac {\left (56 b^{5} x^{5}+210 a \,b^{4} x^{4}+336 a^{2} b^{3} x^{3}+280 a^{3} b^{2} x^{2}+120 a^{4} b x +21 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{168 x^{8} \left (b x +a \right )^{5}}\)	\(74\)

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.49 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=-\frac {56 \, b^{5} x^{5} + 210 \, a b^{4} x^{4} + 336 \, a^{2} b^{3} x^{3} + 280 \, a^{3} b^{2} x^{2} + 120 \, a^{4} b x + 21 \, a^{5}}{168 \, x^{8}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x, algorithm="fricas")

[Out]

-1/168*(56*b^5*x^5 + 210*a*b^4*x^4 + 336*a^2*b^3*x^3 + 280*a^3*b^2*x^2 + 120*a^4*b*x + 21*a^5)/x^8

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{9}}\, dx \]

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/x**9,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x**9, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (77) = 154\).

Time = 0.20 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.19 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{8}}{6 \, a^{8}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{7}}{6 \, a^{7} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{6}}{6 \, a^{8} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{5}}{6 \, a^{7} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{4}}{6 \, a^{6} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{6 \, a^{5} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{6 \, a^{4} x^{6}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{56 \, a^{3} x^{7}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{8 \, a^{2} x^{8}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^8/a^8 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^7/(a^7*x) - 1/6*(b^2*x^2 +
2*a*b*x + a^2)^(7/2)*b^6/(a^8*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^5/(a^7*x^3) - 1/6*(b^2*x^2 + 2*a*b
*x + a^2)^(7/2)*b^4/(a^6*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^3/(a^5*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + a
^2)^(7/2)*b^2/(a^4*x^6) + 9/56*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^7) - 1/8*(b^2*x^2 + 2*a*b*x + a^2)^(7/
2)/(a^2*x^8)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=-\frac {b^{8} \mathrm {sgn}\left (b x + a\right )}{168 \, a^{3}} - \frac {56 \, b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 210 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 336 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 280 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 120 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 21 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{168 \, x^{8}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x, algorithm="giac")

[Out]

-1/168*b^8*sgn(b*x + a)/a^3 - 1/168*(56*b^5*x^5*sgn(b*x + a) + 210*a*b^4*x^4*sgn(b*x + a) + 336*a^2*b^3*x^3*sg
n(b*x + a) + 280*a^3*b^2*x^2*sgn(b*x + a) + 120*a^4*b*x*sgn(b*x + a) + 21*a^5*sgn(b*x + a))/x^8

Mupad [B] (verification not implemented)

Time = 10.32 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.78 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,x^8\,\left (a+b\,x\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^3\,\left (a+b\,x\right )}-\frac {2\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^6\,\left (a+b\,x\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )} \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^9,x)

[Out]

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