\(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{x^{10}} \, dx\) [178]

Optimal result

Integrand size = 24, antiderivative size = 229 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^6 (a+b x)}-\frac {a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{x^5 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{x^5 (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^6 (a+b x)}-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^10,x]

[Out]

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

Rule 45

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 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^{10}} \, 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^{10}}+\frac {5 a^4 b^6}{x^9}+\frac {10 a^3 b^7}{x^8}+\frac {10 a^2 b^8}{x^7}+\frac {5 a b^9}{x^6}+\frac {b^{10}}{x^5}\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}}{9 x^9 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^6 (a+b x)}-\frac {a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{x^5 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (56 a^5+315 a^4 b x+720 a^3 b^2 x^2+840 a^2 b^3 x^3+504 a b^4 x^4+126 b^5 x^5\right )}{504 x^9 (a+b x)} \]

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^10,x]

[Out]

-1/504*(Sqrt[(a + b*x)^2]*(56*a^5 + 315*a^4*b*x + 720*a^3*b^2*x^2 + 840*a^2*b^3*x^3 + 504*a*b^4*x^4 + 126*b^5*
x^5))/(x^9*(a + b*x))

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.32

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {1}{4} b^{5} x^{5}-a \,b^{4} x^{4}-\frac {5}{3} a^{2} b^{3} x^{3}-\frac {10}{7} a^{3} b^{2} x^{2}-\frac {5}{8} a^{4} b x -\frac {1}{9} a^{5}\right )}{\left (b x +a \right ) x^{9}}\)	\(73\)
gosper	\(-\frac {\left (126 b^{5} x^{5}+504 a \,b^{4} x^{4}+840 a^{2} b^{3} x^{3}+720 a^{3} b^{2} x^{2}+315 a^{4} b x +56 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{504 x^{9} \left (b x +a \right )^{5}}\)	\(74\)
default	\(-\frac {\left (126 b^{5} x^{5}+504 a \,b^{4} x^{4}+840 a^{2} b^{3} x^{3}+720 a^{3} b^{2} x^{2}+315 a^{4} b x +56 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{504 x^{9} \left (b x +a \right )^{5}}\)	\(74\)

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^10,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.25 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=-\frac {126 \, b^{5} x^{5} + 504 \, a b^{4} x^{4} + 840 \, a^{2} b^{3} x^{3} + 720 \, a^{3} b^{2} x^{2} + 315 \, a^{4} b x + 56 \, a^{5}}{504 \, x^{9}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^10,x, algorithm="fricas")

[Out]

-1/504*(126*b^5*x^5 + 504*a*b^4*x^4 + 840*a^2*b^3*x^3 + 720*a^3*b^2*x^2 + 315*a^4*b*x + 56*a^5)/x^9

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{10}}\, dx \]

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/x**10,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x**10, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{9}}{6 \, a^{9}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{8}}{6 \, a^{8} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{7}}{6 \, a^{9} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{6}}{6 \, a^{8} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{5}}{6 \, a^{7} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{4}}{6 \, a^{6} x^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{6 \, a^{5} x^{6}} - \frac {83 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{504 \, a^{4} x^{7}} + \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{72 \, a^{3} x^{8}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{9 \, a^{2} x^{9}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^10,x, algorithm="maxima")

[Out]

-1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^9/a^9 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^8/(a^8*x) + 1/6*(b^2*x^2
+ 2*a*b*x + a^2)^(7/2)*b^7/(a^9*x^2) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^6/(a^8*x^3) + 1/6*(b^2*x^2 + 2*a*
b*x + a^2)^(7/2)*b^5/(a^7*x^4) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^4/(a^6*x^5) + 1/6*(b^2*x^2 + 2*a*b*x +
a^2)^(7/2)*b^3/(a^5*x^6) - 83/504*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^2/(a^4*x^7) + 11/72*(b^2*x^2 + 2*a*b*x + a
^2)^(7/2)*b/(a^3*x^8) - 1/9*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/(a^2*x^9)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.47 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=\frac {b^{9} \mathrm {sgn}\left (b x + a\right )}{504 \, a^{4}} - \frac {126 \, b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 504 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 840 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 720 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 315 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 56 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{504 \, x^{9}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^10,x, algorithm="giac")

[Out]

1/504*b^9*sgn(b*x + a)/a^4 - 1/504*(126*b^5*x^5*sgn(b*x + a) + 504*a*b^4*x^4*sgn(b*x + a) + 840*a^2*b^3*x^3*sg
n(b*x + a) + 720*a^3*b^2*x^2*sgn(b*x + a) + 315*a^4*b*x*sgn(b*x + a) + 56*a^5*sgn(b*x + a))/x^9

Mupad [B] (verification not implemented)

Time = 10.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^6\,\left (a+b\,x\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {a\,b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,x^8\,\left (a+b\,x\right )} \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^10,x)

[Out]

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