\(\int \frac {x^5}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [199]

Optimal result

Integrand size = 24, antiderivative size = 205 \[ \int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {10 a^2}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^5}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a^4}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 a^3}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 205, 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} \[ \int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {10 a^2}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^5}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a^4}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 a^3}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

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

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 {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {x^5}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{b^{10}}-\frac {a^5}{b^{10} (a+b x)^5}+\frac {5 a^4}{b^{10} (a+b x)^4}-\frac {10 a^3}{b^{10} (a+b x)^3}+\frac {10 a^2}{b^{10} (a+b x)^2}-\frac {5 a}{b^{10} (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {10 a^2}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^5}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a^4}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 a^3}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.45 \[ \int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-77 a^5-248 a^4 b x-252 a^3 b^2 x^2-48 a^2 b^3 x^3+48 a b^4 x^4+12 b^5 x^5-60 a (a+b x)^4 \log (a+b x)}{12 b^6 (a+b x)^3 \sqrt {(a+b x)^2}} \]

[In]

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

[Out]

(-77*a^5 - 248*a^4*b*x - 252*a^3*b^2*x^2 - 48*a^2*b^3*x^3 + 48*a*b^4*x^4 + 12*b^5*x^5 - 60*a*(a + b*x)^4*Log[a
+ b*x])/(12*b^6*(a + b*x)^3*Sqrt[(a + b*x)^2])

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.52

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, x}{\left (b x +a \right ) b^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-10 a^{2} b^{2} x^{3}-25 a^{3} b \,x^{2}-\frac {65 a^{4} x}{3}-\frac {77 a^{5}}{12 b}\right )}{\left (b x +a \right )^{5} b^{5}}-\frac {5 \sqrt {\left (b x +a \right )^{2}}\, a \ln \left (b x +a \right )}{\left (b x +a \right ) b^{6}}\)	\(106\)
default	\(-\frac {\left (60 \ln \left (b x +a \right ) a \,b^{4} x^{4}-12 b^{5} x^{5}+240 \ln \left (b x +a \right ) a^{2} b^{3} x^{3}-48 a \,b^{4} x^{4}+360 \ln \left (b x +a \right ) a^{3} b^{2} x^{2}+48 a^{2} b^{3} x^{3}+240 \ln \left (b x +a \right ) a^{4} b x +252 a^{3} b^{2} x^{2}+60 a^{5} \ln \left (b x +a \right )+248 a^{4} b x +77 a^{5}\right ) \left (b x +a \right )}{12 b^{6} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(145\)

[In]

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

[Out]

((b*x+a)^2)^(1/2)/(b*x+a)/b^5*x+((b*x+a)^2)^(1/2)/(b*x+a)^5*(-10*a^2*b^2*x^3-25*a^3*b*x^2-65/3*a^4*x-77/12*a^5
/b)/b^5-5*((b*x+a)^2)^(1/2)/(b*x+a)/b^6*a*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.73 \[ \int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {12 \, b^{5} x^{5} + 48 \, a b^{4} x^{4} - 48 \, a^{2} b^{3} x^{3} - 252 \, a^{3} b^{2} x^{2} - 248 \, a^{4} b x - 77 \, a^{5} - 60 \, {\left (a b^{4} x^{4} + 4 \, a^{2} b^{3} x^{3} + 6 \, a^{3} b^{2} x^{2} + 4 \, a^{4} b x + a^{5}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}\right )}} \]

[In]

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

[Out]

1/12*(12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b*x - 77*a^5 - 60*(a*b^4*x^4 + 4*
a^2*b^3*x^3 + 6*a^3*b^2*x^2 + 4*a^4*b*x + a^5)*log(b*x + a))/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b
^7*x + a^4*b^6)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.55 \[ \int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {12 \, b^{5} x^{5} + 48 \, a b^{4} x^{4} - 48 \, a^{2} b^{3} x^{3} - 252 \, a^{3} b^{2} x^{2} - 248 \, a^{4} b x - 77 \, a^{5}}{12 \, {\left (b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}\right )}} - \frac {5 \, a \log \left (b x + a\right )}{b^{6}} \]

[In]

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

[Out]

1/12*(12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b*x - 77*a^5)/(b^10*x^4 + 4*a*b^9
*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^6) - 5*a*log(b*x + a)/b^6

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.44 \[ \int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {x}{b^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {5 \, a \log \left ({\left | b x + a \right |}\right )}{b^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {120 \, a^{2} b^{3} x^{3} + 300 \, a^{3} b^{2} x^{2} + 260 \, a^{4} b x + 77 \, a^{5}}{12 \, {\left (b x + a\right )}^{4} b^{6} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

x/(b^5*sgn(b*x + a)) - 5*a*log(abs(b*x + a))/(b^6*sgn(b*x + a)) - 1/12*(120*a^2*b^3*x^3 + 300*a^3*b^2*x^2 + 26
0*a^4*b*x + 77*a^5)/((b*x + a)^4*b^6*sgn(b*x + a))

Mupad [F(-1)]

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

[In]

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

[Out]

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