\(\int x^5 (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [163]

Optimal result

Integrand size = 24, antiderivative size = 231 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {a^5 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {a b^4 x^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^5 x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)} \]

[Out]

1/6*a^5*x^6*((b*x+a)^2)^(1/2)/(b*x+a)+5/7*a^4*b*x^7*((b*x+a)^2)^(1/2)/(b*x+a)+5/4*a^3*b^2*x^8*((b*x+a)^2)^(1/2
)/(b*x+a)+10/9*a^2*b^3*x^9*((b*x+a)^2)^(1/2)/(b*x+a)+1/2*a*b^4*x^10*((b*x+a)^2)^(1/2)/(b*x+a)+1/11*b^5*x^11*((
b*x+a)^2)^(1/2)/(b*x+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 231, 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} \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {b^5 x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {a b^4 x^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {a^5 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.33 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^6 \sqrt {(a+b x)^2} \left (462 a^5+1980 a^4 b x+3465 a^3 b^2 x^2+3080 a^2 b^3 x^3+1386 a b^4 x^4+252 b^5 x^5\right )}{2772 (a+b x)} \]

[In]

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

[Out]

(x^6*Sqrt[(a + b*x)^2]*(462*a^5 + 1980*a^4*b*x + 3465*a^3*b^2*x^2 + 3080*a^2*b^3*x^3 + 1386*a*b^4*x^4 + 252*b^
5*x^5))/(2772*(a + b*x))

Maple [A] (verified)

Time = 2.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.32

method	result	size
gosper	\(\frac {x^{6} \left (252 b^{5} x^{5}+1386 a \,b^{4} x^{4}+3080 a^{2} b^{3} x^{3}+3465 a^{3} b^{2} x^{2}+1980 a^{4} b x +462 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2772 \left (b x +a \right )^{5}}\)	\(74\)
default	\(\frac {x^{6} \left (252 b^{5} x^{5}+1386 a \,b^{4} x^{4}+3080 a^{2} b^{3} x^{3}+3465 a^{3} b^{2} x^{2}+1980 a^{4} b x +462 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2772 \left (b x +a \right )^{5}}\)	\(74\)
risch	\(\frac {a^{5} x^{6} \sqrt {\left (b x +a \right )^{2}}}{6 b x +6 a}+\frac {5 a^{4} b \,x^{7} \sqrt {\left (b x +a \right )^{2}}}{7 \left (b x +a \right )}+\frac {5 a^{3} b^{2} x^{8} \sqrt {\left (b x +a \right )^{2}}}{4 \left (b x +a \right )}+\frac {10 a^{2} b^{3} x^{9} \sqrt {\left (b x +a \right )^{2}}}{9 \left (b x +a \right )}+\frac {a \,b^{4} x^{10} \sqrt {\left (b x +a \right )^{2}}}{2 b x +2 a}+\frac {b^{5} x^{11} \sqrt {\left (b x +a \right )^{2}}}{11 b x +11 a}\)	\(154\)

[In]

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

[Out]

1/2772*x^6*(252*b^5*x^5+1386*a*b^4*x^4+3080*a^2*b^3*x^3+3465*a^3*b^2*x^2+1980*a^4*b*x+462*a^5)*((b*x+a)^2)^(5/
2)/(b*x+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.25 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{11} \, b^{5} x^{11} + \frac {1}{2} \, a b^{4} x^{10} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{7} \, a^{4} b x^{7} + \frac {1}{6} \, a^{5} x^{6} \]

[In]

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

[Out]

1/11*b^5*x^11 + 1/2*a*b^4*x^10 + 10/9*a^2*b^3*x^9 + 5/4*a^3*b^2*x^8 + 5/7*a^4*b*x^7 + 1/6*a^5*x^6

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.19 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{10}}{2772 b^{6}} + \frac {a^{9} x}{2772 b^{5}} - \frac {a^{8} x^{2}}{2772 b^{4}} + \frac {a^{7} x^{3}}{2772 b^{3}} - \frac {a^{6} x^{4}}{2772 b^{2}} + \frac {a^{5} x^{5}}{2772 b} + \frac {461 a^{4} x^{6}}{2772} + \frac {217 a^{3} b x^{7}}{396} + \frac {139 a^{2} b^{2} x^{8}}{198} + \frac {9 a b^{3} x^{9}}{22} + \frac {b^{4} x^{10}}{11}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{10} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {5 a^{8} \left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9} - \frac {10 a^{6} \left (a^{2} + 2 a b x\right )^{\frac {11}{2}}}{11} + \frac {10 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {13}{2}}}{13} - \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {15}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {17}{2}}}{17}}{32 a^{6} b^{6}} & \text {for}\: a b \neq 0 \\\frac {x^{6} \left (a^{2}\right )^{\frac {5}{2}}}{6} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**10/(2772*b**6) + a**9*x/(2772*b**5) - a**8*x**2/(2772*b**4) +
a**7*x**3/(2772*b**3) - a**6*x**4/(2772*b**2) + a**5*x**5/(2772*b) + 461*a**4*x**6/2772 + 217*a**3*b*x**7/396
+ 139*a**2*b**2*x**8/198 + 9*a*b**3*x**9/22 + b**4*x**10/11), Ne(b**2, 0)), ((-a**10*(a**2 + 2*a*b*x)**(7/2)/
7 + 5*a**8*(a**2 + 2*a*b*x)**(9/2)/9 - 10*a**6*(a**2 + 2*a*b*x)**(11/2)/11 + 10*a**4*(a**2 + 2*a*b*x)**(13/2)/
13 - a**2*(a**2 + 2*a*b*x)**(15/2)/3 + (a**2 + 2*a*b*x)**(17/2)/17)/(32*a**6*b**6), Ne(a*b, 0)), (x**6*(a**2)*
*(5/2)/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.82 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} x^{4}}{11 \, b^{2}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a x^{3}}{22 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{5} x}{6 \, b^{5}} + \frac {31 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a^{2} x^{2}}{198 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{6}}{6 \, b^{6}} - \frac {65 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a^{3} x}{396 \, b^{5}} + \frac {461 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a^{4}}{2772 \, b^{6}} \]

[In]

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

[Out]

1/11*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*x^4/b^2 - 3/22*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*a*x^3/b^3 - 1/6*(b^2*x^2 +
2*a*b*x + a^2)^(5/2)*a^5*x/b^5 + 31/198*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*a^2*x^2/b^4 - 1/6*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*a^6/b^6 - 65/396*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*a^3*x/b^5 + 461/2772*(b^2*x^2 + 2*a*b*x + a^2)^(
7/2)*a^4/b^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.46 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{11} \, b^{5} x^{11} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a b^{4} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, a^{3} b^{2} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, a^{4} b x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, a^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{11} \mathrm {sgn}\left (b x + a\right )}{2772 \, b^{6}} \]

[In]

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

[Out]

1/11*b^5*x^11*sgn(b*x + a) + 1/2*a*b^4*x^10*sgn(b*x + a) + 10/9*a^2*b^3*x^9*sgn(b*x + a) + 5/4*a^3*b^2*x^8*sgn
(b*x + a) + 5/7*a^4*b*x^7*sgn(b*x + a) + 1/6*a^5*x^6*sgn(b*x + a) - 1/2772*a^11*sgn(b*x + a)/b^6

Mupad [F(-1)]

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

[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)