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

Optimal. Leaf size=231 \[ \frac {b^5 x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {a b^4 x^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {a^5 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

[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)

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Rubi [A]  time = 0.06, 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} \[ \frac {b^5 x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {a b^4 x^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {a^5 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)} \]

Antiderivative was successfully verified.

[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 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 x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx &=\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*}

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Mathematica [A]  time = 0.02, size = 77, normalized size = 0.33 \[ \frac {x^6 \sqrt {(a+b x)^2} \left (462 a^5+1980 a^4 b x+3465 a^3 b^2 x^2+3080 a^2 b^3 x^3+1386 a b^4 x^4+252 b^5 x^5\right )}{2772 (a+b x)} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.92, size = 57, normalized size = 0.25 \[ \frac {1}{11} \, b^{5} x^{11} + \frac {1}{2} \, a b^{4} x^{10} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{7} \, a^{4} b x^{7} + \frac {1}{6} \, a^{5} x^{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.16, size = 107, normalized size = 0.46 \[ \frac {1}{11} \, b^{5} x^{11} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a b^{4} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, a^{3} b^{2} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, a^{4} b x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, a^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{11} \mathrm {sgn}\left (b x + a\right )}{2772 \, b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.05, size = 74, normalized size = 0.32 \[ \frac {\left (252 b^{5} x^{5}+1386 a \,b^{4} x^{4}+3080 a^{2} b^{3} x^{3}+3465 a^{3} b^{2} x^{2}+1980 a^{4} b x +462 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x^{6}}{2772 \left (b x +a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [A]  time = 1.46, size = 189, normalized size = 0.82 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} x^{4}}{11 \, b^{2}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a x^{3}}{22 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{5} x}{6 \, b^{5}} + \frac {31 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a^{2} x^{2}}{198 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{6}}{6 \, b^{6}} - \frac {65 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a^{3} x}{396 \, b^{5}} + \frac {461 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a^{4}}{2772 \, b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^5\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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