3.165 \(\int x^3 (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=144 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4}-\frac {3 a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^4}+\frac {3 a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4}-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^4} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 144, 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} \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4}-\frac {3 a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^4}+\frac {3 a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4}-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 77, normalized size = 0.53 \[ \frac {x^4 \sqrt {(a+b x)^2} \left (126 a^5+504 a^4 b x+840 a^3 b^2 x^2+720 a^2 b^3 x^3+315 a b^4 x^4+56 b^5 x^5\right )}{504 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.87, size = 56, normalized size = 0.39 \[ \frac {1}{9} \, b^{5} x^{9} + \frac {5}{8} \, a b^{4} x^{8} + \frac {10}{7} \, a^{2} b^{3} x^{7} + \frac {5}{3} \, a^{3} b^{2} x^{6} + a^{4} b x^{5} + \frac {1}{4} \, a^{5} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 106, normalized size = 0.74 \[ \frac {1}{9} \, b^{5} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{8} \, a b^{4} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, a^{2} b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + a^{4} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a^{5} x^{4} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{9} \mathrm {sgn}\left (b x + a\right )}{504 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/9*b^5*x^9*sgn(b*x + a) + 5/8*a*b^4*x^8*sgn(b*x + a) + 10/7*a^2*b^3*x^7*sgn(b*x + a) + 5/3*a^3*b^2*x^6*sgn(b*
x + a) + a^4*b*x^5*sgn(b*x + a) + 1/4*a^5*x^4*sgn(b*x + a) - 1/504*a^9*sgn(b*x + a)/b^4

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maple [A]  time = 0.05, size = 74, normalized size = 0.51 \[ \frac {\left (56 b^{5} x^{5}+315 a \,b^{4} x^{4}+720 a^{2} b^{3} x^{3}+840 a^{3} b^{2} x^{2}+504 a^{4} b x +126 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x^{4}}{504 \left (b x +a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

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

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maxima [A]  time = 1.35, size = 131, normalized size = 0.91 \[ -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{3} x}{6 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} x^{2}}{9 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{4}}{6 \, b^{4}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a x}{72 \, b^{3}} + \frac {83 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a^{2}}{504 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

int(x^3*(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 \[ \int x^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**3*((a + b*x)**2)**(5/2), x)

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