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

Optimal. Leaf size=167 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7 (A b-3 a B)}{8 b^4}-\frac {a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (2 A b-3 a B)}{7 b^4}+\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^4}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 76} \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7 (A b-3 a B)}{8 b^4}-\frac {a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (2 A b-3 a B)}{7 b^4}+\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^4}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int x^2 (A+B x) \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^2 \left (a b+b^2 x\right )^5 (A+B x) \, 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^2 (-A b+a B) \left (a b+b^2 x\right )^5}{b^3}+\frac {a (-2 A b+3 a B) \left (a b+b^2 x\right )^6}{b^4}+\frac {(A b-3 a B) \left (a b+b^2 x\right )^7}{b^5}+\frac {B \left (a b+b^2 x\right )^8}{b^6}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {a^2 (A b-a B) (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^4}-\frac {a (2 A b-3 a B) (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^4}+\frac {(A b-3 a B) (a+b x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{8 b^4}+\frac {B (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 = 0.04, size = 125, normalized size = 0.75 \[ \frac {x^3 \sqrt {(a+b x)^2} \left (42 a^5 (4 A+3 B x)+126 a^4 b x (5 A+4 B x)+168 a^3 b^2 x^2 (6 A+5 B x)+120 a^2 b^3 x^3 (7 A+6 B x)+45 a b^4 x^4 (8 A+7 B x)+7 b^5 x^5 (9 A+8 B x)\right )}{504 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x)^2]*(42*a^5*(4*A + 3*B*x) + 126*a^4*b*x*(5*A + 4*B*x) + 168*a^3*b^2*x^2*(6*A + 5*B*x) + 120
*a^2*b^3*x^3*(7*A + 6*B*x) + 45*a*b^4*x^4*(8*A + 7*B*x) + 7*b^5*x^5*(9*A + 8*B*x)))/(504*(a + b*x))

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fricas [A]  time = 0.92, size = 118, normalized size = 0.71 \[ \frac {1}{9} \, B b^{5} x^{9} + \frac {1}{3} \, A a^{5} x^{3} + \frac {1}{8} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{8} + \frac {5}{7} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{7} + \frac {5}{3} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} + {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 220, normalized size = 1.32 \[ \frac {1}{9} \, B b^{5} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{8} \, B a b^{4} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{8} \, A b^{5} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, B a^{2} b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, A a b^{4} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, B a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, A a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + B a^{4} b x^{5} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{3} b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, B a^{5} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, A a^{4} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A a^{5} x^{3} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (B a^{9} - 3 \, A a^{8} b\right )} \mathrm {sgn}\left (b x + a\right )}{504 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 140, normalized size = 0.84 \[ \frac {\left (56 B \,b^{5} x^{6}+63 x^{5} A \,b^{5}+315 x^{5} B a \,b^{4}+360 x^{4} A a \,b^{4}+720 x^{4} B \,a^{2} b^{3}+840 A \,a^{2} b^{3} x^{3}+840 B \,a^{3} b^{2} x^{3}+1008 x^{2} A \,a^{3} b^{2}+504 x^{2} B \,a^{4} b +630 x A \,a^{4} b +126 x B \,a^{5}+168 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x^{3}}{504 \left (b x +a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/504*x^3*(56*B*b^5*x^6+63*A*b^5*x^5+315*B*a*b^4*x^5+360*A*a*b^4*x^4+720*B*a^2*b^3*x^4+840*A*a^2*b^3*x^3+840*B
*a^3*b^2*x^3+1008*A*a^3*b^2*x^2+504*B*a^4*b*x^2+630*A*a^4*b*x+126*B*a^5*x+168*A*a^5)*((b*x+a)^2)^(5/2)/(b*x+a)
^5

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maxima [B]  time = 0.61, size = 241, normalized size = 1.44 \[ -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{3} x}{6 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{2} x}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B x^{2}}{9 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{4}}{6 \, b^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{3}}{6 \, b^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a x}{72 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A x}{8 \, b^{2}} + \frac {83 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{2}}{504 \, b^{4}} - \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a}{56 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x+A)*(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)*B*a^3*x/b^3 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^2*x/b^2 + 1/9*(b^2*
x^2 + 2*a*b*x + a^2)^(7/2)*B*x^2/b^2 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^4/b^4 + 1/6*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*A*a^3/b^3 - 11/72*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a*x/b^3 + 1/8*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)
*A*x/b^2 + 83/504*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^2/b^4 - 9/56*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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