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

Optimal. Leaf size=212 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8 (A b-4 a B)}{9 b^5}-\frac {3 a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7 (A b-2 a B)}{8 b^5}+\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (3 A b-4 a B)}{7 b^5}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^9}{10 b^5}-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^5} \]

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Rubi [A]  time = 0.12, antiderivative size = 212, 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} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8 (A b-4 a B)}{9 b^5}-\frac {3 a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7 (A b-2 a B)}{8 b^5}+\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (3 A b-4 a B)}{7 b^5}-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^5}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^9}{10 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(A*b - a*B)*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^5) + (a^2*(3*A*b - 4*a*B)*(a + b*x)^6*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(7*b^5) - (3*a*(A*b - 2*a*B)*(a + b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*b^5) + ((A
*b - 4*a*B)*(a + b*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*b^5) + (B*(a + b*x)^9*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(10*b^5)

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^3 (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^3 \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^3 (-A b+a B) \left (a b+b^2 x\right )^5}{b^4}-\frac {a^2 (-3 A b+4 a B) \left (a b+b^2 x\right )^6}{b^5}+\frac {3 a (-A b+2 a B) \left (a b+b^2 x\right )^7}{b^6}+\frac {(A b-4 a B) \left (a b+b^2 x\right )^8}{b^7}+\frac {B \left (a b+b^2 x\right )^9}{b^8}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac {a^3 (A b-a B) (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^5}+\frac {a^2 (3 A b-4 a B) (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^5}-\frac {3 a (A b-2 a B) (a+b x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{8 b^5}+\frac {(A b-4 a B) (a+b x)^8 \sqrt {a^2+2 a b x+b^2 x^2}}{9 b^5}+\frac {B (a+b x)^9 \sqrt {a^2+2 a b x+b^2 x^2}}{10 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 125, normalized size = 0.59 \begin {gather*} \frac {x^4 \sqrt {(a+b x)^2} \left (126 a^5 (5 A+4 B x)+420 a^4 b x (6 A+5 B x)+600 a^3 b^2 x^2 (7 A+6 B x)+450 a^2 b^3 x^3 (8 A+7 B x)+175 a b^4 x^4 (9 A+8 B x)+28 b^5 x^5 (10 A+9 B x)\right )}{2520 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[(a + b*x)^2]*(126*a^5*(5*A + 4*B*x) + 420*a^4*b*x*(6*A + 5*B*x) + 600*a^3*b^2*x^2*(7*A + 6*B*x) + 45
0*a^2*b^3*x^3*(8*A + 7*B*x) + 175*a*b^4*x^4*(9*A + 8*B*x) + 28*b^5*x^5*(10*A + 9*B*x)))/(2520*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.34, size = 0, normalized size = 0.00 \begin {gather*} \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][x^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2), x]

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fricas [A]  time = 0.42, size = 119, normalized size = 0.56 \begin {gather*} \frac {1}{10} \, B b^{5} x^{10} + \frac {1}{4} \, A a^{5} x^{4} + \frac {1}{9} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{9} + \frac {5}{8} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} + \frac {10}{7} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{7} + \frac {5}{6} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 221, normalized size = 1.04 \begin {gather*} \frac {1}{10} \, B b^{5} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{9} \, B a b^{4} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{9} \, A b^{5} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, B a^{2} b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{8} \, A a b^{4} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, B a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, A a^{2} b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, B a^{4} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, A a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B a^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + A a^{4} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A a^{5} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (2 \, B a^{10} - 5 \, A a^{9} b\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*b^5*x^10*sgn(b*x + a) + 5/9*B*a*b^4*x^9*sgn(b*x + a) + 1/9*A*b^5*x^9*sgn(b*x + a) + 5/4*B*a^2*b^3*x^8*s
gn(b*x + a) + 5/8*A*a*b^4*x^8*sgn(b*x + a) + 10/7*B*a^3*b^2*x^7*sgn(b*x + a) + 10/7*A*a^2*b^3*x^7*sgn(b*x + a)
+ 5/6*B*a^4*b*x^6*sgn(b*x + a) + 5/3*A*a^3*b^2*x^6*sgn(b*x + a) + 1/5*B*a^5*x^5*sgn(b*x + a) + A*a^4*b*x^5*sg
n(b*x + a) + 1/4*A*a^5*x^4*sgn(b*x + a) + 1/2520*(2*B*a^10 - 5*A*a^9*b)*sgn(b*x + a)/b^5

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maple [A]  time = 0.05, size = 140, normalized size = 0.66 \begin {gather*} \frac {\left (252 B \,b^{5} x^{6}+280 x^{5} A \,b^{5}+1400 x^{5} B a \,b^{4}+1575 x^{4} A a \,b^{4}+3150 x^{4} B \,a^{2} b^{3}+3600 A \,a^{2} b^{3} x^{3}+3600 B \,a^{3} b^{2} x^{3}+4200 x^{2} A \,a^{3} b^{2}+2100 x^{2} B \,a^{4} b +2520 x A \,a^{4} b +504 x B \,a^{5}+630 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x^{4}}{2520 \left (b x +a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*x^4*(252*B*b^5*x^6+280*A*b^5*x^5+1400*B*a*b^4*x^5+1575*A*a*b^4*x^4+3150*B*a^2*b^3*x^4+3600*A*a^2*b^3*x^
3+3600*B*a^3*b^2*x^3+4200*A*a^3*b^2*x^2+2100*B*a^4*b*x^2+2520*A*a^4*b*x+504*B*a^5*x+630*A*a^5)*((b*x+a)^2)^(5/
2)/(b*x+a)^5

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maxima [B]  time = 0.75, size = 301, normalized size = 1.42 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B x^{3}}{10 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{4} x}{6 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{3} x}{6 \, b^{3}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a x^{2}}{90 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A x^{2}}{9 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{5}}{6 \, b^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{4}}{6 \, b^{4}} + \frac {29 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{2} x}{180 \, b^{4}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a x}{72 \, b^{3}} - \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{3}}{1260 \, b^{5}} + \frac {83 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{2}}{504 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*x^3/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^4*x/b^4 - 1/6*(b^2*x^
2 + 2*a*b*x + a^2)^(5/2)*A*a^3*x/b^3 - 13/90*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a*x^2/b^3 + 1/9*(b^2*x^2 + 2*a*
b*x + a^2)^(7/2)*A*x^2/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^5/b^5 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/
2)*A*a^4/b^4 + 29/180*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^2*x/b^4 - 11/72*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a*
x/b^3 - 209/1260*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^3/b^5 + 83/504*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^2/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*(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 \begin {gather*} \int x^{3} \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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