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

Optimal. Leaf size=121 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (A b-2 a B)}{7 b^3}-\frac {a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^3}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {770, 76} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (A b-2 a B)}{7 b^3}-\frac {a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^3}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 125, normalized size = 1.03 \begin {gather*} \frac {x^2 \sqrt {(a+b x)^2} \left (28 a^5 (3 A+2 B x)+70 a^4 b x (4 A+3 B x)+84 a^3 b^2 x^2 (5 A+4 B x)+56 a^2 b^3 x^3 (6 A+5 B x)+20 a b^4 x^4 (7 A+6 B x)+3 b^5 x^5 (8 A+7 B x)\right )}{168 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[(a + b*x)^2]*(28*a^5*(3*A + 2*B*x) + 70*a^4*b*x*(4*A + 3*B*x) + 84*a^3*b^2*x^2*(5*A + 4*B*x) + 56*a^
2*b^3*x^3*(6*A + 5*B*x) + 20*a*b^4*x^4*(7*A + 6*B*x) + 3*b^5*x^5*(8*A + 7*B*x)))/(168*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.25, size = 0, normalized size = 0.00 \begin {gather*} \int x (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*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.41, size = 119, normalized size = 0.98 \begin {gather*} \frac {1}{8} \, B b^{5} x^{8} + \frac {1}{2} \, A a^{5} x^{2} + \frac {1}{7} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{7} + \frac {5}{6} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{6} + 2 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{5} + \frac {5}{4} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 221, normalized size = 1.83 \begin {gather*} \frac {1}{8} \, B b^{5} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, B a b^{4} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{7} \, A b^{5} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, B a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, A a b^{4} x^{6} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{3} b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, B a^{4} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, A a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a^{5} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, A a^{4} b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (B a^{8} - 4 \, A a^{7} b\right )} \mathrm {sgn}\left (b x + a\right )}{168 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*b^5*x^8*sgn(b*x + a) + 5/7*B*a*b^4*x^7*sgn(b*x + a) + 1/7*A*b^5*x^7*sgn(b*x + a) + 5/3*B*a^2*b^3*x^6*sgn
(b*x + a) + 5/6*A*a*b^4*x^6*sgn(b*x + a) + 2*B*a^3*b^2*x^5*sgn(b*x + a) + 2*A*a^2*b^3*x^5*sgn(b*x + a) + 5/4*B
*a^4*b*x^4*sgn(b*x + a) + 5/2*A*a^3*b^2*x^4*sgn(b*x + a) + 1/3*B*a^5*x^3*sgn(b*x + a) + 5/3*A*a^4*b*x^3*sgn(b*
x + a) + 1/2*A*a^5*x^2*sgn(b*x + a) + 1/168*(B*a^8 - 4*A*a^7*b)*sgn(b*x + a)/b^3

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maple [A]  time = 0.05, size = 140, normalized size = 1.16 \begin {gather*} \frac {\left (21 B \,b^{5} x^{6}+24 x^{5} A \,b^{5}+120 x^{5} B a \,b^{4}+140 x^{4} A a \,b^{4}+280 x^{4} B \,a^{2} b^{3}+336 A \,a^{2} b^{3} x^{3}+336 B \,a^{3} b^{2} x^{3}+420 x^{2} A \,a^{3} b^{2}+210 x^{2} B \,a^{4} b +280 x A \,a^{4} b +56 x B \,a^{5}+84 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x^{2}}{168 \left (b x +a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*x^2*(21*B*b^5*x^6+24*A*b^5*x^5+120*B*a*b^4*x^5+140*A*a*b^4*x^4+280*B*a^2*b^3*x^4+336*A*a^2*b^3*x^3+336*B
*a^3*b^2*x^3+420*A*a^3*b^2*x^2+210*B*a^4*b*x^2+280*A*a^4*b*x+56*B*a^5*x+84*A*a^5)*((b*x+a)^2)^(5/2)/(b*x+a)^5

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maxima [B]  time = 0.68, size = 183, normalized size = 1.51 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{2} x}{6 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a x}{6 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{3}}{6 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{2}}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B x}{8 \, b^{2}} - \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a}{56 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A}{7 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

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

[Out]

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

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