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

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

[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)

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Rubi [A]  time = 0.0757351, antiderivative size = 121, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (A b-2 a B)}{7 b^3}-\frac{a \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^3}+\frac{B \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \]

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 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]

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])

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*}

Mathematica [A]  time = 0.0397976, size = 125, normalized size = 1.03 \[ \frac{x^2 \sqrt{(a+b x)^2} \left (84 a^3 b^2 x^2 (5 A+4 B x)+56 a^2 b^3 x^3 (6 A+5 B x)+70 a^4 b x (4 A+3 B x)+28 a^5 (3 A+2 B x)+20 a b^4 x^4 (7 A+6 B x)+3 b^5 x^5 (8 A+7 B x)\right )}{168 (a+b x)} \]

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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Maple [A]  time = 0.004, size = 140, normalized size = 1.2 \begin{align*}{\frac{{x}^{2} \left ( 21\,B{b}^{5}{x}^{6}+24\,{x}^{5}A{b}^{5}+120\,{x}^{5}Ba{b}^{4}+140\,{x}^{4}Aa{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\,xA{a}^{4}b+56\,xB{a}^{5}+84\,A{a}^{5} \right ) }{168\, \left ( bx+a \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.31314, size = 258, normalized size = 2.13 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

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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Giac [B]  time = 1.211, size = 298, normalized size = 2.46 \begin{align*} \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{align*}

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