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]

(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)

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Rubi [A]  time = 0.102302, antiderivative size = 167, normalized size of antiderivative = 1., 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 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^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*}

Mathematica [A]  time = 0.0441994, size = 125, normalized size = 0.75 \[ \frac{x^3 \sqrt{(a+b x)^2} \left (168 a^3 b^2 x^2 (6 A+5 B x)+120 a^2 b^3 x^3 (7 A+6 B x)+126 a^4 b x (5 A+4 B x)+42 a^5 (4 A+3 B x)+45 a b^4 x^4 (8 A+7 B x)+7 b^5 x^5 (9 A+8 B x)\right )}{504 (a+b x)} \]

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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Maple [A]  time = 0.006, size = 140, normalized size = 0.8 \begin{align*}{\frac{{x}^{3} \left ( 56\,B{b}^{5}{x}^{6}+63\,{x}^{5}A{b}^{5}+315\,{x}^{5}Ba{b}^{4}+360\,{x}^{4}Aa{b}^{4}+720\,{x}^{4}B{a}^{2}{b}^{3}+840\,{x}^{3}A{a}^{2}{b}^{3}+840\,{x}^{3}B{a}^{3}{b}^{2}+1008\,A{a}^{3}{b}^{2}{x}^{2}+504\,B{a}^{4}b{x}^{2}+630\,xA{a}^{4}b+126\,xB{a}^{5}+168\,A{a}^{5} \right ) }{504\, \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^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 [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^2*(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 = 1.25927, size = 255, normalized size = 1.53 \begin{align*} \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} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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**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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Giac [A]  time = 1.31995, size = 297, normalized size = 1.78 \begin{align*} \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}} \end{align*}

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