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

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

[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)

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Rubi [A]  time = 0.118961, antiderivative size = 212, 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} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^8 (A b-4 a B)}{9 b^5}-\frac{3 a \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7 (A b-2 a B)}{8 b^5}+\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (3 A b-4 a B)}{7 b^5}-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B)}{6 b^5}+\frac{B \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^9}{10 b^5} \]

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 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^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*}

Mathematica [A]  time = 0.040887, size = 125, normalized size = 0.59 \[ \frac{x^4 \sqrt{(a+b x)^2} \left (600 a^3 b^2 x^2 (7 A+6 B x)+450 a^2 b^3 x^3 (8 A+7 B x)+420 a^4 b x (6 A+5 B x)+126 a^5 (5 A+4 B x)+175 a b^4 x^4 (9 A+8 B x)+28 b^5 x^5 (10 A+9 B x)\right )}{2520 (a+b x)} \]

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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Maple [A]  time = 0.005, size = 140, normalized size = 0.7 \begin{align*}{\frac{{x}^{4} \left ( 252\,B{b}^{5}{x}^{6}+280\,{x}^{5}A{b}^{5}+1400\,{x}^{5}Ba{b}^{4}+1575\,{x}^{4}Aa{b}^{4}+3150\,{x}^{4}B{a}^{2}{b}^{3}+3600\,{x}^{3}A{a}^{2}{b}^{3}+3600\,{x}^{3}B{a}^{3}{b}^{2}+4200\,{x}^{2}A{a}^{3}{b}^{2}+2100\,{x}^{2}B{a}^{4}b+2520\,xA{a}^{4}b+504\,xB{a}^{5}+630\,A{a}^{5} \right ) }{2520\, \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^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 [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^3*(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 = 1.38224, size = 265, normalized size = 1.25 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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**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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Giac [A]  time = 1.17422, size = 298, normalized size = 1.41 \begin{align*} \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{align*}

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