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

Optimal. Leaf size=121 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (A b-2 a B)}{5 b^3}-\frac {a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B)}{4 b^3}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

[Out]

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

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Rubi [A]  time = 0.07, 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} \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (A b-2 a B)}{5 b^3}-\frac {a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B)}{4 b^3}+\frac {B \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 87, normalized size = 0.72 \[ \frac {x^2 \sqrt {(a+b x)^2} \left (10 a^3 (3 A+2 B x)+15 a^2 b x (4 A+3 B x)+9 a b^2 x^2 (5 A+4 B x)+2 b^3 x^3 (6 A+5 B x)\right )}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[(a + b*x)^2]*(10*a^3*(3*A + 2*B*x) + 15*a^2*b*x*(4*A + 3*B*x) + 9*a*b^2*x^2*(5*A + 4*B*x) + 2*b^3*x^
3*(6*A + 5*B*x)))/(60*(a + b*x))

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fricas [A]  time = 0.77, size = 73, normalized size = 0.60 \[ \frac {1}{6} \, B b^{3} x^{6} + \frac {1}{2} \, A a^{3} x^{2} + \frac {1}{5} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{5} + \frac {3}{4} \, {\left (B a^{2} b + A a b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3} \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 148, normalized size = 1.22 \[ \frac {1}{6} \, B b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, B a b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a^{2} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, A a b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + A a^{2} b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (B a^{6} - 3 \, A a^{5} b\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{3}} \]

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)^(3/2),x, algorithm="giac")

[Out]

1/6*B*b^3*x^6*sgn(b*x + a) + 3/5*B*a*b^2*x^5*sgn(b*x + a) + 1/5*A*b^3*x^5*sgn(b*x + a) + 3/4*B*a^2*b*x^4*sgn(b
*x + a) + 3/4*A*a*b^2*x^4*sgn(b*x + a) + 1/3*B*a^3*x^3*sgn(b*x + a) + A*a^2*b*x^3*sgn(b*x + a) + 1/2*A*a^3*x^2
*sgn(b*x + a) + 1/60*(B*a^6 - 3*A*a^5*b)*sgn(b*x + a)/b^3

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maple [A]  time = 0.05, size = 92, normalized size = 0.76 \[ \frac {\left (10 b^{3} B \,x^{4}+12 A \,b^{3} x^{3}+36 x^{3} B a \,b^{2}+45 x^{2} A a \,b^{2}+45 B \,a^{2} b \,x^{2}+60 x A \,a^{2} b +20 B \,a^{3} x +30 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{2}}{60 \left (b x +a \right )^{3}} \]

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)^(3/2),x)

[Out]

1/60*x^2*(10*B*b^3*x^4+12*A*b^3*x^3+36*B*a*b^2*x^3+45*A*a*b^2*x^2+45*B*a^2*b*x^2+60*A*a^2*b*x+20*B*a^3*x+30*A*
a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [B]  time = 0.66, size = 183, normalized size = 1.51 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} x}{4 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3}}{4 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x}{6 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a}{30 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{5 \, b^{2}} \]

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)^(3/2),x, algorithm="maxima")

[Out]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^2*x/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a*x/b + 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*B*a^3/b^3 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^2/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2
)^(5/2)*B*x/b^2 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a/b^3 + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]

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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \]

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)**(3/2),x)

[Out]

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

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