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

Optimal. Leaf size=77 \[ \frac {A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {x^4 (A b-a B)}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {769, 646, 37} \begin {gather*} \frac {A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {x^4 (A b-a B)}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

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*x^3)/(3*a^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - ((A*b - a*B)*x^4)/(4*a^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 73, normalized size = 0.95 \begin {gather*} \frac {-3 a^3 B-a^2 b (A+12 B x)-2 a b^2 x (2 A+9 B x)-6 b^3 x^2 (A+2 B x)}{12 b^4 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

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]

(-3*a^3*B - 6*b^3*x^2*(A + 2*B*x) - 2*a*b^2*x*(2*A + 9*B*x) - a^2*b*(A + 12*B*x))/(12*b^4*(a + b*x)^3*Sqrt[(a
+ b*x)^2])

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 1.29, size = 381, normalized size = 4.95 \begin {gather*} \frac {-2 \left (3 a^7 b B-3 a^6 A b^2-3 a^3 b^5 B x^4-a^2 A b^6 x^4-12 a^2 b^6 B x^5-4 a A b^7 x^5-18 a b^7 B x^6-6 A b^8 x^6-12 b^8 B x^7\right )-2 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (3 a^6 B-3 a^5 A b-3 a^5 b B x+3 a^4 A b^2 x+3 a^4 b^2 B x^2-3 a^3 A b^3 x^2-3 a^3 b^3 B x^3+3 a^2 A b^4 x^3+6 a^2 b^4 B x^4-2 a A b^5 x^4+6 a b^5 B x^5+6 A b^6 x^5+12 b^6 B x^6\right )}{3 b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2} \left (-8 a^3 b^5-24 a^2 b^6 x-24 a b^7 x^2-8 b^8 x^3\right )+3 b^4 \sqrt {b^2} x^4 \left (8 a^4 b^4+32 a^3 b^5 x+48 a^2 b^6 x^2+32 a b^7 x^3+8 b^8 x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-3*a^5*A*b + 3*a^6*B + 3*a^4*A*b^2*x - 3*a^5*b*B*x - 3*a^3*A*b^3*
x^2 + 3*a^4*b^2*B*x^2 + 3*a^2*A*b^4*x^3 - 3*a^3*b^3*B*x^3 - 2*a*A*b^5*x^4 + 6*a^2*b^4*B*x^4 + 6*A*b^6*x^5 + 6*
a*b^5*B*x^5 + 12*b^6*B*x^6) - 2*(-3*a^6*A*b^2 + 3*a^7*b*B - a^2*A*b^6*x^4 - 3*a^3*b^5*B*x^4 - 4*a*A*b^7*x^5 -
12*a^2*b^6*B*x^5 - 6*A*b^8*x^6 - 18*a*b^7*B*x^6 - 12*b^8*B*x^7))/(3*b^4*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*
a^3*b^5 - 24*a^2*b^6*x - 24*a*b^7*x^2 - 8*b^8*x^3) + 3*b^4*Sqrt[b^2]*x^4*(8*a^4*b^4 + 32*a^3*b^5*x + 48*a^2*b^
6*x^2 + 32*a*b^7*x^3 + 8*b^8*x^4))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 105, normalized size = 1.36 \begin {gather*} -\frac {12 \, B b^{3} x^{3} + 3 \, B a^{3} + A a^{2} b + 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 4 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x}{12 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \end {gather*}

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]

-1/12*(12*B*b^3*x^3 + 3*B*a^3 + A*a^2*b + 6*(3*B*a*b^2 + A*b^3)*x^2 + 4*(3*B*a^2*b + A*a*b^2)*x)/(b^8*x^4 + 4*
a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^4)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

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]

sage0*x

________________________________________________________________________________________

maple [A]  time = 0.05, size = 77, normalized size = 1.00 \begin {gather*} -\frac {\left (b x +a \right ) \left (12 B \,b^{3} x^{3}+6 A \,b^{3} x^{2}+18 B a \,b^{2} x^{2}+4 A a \,b^{2} x +12 B \,a^{2} b x +A \,a^{2} b +3 B \,a^{3}\right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{4}} \end {gather*}

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]

-1/12*(b*x+a)*(12*B*b^3*x^3+6*A*b^3*x^2+18*B*a*b^2*x^2+4*A*a*b^2*x+12*B*a^2*b*x+A*a^2*b+3*B*a^3)/b^4/((b*x+a)^
2)^(5/2)

________________________________________________________________________________________

maxima [B]  time = 0.52, size = 156, normalized size = 2.03 \begin {gather*} -\frac {B x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, B a^{2}}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} - \frac {B a}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, B a^{2}}{3 \, b^{7} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {2 \, A a}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {B a^{3}}{4 \, b^{8} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {A a^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

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]

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

________________________________________________________________________________________

mupad [B]  time = 1.27, size = 201, normalized size = 2.61 \begin {gather*} -\frac {\left (\frac {B\,a^2-A\,a\,b}{3\,b^4}-\frac {a\,\left (\frac {A\,b^2-B\,a\,b}{3\,b^4}-\frac {B\,a}{3\,b^3}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^4}-\frac {\left (\frac {A\,b-2\,B\,a}{2\,b^4}-\frac {B\,a}{2\,b^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^3}-\frac {B\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^2}-\frac {a^2\,\left (\frac {A}{4\,b}-\frac {B\,a}{4\,b^2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^2\,{\left (a+b\,x\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \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**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)

________________________________________________________________________________________