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

Optimal. Leaf size=202 \[ \frac{a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a^2 (3 A b-4 a B)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x (a+b x) (A b-3 a B)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a (a+b x) (A b-2 a B) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-((a^2*(3*A*b - 4*a*B))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (a^3*(A*b - a*B))/(2*b^5*(a + b*x)*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]) + ((A*b - 3*a*B)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*x^2*(a + b*x))/(2*b^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*(A*b - 2*a*B)*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])

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Rubi [A]  time = 0.137825, antiderivative size = 202, 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, 77} \[ \frac{a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a^2 (3 A b-4 a B)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x (a+b x) (A b-3 a B)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a (a+b x) (A b-2 a B) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(3*A*b - 4*a*B))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (a^3*(A*b - a*B))/(2*b^5*(a + b*x)*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]) + ((A*b - 3*a*B)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*x^2*(a + b*x))/(2*b^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*(A*b - 2*a*B)*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{x^3 (A+B x)}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac{A b-3 a B}{b^7}+\frac{B x}{b^6}+\frac{a^3 (-A b+a B)}{b^7 (a+b x)^3}-\frac{a^2 (-3 A b+4 a B)}{b^7 (a+b x)^2}+\frac{3 a (-A b+2 a B)}{b^7 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a^2 (3 A b-4 a B)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-3 a B) x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a (A b-2 a B) (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0546608, size = 117, normalized size = 0.58 \[ \frac{-a^2 b^2 x (4 A+11 B x)+a^3 (2 b B x-5 A b)+7 a^4 B+4 a b^3 x^2 (A-B x)+6 a (a+b x)^2 (2 a B-A b) \log (a+b x)+b^4 x^3 (2 A+B x)}{2 b^5 (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^4*B + 4*a*b^3*x^2*(A - B*x) + b^4*x^3*(2*A + B*x) - a^2*b^2*x*(4*A + 11*B*x) + a^3*(-5*A*b + 2*b*B*x) + 6
*a*(-(A*b) + 2*a*B)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.014, size = 191, normalized size = 1. \begin{align*} -{\frac{ \left ( -{b}^{4}B{x}^{4}+6\,A\ln \left ( bx+a \right ){x}^{2}a{b}^{3}-2\,A{x}^{3}{b}^{4}-12\,B\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}+4\,B{x}^{3}a{b}^{3}+12\,A\ln \left ( bx+a \right ) x{a}^{2}{b}^{2}-4\,A{x}^{2}a{b}^{3}-24\,B\ln \left ( bx+a \right ) x{a}^{3}b+11\,B{x}^{2}{a}^{2}{b}^{2}+6\,A\ln \left ( bx+a \right ){a}^{3}b+4\,A{a}^{2}{b}^{2}x-12\,B\ln \left ( bx+a \right ){a}^{4}-2\,B{a}^{3}bx+5\,A{a}^{3}b-7\,B{a}^{4} \right ) \left ( bx+a \right ) }{2\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

-1/2*(-b^4*B*x^4+6*A*ln(b*x+a)*x^2*a*b^3-2*A*x^3*b^4-12*B*ln(b*x+a)*x^2*a^2*b^2+4*B*x^3*a*b^3+12*A*ln(b*x+a)*x
*a^2*b^2-4*A*x^2*a*b^3-24*B*ln(b*x+a)*x*a^3*b+11*B*x^2*a^2*b^2+6*A*ln(b*x+a)*a^3*b+4*A*a^2*b^2*x-12*B*ln(b*x+a
)*a^4-2*B*a^3*b*x+5*A*a^3*b-7*B*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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Maxima [B]  time = 1.02563, size = 419, normalized size = 2.07 \begin{align*} \frac{B x^{3}}{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac{5 \, B a x^{2}}{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac{A x^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{6 \, B a^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{3 \, A a \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b} + \frac{9 \, B a^{4}}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{9 \, A a^{3} b}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{6 \, A a^{2} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{12 \, B a^{3} x}{{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{5 \, B a^{3}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac{2 \, A a^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac{5 \, B a^{4}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}} b^{4}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{A a^{3}}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{2}} \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)^(3/2),x, algorithm="maxima")

[Out]

1/2*B*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*B*a*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + A*x^2/(sqrt(
b^2*x^2 + 2*a*b*x + a^2)*b^2) + 6*B*a^2*log(x + a/b)/((b^2)^(3/2)*b^2) - 3*A*a*log(x + a/b)/((b^2)^(3/2)*b) +
9*B*a^4/((b^2)^(7/2)*(x + a/b)^2) - 9/2*A*a^3*b/((b^2)^(7/2)*(x + a/b)^2) - 6*A*a^2*x/((b^2)^(5/2)*(x + a/b)^2
) + 12*B*a^3*x/((b^2)^(5/2)*b*(x + a/b)^2) - 5*B*a^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) + 2*A*a^2/(sqrt(b^2*x
^2 + 2*a*b*x + a^2)*b^4) + 5/2*B*a^4/((b^2)^(3/2)*b^4*(x + a/b)^2) - A*a^3/((b^2)^(3/2)*b^3*(x + a/b)^2)

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Fricas [A]  time = 1.57715, size = 350, normalized size = 1.73 \begin{align*} \frac{B b^{4} x^{4} + 7 \, B a^{4} - 5 \, A a^{3} b - 2 \,{\left (2 \, B a b^{3} - A b^{4}\right )} x^{3} -{\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \,{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x + 6 \,{\left (2 \, B a^{4} - A a^{3} b +{\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 2 \,{\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \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)^(3/2),x, algorithm="fricas")

[Out]

1/2*(B*b^4*x^4 + 7*B*a^4 - 5*A*a^3*b - 2*(2*B*a*b^3 - A*b^4)*x^3 - (11*B*a^2*b^2 - 4*A*a*b^3)*x^2 + 2*(B*a^3*b
- 2*A*a^2*b^2)*x + 6*(2*B*a^4 - A*a^3*b + (2*B*a^2*b^2 - A*a*b^3)*x^2 + 2*(2*B*a^3*b - A*a^2*b^2)*x)*log(b*x
+ a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{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)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \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)^(3/2),x, algorithm="giac")

[Out]

sage0*x