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

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

[Out]

(a^2*(A*b - a*B)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a*(A*b - a*B)*x^2*(a + b*x))/(2*b^3*Sqrt[
a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^3*(a + b*x))/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*x^4*(a + b*
x))/(4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a^3*(A*b - 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 = 0.114633, 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, 77} \[ \frac{x^3 (a+b x) (A b-a B)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a x^2 (a+b x) (A b-a B)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 x (a+b x) (A b-a B)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a^3 (a+b x) (A b-a B) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^4 (a+b x)}{4 b \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(A*b - a*B)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a*(A*b - a*B)*x^2*(a + b*x))/(2*b^3*Sqrt[
a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^3*(a + b*x))/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*x^4*(a + b*
x))/(4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a^3*(A*b - 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)}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{x^3 (A+B x)}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (-\frac{a^2 (-A b+a B)}{b^5}+\frac{a (-A b+a B) x}{b^4}+\frac{(A b-a B) x^2}{b^3}+\frac{B x^3}{b^2}+\frac{a^3 (-A b+a B)}{b^5 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{a^2 (A b-a B) x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a (A b-a B) x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^3 (a+b x)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^4 (a+b x)}{4 b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a^3 (A b-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 = 0.0462312, size = 96, normalized size = 0.45 \[ \frac{(a+b x) \left (b x \left (6 a^2 b (2 A+B x)-12 a^3 B-2 a b^2 x (3 A+2 B x)+b^3 x^2 (4 A+3 B x)\right )+12 a^3 (a B-A b) \log (a+b x)\right )}{12 b^5 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(b*x*(-12*a^3*B + 6*a^2*b*(2*A + B*x) - 2*a*b^2*x*(3*A + 2*B*x) + b^3*x^2*(4*A + 3*B*x)) + 12*a^3*(
-(A*b) + a*B)*Log[a + b*x]))/(12*b^5*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.009, size = 114, normalized size = 0.5 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( -3\,{b}^{4}B{x}^{4}-4\,A{x}^{3}{b}^{4}+4\,B{x}^{3}a{b}^{3}+6\,A{x}^{2}a{b}^{3}-6\,B{x}^{2}{a}^{2}{b}^{2}+12\,A\ln \left ( bx+a \right ){a}^{3}b-12\,A{a}^{2}{b}^{2}x-12\,B\ln \left ( bx+a \right ){a}^{4}+12\,B{a}^{3}bx \right ) }{12\,{b}^{5}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/((b*x+a)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.0387, size = 377, normalized size = 1.78 \begin{align*} \frac{13 \, B a^{4} \log \left (x + \frac{a}{b}\right )}{6 \,{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{5 \, A a^{3} b \log \left (x + \frac{a}{b}\right )}{3 \,{\left (b^{2}\right )}^{\frac{5}{2}}} + \frac{5 \, A a^{2} x}{3 \,{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{13 \, B a^{3} x}{6 \,{\left (b^{2}\right )}^{\frac{3}{2}} b} + \frac{13 \, B a^{2} x^{2}}{12 \, \sqrt{b^{2}} b^{2}} - \frac{5 \, A a x^{2}}{6 \, \sqrt{b^{2}} b} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B x^{3}}{4 \, b^{2}} - \frac{7 \, B a^{4} \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{6 \, b^{4}} + \frac{2 \, A a^{3} \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{3 \, b^{3}} - \frac{7 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B a x^{2}}{12 \, b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} A x^{2}}{3 \, b^{2}} + \frac{7 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3}}{6 \, b^{5}} - \frac{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2}}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

13/6*B*a^4*log(x + a/b)/(b^2)^(5/2) - 5/3*A*a^3*b*log(x + a/b)/(b^2)^(5/2) + 5/3*A*a^2*x/(b^2)^(3/2) - 13/6*B*
a^3*x/((b^2)^(3/2)*b) + 13/12*B*a^2*x^2/(sqrt(b^2)*b^2) - 5/6*A*a*x^2/(sqrt(b^2)*b) + 1/4*sqrt(b^2*x^2 + 2*a*b
*x + a^2)*B*x^3/b^2 - 7/6*B*a^4*sqrt(b^(-2))*log(x + a/b)/b^4 + 2/3*A*a^3*sqrt(b^(-2))*log(x + a/b)/b^3 - 7/12
*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*x^2/b^3 + 1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*x^2/b^2 + 7/6*sqrt(b^2*x^2 +
2*a*b*x + a^2)*B*a^3/b^5 - 2/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^2/b^4

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Fricas [A]  time = 1.57808, size = 196, normalized size = 0.92 \begin{align*} \frac{3 \, B b^{4} x^{4} - 4 \,{\left (B a b^{3} - A b^{4}\right )} x^{3} + 6 \,{\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 12 \,{\left (B a^{3} b - A a^{2} b^{2}\right )} x + 12 \,{\left (B a^{4} - A a^{3} b\right )} \log \left (b x + a\right )}{12 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.450113, size = 78, normalized size = 0.37 \begin{align*} \frac{B x^{4}}{4 b} + \frac{a^{3} \left (- A b + B a\right ) \log{\left (a + b x \right )}}{b^{5}} - \frac{x^{3} \left (- A b + B a\right )}{3 b^{2}} + \frac{x^{2} \left (- A a b + B a^{2}\right )}{2 b^{3}} - \frac{x \left (- A a^{2} b + B a^{3}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)/((b*x+a)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.32996, size = 200, normalized size = 0.94 \begin{align*} \frac{3 \, B b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) - 4 \, B a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 4 \, A b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \, B a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) - 6 \, A a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) - 12 \, B a^{3} x \mathrm{sgn}\left (b x + a\right ) + 12 \, A a^{2} b x \mathrm{sgn}\left (b x + a\right )}{12 \, b^{4}} + \frac{{\left (B a^{4} \mathrm{sgn}\left (b x + a\right ) - A a^{3} b \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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