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 \[ \frac {x^3 (a+b x) (A b-a B)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x (a+b x) (A b-a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^2 (a+b x) (A b-a B)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^4 (a+b x)}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 (a+b x) (A b-a B) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 212, normalized size of antiderivative = 1.00, 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 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]))))

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 \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*}

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Mathematica [A]  time = 0.04, size = 96, normalized size = 0.45 \[ \frac {(a+b x) \left (12 a^3 (a B-A b) \log (a+b x)+b x \left (-12 a^3 B+6 a^2 b (2 A+B x)-2 a b^2 x (3 A+2 B x)+b^3 x^2 (4 A+3 B x)\right )\right )}{12 b^5 \sqrt {(a+b x)^2}} \]

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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fricas [A]  time = 0.85, size = 94, normalized size = 0.44 \[ \frac {3 \, B b^{4} x^{4} - 4 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 6 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 12 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x + 12 \, {\left (B a^{4} - A a^{3} b\right )} \log \left (b x + a\right )}{12 \, b^{5}} \]

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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giac [A]  time = 0.16, size = 148, normalized size = 0.70 \[ \frac {3 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) - 12 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 12 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right )}{12 \, b^{4}} + \frac {{\left (B a^{4} \mathrm {sgn}\left (b x + a\right ) - A a^{3} b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \]

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

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maple [A]  time = 0.05, size = 114, normalized size = 0.54 \[ -\frac {\left (b x +a \right ) \left (-3 B \,b^{4} x^{4}-4 A \,b^{4} x^{3}+4 B a \,b^{3} x^{3}+6 A a \,b^{3} x^{2}-6 B \,a^{2} b^{2} x^{2}+12 A \,a^{3} b \ln \left (b x +a \right )-12 A \,a^{2} b^{2} x -12 B \,a^{4} \ln \left (b x +a \right )+12 B \,a^{3} b x \right )}{12 \sqrt {\left (b x +a \right )^{2}}\, b^{5}} \]

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

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maxima [A]  time = 0.75, size = 212, normalized size = 1.00 \[ \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B x^{3}}{4 \, b^{2}} + \frac {13 \, B a^{2} x^{2}}{12 \, b^{3}} - \frac {5 \, A a x^{2}}{6 \, b^{2}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a x^{2}}{12 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A x^{2}}{3 \, b^{2}} - \frac {13 \, B a^{3} x}{6 \, b^{4}} + \frac {5 \, A a^{2} x}{3 \, b^{3}} + \frac {B a^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {A a^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3}}{6 \, b^{5}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2}}{3 \, b^{4}} \]

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]

1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*x^3/b^2 + 13/12*B*a^2*x^2/b^3 - 5/6*A*a*x^2/b^2 - 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 - 13/6*B*a^3*x/b^4 + 5/3*A*a^2*x/b^3 + B*
a^4*log(x + a/b)/b^5 - A*a^3*log(x + a/b)/b^4 + 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,\left (A+B\,x\right )}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.27, size = 85, normalized size = 0.40 \[ \frac {B x^{4}}{4 b} + \frac {a^{3} \left (- A b + B a\right ) \log {\left (a + b x \right )}}{b^{5}} + x^{3} \left (\frac {A}{3 b} - \frac {B a}{3 b^{2}}\right ) + x^{2} \left (- \frac {A a}{2 b^{2}} + \frac {B a^{2}}{2 b^{3}}\right ) + x \left (\frac {A a^{2}}{b^{3}} - \frac {B a^{3}}{b^{4}}\right ) \]

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

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