3.2.85 \(\int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx\) [185]

Optimal. Leaf size=90 \[ -\frac {a (2 A b-3 a B) x}{b^4}+\frac {(A b-2 a B) x^2}{2 b^3}+\frac {B x^3}{3 b^2}+\frac {a^3 (A b-a B)}{b^5 (a+b x)}+\frac {a^2 (3 A b-4 a B) \log (a+b x)}{b^5} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \begin {gather*} \frac {a^3 (A b-a B)}{b^5 (a+b x)}+\frac {a^2 (3 A b-4 a B) \log (a+b x)}{b^5}-\frac {a x (2 A b-3 a B)}{b^4}+\frac {x^2 (A b-2 a B)}{2 b^3}+\frac {B x^3}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(a + b*x)^2,x]

[Out]

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

Rule 78

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

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Mathematica [A]
time = 0.04, size = 87, normalized size = 0.97 \begin {gather*} \frac {6 a b (-2 A b+3 a B) x+3 b^2 (A b-2 a B) x^2+2 b^3 B x^3+\frac {6 a^3 (A b-a B)}{a+b x}+6 a^2 (3 A b-4 a B) \log (a+b x)}{6 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 90, normalized size = 1.00

method	result	size
default	\(-\frac {-\frac {1}{3} b^{2} B \,x^{3}-\frac {1}{2} A \,b^{2} x^{2}+B a b \,x^{2}+2 a A b x -3 a^{2} B x}{b^{4}}+\frac {a^{2} \left (3 A b -4 B a \right ) \ln \left (b x +a \right )}{b^{5}}+\frac {a^{3} \left (A b -B a \right )}{b^{5} \left (b x +a \right )}\)	\(90\)
norman	\(\frac {\frac {a \left (3 a^{2} b A -4 a^{3} B \right )}{b^{5}}+\frac {B \,x^{4}}{3 b}+\frac {\left (3 A b -4 B a \right ) x^{3}}{6 b^{2}}-\frac {a \left (3 A b -4 B a \right ) x^{2}}{2 b^{3}}}{b x +a}+\frac {a^{2} \left (3 A b -4 B a \right ) \ln \left (b x +a \right )}{b^{5}}\)	\(96\)
risch	\(\frac {B \,x^{3}}{3 b^{2}}+\frac {A \,x^{2}}{2 b^{2}}-\frac {B a \,x^{2}}{b^{3}}-\frac {2 a A x}{b^{3}}+\frac {3 a^{2} B x}{b^{4}}+\frac {a^{3} A}{b^{4} \left (b x +a \right )}-\frac {a^{4} B}{b^{5} \left (b x +a \right )}+\frac {3 a^{2} \ln \left (b x +a \right ) A}{b^{4}}-\frac {4 a^{3} \ln \left (b x +a \right ) B}{b^{5}}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 101, normalized size = 1.12 \begin {gather*} -\frac {B a^{4} - A a^{3} b}{b^{6} x + a b^{5}} + \frac {2 \, B b^{2} x^{3} - 3 \, {\left (2 \, B a b - A b^{2}\right )} x^{2} + 6 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} x}{6 \, b^{4}} - \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x + a\right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.92, size = 140, normalized size = 1.56 \begin {gather*} \frac {2 \, B b^{4} x^{4} - 6 \, B a^{4} + 6 \, A a^{3} b - {\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} x^{3} + 3 \, {\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{2} + 6 \, {\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} x - 6 \, {\left (4 \, B a^{4} - 3 \, A a^{3} b + {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{6} x + a b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.21, size = 92, normalized size = 1.02 \begin {gather*} \frac {B x^{3}}{3 b^{2}} - \frac {a^{2} \left (- 3 A b + 4 B a\right ) \log {\left (a + b x \right )}}{b^{5}} + x^{2} \left (\frac {A}{2 b^{2}} - \frac {B a}{b^{3}}\right ) + x \left (- \frac {2 A a}{b^{3}} + \frac {3 B a^{2}}{b^{4}}\right ) + \frac {A a^{3} b - B a^{4}}{a b^{5} + b^{6} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.43, size = 144, normalized size = 1.60 \begin {gather*} \frac {{\left (b x + a\right )}^{3} {\left (2 \, B - \frac {3 \, {\left (4 \, B a b - A b^{2}\right )}}{{\left (b x + a\right )} b} + \frac {18 \, {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )}}{{\left (b x + a\right )}^{2} b^{2}}\right )}}{6 \, b^{5}} + \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{5}} - \frac {\frac {B a^{4} b^{3}}{b x + a} - \frac {A a^{3} b^{4}}{b x + a}}{b^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x + a)^3*(2*B - 3*(4*B*a*b - A*b^2)/((b*x + a)*b) + 18*(2*B*a^2*b^2 - A*a*b^3)/((b*x + a)^2*b^2))/b^5 +
(4*B*a^3 - 3*A*a^2*b)*log(abs(b*x + a)/((b*x + a)^2*abs(b)))/b^5 - (B*a^4*b^3/(b*x + a) - A*a^3*b^4/(b*x + a)
)/b^8

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Mupad [B]
time = 0.05, size = 115, normalized size = 1.28 \begin {gather*} x^2\,\left (\frac {A}{2\,b^2}-\frac {B\,a}{b^3}\right )-x\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )-\frac {\ln \left (a+b\,x\right )\,\left (4\,B\,a^3-3\,A\,a^2\,b\right )}{b^5}+\frac {B\,x^3}{3\,b^2}-\frac {B\,a^4-A\,a^3\,b}{b\,\left (x\,b^5+a\,b^4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(A + B*x))/(a + b*x)^2,x)

[Out]

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

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