\(\int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx\) [1765]

Optimal result

Integrand size = 27, antiderivative size = 32 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=-\frac {b c-a d}{b^2 (a+b x)}+\frac {d \log (a+b x)}{b^2} \]

[Out]

(a*d-b*c)/b^2/(b*x+a)+d*ln(b*x+a)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {24, 45} \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=\frac {d \log (a+b x)}{b^2}-\frac {b c-a d}{b^2 (a+b x)} \]

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)/(a + b*x)^3,x]

[Out]

-((b*c - a*d)/(b^2*(a + b*x))) + (d*Log[a + b*x])/b^2

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {b^2 c+b^2 d x}{(a+b x)^2} \, dx}{b^2} \\ & = \frac {\int \left (\frac {b (b c-a d)}{(a+b x)^2}+\frac {b d}{a+b x}\right ) \, dx}{b^2} \\ & = -\frac {b c-a d}{b^2 (a+b x)}+\frac {d \log (a+b x)}{b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=\frac {-b c+a d}{b^2 (a+b x)}+\frac {d \log (a+b x)}{b^2} \]

[In]

Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)/(a + b*x)^3,x]

[Out]

(-(b*c) + a*d)/(b^2*(a + b*x)) + (d*Log[a + b*x])/b^2

Maple [A] (verified)

Time = 2.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {d \ln \left (b x +a \right )}{b^{2}}-\frac {-a d +b c}{b^{2} \left (b x +a \right )}\)	\(33\)
risch	\(\frac {a d}{b^{2} \left (b x +a \right )}-\frac {c}{b \left (b x +a \right )}+\frac {d \ln \left (b x +a \right )}{b^{2}}\)	\(39\)
parallelrisch	\(\frac {\ln \left (b x +a \right ) x b d +\ln \left (b x +a \right ) a d +a d -b c}{b^{2} \left (b x +a \right )}\)	\(39\)
norman	\(\frac {\frac {a \left (a d -b c \right )}{b^{2}}+\frac {\left (a d -b c \right ) x}{b}}{\left (b x +a \right )^{2}}+\frac {d \ln \left (b x +a \right )}{b^{2}}\)	\(48\)

[In]

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

[Out]

d*ln(b*x+a)/b^2-(-a*d+b*c)/b^2/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=-\frac {b c - a d - {\left (b d x + a d\right )} \log \left (b x + a\right )}{b^{3} x + a b^{2}} \]

[In]

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

[Out]

-(b*c - a*d - (b*d*x + a*d)*log(b*x + a))/(b^3*x + a*b^2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=\frac {a d - b c}{a b^{2} + b^{3} x} + \frac {d \log {\left (a + b x \right )}}{b^{2}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x**2)/(b*x+a)**3,x)

[Out]

(a*d - b*c)/(a*b**2 + b**3*x) + d*log(a + b*x)/b**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=-\frac {b c - a d}{b^{3} x + a b^{2}} + \frac {d \log \left (b x + a\right )}{b^{2}} \]

[In]

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

[Out]

-(b*c - a*d)/(b^3*x + a*b^2) + d*log(b*x + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=\frac {d \log \left ({\left | b x + a \right |}\right )}{b^{2}} - \frac {b c - a d}{{\left (b x + a\right )} b^{2}} \]

[In]

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

[Out]

d*log(abs(b*x + a))/b^2 - (b*c - a*d)/((b*x + a)*b^2)

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {a c+(b c+a d) x+b d x^2}{(a+b x)^3} \, dx=\frac {a\,d-b\,c}{b^2\,\left (a+b\,x\right )}+\frac {d\,\ln \left (a+b\,x\right )}{b^2} \]

[In]

int((a*c + x*(a*d + b*c) + b*d*x^2)/(a + b*x)^3,x)

[Out]

(a*d - b*c)/(b^2*(a + b*x)) + (d*log(a + b*x))/b^2