\(\int \frac {(a+b x)^3}{(a c+(b c+a d) x+b d x^2)^2} \, dx\) [1813]

Optimal result

Integrand size = 29, antiderivative size = 31 \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b c-a d}{d^2 (c+d x)}+\frac {b \log (c+d x)}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {640, 45} \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b c-a d}{d^2 (c+d x)}+\frac {b \log (c+d x)}{d^2} \]

[In]

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

[Out]

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

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])

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b c-a d}{d^2 (c+d x)}+\frac {b \log (c+d x)}{d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {b \ln \left (d x +c \right )}{d^{2}}-\frac {a d -b c}{d^{2} \left (d x +c \right )}\)	\(33\)
risch	\(\frac {b \ln \left (d x +c \right )}{d^{2}}-\frac {a}{d \left (d x +c \right )}+\frac {b c}{d^{2} \left (d x +c \right )}\)	\(39\)
parallelrisch	\(\frac {\ln \left (d x +c \right ) x b d +\ln \left (d x +c \right ) b c -a d +b c}{\left (d x +c \right ) d^{2}}\)	\(39\)
norman	\(\frac {-\frac {a \left (a b d -b^{2} c \right )}{d^{2} b}-\frac {\left (a \,b^{2} d -b^{3} c \right ) x}{d^{2} b}}{\left (b x +a \right ) \left (d x +c \right )}+\frac {b \ln \left (d x +c \right )}{d^{2}}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b c - a d + {\left (b d x + b c\right )} \log \left (d x + c\right )}{d^{3} x + c d^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b \log {\left (c + d x \right )}}{d^{2}} + \frac {- a d + b c}{c d^{2} + d^{3} x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b c - a d}{d^{3} x + c d^{2}} + \frac {b \log \left (d x + c\right )}{d^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b \log \left ({\left | d x + c \right |}\right )}{d^{2}} + \frac {b c - a d}{{\left (d x + c\right )} d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^3}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b\,\ln \left (c+d\,x\right )}{d^2}-\frac {a\,d-b\,c}{d^2\,\left (c+d\,x\right )} \]

[In]

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

[Out]

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