\(\int \frac {a+b x}{(a c+(b c+a d) x+b d x^2)^2} \, dx\) [1815]

Optimal result

Integrand size = 27, antiderivative size = 56 \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {1}{(b c-a d) (c+d x)}+\frac {b \log (a+b x)}{(b c-a d)^2}-\frac {b \log (c+d x)}{(b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, 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 = {640, 46} \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {1}{(c+d x) (b c-a d)}+\frac {b \log (a+b x)}{(b c-a d)^2}-\frac {b \log (c+d x)}{(b c-a d)^2} \]

[In]

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

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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 {1}{(a+b x) (c+d x)^2} \, dx \\ & = \int \left (\frac {b^2}{(b c-a d)^2 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^2}-\frac {b d}{(b c-a d)^2 (c+d x)}\right ) \, dx \\ & = \frac {1}{(b c-a d) (c+d x)}+\frac {b \log (a+b x)}{(b c-a d)^2}-\frac {b \log (c+d x)}{(b c-a d)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b c-a d+b (c+d x) \log (a+b x)-b (c+d x) \log (c+d x)}{(b c-a d)^2 (c+d x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {1}{\left (a d -b c \right ) \left (d x +c \right )}-\frac {b \ln \left (d x +c \right )}{\left (a d -b c \right )^{2}}+\frac {b \ln \left (b x +a \right )}{\left (a d -b c \right )^{2}}\)	\(58\)
risch	\(-\frac {1}{\left (a d -b c \right ) \left (d x +c \right )}+\frac {b \ln \left (-b x -a \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}-\frac {b \ln \left (d x +c \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}\)	\(87\)
parallelrisch	\(\frac {\ln \left (b x +a \right ) x b c d -\ln \left (d x +c \right ) x b c d +\ln \left (b x +a \right ) b \,c^{2}-\ln \left (d x +c \right ) b \,c^{2}+x a \,d^{2}-b c d x}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (d x +c \right ) c}\)	\(94\)
norman	\(\frac {-\frac {a}{a d -b c}-\frac {b x}{a d -b c}}{\left (b x +a \right ) \left (d x +c \right )}+\frac {b \ln \left (b x +a \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}-\frac {b \ln \left (d x +c \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}\)	\(108\)

[In]

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

[Out]

-1/(a*d-b*c)/(d*x+c)-1/(a*d-b*c)^2*b*ln(d*x+c)+1/(a*d-b*c)^2*b*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.64 \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b c - a d + {\left (b d x + b c\right )} \log \left (b x + a\right ) - {\left (b d x + b c\right )} \log \left (d x + c\right )}{b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x} \]

[In]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (46) = 92\).

Time = 0.38 (sec) , antiderivative size = 233, normalized size of antiderivative = 4.16 \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=- \frac {b \log {\left (x + \frac {- \frac {a^{3} b d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} - \frac {3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d + \frac {b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{\left (a d - b c\right )^{2}} + \frac {b \log {\left (x + \frac {\frac {a^{3} b d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} + \frac {3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d - \frac {b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{\left (a d - b c\right )^{2}} - \frac {1}{a c d - b c^{2} + x \left (a d^{2} - b c d\right )} \]

[In]

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

[Out]

-b*log(x + (-a**3*b*d**3/(a*d - b*c)**2 + 3*a**2*b**2*c*d**2/(a*d - b*c)**2 - 3*a*b**3*c**2*d/(a*d - b*c)**2 +
a*b*d + b**4*c**3/(a*d - b*c)**2 + b**2*c)/(2*b**2*d))/(a*d - b*c)**2 + b*log(x + (a**3*b*d**3/(a*d - b*c)**2
- 3*a**2*b**2*c*d**2/(a*d - b*c)**2 + 3*a*b**3*c**2*d/(a*d - b*c)**2 + a*b*d - b**4*c**3/(a*d - b*c)**2 + b**
2*c)/(2*b**2*d))/(a*d - b*c)**2 - 1/(a*c*d - b*c**2 + x*(a*d**2 - b*c*d))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.61 \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b \log \left (b x + a\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac {b \log \left (d x + c\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} + \frac {1}{b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x} \]

[In]

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

[Out]

b*log(b*x + a)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2) - b*log(d*x + c)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2) + 1/(b*c^2 - a
*c*d + (b*c*d - a*d^2)*x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.66 \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} - \frac {b d \log \left ({\left | d x + c \right |}\right )}{b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}} + \frac {1}{{\left (b c - a d\right )} {\left (d x + c\right )}} \]

[In]

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

[Out]

b^2*log(abs(b*x + a))/(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2) - b*d*log(abs(d*x + c))/(b^2*c^2*d - 2*a*b*c*d^2 + a
^2*d^3) + 1/((b*c - a*d)*(d*x + c))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.38 \[ \int \frac {a+b x}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {a^2\,d^2-b^2\,c^2}{{\left (a\,d-b\,c\right )}^2}+\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{{\left (a\,d-b\,c\right )}^2}-\frac {1}{\left (a\,d-b\,c\right )\,\left (c+d\,x\right )} \]

[In]

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

[Out]

(2*b*atanh((a^2*d^2 - b^2*c^2)/(a*d - b*c)^2 + (2*b*d*x)/(a*d - b*c)))/(a*d - b*c)^2 - 1/((a*d - b*c)*(c + d*x
))