3.1806 \(\int \frac {1}{(a+b x) (a c+(b c+a d) x+b d x^2)} \, dx\)

Optimal. Leaf size=57 \[ -\frac {1}{(a+b x) (b c-a d)}-\frac {d \log (a+b x)}{(b c-a d)^2}+\frac {d \log (c+d x)}{(b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, 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 = {626, 44} \[ -\frac {1}{(a+b x) (b c-a d)}-\frac {d \log (a+b x)}{(b c-a d)^2}+\frac {d \log (c+d x)}{(b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(a*c + (b*c + a*d)*x + b*d*x^2)),x]

[Out]

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

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^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*} \int \frac {1}{(a+b x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx &=\int \frac {1}{(a+b x)^2 (c+d x)} \, dx\\ &=\int \left (\frac {b}{(b c-a d) (a+b x)^2}-\frac {b d}{(b c-a d)^2 (a+b x)}+\frac {d^2}{(b c-a d)^2 (c+d x)}\right ) \, dx\\ &=-\frac {1}{(b c-a d) (a+b x)}-\frac {d \log (a+b x)}{(b c-a d)^2}+\frac {d \log (c+d x)}{(b c-a d)^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 0.93 \[ \frac {d (a+b x) \log (c+d x)-d (a+b x) \log (a+b x)+a d-b c}{(a+b x) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(a*c + (b*c + a*d)*x + b*d*x^2)),x]

[Out]

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

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fricas [A]  time = 0.93, size = 93, normalized size = 1.63 \[ -\frac {b c - a d + {\left (b d x + a d\right )} \log \left (b x + a\right ) - {\left (b d x + a d\right )} \log \left (d x + c\right )}{a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 94, normalized size = 1.65 \[ -\frac {b d \log \left ({\left | b x + a \right |}\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} + \frac {d^{2} \log \left ({\left | d x + c \right |}\right )}{b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}} - \frac {1}{{\left (b c - a d\right )} {\left (b x + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 57, normalized size = 1.00 \[ -\frac {d \ln \left (b x +a \right )}{\left (a d -b c \right )^{2}}+\frac {d \ln \left (d x +c \right )}{\left (a d -b c \right )^{2}}+\frac {1}{\left (a d -b c \right ) \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(a*c+(a*d+b*c)*x+b*d*x^2),x)

[Out]

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

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maxima [A]  time = 1.01, size = 92, normalized size = 1.61 \[ -\frac {d \log \left (b x + a\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} + \frac {d \log \left (d x + c\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac {1}{a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.65, size = 76, normalized size = 1.33 \[ \frac {1}{\left (a\,d-b\,c\right )\,\left (a+b\,x\right )}-\frac {2\,d\,\mathrm {atanh}\left (\frac {a^2\,d^2-b^2\,c^2}{{\left (a\,d-b\,c\right )}^2}+\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{{\left (a\,d-b\,c\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)*(a*c + x*(a*d + b*c) + b*d*x^2)),x)

[Out]

1/((a*d - b*c)*(a + b*x)) - (2*d*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

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sympy [B]  time = 0.81, size = 233, normalized size = 4.09 \[ \frac {d \log {\left (x + \frac {- \frac {a^{3} d^{4}}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b c d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 a b^{2} c^{2} d^{2}}{\left (a d - b c\right )^{2}} + a d^{2} + \frac {b^{3} c^{3} d}{\left (a d - b c\right )^{2}} + b c d}{2 b d^{2}} \right )}}{\left (a d - b c\right )^{2}} - \frac {d \log {\left (x + \frac {\frac {a^{3} d^{4}}{\left (a d - b c\right )^{2}} - \frac {3 a^{2} b c d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a b^{2} c^{2} d^{2}}{\left (a d - b c\right )^{2}} + a d^{2} - \frac {b^{3} c^{3} d}{\left (a d - b c\right )^{2}} + b c d}{2 b d^{2}} \right )}}{\left (a d - b c\right )^{2}} + \frac {1}{a^{2} d - a b c + x \left (a b d - b^{2} c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(a*c+(a*d+b*c)*x+b*d*x**2),x)

[Out]

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

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