\(\int \frac {(a c+(b c+a d) x+b d x^2)^2}{(a+b x)^4} \, dx\) [1776]

Optimal result

Integrand size = 29, antiderivative size = 51 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^2 x}{b^2}-\frac {(b c-a d)^2}{b^3 (a+b x)}+\frac {2 d (b c-a d) \log (a+b x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, 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} \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=-\frac {(b c-a d)^2}{b^3 (a+b x)}+\frac {2 d (b c-a d) \log (a+b x)}{b^3}+\frac {d^2 x}{b^2} \]

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)^2/(a + b*x)^4,x]

[Out]

(d^2*x)/b^2 - (b*c - a*d)^2/(b^3*(a + b*x)) + (2*d*(b*c - a*d)*Log[a + b*x])/b^3

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {b d^2 x-\frac {(b c-a d)^2}{a+b x}+2 d (b c-a d) \log (a+b x)}{b^3} \]

[In]

Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)^2/(a + b*x)^4,x]

[Out]

(b*d^2*x - (b*c - a*d)^2/(a + b*x) + 2*d*(b*c - a*d)*Log[a + b*x])/b^3

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {d^{2} x}{b^{2}}-\frac {2 d \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{3}}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{b^{3} \left (b x +a \right )}\)	\(63\)
risch	\(\frac {d^{2} x}{b^{2}}-\frac {2 d^{2} \ln \left (b x +a \right ) a}{b^{3}}+\frac {2 d \ln \left (b x +a \right ) c}{b^{2}}-\frac {a^{2} d^{2}}{b^{3} \left (b x +a \right )}+\frac {2 a c d}{b^{2} \left (b x +a \right )}-\frac {c^{2}}{b \left (b x +a \right )}\)	\(86\)
parallelrisch	\(-\frac {2 \ln \left (b x +a \right ) x a b \,d^{2}-2 \ln \left (b x +a \right ) x \,b^{2} c d -d^{2} x^{2} b^{2}+2 \ln \left (b x +a \right ) a^{2} d^{2}-2 \ln \left (b x +a \right ) a b c d +2 a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{b^{3} \left (b x +a \right )}\)	\(100\)
norman	\(\frac {b \,d^{2} x^{4}-\frac {a^{2} \left (4 a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{b^{3}}-\frac {\left (7 a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{b}-\frac {a \left (10 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) x}{b^{2}}}{\left (b x +a \right )^{3}}-\frac {2 d \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{3}}\)	\(129\)

[In]

int((b*d*x^2+(a*d+b*c)*x+a*c)^2/(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {b^{2} d^{2} x^{2} + a b d^{2} x - b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} + 2 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^2/(b*x+a)^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {- a^{2} d^{2} + 2 a b c d - b^{2} c^{2}}{a b^{3} + b^{4} x} + \frac {d^{2} x}{b^{2}} - \frac {2 d \left (a d - b c\right ) \log {\left (a + b x \right )}}{b^{3}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x**2)**2/(b*x+a)**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^{2} x}{b^{2}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{b^{4} x + a b^{3}} + \frac {2 \, {\left (b c d - a d^{2}\right )} \log \left (b x + a\right )}{b^{3}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^2/(b*x+a)^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^{2} x}{b^{2}} + \frac {2 \, {\left (b c d - a d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{{\left (b x + a\right )} b^{3}} \]

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^2/(b*x+a)^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^2\,x}{b^2}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{b\,\left (x\,b^3+a\,b^2\right )}-\frac {\ln \left (a+b\,x\right )\,\left (2\,a\,d^2-2\,b\,c\,d\right )}{b^3} \]

[In]

int((a*c + x*(a*d + b*c) + b*d*x^2)^2/(a + b*x)^4,x)

[Out]

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