\(\int \frac {(a+b x)^2}{c+d x} \, dx\) [1337]

Optimal result

Integrand size = 15, antiderivative size = 50 \[ \int \frac {(a+b x)^2}{c+d x} \, dx=-\frac {b (b c-a d) x}{d^2}+\frac {(a+b x)^2}{2 d}+\frac {(b c-a d)^2 \log (c+d x)}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {(a+b x)^2}{c+d x} \, dx=\frac {(b c-a d)^2 \log (c+d x)}{d^3}-\frac {b x (b c-a d)}{d^2}+\frac {(a+b x)^2}{2 d} \]

[In]

Int[(a + b*x)^2/(c + d*x),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^2}{c+d x} \, dx=\frac {b d x (-2 b c+4 a d+b d x)+2 (b c-a d)^2 \log (c+d x)}{2 d^3} \]

[In]

Integrate[(a + b*x)^2/(c + d*x),x]

[Out]

(b*d*x*(-2*b*c + 4*a*d + b*d*x) + 2*(b*c - a*d)^2*Log[c + d*x])/(2*d^3)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {b \left (\frac {1}{2} b d \,x^{2}+2 a d x -b c x \right )}{d^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (d x +c \right )}{d^{3}}\)	\(56\)
norman	\(\frac {b \left (2 a d -b c \right ) x}{d^{2}}+\frac {b^{2} x^{2}}{2 d}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (d x +c \right )}{d^{3}}\)	\(59\)
parallelrisch	\(\frac {d^{2} x^{2} b^{2}+2 \ln \left (d x +c \right ) a^{2} d^{2}-4 \ln \left (d x +c \right ) a b c d +2 \ln \left (d x +c \right ) b^{2} c^{2}+4 x a b \,d^{2}-2 x \,b^{2} c d}{2 d^{3}}\)	\(73\)
risch	\(\frac {b^{2} x^{2}}{2 d}+\frac {2 b a x}{d}-\frac {b^{2} c x}{d^{2}}+\frac {\ln \left (d x +c \right ) a^{2}}{d}-\frac {2 \ln \left (d x +c \right ) a b c}{d^{2}}+\frac {\ln \left (d x +c \right ) b^{2} c^{2}}{d^{3}}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^2}{c+d x} \, dx=\frac {b^{2} d^{2} x^{2} - 2 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d x + c\right )}{2 \, d^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^2}{c+d x} \, dx=\frac {b^{2} x^{2}}{2 d} + x \left (\frac {2 a b}{d} - \frac {b^{2} c}{d^{2}}\right ) + \frac {\left (a d - b c\right )^{2} \log {\left (c + d x \right )}}{d^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^2}{c+d x} \, dx=\frac {b^{2} d x^{2} - 2 \, {\left (b^{2} c - 2 \, a b d\right )} x}{2 \, d^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d x + c\right )}{d^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^2}{c+d x} \, dx=\frac {b^{2} d x^{2} - 2 \, b^{2} c x + 4 \, a b d x}{2 \, d^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | d x + c \right |}\right )}{d^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^2}{c+d x} \, dx=\frac {\ln \left (c+d\,x\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{d^3}-x\,\left (\frac {b^2\,c}{d^2}-\frac {2\,a\,b}{d}\right )+\frac {b^2\,x^2}{2\,d} \]

[In]

int((a + b*x)^2/(c + d*x),x)

[Out]

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