\(\int \frac {a+c x^2}{(d+e x)^2} \, dx\) [456]

Optimal result

Integrand size = 15, antiderivative size = 43 \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=\frac {c x}{e^2}-\frac {c d^2+a e^2}{e^3 (d+e x)}-\frac {2 c d \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, 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 = {711} \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=-\frac {a e^2+c d^2}{e^3 (d+e x)}-\frac {2 c d \log (d+e x)}{e^3}+\frac {c x}{e^2} \]

[In]

Int[(a + c*x^2)/(d + e*x)^2,x]

[Out]

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

Rule 711

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=\frac {c e x-\frac {c d^2+a e^2}{d+e x}-2 c d \log (d+e x)}{e^3} \]

[In]

Integrate[(a + c*x^2)/(d + e*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {c x}{e^{2}}-\frac {e^{2} a +c \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(44\)
norman	\(\frac {\frac {c \,x^{2}}{e}-\frac {e^{2} a +2 c \,d^{2}}{e^{3}}}{e x +d}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(49\)
risch	\(\frac {c x}{e^{2}}-\frac {a}{e \left (e x +d \right )}-\frac {c \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(50\)
parallelrisch	\(-\frac {2 \ln \left (e x +d \right ) x c d e -c \,x^{2} e^{2}+2 \ln \left (e x +d \right ) c \,d^{2}+e^{2} a +2 c \,d^{2}}{e^{3} \left (e x +d \right )}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.37 \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=\frac {c e^{2} x^{2} + c d e x - c d^{2} - a e^{2} - 2 \, {\left (c d e x + c d^{2}\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=- \frac {2 c d \log {\left (d + e x \right )}}{e^{3}} + \frac {c x}{e^{2}} + \frac {- a e^{2} - c d^{2}}{d e^{3} + e^{4} x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=-\frac {c d^{2} + a e^{2}}{e^{4} x + d e^{3}} + \frac {c x}{e^{2}} - \frac {2 \, c d \log \left (e x + d\right )}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.53 \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=c {\left (\frac {2 \, d \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{3}} + \frac {e x + d}{e^{3}} - \frac {d^{2}}{{\left (e x + d\right )} e^{3}}\right )} - \frac {a}{{\left (e x + d\right )} e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.14 \[ \int \frac {a+c x^2}{(d+e x)^2} \, dx=\frac {c\,x}{e^2}-\frac {c\,d^2+a\,e^2}{e\,\left (x\,e^3+d\,e^2\right )}-\frac {2\,c\,d\,\ln \left (d+e\,x\right )}{e^3} \]

[In]

int((a + c*x^2)/(d + e*x)^2,x)

[Out]

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