### 3.456 $$\int \frac{a+c x^2}{(d+e x)^2} \, dx$$

Optimal. Leaf size=43 $-\frac{a e^2+c d^2}{e^3 (d+e x)}-\frac{2 c d \log (d+e x)}{e^3}+\frac{c x}{e^2}$

[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

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Rubi [A]  time = 0.0282407, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.067, Rules used = {697} $-\frac{a e^2+c d^2}{e^3 (d+e x)}-\frac{2 c d \log (d+e x)}{e^3}+\frac{c x}{e^2}$

Antiderivative was successfully veriﬁed.

[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 697

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*} \int \frac{a+c x^2}{(d+e x)^2} \, dx &=\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]  time = 0.0208162, size = 39, normalized size = 0.91 $\frac{-\frac{a e^2+c d^2}{d+e x}-2 c d \log (d+e x)+c e x}{e^3}$

Antiderivative was successfully veriﬁed.

[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

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Maple [A]  time = 0.054, size = 50, normalized size = 1.2 \begin{align*}{\frac{cx}{{e}^{2}}}-2\,{\frac{cd\ln \left ( ex+d \right ) }{{e}^{3}}}-{\frac{a}{e \left ( ex+d \right ) }}-{\frac{c{d}^{2}}{{e}^{3} \left ( ex+d \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.17228, size = 62, normalized size = 1.44 \begin{align*} -\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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.80624, size = 122, normalized size = 2.84 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.516944, size = 41, normalized size = 0.95 \begin{align*} - \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.3839, size = 88, normalized size = 2.05 \begin{align*}{\left (2 \, d e^{\left (-3\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) +{\left (x e + d\right )} e^{\left (-3\right )} - \frac{d^{2} e^{\left (-3\right )}}{x e + d}\right )} c - \frac{a e^{\left (-1\right )}}{x e + d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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