### 3.457 $$\int \frac{a+c x^2}{(d+e x)^3} \, dx$$

Optimal. Leaf size=51 $-\frac{a e^2+c d^2}{2 e^3 (d+e x)^2}+\frac{2 c d}{e^3 (d+e x)}+\frac{c \log (d+e x)}{e^3}$

[Out]

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

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Rubi [A]  time = 0.0324775, antiderivative size = 51, 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} $-\frac{a e^2+c d^2}{2 e^3 (d+e x)^2}+\frac{2 c d}{e^3 (d+e x)}+\frac{c \log (d+e x)}{e^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + c*x^2)/(d + e*x)^3,x]

[Out]

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

Mathematica [A]  time = 0.015273, size = 48, normalized size = 0.94 $\frac{-a e^2+c d (3 d+4 e x)+2 c (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + c*x^2)/(d + e*x)^3,x]

[Out]

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

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Maple [A]  time = 0.052, size = 56, normalized size = 1.1 \begin{align*}{\frac{c\ln \left ( ex+d \right ) }{{e}^{3}}}+2\,{\frac{cd}{{e}^{3} \left ( ex+d \right ) }}-{\frac{a}{2\,e \left ( ex+d \right ) ^{2}}}-{\frac{c{d}^{2}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)/(e*x+d)^3,x)

[Out]

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

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Maxima [A]  time = 1.16519, size = 77, normalized size = 1.51 \begin{align*} \frac{4 \, c d e x + 3 \, c d^{2} - a e^{2}}{2 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac{c \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)^3,x, algorithm="maxima")

[Out]

1/2*(4*c*d*e*x + 3*c*d^2 - a*e^2)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) + c*log(e*x + d)/e^3

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Fricas [A]  time = 1.78359, size = 157, normalized size = 3.08 \begin{align*} \frac{4 \, c d e x + 3 \, c d^{2} - a e^{2} + 2 \,{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )} \log \left (e x + d\right )}{2 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*(4*c*d*e*x + 3*c*d^2 - a*e^2 + 2*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*log(e*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*
e^3)

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Sympy [A]  time = 0.613881, size = 56, normalized size = 1.1 \begin{align*} \frac{c \log{\left (d + e x \right )}}{e^{3}} + \frac{- a e^{2} + 3 c d^{2} + 4 c d e x}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)/(e*x+d)**3,x)

[Out]

c*log(d + e*x)/e**3 + (-a*e**2 + 3*c*d**2 + 4*c*d*e*x)/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2)

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Giac [A]  time = 1.41966, size = 62, normalized size = 1.22 \begin{align*} c e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{{\left (4 \, c d x +{\left (3 \, c d^{2} - a e^{2}\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)/(e*x+d)^3,x, algorithm="giac")

[Out]

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