### 3.455 $$\int \frac{a+c x^2}{d+e x} \, dx$$

Optimal. Leaf size=41 $\frac{\left (a e^2+c d^2\right ) \log (d+e x)}{e^3}-\frac{c d x}{e^2}+\frac{c x^2}{2 e}$

[Out]

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

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Rubi [A]  time = 0.0300263, antiderivative size = 41, 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} $\frac{\left (a e^2+c d^2\right ) \log (d+e x)}{e^3}-\frac{c d x}{e^2}+\frac{c x^2}{2 e}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + c*x^2)/(d + e*x),x]

[Out]

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

Mathematica [A]  time = 0.0112357, size = 38, normalized size = 0.93 $\frac{2 \left (a e^2+c d^2\right ) \log (d+e x)+c e x (e x-2 d)}{2 e^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + c*x^2)/(d + e*x),x]

[Out]

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

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Maple [A]  time = 0.044, size = 44, normalized size = 1.1 \begin{align*}{\frac{c{x}^{2}}{2\,e}}-{\frac{cdx}{{e}^{2}}}+{\frac{\ln \left ( ex+d \right ) a}{e}}+{\frac{\ln \left ( ex+d \right ) c{d}^{2}}{{e}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1435, size = 53, normalized size = 1.29 \begin{align*} \frac{c e x^{2} - 2 \, c d x}{2 \, e^{2}} + \frac{{\left (c d^{2} + a e^{2}\right )} \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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.12247, size = 89, normalized size = 2.17 \begin{align*} \frac{c e^{2} x^{2} - 2 \, c d e x + 2 \,{\left (c d^{2} + a e^{2}\right )} \log \left (e x + d\right )}{2 \, e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.354507, size = 36, normalized size = 0.88 \begin{align*} - \frac{c d x}{e^{2}} + \frac{c x^{2}}{2 e} + \frac{\left (a e^{2} + c d^{2}\right ) \log{\left (d + e x \right )}}{e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24533, size = 53, normalized size = 1.29 \begin{align*}{\left (c d^{2} + a e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{2} \,{\left (c x^{2} e - 2 \, c d x\right )} e^{\left (-2\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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