### 3.500 $$\int \frac{(d+e x)^2}{a+c x^2} \, dx$$

Optimal. Leaf size=59 $\frac{\left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{d e \log \left (a+c x^2\right )}{c}+\frac{e^2 x}{c}$

[Out]

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

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Rubi [A]  time = 0.0509034, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {702, 635, 205, 260} $\frac{\left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{d e \log \left (a+c x^2\right )}{c}+\frac{e^2 x}{c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^2/(a + c*x^2),x]

[Out]

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

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{a+c x^2} \, dx &=\int \left (\frac{e^2}{c}+\frac{c d^2-a e^2+2 c d e x}{c \left (a+c x^2\right )}\right ) \, dx\\ &=\frac{e^2 x}{c}+\frac{\int \frac{c d^2-a e^2+2 c d e x}{a+c x^2} \, dx}{c}\\ &=\frac{e^2 x}{c}+(2 d e) \int \frac{x}{a+c x^2} \, dx+\frac{\left (c d^2-a e^2\right ) \int \frac{1}{a+c x^2} \, dx}{c}\\ &=\frac{e^2 x}{c}+\frac{\left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{d e \log \left (a+c x^2\right )}{c}\\ \end{align*}

Mathematica [A]  time = 0.0400094, size = 56, normalized size = 0.95 $\frac{\left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{e \left (d \log \left (a+c x^2\right )+e x\right )}{c}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^2/(a + c*x^2),x]

[Out]

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

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Maple [A]  time = 0.046, size = 65, normalized size = 1.1 \begin{align*}{\frac{{e}^{2}x}{c}}+{\frac{de\ln \left ( c{x}^{2}+a \right ) }{c}}-{\frac{a{e}^{2}}{c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{{d}^{2}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.91244, size = 309, normalized size = 5.24 \begin{align*} \left [\frac{2 \, a c e^{2} x + 2 \, a c d e \log \left (c x^{2} + a\right ) +{\left (c d^{2} - a e^{2}\right )} \sqrt{-a c} \log \left (\frac{c x^{2} + 2 \, \sqrt{-a c} x - a}{c x^{2} + a}\right )}{2 \, a c^{2}}, \frac{a c e^{2} x + a c d e \log \left (c x^{2} + a\right ) +{\left (c d^{2} - a e^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right )}{a c^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a*c*e^2*x + 2*a*c*d*e*log(c*x^2 + a) + (c*d^2 - a*e^2)*sqrt(-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*
x^2 + a)))/(a*c^2), (a*c*e^2*x + a*c*d*e*log(c*x^2 + a) + (c*d^2 - a*e^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a))/(a*
c^2)]

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Sympy [B]  time = 0.700622, size = 185, normalized size = 3.14 \begin{align*} \left (\frac{d e}{c} - \frac{\sqrt{- a c^{3}} \left (a e^{2} - c d^{2}\right )}{2 a c^{3}}\right ) \log{\left (x + \frac{- 2 a c \left (\frac{d e}{c} - \frac{\sqrt{- a c^{3}} \left (a e^{2} - c d^{2}\right )}{2 a c^{3}}\right ) + 2 a d e}{a e^{2} - c d^{2}} \right )} + \left (\frac{d e}{c} + \frac{\sqrt{- a c^{3}} \left (a e^{2} - c d^{2}\right )}{2 a c^{3}}\right ) \log{\left (x + \frac{- 2 a c \left (\frac{d e}{c} + \frac{\sqrt{- a c^{3}} \left (a e^{2} - c d^{2}\right )}{2 a c^{3}}\right ) + 2 a d e}{a e^{2} - c d^{2}} \right )} + \frac{e^{2} x}{c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.42494, size = 70, normalized size = 1.19 \begin{align*} \frac{d e \log \left (c x^{2} + a\right )}{c} + \frac{x e^{2}}{c} + \frac{{\left (c d^{2} - a e^{2}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{\sqrt{a c} c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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