3.4.10 \(\int \frac {d+e x}{(a^2-c^2 x^2)^2} \, dx\)

Optimal. Leaf size=55 \[ \frac {d \tanh ^{-1}\left (\frac {c x}{a}\right )}{2 a^3 c}+\frac {a^2 e+c^2 d x}{2 a^2 c^2 \left (a^2-c^2 x^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {639, 208} \begin {gather*} \frac {a^2 e+c^2 d x}{2 a^2 c^2 \left (a^2-c^2 x^2\right )}+\frac {d \tanh ^{-1}\left (\frac {c x}{a}\right )}{2 a^3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a^2 - c^2*x^2)^2,x]

[Out]

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

Rule 208

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

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 58, normalized size = 1.05 \begin {gather*} \frac {d \tanh ^{-1}\left (\frac {c x}{a}\right )}{2 a^3 c}+\frac {a^2 (-e)-c^2 d x}{2 a^2 c^2 \left (c^2 x^2-a^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(a^2 - c^2*x^2)^2,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d+e x}{\left (a^2-c^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x)/(a^2 - c^2*x^2)^2,x]

[Out]

IntegrateAlgebraic[(d + e*x)/(a^2 - c^2*x^2)^2, x]

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fricas [A]  time = 0.45, size = 87, normalized size = 1.58 \begin {gather*} -\frac {2 \, a c^{2} d x + 2 \, a^{3} e - {\left (c^{3} d x^{2} - a^{2} c d\right )} \log \left (c x + a\right ) + {\left (c^{3} d x^{2} - a^{2} c d\right )} \log \left (c x - a\right )}{4 \, {\left (a^{3} c^{4} x^{2} - a^{5} c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 71, normalized size = 1.29 \begin {gather*} \frac {d \log \left ({\left | c x + a \right |}\right )}{4 \, a^{3} c} - \frac {d \log \left ({\left | c x - a \right |}\right )}{4 \, a^{3} c} - \frac {c^{2} d x + a^{2} e}{2 \, {\left (c^{2} x^{2} - a^{2}\right )} a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 102, normalized size = 1.85 \begin {gather*} \frac {e}{4 \left (c x +a \right ) a \,c^{2}}-\frac {e}{4 \left (c x -a \right ) a \,c^{2}}-\frac {d}{4 \left (c x +a \right ) a^{2} c}-\frac {d}{4 \left (c x -a \right ) a^{2} c}-\frac {d \ln \left (c x -a \right )}{4 a^{3} c}+\frac {d \ln \left (c x +a \right )}{4 a^{3} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 68, normalized size = 1.24 \begin {gather*} -\frac {c^{2} d x + a^{2} e}{2 \, {\left (a^{2} c^{4} x^{2} - a^{4} c^{2}\right )}} + \frac {d \log \left (c x + a\right )}{4 \, a^{3} c} - \frac {d \log \left (c x - a\right )}{4 \, a^{3} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.07, size = 46, normalized size = 0.84 \begin {gather*} \frac {\frac {e}{2\,c^2}+\frac {d\,x}{2\,a^2}}{a^2-c^2\,x^2}+\frac {d\,\mathrm {atanh}\left (\frac {c\,x}{a}\right )}{2\,a^3\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(a^2 - c^2*x^2)^2,x)

[Out]

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

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sympy [A]  time = 0.33, size = 56, normalized size = 1.02 \begin {gather*} \frac {- a^{2} e - c^{2} d x}{- 2 a^{4} c^{2} + 2 a^{2} c^{4} x^{2}} + \frac {d \left (- \frac {\log {\left (- \frac {a}{c} + x \right )}}{4} + \frac {\log {\left (\frac {a}{c} + x \right )}}{4}\right )}{a^{3} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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