3.4.56 \(\int \frac {(f+g x)^2}{(d+e x)^2 (d^2-e^2 x^2)} \, dx\)

Optimal. Leaf size=87 \[ \frac {(d g+e f)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{4 d^3 e^3}-\frac {(3 d g+e f) (e f-d g)}{4 d^2 e^3 (d+e x)}-\frac {(e f-d g)^2}{4 d e^3 (d+e x)^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {848, 88, 208} \begin {gather*} -\frac {(3 d g+e f) (e f-d g)}{4 d^2 e^3 (d+e x)}+\frac {(d g+e f)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{4 d^3 e^3}-\frac {(e f-d g)^2}{4 d e^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^2/((d + e*x)^2*(d^2 - e^2*x^2)),x]

[Out]

-(e*f - d*g)^2/(4*d*e^3*(d + e*x)^2) - ((e*f - d*g)*(e*f + 3*d*g))/(4*d^2*e^3*(d + e*x)) + ((e*f + d*g)^2*ArcT
anh[(e*x)/d])/(4*d^3*e^3)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rubi steps

\begin {align*} \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )} \, dx &=\int \frac {(f+g x)^2}{(d-e x) (d+e x)^3} \, dx\\ &=\int \left (\frac {(-e f+d g)^2}{2 d e^2 (d+e x)^3}+\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^2 (d+e x)^2}+\frac {(e f+d g)^2}{4 d^2 e^2 \left (d^2-e^2 x^2\right )}\right ) \, dx\\ &=-\frac {(e f-d g)^2}{4 d e^3 (d+e x)^2}-\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^3 (d+e x)}+\frac {(e f+d g)^2 \int \frac {1}{d^2-e^2 x^2} \, dx}{4 d^2 e^2}\\ &=-\frac {(e f-d g)^2}{4 d e^3 (d+e x)^2}-\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^3 (d+e x)}+\frac {(e f+d g)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{4 d^3 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 87, normalized size = 1.00 \begin {gather*} \frac {\frac {2 d (d g-e f) \left (2 d^2 g+d e (2 f+3 g x)+e^2 f x\right )}{(d+e x)^2}+(d g+e f)^2 (-\log (d-e x))+(d g+e f)^2 \log (d+e x)}{8 d^3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^2/((d + e*x)^2*(d^2 - e^2*x^2)),x]

[Out]

((2*d*(-(e*f) + d*g)*(2*d^2*g + e^2*f*x + d*e*(2*f + 3*g*x)))/(d + e*x)^2 - (e*f + d*g)^2*Log[d - e*x] + (e*f
+ d*g)^2*Log[d + e*x])/(8*d^3*e^3)

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(f + g*x)^2/((d + e*x)^2*(d^2 - e^2*x^2)),x]

[Out]

IntegrateAlgebraic[(f + g*x)^2/((d + e*x)^2*(d^2 - e^2*x^2)), x]

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fricas [B]  time = 0.41, size = 271, normalized size = 3.11 \begin {gather*} -\frac {4 \, d^{2} e^{2} f^{2} - 4 \, d^{4} g^{2} + 2 \, {\left (d e^{3} f^{2} + 2 \, d^{2} e^{2} f g - 3 \, d^{3} e g^{2}\right )} x - {\left (d^{2} e^{2} f^{2} + 2 \, d^{3} e f g + d^{4} g^{2} + {\left (e^{4} f^{2} + 2 \, d e^{3} f g + d^{2} e^{2} g^{2}\right )} x^{2} + 2 \, {\left (d e^{3} f^{2} + 2 \, d^{2} e^{2} f g + d^{3} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + {\left (d^{2} e^{2} f^{2} + 2 \, d^{3} e f g + d^{4} g^{2} + {\left (e^{4} f^{2} + 2 \, d e^{3} f g + d^{2} e^{2} g^{2}\right )} x^{2} + 2 \, {\left (d e^{3} f^{2} + 2 \, d^{2} e^{2} f g + d^{3} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{8 \, {\left (d^{3} e^{5} x^{2} + 2 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*d^2*e^2*f^2 - 4*d^4*g^2 + 2*(d*e^3*f^2 + 2*d^2*e^2*f*g - 3*d^3*e*g^2)*x - (d^2*e^2*f^2 + 2*d^3*e*f*g +
d^4*g^2 + (e^4*f^2 + 2*d*e^3*f*g + d^2*e^2*g^2)*x^2 + 2*(d*e^3*f^2 + 2*d^2*e^2*f*g + d^3*e*g^2)*x)*log(e*x +
d) + (d^2*e^2*f^2 + 2*d^3*e*f*g + d^4*g^2 + (e^4*f^2 + 2*d*e^3*f*g + d^2*e^2*g^2)*x^2 + 2*(d*e^3*f^2 + 2*d^2*e
^2*f*g + d^3*e*g^2)*x)*log(e*x - d))/(d^3*e^5*x^2 + 2*d^4*e^4*x + d^5*e^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -(-g^2*d^2*exp(1)+g*d*exp(1)^2*f+g*d*f*e
xp(2)-exp(1)*f^2*exp(2))/(d^3*exp(1)^4-2*d^3*exp(1)^2*exp(2)+d^3*exp(2)^2)*ln(abs(-(-(exp(1)*x+d)^-1/exp(1))^2
*d^2*exp(1)^4+(-(exp(1)*x+d)^-1/exp(1))^2*d^2*exp(1)^2*exp(2)-2*(exp(1)*x+d)^-1/exp(1)*d*exp(1)*exp(2)+exp(2))
)-(g^2*d^2*exp(1)^4+g^2*d^2*exp(1)^2*exp(2)-4*g*d*exp(1)^3*f*exp(2)+exp(1)^4*f^2*exp(2)+exp(1)^2*f^2*exp(2)^2)
/2/(d^2*exp(1)^4-2*d^2*exp(1)^2*exp(2)+d^2*exp(2)^2)/exp(1)/abs(d)/exp(1)^2*ln(abs(2*(exp(1)*x+d)^-1/exp(1)*d^
2*exp(1)^4-2*(exp(1)*x+d)^-1/exp(1)*d^2*exp(1)^2*exp(2)+2*d*exp(1)*exp(2)-2*exp(1)*abs(d)*exp(1)^2)/abs(2*(exp
(1)*x+d)^-1/exp(1)*d^2*exp(1)^4-2*(exp(1)*x+d)^-1/exp(1)*d^2*exp(1)^2*exp(2)+2*d*exp(1)*exp(2)+2*exp(1)*abs(d)
*exp(1)^2))-((exp(1)*x+d)^-1/exp(1)*g^2*d^2*exp(1)^2-2*(exp(1)*x+d)^-1/exp(1)*g*d*exp(1)^3*f+(exp(1)*x+d)^-1/e
xp(1)*exp(1)^4*f^2)/(d^2*exp(1)^4-d^2*exp(1)^2*exp(2))

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maple [B]  time = 0.01, size = 206, normalized size = 2.37 \begin {gather*} -\frac {d \,g^{2}}{4 \left (e x +d \right )^{2} e^{3}}-\frac {f^{2}}{4 \left (e x +d \right )^{2} d e}+\frac {f g}{2 \left (e x +d \right )^{2} e^{2}}-\frac {f g}{2 \left (e x +d \right ) d \,e^{2}}-\frac {g^{2} \ln \left (e x -d \right )}{8 d \,e^{3}}+\frac {g^{2} \ln \left (e x +d \right )}{8 d \,e^{3}}-\frac {f^{2}}{4 \left (e x +d \right ) d^{2} e}-\frac {f g \ln \left (e x -d \right )}{4 d^{2} e^{2}}+\frac {f g \ln \left (e x +d \right )}{4 d^{2} e^{2}}-\frac {f^{2} \ln \left (e x -d \right )}{8 d^{3} e}+\frac {f^{2} \ln \left (e x +d \right )}{8 d^{3} e}+\frac {3 g^{2}}{4 \left (e x +d \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^2/(e*x+d)^2/(-e^2*x^2+d^2),x)

[Out]

-1/8/e^3/d*ln(e*x-d)*g^2-1/4/e^2/d^2*ln(e*x-d)*f*g-1/8/e/d^3*ln(e*x-d)*f^2+3/4/e^3/(e*x+d)*g^2-1/2/d/e^2/(e*x+
d)*f*g-1/4/d^2/e/(e*x+d)*f^2-1/4/e^3*d/(e*x+d)^2*g^2+1/2/e^2/(e*x+d)^2*f*g-1/4/e/d/(e*x+d)^2*f^2+1/8/e^3/d*ln(
e*x+d)*g^2+1/4/e^2/d^2*ln(e*x+d)*f*g+1/8/e/d^3*ln(e*x+d)*f^2

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maxima [A]  time = 0.45, size = 149, normalized size = 1.71 \begin {gather*} -\frac {2 \, d e^{2} f^{2} - 2 \, d^{3} g^{2} + {\left (e^{3} f^{2} + 2 \, d e^{2} f g - 3 \, d^{2} e g^{2}\right )} x}{4 \, {\left (d^{2} e^{5} x^{2} + 2 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} + \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{8 \, d^{3} e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{8 \, d^{3} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*d*e^2*f^2 - 2*d^3*g^2 + (e^3*f^2 + 2*d*e^2*f*g - 3*d^2*e*g^2)*x)/(d^2*e^5*x^2 + 2*d^3*e^4*x + d^4*e^3)
+ 1/8*(e^2*f^2 + 2*d*e*f*g + d^2*g^2)*log(e*x + d)/(d^3*e^3) - 1/8*(e^2*f^2 + 2*d*e*f*g + d^2*g^2)*log(e*x -
d)/(d^3*e^3)

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mupad [B]  time = 0.13, size = 100, normalized size = 1.15 \begin {gather*} \frac {\frac {d^2\,g^2-e^2\,f^2}{2\,d\,e^3}-\frac {x\,\left (-3\,d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^2\,e^2}}{d^2+2\,d\,e\,x+e^2\,x^2}+\frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,{\left (d\,g+e\,f\right )}^2}{4\,d^3\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)^2/((d^2 - e^2*x^2)*(d + e*x)^2),x)

[Out]

((d^2*g^2 - e^2*f^2)/(2*d*e^3) - (x*(e^2*f^2 - 3*d^2*g^2 + 2*d*e*f*g))/(4*d^2*e^2))/(d^2 + e^2*x^2 + 2*d*e*x)
+ (atanh((e*x)/d)*(d*g + e*f)^2)/(4*d^3*e^3)

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sympy [B]  time = 1.03, size = 185, normalized size = 2.13 \begin {gather*} - \frac {- 2 d^{3} g^{2} + 2 d e^{2} f^{2} + x \left (- 3 d^{2} e g^{2} + 2 d e^{2} f g + e^{3} f^{2}\right )}{4 d^{4} e^{3} + 8 d^{3} e^{4} x + 4 d^{2} e^{5} x^{2}} - \frac {\left (d g + e f\right )^{2} \log {\left (- \frac {d \left (d g + e f\right )^{2}}{e \left (d^{2} g^{2} + 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{8 d^{3} e^{3}} + \frac {\left (d g + e f\right )^{2} \log {\left (\frac {d \left (d g + e f\right )^{2}}{e \left (d^{2} g^{2} + 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{8 d^{3} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2/(e*x+d)**2/(-e**2*x**2+d**2),x)

[Out]

-(-2*d**3*g**2 + 2*d*e**2*f**2 + x*(-3*d**2*e*g**2 + 2*d*e**2*f*g + e**3*f**2))/(4*d**4*e**3 + 8*d**3*e**4*x +
4*d**2*e**5*x**2) - (d*g + e*f)**2*log(-d*(d*g + e*f)**2/(e*(d**2*g**2 + 2*d*e*f*g + e**2*f**2)) + x)/(8*d**3
*e**3) + (d*g + e*f)**2*log(d*(d*g + e*f)**2/(e*(d**2*g**2 + 2*d*e*f*g + e**2*f**2)) + x)/(8*d**3*e**3)

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