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

Optimal. Leaf size=86 \[ \frac {(3 d g+e f) (e f-d g) \log (d+e x)}{4 d^2 e^3}-\frac {(d g+e f)^2 \log (d-e x)}{4 d^2 e^3}-\frac {(e f-d g)^2}{2 d e^3 (d+e x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {848, 88} \begin {gather*} \frac {(3 d g+e f) (e f-d g) \log (d+e x)}{4 d^2 e^3}-\frac {(d g+e f)^2 \log (d-e x)}{4 d^2 e^3}-\frac {(e f-d g)^2}{2 d e^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*f - d*g)^2/(2*d*e^3*(d + e*x)) - ((e*f + d*g)^2*Log[d - e*x])/(4*d^2*e^3) + ((e*f - d*g)*(e*f + 3*d*g)*Log
[d + e*x])/(4*d^2*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 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) \left (d^2-e^2 x^2\right )} \, dx &=\int \frac {(f+g x)^2}{(d-e x) (d+e x)^2} \, dx\\ &=\int \left (\frac {(e f+d g)^2}{4 d^2 e^2 (d-e x)}+\frac {(-e f+d g)^2}{2 d e^2 (d+e x)^2}+\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^2 (d+e x)}\right ) \, dx\\ &=-\frac {(e f-d g)^2}{2 d e^3 (d+e x)}-\frac {(e f+d g)^2 \log (d-e x)}{4 d^2 e^3}+\frac {(e f-d g) (e f+3 d g) \log (d+e x)}{4 d^2 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 82, normalized size = 0.95 \begin {gather*} \frac {(e f-d g) ((d+e x) (3 d g+e f) \log (d+e x)+2 d (d g-e f))-(d+e x) (d g+e f)^2 \log (d-e x)}{4 d^2 e^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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) \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)*(d^2 - e^2*x^2)),x]

[Out]

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

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fricas [B]  time = 0.40, size = 165, normalized size = 1.92 \begin {gather*} -\frac {2 \, d e^{2} f^{2} - 4 \, d^{2} e f g + 2 \, d^{3} g^{2} - {\left (d e^{2} f^{2} + 2 \, d^{2} e f g - 3 \, d^{3} g^{2} + {\left (e^{3} f^{2} + 2 \, d e^{2} f g - 3 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2} + {\left (e^{3} f^{2} + 2 \, d e^{2} f g + d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{4 \, {\left (d^{2} e^{4} x + d^{3} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*d*e^2*f^2 - 4*d^2*e*f*g + 2*d^3*g^2 - (d*e^2*f^2 + 2*d^2*e*f*g - 3*d^3*g^2 + (e^3*f^2 + 2*d*e^2*f*g -
3*d^2*e*g^2)*x)*log(e*x + d) + (d*e^2*f^2 + 2*d^2*e*f*g + d^3*g^2 + (e^3*f^2 + 2*d*e^2*f*g + d^2*e*g^2)*x)*log
(e*x - d))/(d^2*e^4*x + d^3*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)/(-e^2*x^2+d^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 149, normalized size = 1.73 \begin {gather*} -\frac {d \,g^{2}}{2 \left (e x +d \right ) e^{3}}-\frac {f^{2}}{2 \left (e x +d \right ) d e}-\frac {f g \ln \left (e x -d \right )}{2 d \,e^{2}}+\frac {f g \ln \left (e x +d \right )}{2 d \,e^{2}}-\frac {f^{2} \ln \left (e x -d \right )}{4 d^{2} e}+\frac {f^{2} \ln \left (e x +d \right )}{4 d^{2} e}+\frac {f g}{\left (e x +d \right ) e^{2}}-\frac {g^{2} \ln \left (e x -d \right )}{4 e^{3}}-\frac {3 g^{2} \ln \left (e x +d \right )}{4 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 113, normalized size = 1.31 \begin {gather*} -\frac {e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}}{2 \, {\left (d e^{4} x + d^{2} e^{3}\right )}} + \frac {{\left (e^{2} f^{2} + 2 \, d e f g - 3 \, d^{2} g^{2}\right )} \log \left (e x + d\right )}{4 \, d^{2} e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{4 \, d^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.70, size = 109, normalized size = 1.27 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (-3\,d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^2\,e^3}-\frac {\ln \left (d-e\,x\right )\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^2\,e^3}-\frac {d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2}{2\,d\,e^3\,\left (d+e\,x\right )} \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)),x)

[Out]

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

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sympy [B]  time = 1.00, size = 182, normalized size = 2.12 \begin {gather*} - \frac {d^{2} g^{2} - 2 d e f g + e^{2} f^{2}}{2 d^{2} e^{3} + 2 d e^{4} x} - \frac {\left (d g - e f\right ) \left (3 d g + e f\right ) \log {\left (x + \frac {- 2 d^{3} g^{2} + d \left (d g - e f\right ) \left (3 d g + e f\right )}{d^{2} e g^{2} - 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} - \frac {\left (d g + e f\right )^{2} \log {\left (x + \frac {- 2 d^{3} g^{2} + d \left (d g + e f\right )^{2}}{d^{2} e g^{2} - 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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