3.6.76 \(\int \frac {(f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [576]

Optimal. Leaf size=127 \[ \frac {\left (d^2 g+e^2 f x\right ) (f+g x)}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}+\frac {2 d^2 f g+\left (3 e^2 f^2-d^2 g^2\right ) x}{8 d^4 e^2 \left (d^2-e^2 x^2\right )}+\frac {\left (3 e^2 f^2-d^2 g^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{8 d^5 e^3} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {753, 653, 214} \begin {gather*} \frac {(f+g x) \left (d^2 g+e^2 f x\right )}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}+\frac {\left (3 e^2 f^2-d^2 g^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{8 d^5 e^3}+\frac {x \left (3 e^2 f^2-d^2 g^2\right )+2 d^2 f g}{8 d^4 e^2 \left (d^2-e^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^2/(d^2 - e^2*x^2)^3,x]

[Out]

((d^2*g + e^2*f*x)*(f + g*x))/(4*d^2*e^2*(d^2 - e^2*x^2)^2) + (2*d^2*f*g + (3*e^2*f^2 - d^2*g^2)*x)/(8*d^4*e^2
*(d^2 - e^2*x^2)) + ((3*e^2*f^2 - d^2*g^2)*ArcTanh[(e*x)/d])/(8*d^5*e^3)

Rule 214

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

Rule 653

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)/(2*a*c*(p + 1)))*(a + c*x
^2)^(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]

Rule 753

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rubi steps

\begin {align*} \int \frac {(f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx &=\frac {\left (d^2 g+e^2 f x\right ) (f+g x)}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}-\frac {\int \frac {-3 e^2 f^2+d^2 g^2-2 e^2 f g x}{\left (d^2-e^2 x^2\right )^2} \, dx}{4 d^2 e^2}\\ &=\frac {\left (d^2 g+e^2 f x\right ) (f+g x)}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}+\frac {2 d^2 f g+\left (3 e^2 f^2-d^2 g^2\right ) x}{8 d^4 e^2 \left (d^2-e^2 x^2\right )}-\frac {\left (-\frac {3 e^2 f^2}{d^2}+g^2\right ) \int \frac {1}{d^2-e^2 x^2} \, dx}{8 d^2 e^2}\\ &=\frac {\left (d^2 g+e^2 f x\right ) (f+g x)}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}+\frac {2 d^2 f g+\left (3 e^2 f^2-d^2 g^2\right ) x}{8 d^4 e^2 \left (d^2-e^2 x^2\right )}+\frac {\left (3 e^2 f^2-d^2 g^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{8 d^5 e^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 110, normalized size = 0.87 \begin {gather*} \frac {-3 d e^5 f^2 x^3+d^5 e g (4 f+g x)+d^3 e^3 x \left (5 f^2+g^2 x^2\right )+\left (3 e^2 f^2-d^2 g^2\right ) \left (d^2-e^2 x^2\right )^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{8 d^5 e^3 \left (d^2-e^2 x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*e^5*f^2*x^3 + d^5*e*g*(4*f + g*x) + d^3*e^3*x*(5*f^2 + g^2*x^2) + (3*e^2*f^2 - d^2*g^2)*(d^2 - e^2*x^2)^
2*ArcTanh[(e*x)/d])/(8*d^5*e^3*(d^2 - e^2*x^2)^2)

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Maple [A]
time = 0.08, size = 216, normalized size = 1.70

method	result	size
norman	\(\frac {\frac {f g}{2 e^{2}}+\frac {\left (d^{2} g^{2}-3 e^{2} f^{2}\right ) x^{3}}{8 d^{4}}+\frac {\left (d^{2} g^{2}+5 e^{2} f^{2}\right ) x}{8 e^{2} d^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}+\frac {\left (d^{2} g^{2}-3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 d^{5} e^{3}}-\frac {\left (d^{2} g^{2}-3 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 d^{5} e^{3}}\)	\(135\)
risch	\(\frac {\frac {f g}{2 e^{2}}+\frac {\left (d^{2} g^{2}-3 e^{2} f^{2}\right ) x^{3}}{8 d^{4}}+\frac {\left (d^{2} g^{2}+5 e^{2} f^{2}\right ) x}{8 e^{2} d^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {\ln \left (-e x -d \right ) g^{2}}{16 d^{3} e^{3}}+\frac {3 \ln \left (-e x -d \right ) f^{2}}{16 d^{5} e}+\frac {\ln \left (e x -d \right ) g^{2}}{16 d^{3} e^{3}}-\frac {3 \ln \left (e x -d \right ) f^{2}}{16 d^{5} e}\)	\(152\)
default	\(\frac {\left (-d^{2} g^{2}+3 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 e^{3} d^{5}}-\frac {-d^{2} g^{2}-2 d e f g +3 e^{2} f^{2}}{16 e^{3} d^{4} \left (e x +d \right )}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{16 e^{3} d^{3} \left (e x +d \right )^{2}}+\frac {\left (d^{2} g^{2}-3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 d^{5} e^{3}}-\frac {-d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{16 e^{3} d^{3} \left (-e x +d \right )^{2}}+\frac {-d^{2} g^{2}+2 d e f g +3 e^{2} f^{2}}{16 e^{3} d^{4} \left (-e x +d \right )}\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^2/(-e^2*x^2+d^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/16*(-d^2*g^2+3*e^2*f^2)/e^3/d^5*ln(e*x+d)-1/16*(-d^2*g^2-2*d*e*f*g+3*e^2*f^2)/e^3/d^4/(e*x+d)-1/16*(d^2*g^2-
2*d*e*f*g+e^2*f^2)/e^3/d^3/(e*x+d)^2+1/16*(d^2*g^2-3*e^2*f^2)/d^5/e^3*ln(-e*x+d)-1/16*(-d^2*g^2-2*d*e*f*g-e^2*
f^2)/e^3/d^3/(-e*x+d)^2+1/16*(-d^2*g^2+2*d*e*f*g+3*e^2*f^2)/e^3/d^4/(-e*x+d)

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Maxima [A]
time = 0.29, size = 140, normalized size = 1.10 \begin {gather*} \frac {4 \, d^{4} f g + {\left (d^{2} g^{2} e^{2} - 3 \, f^{2} e^{4}\right )} x^{3} + {\left (d^{4} g^{2} + 5 \, d^{2} f^{2} e^{2}\right )} x}{8 \, {\left (d^{4} x^{4} e^{6} - 2 \, d^{6} x^{2} e^{4} + d^{8} e^{2}\right )}} - \frac {{\left (d^{2} g^{2} - 3 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e + d\right )}{16 \, d^{5}} + \frac {{\left (d^{2} g^{2} - 3 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right )}{16 \, d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

1/8*(4*d^4*f*g + (d^2*g^2*e^2 - 3*f^2*e^4)*x^3 + (d^4*g^2 + 5*d^2*f^2*e^2)*x)/(d^4*x^4*e^6 - 2*d^6*x^2*e^4 + d
^8*e^2) - 1/16*(d^2*g^2 - 3*f^2*e^2)*e^(-3)*log(x*e + d)/d^5 + 1/16*(d^2*g^2 - 3*f^2*e^2)*e^(-3)*log(x*e - d)/
d^5

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Fricas [A]
time = 2.49, size = 188, normalized size = 1.48 \begin {gather*} -\frac {6 \, d f^{2} x^{3} e^{6} + {\left (d^{6} g^{2} - 3 \, f^{2} x^{4} e^{6} + {\left (d^{2} g^{2} x^{4} + 6 \, d^{2} f^{2} x^{2}\right )} e^{4} - {\left (2 \, d^{4} g^{2} x^{2} + 3 \, d^{4} f^{2}\right )} e^{2}\right )} e \log \left (\frac {x^{2} e^{2} + 2 \, d x e + d^{2}}{x^{2} e^{2} - d^{2}}\right ) - 2 \, {\left (d^{3} g^{2} x^{3} + 5 \, d^{3} f^{2} x\right )} e^{4} - 2 \, {\left (d^{5} g^{2} x + 4 \, d^{5} f g\right )} e^{2}}{16 \, {\left (d^{5} x^{4} e^{8} - 2 \, d^{7} x^{2} e^{6} + d^{9} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

-1/16*(6*d*f^2*x^3*e^6 + (d^6*g^2 - 3*f^2*x^4*e^6 + (d^2*g^2*x^4 + 6*d^2*f^2*x^2)*e^4 - (2*d^4*g^2*x^2 + 3*d^4
*f^2)*e^2)*e*log((x^2*e^2 + 2*d*x*e + d^2)/(x^2*e^2 - d^2)) - 2*(d^3*g^2*x^3 + 5*d^3*f^2*x)*e^4 - 2*(d^5*g^2*x
+ 4*d^5*f*g)*e^2)/(d^5*x^4*e^8 - 2*d^7*x^2*e^6 + d^9*e^4)

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Sympy [A]
time = 0.48, size = 144, normalized size = 1.13 \begin {gather*} - \frac {- 4 d^{4} f g + x^{3} \left (- d^{2} e^{2} g^{2} + 3 e^{4} f^{2}\right ) + x \left (- d^{4} g^{2} - 5 d^{2} e^{2} f^{2}\right )}{8 d^{8} e^{2} - 16 d^{6} e^{4} x^{2} + 8 d^{4} e^{6} x^{4}} + \frac {\left (d^{2} g^{2} - 3 e^{2} f^{2}\right ) \log {\left (- \frac {d}{e} + x \right )}}{16 d^{5} e^{3}} - \frac {\left (d^{2} g^{2} - 3 e^{2} f^{2}\right ) \log {\left (\frac {d}{e} + x \right )}}{16 d^{5} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2/(-e**2*x**2+d**2)**3,x)

[Out]

-(-4*d**4*f*g + x**3*(-d**2*e**2*g**2 + 3*e**4*f**2) + x*(-d**4*g**2 - 5*d**2*e**2*f**2))/(8*d**8*e**2 - 16*d*
*6*e**4*x**2 + 8*d**4*e**6*x**4) + (d**2*g**2 - 3*e**2*f**2)*log(-d/e + x)/(16*d**5*e**3) - (d**2*g**2 - 3*e**
2*f**2)*log(d/e + x)/(16*d**5*e**3)

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Giac [A]
time = 2.88, size = 127, normalized size = 1.00 \begin {gather*} \frac {{\left (d^{2} g^{2} - 3 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (\frac {{\left | 2 \, x e^{2} - 2 \, {\left | d \right |} e \right |}}{{\left | 2 \, x e^{2} + 2 \, {\left | d \right |} e \right |}}\right )}{16 \, d^{4} {\left | d \right |}} + \frac {{\left (d^{2} g^{2} x^{3} e^{2} + d^{4} g^{2} x + 4 \, d^{4} f g - 3 \, f^{2} x^{3} e^{4} + 5 \, d^{2} f^{2} x e^{2}\right )} e^{\left (-2\right )}}{8 \, {\left (x^{2} e^{2} - d^{2}\right )}^{2} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(d^2*g^2 - 3*f^2*e^2)*e^(-3)*log(abs(2*x*e^2 - 2*abs(d)*e)/abs(2*x*e^2 + 2*abs(d)*e))/(d^4*abs(d)) + 1/8*
(d^2*g^2*x^3*e^2 + d^4*g^2*x + 4*d^4*f*g - 3*f^2*x^3*e^4 + 5*d^2*f^2*x*e^2)*e^(-2)/((x^2*e^2 - d^2)^2*d^4)

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Mupad [B]
time = 0.10, size = 114, normalized size = 0.90 \begin {gather*} \frac {\frac {x^3\,\left (d^2\,g^2-3\,e^2\,f^2\right )}{8\,d^4}+\frac {f\,g}{2\,e^2}+\frac {x\,\left (d^2\,g^2+5\,e^2\,f^2\right )}{8\,d^2\,e^2}}{d^4-2\,d^2\,e^2\,x^2+e^4\,x^4}-\frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,\left (d^2\,g^2-3\,e^2\,f^2\right )}{8\,d^5\,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)^3,x)

[Out]

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

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