3.6.56 \(\int \frac {(f+g x)^2}{(d+e x)^4 (d^2-e^2 x^2)} \, dx\) [556]

Optimal. Leaf size=139 \[ -\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4}-\frac {(e f-d g) (e f+3 d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(e f+d g)^2}{16 d^3 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{16 d^4 e^3 (d+e x)}+\frac {(e f+d g)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^5 e^3} \]

[Out]

-1/8*(-d*g+e*f)^2/d/e^3/(e*x+d)^4-1/12*(-d*g+e*f)*(3*d*g+e*f)/d^2/e^3/(e*x+d)^3-1/16*(d*g+e*f)^2/d^3/e^3/(e*x+
d)^2-1/16*(d*g+e*f)^2/d^4/e^3/(e*x+d)+1/16*(d*g+e*f)^2*arctanh(e*x/d)/d^5/e^3

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Rubi [A]
time = 0.09, antiderivative size = 139, 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 = {862, 90, 214} \begin {gather*} \frac {(d g+e f)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^5 e^3}-\frac {(d g+e f)^2}{16 d^4 e^3 (d+e x)}-\frac {(d g+e f)^2}{16 d^3 e^3 (d+e x)^2}-\frac {(3 d g+e f) (e f-d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(e*f - d*g)^2/(d*e^3*(d + e*x)^4) - ((e*f - d*g)*(e*f + 3*d*g))/(12*d^2*e^3*(d + e*x)^3) - (e*f + d*g)^2/
(16*d^3*e^3*(d + e*x)^2) - (e*f + d*g)^2/(16*d^4*e^3*(d + e*x)) + ((e*f + d*g)^2*ArcTanh[(e*x)/d])/(16*d^5*e^3
)

Rule 90

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 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 862

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/e)*x)^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)^4 \left (d^2-e^2 x^2\right )} \, dx &=\int \frac {(f+g x)^2}{(d-e x) (d+e x)^5} \, dx\\ &=\int \left (\frac {(-e f+d g)^2}{2 d e^2 (d+e x)^5}+\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^2 (d+e x)^4}+\frac {(e f+d g)^2}{8 d^3 e^2 (d+e x)^3}+\frac {(e f+d g)^2}{16 d^4 e^2 (d+e x)^2}+\frac {(e f+d g)^2}{16 d^4 e^2 \left (d^2-e^2 x^2\right )}\right ) \, dx\\ &=-\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4}-\frac {(e f-d g) (e f+3 d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(e f+d g)^2}{16 d^3 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{16 d^4 e^3 (d+e x)}+\frac {(e f+d g)^2 \int \frac {1}{d^2-e^2 x^2} \, dx}{16 d^4 e^2}\\ &=-\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4}-\frac {(e f-d g) (e f+3 d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(e f+d g)^2}{16 d^3 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{16 d^4 e^3 (d+e x)}+\frac {(e f+d g)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^5 e^3}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 142, normalized size = 1.02 \begin {gather*} -\frac {\frac {12 d^4 (e f-d g)^2}{(d+e x)^4}+\frac {8 d^3 \left (e^2 f^2+2 d e f g-3 d^2 g^2\right )}{(d+e x)^3}+\frac {6 d^2 (e f+d g)^2}{(d+e x)^2}+\frac {6 d (e f+d g)^2}{d+e x}+3 (e f+d g)^2 \log (d-e x)-3 (e f+d g)^2 \log (d+e x)}{96 d^5 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*((12*d^4*(e*f - d*g)^2)/(d + e*x)^4 + (8*d^3*(e^2*f^2 + 2*d*e*f*g - 3*d^2*g^2))/(d + e*x)^3 + (6*d^2*(e*
f + d*g)^2)/(d + e*x)^2 + (6*d*(e*f + d*g)^2)/(d + e*x) + 3*(e*f + d*g)^2*Log[d - e*x] - 3*(e*f + d*g)^2*Log[d
+ e*x])/(d^5*e^3)

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Maple [A]
time = 0.10, size = 220, normalized size = 1.58

method	result	size
norman	\(\frac {-\frac {\left (3 d^{2} g^{2}-26 d e f g -61 e^{2} f^{2}\right ) x^{3}}{48 d^{4}}-\frac {\left (d^{2} g^{2}-2 d e f g -7 e^{2} f^{2}\right ) x^{2}}{4 e \,d^{3}}+\frac {e^{2} \left (d f g +2 e \,f^{2}\right ) x^{4}}{6 d^{5}}-\frac {\left (d^{2} g^{2}+2 d e f g -15 e^{2} f^{2}\right ) x}{16 d^{2} e^{2}}}{\left (e x +d \right )^{4}}-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{32 e^{3} d^{5}}+\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{32 e^{3} d^{5}}\)	\(199\)
default	\(-\frac {-3 d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{12 d^{2} e^{3} \left (e x +d \right )^{3}}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{8 e^{3} d \left (e x +d \right )^{4}}+\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{32 e^{3} d^{5}}-\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{16 e^{3} d^{4} \left (e x +d \right )}-\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{16 e^{3} d^{3} \left (e x +d \right )^{2}}+\frac {\left (-d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{32 e^{3} d^{5}}\)	\(220\)
risch	\(\frac {-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) x^{3}}{16 d^{4}}-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) x^{2}}{4 d^{3} e}-\frac {\left (3 d^{2} g^{2}+38 d e f g +19 e^{2} f^{2}\right ) x}{48 d^{2} e^{2}}-\frac {f \left (d g +2 e f \right )}{6 e^{2} d}}{\left (e x +d \right )^{4}}-\frac {\ln \left (-e x +d \right ) g^{2}}{32 e^{3} d^{3}}-\frac {\ln \left (-e x +d \right ) f g}{16 e^{2} d^{4}}-\frac {\ln \left (-e x +d \right ) f^{2}}{32 e \,d^{5}}+\frac {\ln \left (e x +d \right ) g^{2}}{32 e^{3} d^{3}}+\frac {\ln \left (e x +d \right ) f g}{16 e^{2} d^{4}}+\frac {\ln \left (e x +d \right ) f^{2}}{32 e \,d^{5}}\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^2/(e*x+d)^4/(-e^2*x^2+d^2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(-3*d^2*g^2+2*d*e*f*g+e^2*f^2)/d^2/e^3/(e*x+d)^3-1/8/e^3*(d^2*g^2-2*d*e*f*g+e^2*f^2)/d/(e*x+d)^4+1/32*(d
^2*g^2+2*d*e*f*g+e^2*f^2)/e^3/d^5*ln(e*x+d)-1/16*(d^2*g^2+2*d*e*f*g+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+1/32*(-d^2*g^2-2*d*e*f*g-e^2*f^2)/e^3/d^5*ln(-e*x+d)

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Maxima [A]
time = 0.30, size = 228, normalized size = 1.64 \begin {gather*} -\frac {8 \, d^{4} f g + 16 \, d^{3} f^{2} e + 3 \, {\left (d^{2} g^{2} e^{2} + 2 \, d f g e^{3} + f^{2} e^{4}\right )} x^{3} + 12 \, {\left (d^{3} g^{2} e + 2 \, d^{2} f g e^{2} + d f^{2} e^{3}\right )} x^{2} + {\left (3 \, d^{4} g^{2} + 38 \, d^{3} f g e + 19 \, d^{2} f^{2} e^{2}\right )} x}{48 \, {\left (d^{4} x^{4} e^{6} + 4 \, d^{5} x^{3} e^{5} + 6 \, d^{6} x^{2} e^{4} + 4 \, d^{7} x e^{3} + d^{8} e^{2}\right )}} + \frac {{\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e + d\right )}{32 \, d^{5}} - \frac {{\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right )}{32 \, d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(e*x+d)^4/(-e^2*x^2+d^2),x, algorithm="maxima")

[Out]

-1/48*(8*d^4*f*g + 16*d^3*f^2*e + 3*(d^2*g^2*e^2 + 2*d*f*g*e^3 + f^2*e^4)*x^3 + 12*(d^3*g^2*e + 2*d^2*f*g*e^2
+ d*f^2*e^3)*x^2 + (3*d^4*g^2 + 38*d^3*f*g*e + 19*d^2*f^2*e^2)*x)/(d^4*x^4*e^6 + 4*d^5*x^3*e^5 + 6*d^6*x^2*e^4
+ 4*d^7*x*e^3 + d^8*e^2) + 1/32*(d^2*g^2 + 2*d*f*g*e + f^2*e^2)*e^(-3)*log(x*e + d)/d^5 - 1/32*(d^2*g^2 + 2*d
*f*g*e + f^2*e^2)*e^(-3)*log(x*e - d)/d^5

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (135) = 270\).
time = 3.30, size = 505, normalized size = 3.63 \begin {gather*} -\frac {6 \, d f^{2} x^{3} e^{5} + 12 \, {\left (d^{2} f g x^{3} + 2 \, d^{2} f^{2} x^{2}\right )} e^{4} + 2 \, {\left (3 \, d^{3} g^{2} x^{3} + 24 \, d^{3} f g x^{2} + 19 \, d^{3} f^{2} x\right )} e^{3} + 4 \, {\left (6 \, d^{4} g^{2} x^{2} + 19 \, d^{4} f g x + 8 \, d^{4} f^{2}\right )} e^{2} + 2 \, {\left (3 \, d^{5} g^{2} x + 8 \, d^{5} f g\right )} e - 3 \, {\left (d^{6} g^{2} + f^{2} x^{4} e^{6} + 2 \, {\left (d f g x^{4} + 2 \, d f^{2} x^{3}\right )} e^{5} + {\left (d^{2} g^{2} x^{4} + 8 \, d^{2} f g x^{3} + 6 \, d^{2} f^{2} x^{2}\right )} e^{4} + 4 \, {\left (d^{3} g^{2} x^{3} + 3 \, d^{3} f g x^{2} + d^{3} f^{2} x\right )} e^{3} + {\left (6 \, d^{4} g^{2} x^{2} + 8 \, d^{4} f g x + d^{4} f^{2}\right )} e^{2} + 2 \, {\left (2 \, d^{5} g^{2} x + d^{5} f g\right )} e\right )} \log \left (x e + d\right ) + 3 \, {\left (d^{6} g^{2} + f^{2} x^{4} e^{6} + 2 \, {\left (d f g x^{4} + 2 \, d f^{2} x^{3}\right )} e^{5} + {\left (d^{2} g^{2} x^{4} + 8 \, d^{2} f g x^{3} + 6 \, d^{2} f^{2} x^{2}\right )} e^{4} + 4 \, {\left (d^{3} g^{2} x^{3} + 3 \, d^{3} f g x^{2} + d^{3} f^{2} x\right )} e^{3} + {\left (6 \, d^{4} g^{2} x^{2} + 8 \, d^{4} f g x + d^{4} f^{2}\right )} e^{2} + 2 \, {\left (2 \, d^{5} g^{2} x + d^{5} f g\right )} e\right )} \log \left (x e - d\right )}{96 \, {\left (d^{5} x^{4} e^{7} + 4 \, d^{6} x^{3} e^{6} + 6 \, d^{7} x^{2} e^{5} + 4 \, d^{8} x e^{4} + d^{9} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(e*x+d)^4/(-e^2*x^2+d^2),x, algorithm="fricas")

[Out]

-1/96*(6*d*f^2*x^3*e^5 + 12*(d^2*f*g*x^3 + 2*d^2*f^2*x^2)*e^4 + 2*(3*d^3*g^2*x^3 + 24*d^3*f*g*x^2 + 19*d^3*f^2
*x)*e^3 + 4*(6*d^4*g^2*x^2 + 19*d^4*f*g*x + 8*d^4*f^2)*e^2 + 2*(3*d^5*g^2*x + 8*d^5*f*g)*e - 3*(d^6*g^2 + f^2*
x^4*e^6 + 2*(d*f*g*x^4 + 2*d*f^2*x^3)*e^5 + (d^2*g^2*x^4 + 8*d^2*f*g*x^3 + 6*d^2*f^2*x^2)*e^4 + 4*(d^3*g^2*x^3
+ 3*d^3*f*g*x^2 + d^3*f^2*x)*e^3 + (6*d^4*g^2*x^2 + 8*d^4*f*g*x + d^4*f^2)*e^2 + 2*(2*d^5*g^2*x + d^5*f*g)*e)
*log(x*e + d) + 3*(d^6*g^2 + f^2*x^4*e^6 + 2*(d*f*g*x^4 + 2*d*f^2*x^3)*e^5 + (d^2*g^2*x^4 + 8*d^2*f*g*x^3 + 6*
d^2*f^2*x^2)*e^4 + 4*(d^3*g^2*x^3 + 3*d^3*f*g*x^2 + d^3*f^2*x)*e^3 + (6*d^4*g^2*x^2 + 8*d^4*f*g*x + d^4*f^2)*e
^2 + 2*(2*d^5*g^2*x + d^5*f*g)*e)*log(x*e - d))/(d^5*x^4*e^7 + 4*d^6*x^3*e^6 + 6*d^7*x^2*e^5 + 4*d^8*x*e^4 + d
^9*e^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (122) = 244\).
time = 0.74, size = 282, normalized size = 2.03 \begin {gather*} - \frac {8 d^{4} f g + 16 d^{3} e f^{2} + x^{3} \cdot \left (3 d^{2} e^{2} g^{2} + 6 d e^{3} f g + 3 e^{4} f^{2}\right ) + x^{2} \cdot \left (12 d^{3} e g^{2} + 24 d^{2} e^{2} f g + 12 d e^{3} f^{2}\right ) + x \left (3 d^{4} g^{2} + 38 d^{3} e f g + 19 d^{2} e^{2} f^{2}\right )}{48 d^{8} e^{2} + 192 d^{7} e^{3} x + 288 d^{6} e^{4} x^{2} + 192 d^{5} e^{5} x^{3} + 48 d^{4} e^{6} x^{4}} - \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 )}}{32 d^{5} 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 )}}{32 d^{5} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2/(e*x+d)**4/(-e**2*x**2+d**2),x)

[Out]

-(8*d**4*f*g + 16*d**3*e*f**2 + x**3*(3*d**2*e**2*g**2 + 6*d*e**3*f*g + 3*e**4*f**2) + x**2*(12*d**3*e*g**2 +
24*d**2*e**2*f*g + 12*d*e**3*f**2) + x*(3*d**4*g**2 + 38*d**3*e*f*g + 19*d**2*e**2*f**2))/(48*d**8*e**2 + 192*
d**7*e**3*x + 288*d**6*e**4*x**2 + 192*d**5*e**5*x**3 + 48*d**4*e**6*x**4) - (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)/(32*d**5*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)/(32*d**5*e**3)

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Giac [A]
time = 1.25, size = 206, normalized size = 1.48 \begin {gather*} \frac {{\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right )}{32 \, d^{5}} - \frac {{\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right )}{32 \, d^{5}} - \frac {{\left (8 \, d^{5} f g e + 16 \, d^{4} f^{2} e^{2} + 3 \, {\left (d^{3} g^{2} e^{3} + 2 \, d^{2} f g e^{4} + d f^{2} e^{5}\right )} x^{3} + 12 \, {\left (d^{4} g^{2} e^{2} + 2 \, d^{3} f g e^{3} + d^{2} f^{2} e^{4}\right )} x^{2} + {\left (3 \, d^{5} g^{2} e + 38 \, d^{4} f g e^{2} + 19 \, d^{3} f^{2} e^{3}\right )} x\right )} e^{\left (-3\right )}}{48 \, {\left (x e + d\right )}^{4} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(d^2*g^2 + 2*d*f*g*e + f^2*e^2)*e^(-3)*log(abs(x*e + d))/d^5 - 1/32*(d^2*g^2 + 2*d*f*g*e + f^2*e^2)*e^(-3
)*log(abs(x*e - d))/d^5 - 1/48*(8*d^5*f*g*e + 16*d^4*f^2*e^2 + 3*(d^3*g^2*e^3 + 2*d^2*f*g*e^4 + d*f^2*e^5)*x^3
+ 12*(d^4*g^2*e^2 + 2*d^3*f*g*e^3 + d^2*f^2*e^4)*x^2 + (3*d^5*g^2*e + 38*d^4*f*g*e^2 + 19*d^3*f^2*e^3)*x)*e^(
-3)/((x*e + d)^4*d^5)

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Mupad [B]
time = 0.15, size = 180, normalized size = 1.29 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,{\left (d\,g+e\,f\right )}^2}{16\,d^5\,e^3}-\frac {\frac {x^3\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{16\,d^4}+\frac {2\,e\,f^2+d\,g\,f}{6\,d\,e^2}+\frac {x\,\left (3\,d^2\,g^2+38\,d\,e\,f\,g+19\,e^2\,f^2\right )}{48\,d^2\,e^2}+\frac {x^2\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^3\,e}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \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)^4),x)

[Out]

(atanh((e*x)/d)*(d*g + e*f)^2)/(16*d^5*e^3) - ((x^3*(d^2*g^2 + e^2*f^2 + 2*d*e*f*g))/(16*d^4) + (2*e*f^2 + d*f
*g)/(6*d*e^2) + (x*(3*d^2*g^2 + 19*e^2*f^2 + 38*d*e*f*g))/(48*d^2*e^2) + (x^2*(d^2*g^2 + e^2*f^2 + 2*d*e*f*g))
/(4*d^3*e))/(d^4 + e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x)

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