3.4.69 \(\int \frac {(f+g x)^2}{(d+e x)^3 (d^2-e^2 x^2)^2} \, dx\)

Optimal. Leaf size=178 \[ \frac {(d g+e f) (d g+5 e f) \tanh ^{-1}\left (\frac {e x}{d}\right )}{32 d^6 e^3}+\frac {(d g+e f)^2}{32 d^5 e^3 (d-e x)}-\frac {f (d g+e f)}{8 d^5 e^2 (d+e x)}-\frac {(3 e f-d g) (d g+e f)}{32 d^4 e^3 (d+e x)^2}-\frac {(e f-d g)^2}{16 d^2 e^3 (d+e x)^4}-\frac {e^2 f^2-d^2 g^2}{12 d^3 e^3 (d+e x)^3} \]

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Rubi [A]  time = 0.20, antiderivative size = 178, 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 {e^2 f^2-d^2 g^2}{12 d^3 e^3 (d+e x)^3}-\frac {(e f-d g)^2}{16 d^2 e^3 (d+e x)^4}+\frac {(d g+e f)^2}{32 d^5 e^3 (d-e x)}-\frac {f (d g+e f)}{8 d^5 e^2 (d+e x)}-\frac {(3 e f-d g) (d g+e f)}{32 d^4 e^3 (d+e x)^2}+\frac {(d g+e f) (d g+5 e f) \tanh ^{-1}\left (\frac {e x}{d}\right )}{32 d^6 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*f + d*g)^2/(32*d^5*e^3*(d - e*x)) - (e*f - d*g)^2/(16*d^2*e^3*(d + e*x)^4) - (e^2*f^2 - d^2*g^2)/(12*d^3*e^
3*(d + e*x)^3) - ((3*e*f - d*g)*(e*f + d*g))/(32*d^4*e^3*(d + e*x)^2) - (f*(e*f + d*g))/(8*d^5*e^2*(d + e*x))
+ ((e*f + d*g)*(5*e*f + d*g)*ArcTanh[(e*x)/d])/(32*d^6*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)^3 \left (d^2-e^2 x^2\right )^2} \, dx &=\int \frac {(f+g x)^2}{(d-e x)^2 (d+e x)^5} \, dx\\ &=\int \left (\frac {(e f+d g)^2}{32 d^5 e^2 (d-e x)^2}+\frac {(-e f+d g)^2}{4 d^2 e^2 (d+e x)^5}+\frac {e^2 f^2-d^2 g^2}{4 d^3 e^2 (d+e x)^4}+\frac {(3 e f-d g) (e f+d g)}{16 d^4 e^2 (d+e x)^3}+\frac {f (e f+d g)}{8 d^5 e (d+e x)^2}+\frac {(e f+d g) (5 e f+d g)}{32 d^5 e^2 \left (d^2-e^2 x^2\right )}\right ) \, dx\\ &=\frac {(e f+d g)^2}{32 d^5 e^3 (d-e x)}-\frac {(e f-d g)^2}{16 d^2 e^3 (d+e x)^4}-\frac {e^2 f^2-d^2 g^2}{12 d^3 e^3 (d+e x)^3}-\frac {(3 e f-d g) (e f+d g)}{32 d^4 e^3 (d+e x)^2}-\frac {f (e f+d g)}{8 d^5 e^2 (d+e x)}+\frac {((e f+d g) (5 e f+d g)) \int \frac {1}{d^2-e^2 x^2} \, dx}{32 d^5 e^2}\\ &=\frac {(e f+d g)^2}{32 d^5 e^3 (d-e x)}-\frac {(e f-d g)^2}{16 d^2 e^3 (d+e x)^4}-\frac {e^2 f^2-d^2 g^2}{12 d^3 e^3 (d+e x)^3}-\frac {(3 e f-d g) (e f+d g)}{32 d^4 e^3 (d+e x)^2}-\frac {f (e f+d g)}{8 d^5 e^2 (d+e x)}+\frac {(e f+d g) (5 e f+d g) \tanh ^{-1}\left (\frac {e x}{d}\right )}{32 d^6 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 195, normalized size = 1.10 \begin {gather*} \frac {-\frac {12 d^4 (e f-d g)^2}{(d+e x)^4}+\frac {6 d^2 \left (d^2 g^2-2 d e f g-3 e^2 f^2\right )}{(d+e x)^2}-3 \left (d^2 g^2+6 d e f g+5 e^2 f^2\right ) \log (d-e x)+3 \left (d^2 g^2+6 d e f g+5 e^2 f^2\right ) \log (d+e x)+\frac {16 d^3 \left (d^2 g^2-e^2 f^2\right )}{(d+e x)^3}+\frac {6 d (d g+e f)^2}{d-e x}-\frac {24 d e f (d g+e f)}{d+e x}}{192 d^6 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((6*d*(e*f + d*g)^2)/(d - e*x) - (12*d^4*(e*f - d*g)^2)/(d + e*x)^4 + (16*d^3*(-(e^2*f^2) + d^2*g^2))/(d + e*x
)^3 + (6*d^2*(-3*e^2*f^2 - 2*d*e*f*g + d^2*g^2))/(d + e*x)^2 - (24*d*e*f*(e*f + d*g))/(d + e*x) - 3*(5*e^2*f^2
+ 6*d*e*f*g + d^2*g^2)*Log[d - e*x] + 3*(5*e^2*f^2 + 6*d*e*f*g + d^2*g^2)*Log[d + e*x])/(192*d^6*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)^3 \left (d^2-e^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.40, size = 648, normalized size = 3.64 \begin {gather*} \frac {64 \, d^{5} e^{2} f^{2} - 16 \, d^{7} g^{2} - 6 \, {\left (5 \, d e^{6} f^{2} + 6 \, d^{2} e^{5} f g + d^{3} e^{4} g^{2}\right )} x^{4} - 18 \, {\left (5 \, d^{2} e^{5} f^{2} + 6 \, d^{3} e^{4} f g + d^{4} e^{3} g^{2}\right )} x^{3} - 14 \, {\left (5 \, d^{3} e^{4} f^{2} + 6 \, d^{4} e^{3} f g + d^{5} e^{2} g^{2}\right )} x^{2} + 6 \, {\left (5 \, d^{4} e^{3} f^{2} + 6 \, d^{5} e^{2} f g - 7 \, d^{6} e g^{2}\right )} x - 3 \, {\left (5 \, d^{5} e^{2} f^{2} + 6 \, d^{6} e f g + d^{7} g^{2} - {\left (5 \, e^{7} f^{2} + 6 \, d e^{6} f g + d^{2} e^{5} g^{2}\right )} x^{5} - 3 \, {\left (5 \, d e^{6} f^{2} + 6 \, d^{2} e^{5} f g + d^{3} e^{4} g^{2}\right )} x^{4} - 2 \, {\left (5 \, d^{2} e^{5} f^{2} + 6 \, d^{3} e^{4} f g + d^{4} e^{3} g^{2}\right )} x^{3} + 2 \, {\left (5 \, d^{3} e^{4} f^{2} + 6 \, d^{4} e^{3} f g + d^{5} e^{2} g^{2}\right )} x^{2} + 3 \, {\left (5 \, d^{4} e^{3} f^{2} + 6 \, d^{5} e^{2} f g + d^{6} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + 3 \, {\left (5 \, d^{5} e^{2} f^{2} + 6 \, d^{6} e f g + d^{7} g^{2} - {\left (5 \, e^{7} f^{2} + 6 \, d e^{6} f g + d^{2} e^{5} g^{2}\right )} x^{5} - 3 \, {\left (5 \, d e^{6} f^{2} + 6 \, d^{2} e^{5} f g + d^{3} e^{4} g^{2}\right )} x^{4} - 2 \, {\left (5 \, d^{2} e^{5} f^{2} + 6 \, d^{3} e^{4} f g + d^{4} e^{3} g^{2}\right )} x^{3} + 2 \, {\left (5 \, d^{3} e^{4} f^{2} + 6 \, d^{4} e^{3} f g + d^{5} e^{2} g^{2}\right )} x^{2} + 3 \, {\left (5 \, d^{4} e^{3} f^{2} + 6 \, d^{5} e^{2} f g + d^{6} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{192 \, {\left (d^{6} e^{8} x^{5} + 3 \, d^{7} e^{7} x^{4} + 2 \, d^{8} e^{6} x^{3} - 2 \, d^{9} e^{5} x^{2} - 3 \, d^{10} e^{4} x - d^{11} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(64*d^5*e^2*f^2 - 16*d^7*g^2 - 6*(5*d*e^6*f^2 + 6*d^2*e^5*f*g + d^3*e^4*g^2)*x^4 - 18*(5*d^2*e^5*f^2 + 6
*d^3*e^4*f*g + d^4*e^3*g^2)*x^3 - 14*(5*d^3*e^4*f^2 + 6*d^4*e^3*f*g + d^5*e^2*g^2)*x^2 + 6*(5*d^4*e^3*f^2 + 6*
d^5*e^2*f*g - 7*d^6*e*g^2)*x - 3*(5*d^5*e^2*f^2 + 6*d^6*e*f*g + d^7*g^2 - (5*e^7*f^2 + 6*d*e^6*f*g + d^2*e^5*g
^2)*x^5 - 3*(5*d*e^6*f^2 + 6*d^2*e^5*f*g + d^3*e^4*g^2)*x^4 - 2*(5*d^2*e^5*f^2 + 6*d^3*e^4*f*g + d^4*e^3*g^2)*
x^3 + 2*(5*d^3*e^4*f^2 + 6*d^4*e^3*f*g + d^5*e^2*g^2)*x^2 + 3*(5*d^4*e^3*f^2 + 6*d^5*e^2*f*g + d^6*e*g^2)*x)*l
og(e*x + d) + 3*(5*d^5*e^2*f^2 + 6*d^6*e*f*g + d^7*g^2 - (5*e^7*f^2 + 6*d*e^6*f*g + d^2*e^5*g^2)*x^5 - 3*(5*d*
e^6*f^2 + 6*d^2*e^5*f*g + d^3*e^4*g^2)*x^4 - 2*(5*d^2*e^5*f^2 + 6*d^3*e^4*f*g + d^4*e^3*g^2)*x^3 + 2*(5*d^3*e^
4*f^2 + 6*d^4*e^3*f*g + d^5*e^2*g^2)*x^2 + 3*(5*d^4*e^3*f^2 + 6*d^5*e^2*f*g + d^6*e*g^2)*x)*log(e*x - d))/(d^6
*e^8*x^5 + 3*d^7*e^7*x^4 + 2*d^8*e^6*x^3 - 2*d^9*e^5*x^2 - 3*d^10*e^4*x - d^11*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)^3/(-e^2*x^2+d^2)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-3*exp(2)^2*d^2*exp(1)*g^2+12*exp(2)^2*
d*exp(1)^2*g*f-10*exp(2)^2*exp(1)^3*f^2-8*exp(2)*d^2*exp(1)^3*g^2+12*exp(2)*d*exp(1)^4*g*f-2*exp(2)*exp(1)^5*f
^2-d^2*exp(1)^5*g^2)/(2*exp(2)^4*d^6-8*exp(2)^3*d^6*exp(1)^2+12*exp(2)^2*d^6*exp(1)^4-8*exp(2)*d^6*exp(1)^6+2*
d^6*exp(1)^8)*ln(abs(-x^2*exp(2)+d^2))+(exp(2)^4*f^2-exp(2)^3*d^2*g^2+6*exp(2)^3*d*exp(1)*g*f-10*exp(2)^3*exp(
1)^2*f^2-14*exp(2)^2*d^2*exp(1)^2*g^2+36*exp(2)^2*d*exp(1)^3*g*f-15*exp(2)^2*exp(1)^4*f^2-9*exp(2)*d^2*exp(1)^
4*g^2+6*exp(2)*d*exp(1)^5*g*f)*1/2/(2*exp(2)^4*d^5-8*exp(2)^3*d^5*exp(1)^2+12*exp(2)^2*d^5*exp(1)^4-8*exp(2)*d
^5*exp(1)^6+2*d^5*exp(1)^8)/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))
)+(3*exp(2)^2*d^2*exp(1)^2*g^2-12*exp(2)^2*d*exp(1)^3*g*f+10*exp(2)^2*exp(1)^4*f^2+8*exp(2)*d^2*exp(1)^4*g^2-1
2*exp(2)*d*exp(1)^5*g*f+2*exp(2)*exp(1)^6*f^2+d^2*exp(1)^6*g^2)/(exp(2)^4*d^6*exp(1)-4*exp(2)^3*d^6*exp(1)^3+6
*exp(2)^2*d^6*exp(1)^5-4*exp(2)*d^6*exp(1)^7+d^6*exp(1)^9)*ln(abs(x*exp(1)+d))-((-exp(2)^4*d*exp(1)^2*f^2-5*ex
p(2)^3*d^3*exp(1)^2*g^2+18*exp(2)^3*d^2*exp(1)^3*g*f-10*exp(2)^3*d*exp(1)^4*f^2-2*exp(2)^2*d^3*exp(1)^4*g^2-12
*exp(2)^2*d^2*exp(1)^5*g*f+11*exp(2)^2*d*exp(1)^6*f^2+7*exp(2)*d^3*exp(1)^6*g^2-6*exp(2)*d^2*exp(1)^7*g*f)*x^3
+(-2*exp(2)^4*d^2*exp(1)*f^2-7*exp(2)^3*d^4*exp(1)*g^2+24*exp(2)^3*d^3*exp(1)^2*g*f-10*exp(2)^3*d^2*exp(1)^3*f
^2+exp(2)^2*d^4*exp(1)^3*g^2-24*exp(2)^2*d^3*exp(1)^4*g*f+14*exp(2)^2*d^2*exp(1)^5*f^2+7*exp(2)*d^4*exp(1)^5*g
^2-2*exp(2)*d^2*exp(1)^7*f^2-d^4*exp(1)^7*g^2)*x^2+(-exp(2)^4*d^3*f^2-exp(2)^3*d^5*g^2+2*exp(2)^3*d^4*exp(1)*g
*f+4*exp(2)^3*d^3*exp(1)^2*f^2+8*exp(2)^2*d^5*exp(1)^2*g^2-24*exp(2)^2*d^4*exp(1)^3*g*f+7*exp(2)^2*d^3*exp(1)^
4*f^2-exp(2)*d^5*exp(1)^4*g^2+18*exp(2)*d^4*exp(1)^5*g*f-10*exp(2)*d^3*exp(1)^6*f^2-6*d^5*exp(1)^6*g^2+4*d^4*e
xp(1)^7*g*f)*x-2*exp(2)^3*d^5*g*f+3*exp(2)^3*d^4*exp(1)*f^2+8*exp(2)^2*d^6*exp(1)*g^2-18*exp(2)^2*d^5*exp(1)^2
*g*f+7*exp(2)^2*d^4*exp(1)^3*f^2-4*exp(2)*d^6*exp(1)^3*g^2+18*exp(2)*d^5*exp(1)^4*g*f-11*exp(2)*d^4*exp(1)^5*f
^2-4*d^6*exp(1)^5*g^2+2*d^5*exp(1)^6*g*f+d^4*exp(1)^7*f^2)/2/d^6/(exp(2)-exp(1)^2)^4/(-x*exp(1)-d)^2/(-x^2*exp
(2)+d^2)

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maple [B]  time = 0.02, size = 341, normalized size = 1.92 \begin {gather*} \frac {f g}{8 \left (e x +d \right )^{4} d \,e^{2}}-\frac {f^{2}}{16 \left (e x +d \right )^{4} d^{2} e}-\frac {g^{2}}{16 \left (e x +d \right )^{4} e^{3}}+\frac {g^{2}}{12 \left (e x +d \right )^{3} d \,e^{3}}-\frac {f^{2}}{12 \left (e x +d \right )^{3} d^{3} e}+\frac {g^{2}}{32 \left (e x +d \right )^{2} d^{2} e^{3}}-\frac {f g}{16 \left (e x +d \right )^{2} d^{3} e^{2}}-\frac {3 f^{2}}{32 \left (e x +d \right )^{2} d^{4} e}-\frac {g^{2}}{32 \left (e x -d \right ) d^{3} e^{3}}-\frac {f g}{16 \left (e x -d \right ) d^{4} e^{2}}-\frac {f g}{8 \left (e x +d \right ) d^{4} e^{2}}-\frac {g^{2} \ln \left (e x -d \right )}{64 d^{4} e^{3}}+\frac {g^{2} \ln \left (e x +d \right )}{64 d^{4} e^{3}}-\frac {f^{2}}{32 \left (e x -d \right ) d^{5} e}-\frac {f^{2}}{8 \left (e x +d \right ) d^{5} e}-\frac {3 f g \ln \left (e x -d \right )}{32 d^{5} e^{2}}+\frac {3 f g \ln \left (e x +d \right )}{32 d^{5} e^{2}}-\frac {5 f^{2} \ln \left (e x -d \right )}{64 d^{6} e}+\frac {5 f^{2} \ln \left (e x +d \right )}{64 d^{6} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64/e^3/d^4*ln(e*x-d)*g^2-3/32/e^2/d^5*ln(e*x-d)*f*g-5/64/e/d^6*ln(e*x-d)*f^2-1/32/e^3/d^3/(e*x-d)*g^2-1/16/
e^2/d^4/(e*x-d)*f*g-1/32/e/d^5/(e*x-d)*f^2+1/64/e^3/d^4*ln(e*x+d)*g^2+3/32/e^2/d^5*ln(e*x+d)*f*g+5/64/e/d^6*ln
(e*x+d)*f^2+1/12/e^3/d/(e*x+d)^3*g^2-1/12/e/d^3/(e*x+d)^3*f^2+1/32/e^3/d^2/(e*x+d)^2*g^2-1/16/e^2/d^3/(e*x+d)^
2*f*g-3/32/e/d^4/(e*x+d)^2*f^2-1/16/e^3/(e*x+d)^4*g^2+1/8/e^2/d/(e*x+d)^4*f*g-1/16/e/d^2/(e*x+d)^4*f^2-1/8/e^2
*f/d^4/(e*x+d)*g-1/8/e*f^2/d^5/(e*x+d)

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maxima [A]  time = 0.50, size = 298, normalized size = 1.67 \begin {gather*} \frac {32 \, d^{4} e^{2} f^{2} - 8 \, d^{6} g^{2} - 3 \, {\left (5 \, e^{6} f^{2} + 6 \, d e^{5} f g + d^{2} e^{4} g^{2}\right )} x^{4} - 9 \, {\left (5 \, d e^{5} f^{2} + 6 \, d^{2} e^{4} f g + d^{3} e^{3} g^{2}\right )} x^{3} - 7 \, {\left (5 \, d^{2} e^{4} f^{2} + 6 \, d^{3} e^{3} f g + d^{4} e^{2} g^{2}\right )} x^{2} + 3 \, {\left (5 \, d^{3} e^{3} f^{2} + 6 \, d^{4} e^{2} f g - 7 \, d^{5} e g^{2}\right )} x}{96 \, {\left (d^{5} e^{8} x^{5} + 3 \, d^{6} e^{7} x^{4} + 2 \, d^{7} e^{6} x^{3} - 2 \, d^{8} e^{5} x^{2} - 3 \, d^{9} e^{4} x - d^{10} e^{3}\right )}} + \frac {{\left (5 \, e^{2} f^{2} + 6 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{64 \, d^{6} e^{3}} - \frac {{\left (5 \, e^{2} f^{2} + 6 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{64 \, d^{6} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(32*d^4*e^2*f^2 - 8*d^6*g^2 - 3*(5*e^6*f^2 + 6*d*e^5*f*g + d^2*e^4*g^2)*x^4 - 9*(5*d*e^5*f^2 + 6*d^2*e^4*
f*g + d^3*e^3*g^2)*x^3 - 7*(5*d^2*e^4*f^2 + 6*d^3*e^3*f*g + d^4*e^2*g^2)*x^2 + 3*(5*d^3*e^3*f^2 + 6*d^4*e^2*f*
g - 7*d^5*e*g^2)*x)/(d^5*e^8*x^5 + 3*d^6*e^7*x^4 + 2*d^7*e^6*x^3 - 2*d^8*e^5*x^2 - 3*d^9*e^4*x - d^10*e^3) + 1
/64*(5*e^2*f^2 + 6*d*e*f*g + d^2*g^2)*log(e*x + d)/(d^6*e^3) - 1/64*(5*e^2*f^2 + 6*d*e*f*g + d^2*g^2)*log(e*x
- d)/(d^6*e^3)

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mupad [B]  time = 2.70, size = 274, normalized size = 1.54 \begin {gather*} \frac {\frac {d^2\,g^2-4\,e^2\,f^2}{12\,d\,e^3}+\frac {3\,x^3\,\left (d^2\,g^2+6\,d\,e\,f\,g+5\,e^2\,f^2\right )}{32\,d^4}+\frac {e\,x^4\,\left (d^2\,g^2+6\,d\,e\,f\,g+5\,e^2\,f^2\right )}{32\,d^5}-\frac {x\,\left (-7\,d^2\,g^2+6\,d\,e\,f\,g+5\,e^2\,f^2\right )}{32\,d^2\,e^2}+\frac {7\,x^2\,\left (d^2\,g^2+6\,d\,e\,f\,g+5\,e^2\,f^2\right )}{96\,d^3\,e}}{d^5+3\,d^4\,e\,x+2\,d^3\,e^2\,x^2-2\,d^2\,e^3\,x^3-3\,d\,e^4\,x^4-e^5\,x^5}+\frac {\mathrm {atanh}\left (\frac {e\,x\,\left (d\,g+e\,f\right )\,\left (d\,g+5\,e\,f\right )}{d\,\left (d^2\,g^2+6\,d\,e\,f\,g+5\,e^2\,f^2\right )}\right )\,\left (d\,g+e\,f\right )\,\left (d\,g+5\,e\,f\right )}{32\,d^6\,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)^2*(d + e*x)^3),x)

[Out]

((d^2*g^2 - 4*e^2*f^2)/(12*d*e^3) + (3*x^3*(d^2*g^2 + 5*e^2*f^2 + 6*d*e*f*g))/(32*d^4) + (e*x^4*(d^2*g^2 + 5*e
^2*f^2 + 6*d*e*f*g))/(32*d^5) - (x*(5*e^2*f^2 - 7*d^2*g^2 + 6*d*e*f*g))/(32*d^2*e^2) + (7*x^2*(d^2*g^2 + 5*e^2
*f^2 + 6*d*e*f*g))/(96*d^3*e))/(d^5 - e^5*x^5 - 3*d*e^4*x^4 + 2*d^3*e^2*x^2 - 2*d^2*e^3*x^3 + 3*d^4*e*x) + (at
anh((e*x*(d*g + e*f)*(d*g + 5*e*f))/(d*(d^2*g^2 + 5*e^2*f^2 + 6*d*e*f*g)))*(d*g + e*f)*(d*g + 5*e*f))/(32*d^6*
e^3)

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sympy [B]  time = 1.93, size = 376, normalized size = 2.11 \begin {gather*} \frac {- 8 d^{6} g^{2} + 32 d^{4} e^{2} f^{2} + x^{4} \left (- 3 d^{2} e^{4} g^{2} - 18 d e^{5} f g - 15 e^{6} f^{2}\right ) + x^{3} \left (- 9 d^{3} e^{3} g^{2} - 54 d^{2} e^{4} f g - 45 d e^{5} f^{2}\right ) + x^{2} \left (- 7 d^{4} e^{2} g^{2} - 42 d^{3} e^{3} f g - 35 d^{2} e^{4} f^{2}\right ) + x \left (- 21 d^{5} e g^{2} + 18 d^{4} e^{2} f g + 15 d^{3} e^{3} f^{2}\right )}{- 96 d^{10} e^{3} - 288 d^{9} e^{4} x - 192 d^{8} e^{5} x^{2} + 192 d^{7} e^{6} x^{3} + 288 d^{6} e^{7} x^{4} + 96 d^{5} e^{8} x^{5}} - \frac {\left (d g + e f\right ) \left (d g + 5 e f\right ) \log {\left (- \frac {d \left (d g + e f\right ) \left (d g + 5 e f\right )}{e \left (d^{2} g^{2} + 6 d e f g + 5 e^{2} f^{2}\right )} + x \right )}}{64 d^{6} e^{3}} + \frac {\left (d g + e f\right ) \left (d g + 5 e f\right ) \log {\left (\frac {d \left (d g + e f\right ) \left (d g + 5 e f\right )}{e \left (d^{2} g^{2} + 6 d e f g + 5 e^{2} f^{2}\right )} + x \right )}}{64 d^{6} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-8*d**6*g**2 + 32*d**4*e**2*f**2 + x**4*(-3*d**2*e**4*g**2 - 18*d*e**5*f*g - 15*e**6*f**2) + x**3*(-9*d**3*e*
*3*g**2 - 54*d**2*e**4*f*g - 45*d*e**5*f**2) + x**2*(-7*d**4*e**2*g**2 - 42*d**3*e**3*f*g - 35*d**2*e**4*f**2)
+ x*(-21*d**5*e*g**2 + 18*d**4*e**2*f*g + 15*d**3*e**3*f**2))/(-96*d**10*e**3 - 288*d**9*e**4*x - 192*d**8*e*
*5*x**2 + 192*d**7*e**6*x**3 + 288*d**6*e**7*x**4 + 96*d**5*e**8*x**5) - (d*g + e*f)*(d*g + 5*e*f)*log(-d*(d*g
+ e*f)*(d*g + 5*e*f)/(e*(d**2*g**2 + 6*d*e*f*g + 5*e**2*f**2)) + x)/(64*d**6*e**3) + (d*g + e*f)*(d*g + 5*e*f
)*log(d*(d*g + e*f)*(d*g + 5*e*f)/(e*(d**2*g**2 + 6*d*e*f*g + 5*e**2*f**2)) + x)/(64*d**6*e**3)

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