### 3.511 $$\int \frac{1}{(d+e x)^2 (a+c x^2)^2} \, dx$$

Optimal. Leaf size=205 $\frac{\sqrt{c} \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^3}+\frac{a e+c d x}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}-\frac{2 c d e^3 \log \left (a+c x^2\right )}{\left (a e^2+c d^2\right )^3}+\frac{e \left (c d^2-3 a e^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^2}+\frac{4 c d e^3 \log (d+e x)}{\left (a e^2+c d^2\right )^3}$

[Out]

(e*(c*d^2 - 3*a*e^2))/(2*a*(c*d^2 + a*e^2)^2*(d + e*x)) + (a*e + c*d*x)/(2*a*(c*d^2 + a*e^2)*(d + e*x)*(a + c*
x^2)) + (Sqrt[c]*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)
^3) + (4*c*d*e^3*Log[d + e*x])/(c*d^2 + a*e^2)^3 - (2*c*d*e^3*Log[a + c*x^2])/(c*d^2 + a*e^2)^3

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Rubi [A]  time = 0.189461, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.294, Rules used = {741, 801, 635, 205, 260} $\frac{\sqrt{c} \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^3}+\frac{a e+c d x}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}-\frac{2 c d e^3 \log \left (a+c x^2\right )}{\left (a e^2+c d^2\right )^3}+\frac{e \left (c d^2-3 a e^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^2}+\frac{4 c d e^3 \log (d+e x)}{\left (a e^2+c d^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^2*(a + c*x^2)^2),x]

[Out]

(e*(c*d^2 - 3*a*e^2))/(2*a*(c*d^2 + a*e^2)^2*(d + e*x)) + (a*e + c*d*x)/(2*a*(c*d^2 + a*e^2)*(d + e*x)*(a + c*
x^2)) + (Sqrt[c]*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)
^3) + (4*c*d*e^3*Log[d + e*x])/(c*d^2 + a*e^2)^3 - (2*c*d*e^3*Log[a + c*x^2])/(c*d^2 + a*e^2)^3

Rule 741

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*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx &=\frac{a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac{\int \frac{-c d^2-3 a e^2-2 c d e x}{(d+e x)^2 \left (a+c x^2\right )} \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac{a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac{\int \left (\frac{c d^2 e^2-3 a e^4}{\left (c d^2+a e^2\right ) (d+e x)^2}-\frac{8 a c d e^4}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac{c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4-8 a c d e^3 x\right )}{\left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}\right ) \, dx}{2 a \left (c d^2+a e^2\right )}\\ &=\frac{e \left (c d^2-3 a e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}+\frac{a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}+\frac{4 c d e^3 \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac{c \int \frac{c^2 d^4+6 a c d^2 e^2-3 a^2 e^4-8 a c d e^3 x}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=\frac{e \left (c d^2-3 a e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}+\frac{a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}+\frac{4 c d e^3 \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac{\left (4 c^2 d e^3\right ) \int \frac{x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}+\frac{\left (c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right ) \int \frac{1}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=\frac{e \left (c d^2-3 a e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}+\frac{a e+c d x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}+\frac{\sqrt{c} \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )^3}+\frac{4 c d e^3 \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac{2 c d e^3 \log \left (a+c x^2\right )}{\left (c d^2+a e^2\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.213506, size = 162, normalized size = 0.79 $\frac{\frac{\sqrt{c} \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{a^{3/2}}+\frac{c \left (a e^2+c d^2\right ) \left (a e (2 d-e x)+c d^2 x\right )}{a \left (a+c x^2\right )}-\frac{2 e^3 \left (a e^2+c d^2\right )}{d+e x}-4 c d e^3 \log \left (a+c x^2\right )+8 c d e^3 \log (d+e x)}{2 \left (a e^2+c d^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^2*(a + c*x^2)^2),x]

[Out]

((-2*e^3*(c*d^2 + a*e^2))/(d + e*x) + (c*(c*d^2 + a*e^2)*(c*d^2*x + a*e*(2*d - e*x)))/(a*(a + c*x^2)) + (Sqrt[
c]*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(3/2) + 8*c*d*e^3*Log[d + e*x] - 4*c*d
*e^3*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)

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Maple [A]  time = 0.064, size = 314, normalized size = 1.5 \begin{align*} -{\frac{acx{e}^{4}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) }}+{\frac{{c}^{3}x{d}^{4}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) a}}+{\frac{acd{e}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) }}+{\frac{{c}^{2}{d}^{3}e}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3} \left ( c{x}^{2}+a \right ) }}-2\,{\frac{d{e}^{3}c\ln \left ( c{x}^{2}+a \right ) }{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-{\frac{3\,a{e}^{4}c}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+3\,{\frac{{d}^{2}{e}^{2}{c}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) }+{\frac{{c}^{3}{d}^{4}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}a}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{{e}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}+4\,{\frac{d{e}^{3}c\ln \left ( ex+d \right ) }{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/(c*x^2+a)^2,x)

[Out]

-1/2*c/(a*e^2+c*d^2)^3/(c*x^2+a)*x*a*e^4+1/2*c^3/(a*e^2+c*d^2)^3/(c*x^2+a)*x/a*d^4+c/(a*e^2+c*d^2)^3/(c*x^2+a)
*a*d*e^3+c^2/(a*e^2+c*d^2)^3/(c*x^2+a)*d^3*e-2*c*d*e^3*ln(c*x^2+a)/(a*e^2+c*d^2)^3-3/2*c/(a*e^2+c*d^2)^3*a/(a*
c)^(1/2)*arctan(x*c/(a*c)^(1/2))*e^4+3*c^2/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d^2*e^2+1/2*c^3
/(a*e^2+c*d^2)^3/a/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d^4-e^3/(a*e^2+c*d^2)^2/(e*x+d)+4*c*d*e^3*ln(e*x+d)/(a*
e^2+c*d^2)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 16.3465, size = 2225, normalized size = 10.85 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/4*(4*a*c^2*d^4*e - 4*a^3*e^5 + 2*(c^3*d^4*e - 2*a*c^2*d^2*e^3 - 3*a^2*c*e^5)*x^2 - (a*c^2*d^5 + 6*a^2*c*d^3
*e^2 - 3*a^3*d*e^4 + (c^3*d^4*e + 6*a*c^2*d^2*e^3 - 3*a^2*c*e^5)*x^3 + (c^3*d^5 + 6*a*c^2*d^3*e^2 - 3*a^2*c*d*
e^4)*x^2 + (a*c^2*d^4*e + 6*a^2*c*d^2*e^3 - 3*a^3*e^5)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a) - a)/(c*x^2
+ a)) + 2*(c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x - 8*(a*c^2*d*e^4*x^3 + a*c^2*d^2*e^3*x^2 + a^2*c*d*e^4*
x + a^2*c*d^2*e^3)*log(c*x^2 + a) + 16*(a*c^2*d*e^4*x^3 + a*c^2*d^2*e^3*x^2 + a^2*c*d*e^4*x + a^2*c*d^2*e^3)*l
og(e*x + d))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3*a^4*c*d^3*e^4 + a^5*d*e^6 + (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3
+ 3*a^3*c^2*d^2*e^5 + a^4*c*e^7)*x^3 + (a*c^4*d^7 + 3*a^2*c^3*d^5*e^2 + 3*a^3*c^2*d^3*e^4 + a^4*c*d*e^6)*x^2
+ (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^2*e^5 + a^5*e^7)*x), 1/2*(2*a*c^2*d^4*e - 2*a^3*e^5 + (c^3*d^
4*e - 2*a*c^2*d^2*e^3 - 3*a^2*c*e^5)*x^2 + (a*c^2*d^5 + 6*a^2*c*d^3*e^2 - 3*a^3*d*e^4 + (c^3*d^4*e + 6*a*c^2*d
^2*e^3 - 3*a^2*c*e^5)*x^3 + (c^3*d^5 + 6*a*c^2*d^3*e^2 - 3*a^2*c*d*e^4)*x^2 + (a*c^2*d^4*e + 6*a^2*c*d^2*e^3 -
3*a^3*e^5)*x)*sqrt(c/a)*arctan(x*sqrt(c/a)) + (c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x - 4*(a*c^2*d*e^4*x^
3 + a*c^2*d^2*e^3*x^2 + a^2*c*d*e^4*x + a^2*c*d^2*e^3)*log(c*x^2 + a) + 8*(a*c^2*d*e^4*x^3 + a*c^2*d^2*e^3*x^2
+ a^2*c*d*e^4*x + a^2*c*d^2*e^3)*log(e*x + d))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3*a^4*c*d^3*e^4 + a^5*d*e^6
+ (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3 + 3*a^3*c^2*d^2*e^5 + a^4*c*e^7)*x^3 + (a*c^4*d^7 + 3*a^2*c^3*d^5*e^2 + 3*
a^3*c^2*d^3*e^4 + a^4*c*d*e^6)*x^2 + (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^2*e^5 + a^5*e^7)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(c*x**2+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.25612, size = 521, normalized size = 2.54 \begin{align*} -\frac{2 \, c d e^{3} \log \left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac{{\left (c^{3} d^{4} e^{2} + 6 \, a c^{2} d^{2} e^{4} - 3 \, a^{2} c e^{6}\right )} \arctan \left (\frac{{\left (c d - \frac{c d^{2}}{x e + d} - \frac{a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{a c}}\right ) e^{\left (-2\right )}}{2 \,{\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt{a c}} - \frac{e^{7}}{{\left (c^{2} d^{4} e^{4} + 2 \, a c d^{2} e^{6} + a^{2} e^{8}\right )}{\left (x e + d\right )}} + \frac{\frac{c^{3} d^{3} e - 3 \, a c^{2} d e^{3}}{c d^{2} + a e^{2}} - \frac{{\left (c^{3} d^{4} e^{2} - 6 \, a c^{2} d^{2} e^{4} + a^{2} c e^{6}\right )} e^{\left (-1\right )}}{{\left (c d^{2} + a e^{2}\right )}{\left (x e + d\right )}}}{2 \,{\left (c d^{2} + a e^{2}\right )}^{2} a{\left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="giac")

[Out]

-2*c*d*e^3*log(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2)/(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2
*c*d^2*e^4 + a^3*e^6) + 1/2*(c^3*d^4*e^2 + 6*a*c^2*d^2*e^4 - 3*a^2*c*e^6)*arctan((c*d - c*d^2/(x*e + d) - a*e^
2/(x*e + d))*e^(-1)/sqrt(a*c))*e^(-2)/((a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3*c*d^2*e^4 + a^4*e^6)*sqrt(a*c))
- e^7/((c^2*d^4*e^4 + 2*a*c*d^2*e^6 + a^2*e^8)*(x*e + d)) + 1/2*((c^3*d^3*e - 3*a*c^2*d*e^3)/(c*d^2 + a*e^2) -
(c^3*d^4*e^2 - 6*a*c^2*d^2*e^4 + a^2*c*e^6)*e^(-1)/((c*d^2 + a*e^2)*(x*e + d)))/((c*d^2 + a*e^2)^2*a*(c - 2*c
*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2))