\(\int \frac {1}{(d+e x) (a+c x^2)^2} \, dx\) [510]

Optimal result

Integrand size = 17, antiderivative size = 142 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=\frac {a e+c d x}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\sqrt {c} d \left (c d^2+3 a e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )^2}+\frac {e^3 \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {e^3 \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {755, 815, 649, 211, 266} \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=\frac {\sqrt {c} d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (3 a e^2+c d^2\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^2}+\frac {a e+c d x}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}-\frac {e^3 \log \left (a+c x^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac {e^3 \log (d+e x)}{\left (a e^2+c d^2\right )^2} \]

[In]

Int[1/((d + e*x)*(a + c*x^2)^2),x]

[Out]

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

Rule 211

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

Rule 266

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]

Rule 649

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 755

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 815

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]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=\frac {\sqrt {c} d \left (c d^2+3 a e^2\right ) \left (a+c x^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )+\sqrt {a} \left (\left (c d^2+a e^2\right ) (a e+c d x)+2 a e^3 \left (a+c x^2\right ) \log (d+e x)-a e^3 \left (a+c x^2\right ) \log \left (a+c x^2\right )\right )}{2 a^{3/2} \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {c \left (\frac {\frac {d \left (e^{2} a +c \,d^{2}\right ) x}{2 a}+\frac {e \left (e^{2} a +c \,d^{2}\right )}{2 c}}{c \,x^{2}+a}+\frac {-\frac {a \,e^{3} \ln \left (c \,x^{2}+a \right )}{c}+\frac {\left (3 a d \,e^{2}+d^{3} c \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 a}\right )}{\left (e^{2} a +c \,d^{2}\right )^{2}}+\frac {e^{3} \ln \left (e x +d \right )}{\left (e^{2} a +c \,d^{2}\right )^{2}}\)	\(138\)
risch	\(\text {Expression too large to display}\)	\(1407\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.59 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.11 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=\left [\frac {2 \, a c d^{2} e + 2 \, a^{2} e^{3} + {\left (a c d^{3} + 3 \, a^{2} d e^{2} + {\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} x^{2}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} + 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) + 2 \, {\left (c^{2} d^{3} + a c d e^{2}\right )} x - 2 \, {\left (a c e^{3} x^{2} + a^{2} e^{3}\right )} \log \left (c x^{2} + a\right ) + 4 \, {\left (a c e^{3} x^{2} + a^{2} e^{3}\right )} \log \left (e x + d\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} + {\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{2}\right )}}, \frac {a c d^{2} e + a^{2} e^{3} + {\left (a c d^{3} + 3 \, a^{2} d e^{2} + {\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} x^{2}\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + {\left (c^{2} d^{3} + a c d e^{2}\right )} x - {\left (a c e^{3} x^{2} + a^{2} e^{3}\right )} \log \left (c x^{2} + a\right ) + 2 \, {\left (a c e^{3} x^{2} + a^{2} e^{3}\right )} \log \left (e x + d\right )}{2 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} + {\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/4*(2*a*c*d^2*e + 2*a^2*e^3 + (a*c*d^3 + 3*a^2*d*e^2 + (c^2*d^3 + 3*a*c*d*e^2)*x^2)*sqrt(-c/a)*log((c*x^2 +
2*a*x*sqrt(-c/a) - a)/(c*x^2 + a)) + 2*(c^2*d^3 + a*c*d*e^2)*x - 2*(a*c*e^3*x^2 + a^2*e^3)*log(c*x^2 + a) + 4*
(a*c*e^3*x^2 + a^2*e^3)*log(e*x + d))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4 + (a*c^3*d^4 + 2*a^2*c^2*d^2*e^
2 + a^3*c*e^4)*x^2), 1/2*(a*c*d^2*e + a^2*e^3 + (a*c*d^3 + 3*a^2*d*e^2 + (c^2*d^3 + 3*a*c*d*e^2)*x^2)*sqrt(c/a
)*arctan(x*sqrt(c/a)) + (c^2*d^3 + a*c*d*e^2)*x - (a*c*e^3*x^2 + a^2*e^3)*log(c*x^2 + a) + 2*(a*c*e^3*x^2 + a^
2*e^3)*log(e*x + d))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4 + (a*c^3*d^4 + 2*a^2*c^2*d^2*e^2 + a^3*c*e^4)*x^
2)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=-\frac {e^{3} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {e^{3} \log \left (e x + d\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {{\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} + \frac {c d x + a e}{2 \, {\left (a^{2} c d^{2} + a^{3} e^{2} + {\left (a c^{2} d^{2} + a^{2} c e^{2}\right )} x^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.42 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=\frac {e^{4} \log \left ({\left | e x + d \right |}\right )}{c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}} - \frac {e^{3} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {{\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} + \frac {a c d^{2} e + a^{2} e^{3} + {\left (c^{2} d^{3} + a c d e^{2}\right )} x}{2 \, {\left (c d^{2} + a e^{2}\right )}^{2} {\left (c x^{2} + a\right )} a} \]

[In]

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

[Out]

e^4*log(abs(e*x + d))/(c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5) - 1/2*e^3*log(c*x^2 + a)/(c^2*d^4 + 2*a*c*d^2*e^2
+ a^2*e^4) + 1/2*(c^2*d^3 + 3*a*c*d*e^2)*arctan(c*x/sqrt(a*c))/((a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(a
*c)) + 1/2*(a*c*d^2*e + a^2*e^3 + (c^2*d^3 + a*c*d*e^2)*x)/((c*d^2 + a*e^2)^2*(c*x^2 + a)*a)

Mupad [B] (verification not implemented)

Time = 10.11 (sec) , antiderivative size = 609, normalized size of antiderivative = 4.29 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^2} \, dx=\frac {\frac {e}{2\,\left (c\,d^2+a\,e^2\right )}+\frac {c\,d\,x}{2\,a\,\left (c\,d^2+a\,e^2\right )}}{c\,x^2+a}+\frac {e^3\,\ln \left (d+e\,x\right )}{{\left (c\,d^2+a\,e^2\right )}^2}+\frac {\ln \left (36\,a^7\,e^{10}\,\sqrt {-a^3\,c}+a^3\,c^6\,d^{10}\,x+a^2\,c^5\,d^{10}\,\sqrt {-a^3\,c}-81\,a^3\,d^2\,e^8\,{\left (-a^3\,c\right )}^{3/2}-8\,c^3\,d^8\,e^2\,{\left (-a^3\,c\right )}^{3/2}+36\,a^8\,c\,e^{10}\,x+8\,a^4\,c^5\,d^8\,e^2\,x+22\,a^5\,c^4\,d^6\,e^4\,x+60\,a^6\,c^3\,d^4\,e^6\,x+81\,a^7\,c^2\,d^2\,e^8\,x-22\,a\,c^2\,d^6\,e^4\,{\left (-a^3\,c\right )}^{3/2}-60\,a^2\,c\,d^4\,e^6\,{\left (-a^3\,c\right )}^{3/2}\right )\,\left (c\,d^3\,\sqrt {-a^3\,c}-2\,a^3\,e^3+3\,a\,d\,e^2\,\sqrt {-a^3\,c}\right )}{4\,\left (a^5\,e^4+2\,a^4\,c\,d^2\,e^2+a^3\,c^2\,d^4\right )}-\frac {\ln \left (a^3\,c^6\,d^{10}\,x-36\,a^7\,e^{10}\,\sqrt {-a^3\,c}-a^2\,c^5\,d^{10}\,\sqrt {-a^3\,c}+81\,a^3\,d^2\,e^8\,{\left (-a^3\,c\right )}^{3/2}+8\,c^3\,d^8\,e^2\,{\left (-a^3\,c\right )}^{3/2}+36\,a^8\,c\,e^{10}\,x+8\,a^4\,c^5\,d^8\,e^2\,x+22\,a^5\,c^4\,d^6\,e^4\,x+60\,a^6\,c^3\,d^4\,e^6\,x+81\,a^7\,c^2\,d^2\,e^8\,x+22\,a\,c^2\,d^6\,e^4\,{\left (-a^3\,c\right )}^{3/2}+60\,a^2\,c\,d^4\,e^6\,{\left (-a^3\,c\right )}^{3/2}\right )\,\left (2\,a^3\,e^3+c\,d^3\,\sqrt {-a^3\,c}+3\,a\,d\,e^2\,\sqrt {-a^3\,c}\right )}{4\,\left (a^5\,e^4+2\,a^4\,c\,d^2\,e^2+a^3\,c^2\,d^4\right )} \]

[In]

int(1/((a + c*x^2)^2*(d + e*x)),x)

[Out]

(e/(2*(a*e^2 + c*d^2)) + (c*d*x)/(2*a*(a*e^2 + c*d^2)))/(a + c*x^2) + (e^3*log(d + e*x))/(a*e^2 + c*d^2)^2 + (
log(36*a^7*e^10*(-a^3*c)^(1/2) + a^3*c^6*d^10*x + a^2*c^5*d^10*(-a^3*c)^(1/2) - 81*a^3*d^2*e^8*(-a^3*c)^(3/2)
- 8*c^3*d^8*e^2*(-a^3*c)^(3/2) + 36*a^8*c*e^10*x + 8*a^4*c^5*d^8*e^2*x + 22*a^5*c^4*d^6*e^4*x + 60*a^6*c^3*d^4
*e^6*x + 81*a^7*c^2*d^2*e^8*x - 22*a*c^2*d^6*e^4*(-a^3*c)^(3/2) - 60*a^2*c*d^4*e^6*(-a^3*c)^(3/2))*(c*d^3*(-a^
3*c)^(1/2) - 2*a^3*e^3 + 3*a*d*e^2*(-a^3*c)^(1/2)))/(4*(a^5*e^4 + a^3*c^2*d^4 + 2*a^4*c*d^2*e^2)) - (log(a^3*c
^6*d^10*x - 36*a^7*e^10*(-a^3*c)^(1/2) - a^2*c^5*d^10*(-a^3*c)^(1/2) + 81*a^3*d^2*e^8*(-a^3*c)^(3/2) + 8*c^3*d
^8*e^2*(-a^3*c)^(3/2) + 36*a^8*c*e^10*x + 8*a^4*c^5*d^8*e^2*x + 22*a^5*c^4*d^6*e^4*x + 60*a^6*c^3*d^4*e^6*x +
81*a^7*c^2*d^2*e^8*x + 22*a*c^2*d^6*e^4*(-a^3*c)^(3/2) + 60*a^2*c*d^4*e^6*(-a^3*c)^(3/2))*(2*a^3*e^3 + c*d^3*(
-a^3*c)^(1/2) + 3*a*d*e^2*(-a^3*c)^(1/2)))/(4*(a^5*e^4 + a^3*c^2*d^4 + 2*a^4*c*d^2*e^2))