\(\int \frac {d+e x}{(a+c x^2)^3} \, dx\) [516]

Optimal result

Integrand size = 15, antiderivative size = 75 \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=\frac {-a e+c d x}{4 a c \left (a+c x^2\right )^2}+\frac {3 d x}{8 a^2 \left (a+c x^2\right )}+\frac {3 d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {c}} \]

[Out]

1/4*(c*d*x-a*e)/a/c/(c*x^2+a)^2+3/8*d*x/a^2/(c*x^2+a)+3/8*d*arctan(x*c^(1/2)/a^(1/2))/a^(5/2)/c^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {653, 205, 211} \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=\frac {3 d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {c}}+\frac {3 d x}{8 a^2 \left (a+c x^2\right )}-\frac {a e-c d x}{4 a c \left (a+c x^2\right )^2} \]

[In]

Int[(d + e*x)/(a + c*x^2)^3,x]

[Out]

-1/4*(a*e - c*d*x)/(a*c*(a + c*x^2)^2) + (3*d*x)/(8*a^2*(a + c*x^2)) + (3*d*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^
(5/2)*Sqrt[c])

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 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]

Rubi steps \begin{align*} \text {integral}& = -\frac {a e-c d x}{4 a c \left (a+c x^2\right )^2}+\frac {(3 d) \int \frac {1}{\left (a+c x^2\right )^2} \, dx}{4 a} \\ & = -\frac {a e-c d x}{4 a c \left (a+c x^2\right )^2}+\frac {3 d x}{8 a^2 \left (a+c x^2\right )}+\frac {(3 d) \int \frac {1}{a+c x^2} \, dx}{8 a^2} \\ & = -\frac {a e-c d x}{4 a c \left (a+c x^2\right )^2}+\frac {3 d x}{8 a^2 \left (a+c x^2\right )}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=\frac {\frac {\sqrt {a} \left (-2 a^2 e+5 a c d x+3 c^2 d x^3\right )}{\left (a+c x^2\right )^2}+3 \sqrt {c} d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c} \]

[In]

Integrate[(d + e*x)/(a + c*x^2)^3,x]

[Out]

((Sqrt[a]*(-2*a^2*e + 5*a*c*d*x + 3*c^2*d*x^3))/(a + c*x^2)^2 + 3*Sqrt[c]*d*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^
(5/2)*c)

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {2 c d x -2 a e}{8 a c \left (c \,x^{2}+a \right )^{2}}+\frac {3 d \left (\frac {x}{2 a \left (c \,x^{2}+a \right )}+\frac {\arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 a \sqrt {a c}}\right )}{4 a}\)	\(70\)
risch	\(\frac {\frac {3 d c \,x^{3}}{8 a^{2}}+\frac {5 x d}{8 a}-\frac {e}{4 c}}{\left (c \,x^{2}+a \right )^{2}}-\frac {3 d \ln \left (c x +\sqrt {-a c}\right )}{16 \sqrt {-a c}\, a^{2}}+\frac {3 d \ln \left (-c x +\sqrt {-a c}\right )}{16 \sqrt {-a c}\, a^{2}}\)	\(83\)

[In]

int((e*x+d)/(c*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.83 \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=\left [\frac {6 \, a c^{2} d x^{3} + 10 \, a^{2} c d x - 4 \, a^{3} e - 3 \, {\left (c^{2} d x^{4} + 2 \, a c d x^{2} + a^{2} d\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right )}{16 \, {\left (a^{3} c^{3} x^{4} + 2 \, a^{4} c^{2} x^{2} + a^{5} c\right )}}, \frac {3 \, a c^{2} d x^{3} + 5 \, a^{2} c d x - 2 \, a^{3} e + 3 \, {\left (c^{2} d x^{4} + 2 \, a c d x^{2} + a^{2} d\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right )}{8 \, {\left (a^{3} c^{3} x^{4} + 2 \, a^{4} c^{2} x^{2} + a^{5} c\right )}}\right ] \]

[In]

integrate((e*x+d)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(6*a*c^2*d*x^3 + 10*a^2*c*d*x - 4*a^3*e - 3*(c^2*d*x^4 + 2*a*c*d*x^2 + a^2*d)*sqrt(-a*c)*log((c*x^2 - 2*
sqrt(-a*c)*x - a)/(c*x^2 + a)))/(a^3*c^3*x^4 + 2*a^4*c^2*x^2 + a^5*c), 1/8*(3*a*c^2*d*x^3 + 5*a^2*c*d*x - 2*a^
3*e + 3*(c^2*d*x^4 + 2*a*c*d*x^2 + a^2*d)*sqrt(a*c)*arctan(sqrt(a*c)*x/a))/(a^3*c^3*x^4 + 2*a^4*c^2*x^2 + a^5*
c)]

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.65 \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=d \left (- \frac {3 \sqrt {- \frac {1}{a^{5} c}} \log {\left (- a^{3} \sqrt {- \frac {1}{a^{5} c}} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{5} c}} \log {\left (a^{3} \sqrt {- \frac {1}{a^{5} c}} + x \right )}}{16}\right ) + \frac {- 2 a^{2} e + 5 a c d x + 3 c^{2} d x^{3}}{8 a^{4} c + 16 a^{3} c^{2} x^{2} + 8 a^{2} c^{3} x^{4}} \]

[In]

integrate((e*x+d)/(c*x**2+a)**3,x)

[Out]

d*(-3*sqrt(-1/(a**5*c))*log(-a**3*sqrt(-1/(a**5*c)) + x)/16 + 3*sqrt(-1/(a**5*c))*log(a**3*sqrt(-1/(a**5*c)) +
x)/16) + (-2*a**2*e + 5*a*c*d*x + 3*c**2*d*x**3)/(8*a**4*c + 16*a**3*c**2*x**2 + 8*a**2*c**3*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=\frac {3 \, c^{2} d x^{3} + 5 \, a c d x - 2 \, a^{2} e}{8 \, {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} + \frac {3 \, d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2}} \]

[In]

integrate((e*x+d)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

1/8*(3*c^2*d*x^3 + 5*a*c*d*x - 2*a^2*e)/(a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c) + 3/8*d*arctan(c*x/sqrt(a*c))/(s
qrt(a*c)*a^2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=\frac {3 \, d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2}} + \frac {3 \, c^{2} d x^{3} + 5 \, a c d x - 2 \, a^{2} e}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c} \]

[In]

integrate((e*x+d)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

3/8*d*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2) + 1/8*(3*c^2*d*x^3 + 5*a*c*d*x - 2*a^2*e)/((c*x^2 + a)^2*a^2*c)

Mupad [B] (verification not implemented)

Time = 9.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85 \[ \int \frac {d+e x}{\left (a+c x^2\right )^3} \, dx=\frac {\frac {5\,d\,x}{8\,a}-\frac {e}{4\,c}+\frac {3\,c\,d\,x^3}{8\,a^2}}{a^2+2\,a\,c\,x^2+c^2\,x^4}+\frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {c}} \]

[In]

int((d + e*x)/(a + c*x^2)^3,x)

[Out]

((5*d*x)/(8*a) - e/(4*c) + (3*c*d*x^3)/(8*a^2))/(a^2 + c^2*x^4 + 2*a*c*x^2) + (3*d*atan((c^(1/2)*x)/a^(1/2)))/
(8*a^(5/2)*c^(1/2))