\(\int \frac {d+e x}{(a+c x^2)^4} \, dx\) [522]

Optimal result

Integrand size = 15, antiderivative size = 93 \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=\frac {-a e+c d x}{6 a c \left (a+c x^2\right )^3}+\frac {5 d x}{24 a^2 \left (a+c x^2\right )^2}+\frac {5 d x}{16 a^3 \left (a+c x^2\right )}+\frac {5 d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {c}} \]

[Out]

1/6*(c*d*x-a*e)/a/c/(c*x^2+a)^3+5/24*d*x/a^2/(c*x^2+a)^2+5/16*d*x/a^3/(c*x^2+a)+5/16*d*arctan(x*c^(1/2)/a^(1/2
))/a^(7/2)/c^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {653, 205, 211} \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=\frac {5 d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {c}}+\frac {5 d x}{16 a^3 \left (a+c x^2\right )}+\frac {5 d x}{24 a^2 \left (a+c x^2\right )^2}-\frac {a e-c d x}{6 a c \left (a+c x^2\right )^3} \]

[In]

Int[(d + e*x)/(a + c*x^2)^4,x]

[Out]

-1/6*(a*e - c*d*x)/(a*c*(a + c*x^2)^3) + (5*d*x)/(24*a^2*(a + c*x^2)^2) + (5*d*x)/(16*a^3*(a + c*x^2)) + (5*d*
ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*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}{6 a c \left (a+c x^2\right )^3}+\frac {(5 d) \int \frac {1}{\left (a+c x^2\right )^3} \, dx}{6 a} \\ & = -\frac {a e-c d x}{6 a c \left (a+c x^2\right )^3}+\frac {5 d x}{24 a^2 \left (a+c x^2\right )^2}+\frac {(5 d) \int \frac {1}{\left (a+c x^2\right )^2} \, dx}{8 a^2} \\ & = -\frac {a e-c d x}{6 a c \left (a+c x^2\right )^3}+\frac {5 d x}{24 a^2 \left (a+c x^2\right )^2}+\frac {5 d x}{16 a^3 \left (a+c x^2\right )}+\frac {(5 d) \int \frac {1}{a+c x^2} \, dx}{16 a^3} \\ & = -\frac {a e-c d x}{6 a c \left (a+c x^2\right )^3}+\frac {5 d x}{24 a^2 \left (a+c x^2\right )^2}+\frac {5 d x}{16 a^3 \left (a+c x^2\right )}+\frac {5 d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89 \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=\frac {\frac {\sqrt {a} \left (-8 a^3 e+33 a^2 c d x+40 a c^2 d x^3+15 c^3 d x^5\right )}{\left (a+c x^2\right )^3}+15 \sqrt {c} d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{48 a^{7/2} c} \]

[In]

Integrate[(d + e*x)/(a + c*x^2)^4,x]

[Out]

((Sqrt[a]*(-8*a^3*e + 33*a^2*c*d*x + 40*a*c^2*d*x^3 + 15*c^3*d*x^5))/(a + c*x^2)^3 + 15*Sqrt[c]*d*ArcTan[(Sqrt
[c]*x)/Sqrt[a]])/(48*a^(7/2)*c)

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {2 c d x -2 a e}{12 a c \left (c \,x^{2}+a \right )^{3}}+\frac {5 d \left (\frac {x}{4 a \left (c \,x^{2}+a \right )^{2}}+\frac {\frac {3 x}{8 a \left (c \,x^{2}+a \right )}+\frac {3 \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 a \sqrt {a c}}}{a}\right )}{6 a}\)	\(91\)
risch	\(\frac {\frac {5 d \,c^{2} x^{5}}{16 a^{3}}+\frac {5 d c \,x^{3}}{6 a^{2}}+\frac {11 x d}{16 a}-\frac {e}{6 c}}{\left (c \,x^{2}+a \right )^{3}}-\frac {5 d \ln \left (c x +\sqrt {-a c}\right )}{32 \sqrt {-a c}\, a^{3}}+\frac {5 d \ln \left (-c x +\sqrt {-a c}\right )}{32 \sqrt {-a c}\, a^{3}}\)	\(95\)

[In]

int((e*x+d)/(c*x^2+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/12*(2*c*d*x-2*a*e)/a/c/(c*x^2+a)^3+5/6*d/a*(1/4*x/a/(c*x^2+a)^2+3/4/a*(1/2*x/a/(c*x^2+a)+1/2/a/(a*c)^(1/2)*a
rctan(c*x/(a*c)^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 282, normalized size of antiderivative = 3.03 \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=\left [\frac {30 \, a c^{3} d x^{5} + 80 \, a^{2} c^{2} d x^{3} + 66 \, a^{3} c d x - 16 \, a^{4} e - 15 \, {\left (c^{3} d x^{6} + 3 \, a c^{2} d x^{4} + 3 \, a^{2} c d x^{2} + a^{3} d\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right )}{96 \, {\left (a^{4} c^{4} x^{6} + 3 \, a^{5} c^{3} x^{4} + 3 \, a^{6} c^{2} x^{2} + a^{7} c\right )}}, \frac {15 \, a c^{3} d x^{5} + 40 \, a^{2} c^{2} d x^{3} + 33 \, a^{3} c d x - 8 \, a^{4} e + 15 \, {\left (c^{3} d x^{6} + 3 \, a c^{2} d x^{4} + 3 \, a^{2} c d x^{2} + a^{3} d\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right )}{48 \, {\left (a^{4} c^{4} x^{6} + 3 \, a^{5} c^{3} x^{4} + 3 \, a^{6} c^{2} x^{2} + a^{7} c\right )}}\right ] \]

[In]

integrate((e*x+d)/(c*x^2+a)^4,x, algorithm="fricas")

[Out]

[1/96*(30*a*c^3*d*x^5 + 80*a^2*c^2*d*x^3 + 66*a^3*c*d*x - 16*a^4*e - 15*(c^3*d*x^6 + 3*a*c^2*d*x^4 + 3*a^2*c*d
*x^2 + a^3*d)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)))/(a^4*c^4*x^6 + 3*a^5*c^3*x^4 + 3*a^6*c
^2*x^2 + a^7*c), 1/48*(15*a*c^3*d*x^5 + 40*a^2*c^2*d*x^3 + 33*a^3*c*d*x - 8*a^4*e + 15*(c^3*d*x^6 + 3*a*c^2*d*
x^4 + 3*a^2*c*d*x^2 + a^3*d)*sqrt(a*c)*arctan(sqrt(a*c)*x/a))/(a^4*c^4*x^6 + 3*a^5*c^3*x^4 + 3*a^6*c^2*x^2 + a
^7*c)]

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.61 \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=d \left (- \frac {5 \sqrt {- \frac {1}{a^{7} c}} \log {\left (- a^{4} \sqrt {- \frac {1}{a^{7} c}} + x \right )}}{32} + \frac {5 \sqrt {- \frac {1}{a^{7} c}} \log {\left (a^{4} \sqrt {- \frac {1}{a^{7} c}} + x \right )}}{32}\right ) + \frac {- 8 a^{3} e + 33 a^{2} c d x + 40 a c^{2} d x^{3} + 15 c^{3} d x^{5}}{48 a^{6} c + 144 a^{5} c^{2} x^{2} + 144 a^{4} c^{3} x^{4} + 48 a^{3} c^{4} x^{6}} \]

[In]

integrate((e*x+d)/(c*x**2+a)**4,x)

[Out]

d*(-5*sqrt(-1/(a**7*c))*log(-a**4*sqrt(-1/(a**7*c)) + x)/32 + 5*sqrt(-1/(a**7*c))*log(a**4*sqrt(-1/(a**7*c)) +
x)/32) + (-8*a**3*e + 33*a**2*c*d*x + 40*a*c**2*d*x**3 + 15*c**3*d*x**5)/(48*a**6*c + 144*a**5*c**2*x**2 + 14
4*a**4*c**3*x**4 + 48*a**3*c**4*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04 \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=\frac {15 \, c^{3} d x^{5} + 40 \, a c^{2} d x^{3} + 33 \, a^{2} c d x - 8 \, a^{3} e}{48 \, {\left (a^{3} c^{4} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{5} c^{2} x^{2} + a^{6} c\right )}} + \frac {5 \, d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} \]

[In]

integrate((e*x+d)/(c*x^2+a)^4,x, algorithm="maxima")

[Out]

1/48*(15*c^3*d*x^5 + 40*a*c^2*d*x^3 + 33*a^2*c*d*x - 8*a^3*e)/(a^3*c^4*x^6 + 3*a^4*c^3*x^4 + 3*a^5*c^2*x^2 + a
^6*c) + 5/16*d*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=\frac {5 \, d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3}} + \frac {15 \, c^{3} d x^{5} + 40 \, a c^{2} d x^{3} + 33 \, a^{2} c d x - 8 \, a^{3} e}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c} \]

[In]

integrate((e*x+d)/(c*x^2+a)^4,x, algorithm="giac")

[Out]

5/16*d*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3) + 1/48*(15*c^3*d*x^5 + 40*a*c^2*d*x^3 + 33*a^2*c*d*x - 8*a^3*e)/(
(c*x^2 + a)^3*a^3*c)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94 \[ \int \frac {d+e x}{\left (a+c x^2\right )^4} \, dx=\frac {\frac {11\,d\,x}{16\,a}-\frac {e}{6\,c}+\frac {5\,c^2\,d\,x^5}{16\,a^3}+\frac {5\,c\,d\,x^3}{6\,a^2}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6}+\frac {5\,d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{16\,a^{7/2}\,\sqrt {c}} \]

[In]

int((d + e*x)/(a + c*x^2)^4,x)

[Out]

((11*d*x)/(16*a) - e/(6*c) + (5*c^2*d*x^5)/(16*a^3) + (5*c*d*x^3)/(6*a^2))/(a^3 + c^3*x^6 + 3*a^2*c*x^2 + 3*a*
c^2*x^4) + (5*d*atan((c^(1/2)*x)/a^(1/2)))/(16*a^(7/2)*c^(1/2))