\(\int \frac {a+c x^4}{(d+e x^2)^2} \, dx\) [125]

Optimal result

Integrand size = 17, antiderivative size = 74 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=\frac {c x}{e^2}+\frac {\left (a+\frac {c d^2}{e^2}\right ) x}{2 d \left (d+e x^2\right )}-\frac {\left (3 c d^2-a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1172, 396, 211} \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=-\frac {\left (3 c d^2-a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{5/2}}+\frac {x \left (a+\frac {c d^2}{e^2}\right )}{2 d \left (d+e x^2\right )}+\frac {c x}{e^2} \]

[In]

Int[(a + c*x^4)/(d + e*x^2)^2,x]

[Out]

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

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 1172

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + c*
x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, Simp[(-R)*x*((d + e
*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx +
R*(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=\frac {c x}{e^2}+\frac {\left (c d^2+a e^2\right ) x}{2 d e^2 \left (d+e x^2\right )}-\frac {\left (3 c d^2-a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{5/2}} \]

[In]

Integrate[(a + c*x^4)/(d + e*x^2)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {c x}{e^{2}}+\frac {\frac {\left (a \,e^{2}+c \,d^{2}\right ) x}{2 d \left (e \,x^{2}+d \right )}+\frac {\left (a \,e^{2}-3 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{2 d \sqrt {e d}}}{e^{2}}\)	\(70\)
risch	\(\frac {c x}{e^{2}}+\frac {\left (a \,e^{2}+c \,d^{2}\right ) x}{2 d \,e^{2} \left (e \,x^{2}+d \right )}-\frac {\ln \left (e x +\sqrt {-e d}\right ) a}{4 \sqrt {-e d}\, d}+\frac {3 d \ln \left (e x +\sqrt {-e d}\right ) c}{4 e^{2} \sqrt {-e d}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) a}{4 \sqrt {-e d}\, d}-\frac {3 d \ln \left (-e x +\sqrt {-e d}\right ) c}{4 e^{2} \sqrt {-e d}}\)	\(133\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 222, normalized size of antiderivative = 3.00 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=\left [\frac {4 \, c d^{2} e^{2} x^{3} + {\left (3 \, c d^{3} - a d e^{2} + {\left (3 \, c d^{2} e - a e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (3 \, c d^{3} e + a d e^{3}\right )} x}{4 \, {\left (d^{2} e^{4} x^{2} + d^{3} e^{3}\right )}}, \frac {2 \, c d^{2} e^{2} x^{3} - {\left (3 \, c d^{3} - a d e^{2} + {\left (3 \, c d^{2} e - a e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (3 \, c d^{3} e + a d e^{3}\right )} x}{2 \, {\left (d^{2} e^{4} x^{2} + d^{3} e^{3}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (68) = 136\).

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.86 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=\frac {c x}{e^{2}} + \frac {x \left (a e^{2} + c d^{2}\right )}{2 d^{2} e^{2} + 2 d e^{3} x^{2}} - \frac {\sqrt {- \frac {1}{d^{3} e^{5}}} \left (a e^{2} - 3 c d^{2}\right ) \log {\left (- d^{2} e^{2} \sqrt {- \frac {1}{d^{3} e^{5}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{d^{3} e^{5}}} \left (a e^{2} - 3 c d^{2}\right ) \log {\left (d^{2} e^{2} \sqrt {- \frac {1}{d^{3} e^{5}}} + x \right )}}{4} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=\frac {c x}{e^{2}} - \frac {{\left (3 \, c d^{2} - a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} d e^{2}} + \frac {c d^{2} x + a e^{2} x}{2 \, {\left (e x^{2} + d\right )} d e^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.73 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^2} \, dx=\frac {c\,x}{e^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (a\,e^2-3\,c\,d^2\right )}{2\,d^{3/2}\,e^{5/2}}+\frac {x\,\left (c\,d^2+a\,e^2\right )}{2\,d\,\left (e^3\,x^2+d\,e^2\right )} \]

[In]

int((a + c*x^4)/(d + e*x^2)^2,x)

[Out]

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