\(\int \frac {a+c x^4}{(d+e x^2)^3} \, dx\) [126]

Optimal result

Integrand size = 17, antiderivative size = 93 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\left (a+\frac {c d^2}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}+\frac {\left (\frac {3 a}{d^2}-\frac {5 c}{e^2}\right ) x}{8 \left (d+e x^2\right )}+\frac {3 \left (c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 93, 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, 393, 211} \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {3 \left (a e^2+c d^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}}+\frac {x \left (\frac {3 a}{d^2}-\frac {5 c}{e^2}\right )}{8 \left (d+e x^2\right )}+\frac {x \left (a+\frac {c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2} \]

[In]

Int[(a + c*x^4)/(d + e*x^2)^3,x]

[Out]

((a + (c*d^2)/e^2)*x)/(4*d*(d + e*x^2)^2) + (((3*a)/d^2 - (5*c)/e^2)*x)/(8*(d + e*x^2)) + (3*(c*d^2 + a*e^2)*A
rcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(5/2)*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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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}{4 d \left (d+e x^2\right )^2}-\frac {\int \frac {-3 a+\frac {c d^2}{e^2}-\frac {4 c d x^2}{e}}{\left (d+e x^2\right )^2} \, dx}{4 d} \\ & = \frac {\left (a+\frac {c d^2}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}+\frac {\left (\frac {3 a}{d^2}-\frac {5 c}{e^2}\right ) x}{8 \left (d+e x^2\right )}+\frac {1}{8} \left (3 \left (\frac {a}{d^2}+\frac {c}{e^2}\right )\right ) \int \frac {1}{d+e x^2} \, dx \\ & = \frac {\left (a+\frac {c d^2}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}+\frac {\left (\frac {3 a}{d^2}-\frac {5 c}{e^2}\right ) x}{8 \left (d+e x^2\right )}+\frac {3 \left (c d^2+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {a e^2 x \left (5 d+3 e x^2\right )-c d^2 x \left (3 d+5 e x^2\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac {3 \left (c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \]

[In]

Integrate[(a + c*x^4)/(d + e*x^2)^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\frac {\left (3 a \,e^{2}-5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}+\frac {3 \left (a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 e^{2} d^{2} \sqrt {e d}}\)	\(92\)
risch	\(\frac {\frac {\left (3 a \,e^{2}-5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) a}{16 \sqrt {-e d}\, d^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) c}{16 \sqrt {-e d}\, e^{2}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) a}{16 \sqrt {-e d}\, d^{2}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) c}{16 \sqrt {-e d}\, e^{2}}\)	\(153\)

[In]

int((c*x^4+a)/(e*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.29 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, c d^{3} e^{2} - 3 \, a d e^{4}\right )} x^{3} + 3 \, {\left (c d^{4} + a d^{2} e^{2} + {\left (c d^{2} e^{2} + a e^{4}\right )} x^{4} + 2 \, {\left (c d^{3} e + a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (3 \, c d^{4} e - 5 \, a d^{2} e^{3}\right )} x}{16 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}, -\frac {{\left (5 \, c d^{3} e^{2} - 3 \, a d e^{4}\right )} x^{3} - 3 \, {\left (c d^{4} + a d^{2} e^{2} + {\left (c d^{2} e^{2} + a e^{4}\right )} x^{4} + 2 \, {\left (c d^{3} e + a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (3 \, c d^{4} e - 5 \, a d^{2} e^{3}\right )} x}{8 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}\right ] \]

[In]

integrate((c*x^4+a)/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (90) = 180\).

Time = 0.35 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.35 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=- \frac {3 \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right ) \log {\left (- \frac {3 d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right )}{3 a e^{2} + 3 c d^{2}} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right ) \log {\left (\frac {3 d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right )}{3 a e^{2} + 3 c d^{2}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a e^{3} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{4}} \]

[In]

integrate((c*x**4+a)/(e*x**2+d)**3,x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^4+a)/(e*x^2+d)^3,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 = 0.33 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {3 \, {\left (c d^{2} + a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{2} e^{2}} - \frac {5 \, c d^{2} e x^{3} - 3 \, a e^{3} x^{3} + 3 \, c d^{3} x - 5 \, a d e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} d^{2} e^{2}} \]

[In]

integrate((c*x^4+a)/(e*x^2+d)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.73 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\frac {x^3\,\left (3\,a\,e^2-5\,c\,d^2\right )}{8\,d^2\,e}+\frac {x\,\left (5\,a\,e^2-3\,c\,d^2\right )}{8\,d\,e^2}}{d^2+2\,d\,e\,x^2+e^2\,x^4}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+a\,e^2\right )}{8\,d^{5/2}\,e^{5/2}} \]

[In]

int((a + c*x^4)/(d + e*x^2)^3,x)

[Out]

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