\(\int \frac {a+c x^4}{(d+e x^2)^4} \, dx\) [301]

Optimal result

Integrand size = 17, antiderivative size = 123 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {\left (a+\frac {c d^2}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}+\frac {\left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right ) x}{24 \left (d+e x^2\right )^2}+\frac {\left (\frac {5 a}{d^2}+\frac {c}{e^2}\right ) x}{16 d \left (d+e x^2\right )}+\frac {\left (c d^2+5 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.92 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {x \left (c d^2 \left (-3 d^2-8 d e x^2+3 e^2 x^4\right )+a e^2 \left (33 d^2+40 d e x^2+15 e^2 x^4\right )\right )}{48 d^3 e^2 \left (d+e x^2\right )^3}+\frac {\left (c d^2+5 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \] Input:

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

Output:

(x*(c*d^2*(-3*d^2 - 8*d*e*x^2 + 3*e^2*x^4) + a*e^2*(33*d^2 + 40*d*e*x^2 + 
15*e^2*x^4)))/(48*d^3*e^2*(d + e*x^2)^3) + ((c*d^2 + 5*a*e^2)*ArcTan[(Sqrt 
[e]*x)/Sqrt[d]])/(16*d^(7/2)*e^(5/2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1472, 25, 25, 27, 298, 215, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx\)

\(\Big \downarrow \) 1472

\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3}-\frac {\int -\frac {6 c d x^2+\left (5 a-\frac {c d^2}{e^2}\right ) e}{e \left (e x^2+d\right )^3}dx}{6 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int -\frac {\frac {c d^2}{e}-6 c x^2 d-5 a e}{e \left (e x^2+d\right )^3}dx}{6 d}+\frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3}-\frac {\int \frac {\frac {c d^2}{e}-6 c x^2 d-5 a e}{e \left (e x^2+d\right )^3}dx}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3}-\frac {\int \frac {\frac {c d^2}{e}-6 c x^2 d-5 a e}{\left (e x^2+d\right )^3}dx}{6 d e}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3}-\frac {\frac {x \left (\frac {7 c d}{e}-\frac {5 a e}{d}\right )}{4 \left (d+e x^2\right )^2}-\frac {3}{4} \left (\frac {5 a e}{d}+\frac {c d}{e}\right ) \int \frac {1}{\left (e x^2+d\right )^2}dx}{6 d e}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3}-\frac {\frac {x \left (\frac {7 c d}{e}-\frac {5 a e}{d}\right )}{4 \left (d+e x^2\right )^2}-\frac {3}{4} \left (\frac {5 a e}{d}+\frac {c d}{e}\right ) \left (\frac {\int \frac {1}{e x^2+d}dx}{2 d}+\frac {x}{2 d \left (d+e x^2\right )}\right )}{6 d e}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3}-\frac {\frac {x \left (\frac {7 c d}{e}-\frac {5 a e}{d}\right )}{4 \left (d+e x^2\right )^2}-\frac {3}{4} \left (\frac {5 a e}{d}+\frac {c d}{e}\right ) \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} \sqrt {e}}+\frac {x}{2 d \left (d+e x^2\right )}\right )}{6 d e}\)

Input:

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

Output:

((a + (c*d^2)/e^2)*x)/(6*d*(d + e*x^2)^3) - ((((7*c*d)/e - (5*a*e)/d)*x)/( 
4*(d + e*x^2)^2) - (3*((c*d)/e + (5*a*e)/d)*(x/(2*d*(d + e*x^2)) + ArcTan[ 
(Sqrt[e]*x)/Sqrt[d]]/(2*d^(3/2)*Sqrt[e])))/4)/(6*d*e)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 218

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

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 1472

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wi 
th[{Qx = PolynomialQuotient[(a + c*x^4)^p, d + e*x^2, x], R = Coeff[Polynom 
ialRemainder[(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] + Simp[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]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}-c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}+\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 d^{3} e^{2} \sqrt {d e}}\)	\(113\)
risch	\(\frac {\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}-c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}-\frac {5 \ln \left (e x +\sqrt {-d e}\right ) a}{32 \sqrt {-d e}\, d^{3}}-\frac {\ln \left (e x +\sqrt {-d e}\right ) c}{32 \sqrt {-d e}\, e^{2} d}+\frac {5 \ln \left (-e x +\sqrt {-d e}\right ) a}{32 \sqrt {-d e}\, d^{3}}+\frac {\ln \left (-e x +\sqrt {-d e}\right ) c}{32 \sqrt {-d e}\, e^{2} d}\)	\(179\)

Input:

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

Output:

(1/16*(5*a*e^2+c*d^2)/d^3*x^5+1/6*(5*a*e^2-c*d^2)/d^2/e*x^3+1/16*(11*a*e^2 
-c*d^2)/d/e^2*x)/(e*x^2+d)^3+1/16*(5*a*e^2+c*d^2)/d^3/e^2/(d*e)^(1/2)*arct 
an(e*x/(d*e)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 424, normalized size of antiderivative = 3.45 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\left [\frac {6 \, {\left (c d^{3} e^{3} + 5 \, a d e^{5}\right )} x^{5} - 16 \, {\left (c d^{4} e^{2} - 5 \, a d^{2} e^{4}\right )} x^{3} - 3 \, {\left ({\left (c d^{2} e^{3} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + 5 \, a d^{2} e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - 6 \, {\left (c d^{5} e - 11 \, a d^{3} e^{3}\right )} x}{96 \, {\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}, \frac {3 \, {\left (c d^{3} e^{3} + 5 \, a d e^{5}\right )} x^{5} - 8 \, {\left (c d^{4} e^{2} - 5 \, a d^{2} e^{4}\right )} x^{3} + 3 \, {\left ({\left (c d^{2} e^{3} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + 5 \, a d^{2} e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (c d^{5} e - 11 \, a d^{3} e^{3}\right )} x}{48 \, {\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}\right ] \] Input:

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

Output:

[1/96*(6*(c*d^3*e^3 + 5*a*d*e^5)*x^5 - 16*(c*d^4*e^2 - 5*a*d^2*e^4)*x^3 - 
3*((c*d^2*e^3 + 5*a*e^5)*x^6 + c*d^5 + 5*a*d^3*e^2 + 3*(c*d^3*e^2 + 5*a*d* 
e^4)*x^4 + 3*(c*d^4*e + 5*a*d^2*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(- 
d*e)*x - d)/(e*x^2 + d)) - 6*(c*d^5*e - 11*a*d^3*e^3)*x)/(d^4*e^6*x^6 + 3* 
d^5*e^5*x^4 + 3*d^6*e^4*x^2 + d^7*e^3), 1/48*(3*(c*d^3*e^3 + 5*a*d*e^5)*x^ 
5 - 8*(c*d^4*e^2 - 5*a*d^2*e^4)*x^3 + 3*((c*d^2*e^3 + 5*a*e^5)*x^6 + c*d^5 
 + 5*a*d^3*e^2 + 3*(c*d^3*e^2 + 5*a*d*e^4)*x^4 + 3*(c*d^4*e + 5*a*d^2*e^3) 
*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3*(c*d^5*e - 11*a*d^3*e^3)*x)/(d^4 
*e^6*x^6 + 3*d^5*e^5*x^4 + 3*d^6*e^4*x^2 + d^7*e^3)]
 

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.66 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + c d^{2}\right ) \log {\left (- d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + c d^{2}\right ) \log {\left (d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a e^{4} + 3 c d^{2} e^{2}\right ) + x^{3} \cdot \left (40 a d e^{3} - 8 c d^{3} e\right ) + x \left (33 a d^{2} e^{2} - 3 c d^{4}\right )}{48 d^{6} e^{2} + 144 d^{5} e^{3} x^{2} + 144 d^{4} e^{4} x^{4} + 48 d^{3} e^{5} x^{6}} \] Input:

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {{\left (c d^{2} + 5 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} e^{2}} + \frac {3 \, c d^{2} e^{2} x^{5} + 15 \, a e^{4} x^{5} - 8 \, c d^{3} e x^{3} + 40 \, a d e^{3} x^{3} - 3 \, c d^{4} x + 33 \, a d^{2} e^{2} x}{48 \, {\left (e x^{2} + d\right )}^{3} d^{3} e^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 17.02 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.05 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {\frac {x^5\,\left (c\,d^2+5\,a\,e^2\right )}{16\,d^3}+\frac {x^3\,\left (5\,a\,e^2-c\,d^2\right )}{6\,d^2\,e}+\frac {x\,\left (11\,a\,e^2-c\,d^2\right )}{16\,d\,e^2}}{d^3+3\,d^2\,e\,x^2+3\,d\,e^2\,x^4+e^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+5\,a\,e^2\right )}{16\,d^{7/2}\,e^{5/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.54 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {15 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,d^{3} e^{2}+45 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,d^{2} e^{3} x^{2}+45 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d \,e^{4} x^{4}+15 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,e^{5} x^{6}+3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c \,d^{5}+9 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c \,d^{4} e \,x^{2}+9 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c \,d^{3} e^{2} x^{4}+3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c \,d^{2} e^{3} x^{6}+33 a \,d^{3} e^{3} x +40 a \,d^{2} e^{4} x^{3}+15 a d \,e^{5} x^{5}-3 c \,d^{5} e x -8 c \,d^{4} e^{2} x^{3}+3 c \,d^{3} e^{3} x^{5}}{48 d^{4} e^{3} \left (e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}\right )} \] Input:

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

Output:

(15*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d**3*e**2 + 45*sqrt(e) 
*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d**2*e**3*x**2 + 45*sqrt(e)*sqrt( 
d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d*e**4*x**4 + 15*sqrt(e)*sqrt(d)*atan(( 
e*x)/(sqrt(e)*sqrt(d)))*a*e**5*x**6 + 3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e 
)*sqrt(d)))*c*d**5 + 9*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c*d** 
4*e*x**2 + 9*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c*d**3*e**2*x** 
4 + 3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c*d**2*e**3*x**6 + 33* 
a*d**3*e**3*x + 40*a*d**2*e**4*x**3 + 15*a*d*e**5*x**5 - 3*c*d**5*e*x - 8* 
c*d**4*e**2*x**3 + 3*c*d**3*e**3*x**5)/(48*d**4*e**3*(d**3 + 3*d**2*e*x**2 
 + 3*d*e**2*x**4 + e**3*x**6))