Integrand size = 19, antiderivative size = 83 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {24 e^{\arctan (a x)}}{85 a c^3}+\frac {e^{\arctan (a x)} (1+4 a x)}{17 a c^3 \left (1+a^2 x^2\right )^2}+\frac {12 e^{\arctan (a x)} (1+2 a x)}{85 a c^3 \left (1+a^2 x^2\right )} \] Output:
24/85*exp(arctan(a*x))/a/c^3+1/17*exp(arctan(a*x))*(4*a*x+1)/a/c^3/(a^2*x^ 2+1)^2+12/85*exp(arctan(a*x))*(2*a*x+1)/a/c^3/(a^2*x^2+1)
Result contains complex when optimal does not.
Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {5 e^{\arctan (a x)} (1+4 a x)+12 (1-i a x)^{\frac {i}{2}} (1+i a x)^{-\frac {i}{2}} \left (1+a^2 x^2\right ) \left (3+2 a x+2 a^2 x^2\right )}{85 a c^3 \left (1+a^2 x^2\right )^2} \] Input:
Integrate[E^ArcTan[a*x]/(c + a^2*c*x^2)^3,x]
Output:
(5*E^ArcTan[a*x]*(1 + 4*a*x) + (12*(1 - I*a*x)^(I/2)*(1 + a^2*x^2)*(3 + 2* a*x + 2*a^2*x^2))/(1 + I*a*x)^(I/2))/(85*a*c^3*(1 + a^2*x^2)^2)
Time = 0.65 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5593, 27, 5593, 5594}
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.
\(\displaystyle \int \frac {e^{\arctan (a x)}}{\left (a^2 c x^2+c\right )^3} \, dx\) |
\(\Big \downarrow \) 5593 |
\(\displaystyle \frac {12 \int \frac {e^{\arctan (a x)}}{c^2 \left (a^2 x^2+1\right )^2}dx}{17 c}+\frac {(4 a x+1) e^{\arctan (a x)}}{17 a c^3 \left (a^2 x^2+1\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {12 \int \frac {e^{\arctan (a x)}}{\left (a^2 x^2+1\right )^2}dx}{17 c^3}+\frac {(4 a x+1) e^{\arctan (a x)}}{17 a c^3 \left (a^2 x^2+1\right )^2}\) |
\(\Big \downarrow \) 5593 |
\(\displaystyle \frac {12 \left (\frac {2}{5} \int \frac {e^{\arctan (a x)}}{a^2 x^2+1}dx+\frac {(2 a x+1) e^{\arctan (a x)}}{5 a \left (a^2 x^2+1\right )}\right )}{17 c^3}+\frac {(4 a x+1) e^{\arctan (a x)}}{17 a c^3 \left (a^2 x^2+1\right )^2}\) |
\(\Big \downarrow \) 5594 |
\(\displaystyle \frac {(4 a x+1) e^{\arctan (a x)}}{17 a c^3 \left (a^2 x^2+1\right )^2}+\frac {12 \left (\frac {(2 a x+1) e^{\arctan (a x)}}{5 a \left (a^2 x^2+1\right )}+\frac {2 e^{\arctan (a x)}}{5 a}\right )}{17 c^3}\) |
Input:
Int[E^ArcTan[a*x]/(c + a^2*c*x^2)^3,x]
Output:
(E^ArcTan[a*x]*(1 + 4*a*x))/(17*a*c^3*(1 + a^2*x^2)^2) + (12*((2*E^ArcTan[ a*x])/(5*a) + (E^ArcTan[a*x]*(1 + 2*a*x))/(5*a*(1 + a^2*x^2))))/(17*c^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> S imp[(n - 2*a*(p + 1)*x)*(c + d*x^2)^(p + 1)*(E^(n*ArcTan[a*x])/(a*c*(n^2 + 4*(p + 1)^2))), x] + Simp[2*(p + 1)*((2*p + 3)/(c*(n^2 + 4*(p + 1)^2))) I nt[(c + d*x^2)^(p + 1)*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1] && !IntegerQ[I*n] && NeQ[n^2 + 4*(p + 1)^2, 0] && IntegerQ[2*p]
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E ^(n*ArcTan[a*x])/(a*c*n), x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c]
Time = 11.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.66
method | result | size |
gosper | \(\frac {{\mathrm e}^{\arctan \left (a x \right )} \left (24 a^{4} x^{4}+24 a^{3} x^{3}+60 a^{2} x^{2}+44 a x +41\right )}{85 \left (a^{2} x^{2}+1\right )^{2} a \,c^{3}}\) | \(55\) |
orering | \(\frac {\left (24 a^{4} x^{4}+24 a^{3} x^{3}+60 a^{2} x^{2}+44 a x +41\right ) \left (a^{2} x^{2}+1\right ) {\mathrm e}^{\arctan \left (a x \right )}}{85 a \left (a^{2} c \,x^{2}+c \right )^{3}}\) | \(62\) |
parallelrisch | \(\frac {24 a^{4} {\mathrm e}^{\arctan \left (a x \right )} x^{4}+24 a^{3} {\mathrm e}^{\arctan \left (a x \right )} x^{3}+60 a^{2} {\mathrm e}^{\arctan \left (a x \right )} x^{2}+44 \,{\mathrm e}^{\arctan \left (a x \right )} a x +41 \,{\mathrm e}^{\arctan \left (a x \right )}}{85 c^{3} \left (a^{2} x^{2}+1\right )^{2} a}\) | \(76\) |
Input:
int(exp(arctan(a*x))/(a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/85*exp(arctan(a*x))*(24*a^4*x^4+24*a^3*x^3+60*a^2*x^2+44*a*x+41)/(a^2*x^ 2+1)^2/a/c^3
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {{\left (24 \, a^{4} x^{4} + 24 \, a^{3} x^{3} + 60 \, a^{2} x^{2} + 44 \, a x + 41\right )} e^{\left (\arctan \left (a x\right )\right )}}{85 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \] Input:
integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
1/85*(24*a^4*x^4 + 24*a^3*x^3 + 60*a^2*x^2 + 44*a*x + 41)*e^(arctan(a*x))/ (a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (75) = 150\).
Time = 3.10 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.69 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\begin {cases} \frac {24 a^{4} x^{4} e^{\operatorname {atan}{\left (a x \right )}}}{85 a^{5} c^{3} x^{4} + 170 a^{3} c^{3} x^{2} + 85 a c^{3}} + \frac {24 a^{3} x^{3} e^{\operatorname {atan}{\left (a x \right )}}}{85 a^{5} c^{3} x^{4} + 170 a^{3} c^{3} x^{2} + 85 a c^{3}} + \frac {60 a^{2} x^{2} e^{\operatorname {atan}{\left (a x \right )}}}{85 a^{5} c^{3} x^{4} + 170 a^{3} c^{3} x^{2} + 85 a c^{3}} + \frac {44 a x e^{\operatorname {atan}{\left (a x \right )}}}{85 a^{5} c^{3} x^{4} + 170 a^{3} c^{3} x^{2} + 85 a c^{3}} + \frac {41 e^{\operatorname {atan}{\left (a x \right )}}}{85 a^{5} c^{3} x^{4} + 170 a^{3} c^{3} x^{2} + 85 a c^{3}} & \text {for}\: a \neq 0 \\\frac {x}{c^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(exp(atan(a*x))/(a**2*c*x**2+c)**3,x)
Output:
Piecewise((24*a**4*x**4*exp(atan(a*x))/(85*a**5*c**3*x**4 + 170*a**3*c**3* x**2 + 85*a*c**3) + 24*a**3*x**3*exp(atan(a*x))/(85*a**5*c**3*x**4 + 170*a **3*c**3*x**2 + 85*a*c**3) + 60*a**2*x**2*exp(atan(a*x))/(85*a**5*c**3*x** 4 + 170*a**3*c**3*x**2 + 85*a*c**3) + 44*a*x*exp(atan(a*x))/(85*a**5*c**3* x**4 + 170*a**3*c**3*x**2 + 85*a*c**3) + 41*exp(atan(a*x))/(85*a**5*c**3*x **4 + 170*a**3*c**3*x**2 + 85*a*c**3), Ne(a, 0)), (x/c**3, True))
\[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\int { \frac {e^{\left (\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
integrate(e^(arctan(a*x))/(a^2*c*x^2 + c)^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (74) = 148\).
Time = 0.14 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.45 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {41 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{8} - 88 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{7} + 76 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{6} + 72 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{5} + 150 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{4} - 72 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{3} + 76 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{2} + 88 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right ) + 41 \, e^{\left (\arctan \left (a x\right )\right )}}{85 \, {\left (a c^{3} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{8} + 4 \, a c^{3} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{6} + 6 \, a c^{3} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{4} + 4 \, a c^{3} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{2} + a c^{3}\right )}} \] Input:
integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
1/85*(41*e^(arctan(a*x))*tan(1/2*arctan(a*x))^8 - 88*e^(arctan(a*x))*tan(1 /2*arctan(a*x))^7 + 76*e^(arctan(a*x))*tan(1/2*arctan(a*x))^6 + 72*e^(arct an(a*x))*tan(1/2*arctan(a*x))^5 + 150*e^(arctan(a*x))*tan(1/2*arctan(a*x)) ^4 - 72*e^(arctan(a*x))*tan(1/2*arctan(a*x))^3 + 76*e^(arctan(a*x))*tan(1/ 2*arctan(a*x))^2 + 88*e^(arctan(a*x))*tan(1/2*arctan(a*x)) + 41*e^(arctan( a*x)))/(a*c^3*tan(1/2*arctan(a*x))^8 + 4*a*c^3*tan(1/2*arctan(a*x))^6 + 6* a*c^3*tan(1/2*arctan(a*x))^4 + 4*a*c^3*tan(1/2*arctan(a*x))^2 + a*c^3)
Time = 23.45 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {24\,{\mathrm {e}}^{\mathrm {atan}\left (a\,x\right )}}{85\,a\,c^3}+\frac {12\,{\mathrm {e}}^{\mathrm {atan}\left (a\,x\right )}\,\left (2\,a\,x+1\right )}{85\,a\,c^3\,\left (a^2\,x^2+1\right )}+\frac {{\mathrm {e}}^{\mathrm {atan}\left (a\,x\right )}\,\left (4\,a\,x+1\right )}{17\,a\,c^3\,{\left (a^2\,x^2+1\right )}^2} \] Input:
int(exp(atan(a*x))/(c + a^2*c*x^2)^3,x)
Output:
(24*exp(atan(a*x)))/(85*a*c^3) + (12*exp(atan(a*x))*(2*a*x + 1))/(85*a*c^3 *(a^2*x^2 + 1)) + (exp(atan(a*x))*(4*a*x + 1))/(17*a*c^3*(a^2*x^2 + 1)^2)
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {e^{\mathit {atan} \left (a x \right )} \left (24 a^{4} x^{4}+24 a^{3} x^{3}+60 a^{2} x^{2}+44 a x +41\right )}{85 a \,c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )} \] Input:
int(exp(atan(a*x))/(a^2*c*x^2+c)^3,x)
Output:
(e**atan(a*x)*(24*a**4*x**4 + 24*a**3*x**3 + 60*a**2*x**2 + 44*a*x + 41))/ (85*a*c**3*(a**4*x**4 + 2*a**2*x**2 + 1))