Integrand size = 21, antiderivative size = 108 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\frac {e^{\arctan (a x)} (1+5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}+\frac {e^{\arctan (a x)} (1+3 a x)}{13 a c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac {3 e^{\arctan (a x)} (1+a x)}{13 a c^3 \sqrt {c+a^2 c x^2}} \] Output:
1/26*exp(arctan(a*x))*(5*a*x+1)/a/c/(a^2*c*x^2+c)^(5/2)+1/13*exp(arctan(a* x))*(3*a*x+1)/a/c^2/(a^2*c*x^2+c)^(3/2)+3/13*exp(arctan(a*x))*(a*x+1)/a/c^ 3/(a^2*c*x^2+c)^(1/2)
Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\frac {e^{\arctan (a x)} \left (9+17 a x+14 a^2 x^2+18 a^3 x^3+6 a^4 x^4+6 a^5 x^5\right )}{26 a c^3 \left (1+a^2 x^2\right )^2 \sqrt {c+a^2 c x^2}} \] Input:
Integrate[E^ArcTan[a*x]/(c + a^2*c*x^2)^(7/2),x]
Output:
(E^ArcTan[a*x]*(9 + 17*a*x + 14*a^2*x^2 + 18*a^3*x^3 + 6*a^4*x^4 + 6*a^5*x ^5))/(26*a*c^3*(1 + a^2*x^2)^2*Sqrt[c + a^2*c*x^2])
Time = 0.75 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5593, 5593, 5592}
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.
\(\displaystyle \int \frac {e^{\arctan (a x)}}{\left (a^2 c x^2+c\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 5593 |
\(\displaystyle \frac {10 \int \frac {e^{\arctan (a x)}}{\left (a^2 c x^2+c\right )^{5/2}}dx}{13 c}+\frac {(5 a x+1) e^{\arctan (a x)}}{26 a c \left (a^2 c x^2+c\right )^{5/2}}\) |
\(\Big \downarrow \) 5593 |
\(\displaystyle \frac {10 \left (\frac {3 \int \frac {e^{\arctan (a x)}}{\left (a^2 c x^2+c\right )^{3/2}}dx}{5 c}+\frac {(3 a x+1) e^{\arctan (a x)}}{10 a c \left (a^2 c x^2+c\right )^{3/2}}\right )}{13 c}+\frac {(5 a x+1) e^{\arctan (a x)}}{26 a c \left (a^2 c x^2+c\right )^{5/2}}\) |
\(\Big \downarrow \) 5592 |
\(\displaystyle \frac {10 \left (\frac {3 (a x+1) e^{\arctan (a x)}}{10 a c^2 \sqrt {a^2 c x^2+c}}+\frac {(3 a x+1) e^{\arctan (a x)}}{10 a c \left (a^2 c x^2+c\right )^{3/2}}\right )}{13 c}+\frac {(5 a x+1) e^{\arctan (a x)}}{26 a c \left (a^2 c x^2+c\right )^{5/2}}\) |
Input:
Int[E^ArcTan[a*x]/(c + a^2*c*x^2)^(7/2),x]
Output:
(E^ArcTan[a*x]*(1 + 5*a*x))/(26*a*c*(c + a^2*c*x^2)^(5/2)) + (10*((E^ArcTa n[a*x]*(1 + 3*a*x))/(10*a*c*(c + a^2*c*x^2)^(3/2)) + (3*E^ArcTan[a*x]*(1 + a*x))/(10*a*c^2*Sqrt[c + a^2*c*x^2])))/(13*c)
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(n + a*x)*(E^(n*ArcTan[a*x])/(a*c*(n^2 + 1)*Sqrt[c + d*x^2])), x] /; F reeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && !IntegerQ[I*n]
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]
Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(\frac {\left (a^{2} x^{2}+1\right ) \left (6 a^{5} x^{5}+6 a^{4} x^{4}+18 a^{3} x^{3}+14 a^{2} x^{2}+17 a x +9\right ) {\mathrm e}^{\arctan \left (a x \right )}}{26 a \left (a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}\) | \(70\) |
orering | \(\frac {\left (a^{2} x^{2}+1\right ) \left (6 a^{5} x^{5}+6 a^{4} x^{4}+18 a^{3} x^{3}+14 a^{2} x^{2}+17 a x +9\right ) {\mathrm e}^{\arctan \left (a x \right )}}{26 a \left (a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}\) | \(70\) |
Input:
int(exp(arctan(a*x))/(a^2*c*x^2+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/26*(a^2*x^2+1)*(6*a^5*x^5+6*a^4*x^4+18*a^3*x^3+14*a^2*x^2+17*a*x+9)*exp( arctan(a*x))/a/(a^2*c*x^2+c)^(7/2)
Time = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\frac {{\left (6 \, a^{5} x^{5} + 6 \, a^{4} x^{4} + 18 \, a^{3} x^{3} + 14 \, a^{2} x^{2} + 17 \, a x + 9\right )} \sqrt {a^{2} c x^{2} + c} e^{\left (\arctan \left (a x\right )\right )}}{26 \, {\left (a^{7} c^{4} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} + a c^{4}\right )}} \] Input:
integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^(7/2),x, algorithm="fricas")
Output:
1/26*(6*a^5*x^5 + 6*a^4*x^4 + 18*a^3*x^3 + 14*a^2*x^2 + 17*a*x + 9)*sqrt(a ^2*c*x^2 + c)*e^(arctan(a*x))/(a^7*c^4*x^6 + 3*a^5*c^4*x^4 + 3*a^3*c^4*x^2 + a*c^4)
Timed out. \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(exp(atan(a*x))/(a**2*c*x**2+c)**(7/2),x)
Output:
Timed out
\[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {e^{\left (\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^(7/2),x, algorithm="maxima")
Output:
integrate(e^(arctan(a*x))/(a^2*c*x^2 + c)^(7/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (93) = 186\).
Time = 0.19 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.31 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=-\frac {9 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{10} - 34 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{9} + 11 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{8} - 8 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{7} + 18 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{6} - 108 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{5} - 18 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{4} - 8 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{3} - 11 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{2} - 34 \, e^{\left (\arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right ) - 9 \, e^{\left (\arctan \left (a x\right )\right )}}{26 \, {\left (a c^{\frac {7}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{10} + 5 \, a c^{\frac {7}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{8} + 10 \, a c^{\frac {7}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{6} + 10 \, a c^{\frac {7}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{4} + 5 \, a c^{\frac {7}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{2} + a c^{\frac {7}{2}}\right )}} \] Input:
integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^(7/2),x, algorithm="giac")
Output:
-1/26*(9*e^(arctan(a*x))*tan(1/2*arctan(a*x))^10 - 34*e^(arctan(a*x))*tan( 1/2*arctan(a*x))^9 + 11*e^(arctan(a*x))*tan(1/2*arctan(a*x))^8 - 8*e^(arct an(a*x))*tan(1/2*arctan(a*x))^7 + 18*e^(arctan(a*x))*tan(1/2*arctan(a*x))^ 6 - 108*e^(arctan(a*x))*tan(1/2*arctan(a*x))^5 - 18*e^(arctan(a*x))*tan(1/ 2*arctan(a*x))^4 - 8*e^(arctan(a*x))*tan(1/2*arctan(a*x))^3 - 11*e^(arctan (a*x))*tan(1/2*arctan(a*x))^2 - 34*e^(arctan(a*x))*tan(1/2*arctan(a*x)) - 9*e^(arctan(a*x)))/(a*c^(7/2)*tan(1/2*arctan(a*x))^10 + 5*a*c^(7/2)*tan(1/ 2*arctan(a*x))^8 + 10*a*c^(7/2)*tan(1/2*arctan(a*x))^6 + 10*a*c^(7/2)*tan( 1/2*arctan(a*x))^4 + 5*a*c^(7/2)*tan(1/2*arctan(a*x))^2 + a*c^(7/2))
Time = 23.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\frac {{\mathrm {e}}^{\mathrm {atan}\left (a\,x\right )}\,\left (\frac {9}{26\,a^5\,c^3}+\frac {3\,x^5}{13\,c^3}+\frac {17\,x}{26\,a^4\,c^3}+\frac {3\,x^4}{13\,a\,c^3}+\frac {9\,x^3}{13\,a^2\,c^3}+\frac {7\,x^2}{13\,a^3\,c^3}\right )}{\frac {\sqrt {c\,a^2\,x^2+c}}{a^4}+x^4\,\sqrt {c\,a^2\,x^2+c}+\frac {2\,x^2\,\sqrt {c\,a^2\,x^2+c}}{a^2}} \] Input:
int(exp(atan(a*x))/(c + a^2*c*x^2)^(7/2),x)
Output:
(exp(atan(a*x))*(9/(26*a^5*c^3) + (3*x^5)/(13*c^3) + (17*x)/(26*a^4*c^3) + (3*x^4)/(13*a*c^3) + (9*x^3)/(13*a^2*c^3) + (7*x^2)/(13*a^3*c^3)))/((c + a^2*c*x^2)^(1/2)/a^4 + x^4*(c + a^2*c*x^2)^(1/2) + (2*x^2*(c + a^2*c*x^2)^ (1/2))/a^2)
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\frac {e^{\mathit {atan} \left (a x \right )} \sqrt {c}\, \sqrt {a^{2} x^{2}+1}\, \left (6 a^{5} x^{5}+6 a^{4} x^{4}+18 a^{3} x^{3}+14 a^{2} x^{2}+17 a x +9\right )}{26 a \,c^{4} \left (a^{6} x^{6}+3 a^{4} x^{4}+3 a^{2} x^{2}+1\right )} \] Input:
int(exp(atan(a*x))/(a^2*c*x^2+c)^(7/2),x)
Output:
(e**atan(a*x)*sqrt(c)*sqrt(a**2*x**2 + 1)*(6*a**5*x**5 + 6*a**4*x**4 + 18* a**3*x**3 + 14*a**2*x**2 + 17*a*x + 9))/(26*a*c**4*(a**6*x**6 + 3*a**4*x** 4 + 3*a**2*x**2 + 1))