Integrand size = 23, antiderivative size = 76 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{2 \arctan (a x)} (2+3 a x)}{13 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {6 e^{2 \arctan (a x)} (2+a x)}{65 a c^2 \sqrt {c+a^2 c x^2}} \] Output:
1/13*exp(2*arctan(a*x))*(3*a*x+2)/a/c/(a^2*c*x^2+c)^(3/2)+6/65*exp(2*arcta n(a*x))*(a*x+2)/a/c^2/(a^2*c*x^2+c)^(1/2)
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{2 \arctan (a x)} \left (22+21 a x+12 a^2 x^2+6 a^3 x^3\right )}{65 c^2 \left (a+a^3 x^2\right ) \sqrt {c+a^2 c x^2}} \] Input:
Integrate[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]
Output:
(E^(2*ArcTan[a*x])*(22 + 21*a*x + 12*a^2*x^2 + 6*a^3*x^3))/(65*c^2*(a + a^ 3*x^2)*Sqrt[c + a^2*c*x^2])
Time = 0.54 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {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.
\(\displaystyle \int \frac {e^{2 \arctan (a x)}}{\left (a^2 c x^2+c\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5593 |
\(\displaystyle \frac {6 \int \frac {e^{2 \arctan (a x)}}{\left (a^2 c x^2+c\right )^{3/2}}dx}{13 c}+\frac {(3 a x+2) e^{2 \arctan (a x)}}{13 a c \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5592 |
\(\displaystyle \frac {6 (a x+2) e^{2 \arctan (a x)}}{65 a c^2 \sqrt {a^2 c x^2+c}}+\frac {(3 a x+2) e^{2 \arctan (a x)}}{13 a c \left (a^2 c x^2+c\right )^{3/2}}\) |
Input:
Int[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]
Output:
(E^(2*ArcTan[a*x])*(2 + 3*a*x))/(13*a*c*(c + a^2*c*x^2)^(3/2)) + (6*E^(2*A rcTan[a*x])*(2 + a*x))/(65*a*c^2*Sqrt[c + a^2*c*x^2])
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 = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74
method | result | size |
gosper | \(\frac {\left (a^{2} x^{2}+1\right ) \left (6 a^{3} x^{3}+12 a^{2} x^{2}+21 a x +22\right ) {\mathrm e}^{2 \arctan \left (a x \right )}}{65 a \left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\) | \(56\) |
orering | \(\frac {\left (a^{2} x^{2}+1\right ) \left (6 a^{3} x^{3}+12 a^{2} x^{2}+21 a x +22\right ) {\mathrm e}^{2 \arctan \left (a x \right )}}{65 a \left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\) | \(56\) |
Input:
int(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/65*(a^2*x^2+1)*(6*a^3*x^3+12*a^2*x^2+21*a*x+22)*exp(2*arctan(a*x))/a/(a^ 2*c*x^2+c)^(5/2)
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (6 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 21 \, a x + 22\right )} \sqrt {a^{2} c x^{2} + c} e^{\left (2 \, \arctan \left (a x\right )\right )}}{65 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \] Input:
integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")
Output:
1/65*(6*a^3*x^3 + 12*a^2*x^2 + 21*a*x + 22)*sqrt(a^2*c*x^2 + c)*e^(2*arcta n(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)
\[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {e^{2 \operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(exp(2*atan(a*x))/(a**2*c*x**2+c)**(5/2),x)
Output:
Integral(exp(2*atan(a*x))/(c*(a**2*x**2 + 1))**(5/2), x)
\[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (2 \, \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")
Output:
integrate(e^(2*arctan(a*x))/(a^2*c*x^2 + c)^(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (66) = 132\).
Time = 0.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.24 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (11 \, e^{\left (2 \, \arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{6} - 21 \, e^{\left (2 \, \arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{5} - 9 \, e^{\left (2 \, \arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{4} + 18 \, e^{\left (2 \, \arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{3} + 9 \, e^{\left (2 \, \arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{2} - 21 \, e^{\left (2 \, \arctan \left (a x\right )\right )} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right ) - 11 \, e^{\left (2 \, \arctan \left (a x\right )\right )}\right )}}{65 \, {\left (a c^{\frac {5}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{6} + 3 \, a c^{\frac {5}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{4} + 3 \, a c^{\frac {5}{2}} \tan \left (\frac {1}{2} \, \arctan \left (a x\right )\right )^{2} + a c^{\frac {5}{2}}\right )}} \] Input:
integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")
Output:
-2/65*(11*e^(2*arctan(a*x))*tan(1/2*arctan(a*x))^6 - 21*e^(2*arctan(a*x))* tan(1/2*arctan(a*x))^5 - 9*e^(2*arctan(a*x))*tan(1/2*arctan(a*x))^4 + 18*e ^(2*arctan(a*x))*tan(1/2*arctan(a*x))^3 + 9*e^(2*arctan(a*x))*tan(1/2*arct an(a*x))^2 - 21*e^(2*arctan(a*x))*tan(1/2*arctan(a*x)) - 11*e^(2*arctan(a* x)))/(a*c^(5/2)*tan(1/2*arctan(a*x))^6 + 3*a*c^(5/2)*tan(1/2*arctan(a*x))^ 4 + 3*a*c^(5/2)*tan(1/2*arctan(a*x))^2 + a*c^(5/2))
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\mathrm {e}}^{2\,\mathrm {atan}\left (a\,x\right )}\,\left (\frac {22}{65\,a^3\,c^2}+\frac {6\,x^3}{65\,c^2}+\frac {21\,x}{65\,a^2\,c^2}+\frac {12\,x^2}{65\,a\,c^2}\right )}{\frac {\sqrt {c\,a^2\,x^2+c}}{a^2}+x^2\,\sqrt {c\,a^2\,x^2+c}} \] Input:
int(exp(2*atan(a*x))/(c + a^2*c*x^2)^(5/2),x)
Output:
(exp(2*atan(a*x))*(22/(65*a^3*c^2) + (6*x^3)/(65*c^2) + (21*x)/(65*a^2*c^2 ) + (12*x^2)/(65*a*c^2)))/((c + a^2*c*x^2)^(1/2)/a^2 + x^2*(c + a^2*c*x^2) ^(1/2))
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{2 \mathit {atan} \left (a x \right )} \sqrt {c}\, \sqrt {a^{2} x^{2}+1}\, \left (6 a^{3} x^{3}+12 a^{2} x^{2}+21 a x +22\right )}{65 a \,c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )} \] Input:
int(exp(2*atan(a*x))/(a^2*c*x^2+c)^(5/2),x)
Output:
(e**(2*atan(a*x))*sqrt(c)*sqrt(a**2*x**2 + 1)*(6*a**3*x**3 + 12*a**2*x**2 + 21*a*x + 22))/(65*a*c**3*(a**4*x**4 + 2*a**2*x**2 + 1))