Integrand size = 10, antiderivative size = 51 \[ \int e^{m x} \sec ^3(x) \, dx=\frac {8 e^{(3 i+m) x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} (3-i m),\frac {1}{2} (5-i m),-e^{2 i x}\right )}{3 i+m} \]
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.29 \[ \int e^{m x} \sec ^3(x) \, dx=\frac {1}{2} e^{m x} \left (2 e^{i x} (-i+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {i m}{2},\frac {3}{2}-\frac {i m}{2},-e^{2 i x}\right )+\sec (x) (-m+\tan (x))\right ) \]
(E^(m*x)*(2*E^(I*x)*(-I + m)*Hypergeometric2F1[1, 1/2 - (I/2)*m, 3/2 - (I/ 2)*m, -E^((2*I)*x)] + Sec[x]*(-m + Tan[x])))/2
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.59, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4948, 4951}
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.
\(\displaystyle \int e^{m x} \sec ^3(x) \, dx\) |
\(\Big \downarrow \) 4948 |
\(\displaystyle \frac {1}{2} \left (m^2+1\right ) \int e^{m x} \sec (x)dx-\frac {1}{2} m e^{m x} \sec (x)+\frac {1}{2} e^{m x} \tan (x) \sec (x)\) |
\(\Big \downarrow \) 4951 |
\(\displaystyle \frac {\left (m^2+1\right ) e^{(m+i) x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1-i m),\frac {1}{2} (3-i m),-e^{2 i x}\right )}{m+i}-\frac {1}{2} m e^{m x} \sec (x)+\frac {1}{2} e^{m x} \tan (x) \sec (x)\) |
(E^((I + m)*x)*(1 + m^2)*Hypergeometric2F1[1, (1 - I*m)/2, (3 - I*m)/2, -E ^((2*I)*x)])/(I + m) - (E^(m*x)*m*Sec[x])/2 + (E^(m*x)*Sec[x]*Tan[x])/2
3.6.51.3.1 Defintions of rubi rules used
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_), x_Symbo l] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Sec[d + e*x]^(n - 2)/(e^2*(n - 1) *(n - 2))), x] + (Simp[F^(c*(a + b*x))*Sec[d + e*x]^(n - 1)*(Sin[d + e*x]/( e*(n - 1))), x] + Simp[(e^2*(n - 2)^2 + b^2*c^2*Log[F]^2)/(e^2*(n - 1)*(n - 2)) Int[F^(c*(a + b*x))*Sec[d + e*x]^(n - 2), x], x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[b^2*c^2*Log[F]^2 + e^2*(n - 2)^2, 0] && GtQ[n, 1] && N eQ[n, 2]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symb ol] :> Simp[2^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hy pergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(Log[F]/(2*e )), -E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
\[\int \frac {{\mathrm e}^{m x}}{\cos \left (x \right )^{3}}d x\]
\[ \int e^{m x} \sec ^3(x) \, dx=\int { \frac {e^{\left (m x\right )}}{\cos \left (x\right )^{3}} \,d x } \]
\[ \int e^{m x} \sec ^3(x) \, dx=\int \frac {e^{m x}}{\cos ^{3}{\left (x \right )}}\, dx \]
\[ \int e^{m x} \sec ^3(x) \, dx=\int { \frac {e^{\left (m x\right )}}{\cos \left (x\right )^{3}} \,d x } \]
8*(48*m*cos(x)*e^(m*x) + 6*(m^2 - 15)*e^(m*x)*sin(x) + ((m^3 + 25*m)*cos(3 *x)*e^(m*x) + 48*m*cos(x)*e^(m*x) - 3*(m^2 + 25)*e^(m*x)*sin(3*x) + 6*(m^2 - 15)*e^(m*x)*sin(x))*cos(6*x) + 3*((m^3 + 25*m)*cos(3*x)*e^(m*x) + 48*m* cos(x)*e^(m*x) - 3*(m^2 + 25)*e^(m*x)*sin(3*x) + 6*(m^2 - 15)*e^(m*x)*sin( x))*cos(4*x) + (3*(m^3 + 25*m)*cos(2*x)*e^(m*x) + 9*(m^2 + 25)*e^(m*x)*sin (2*x) + (m^3 + 25*m)*e^(m*x))*cos(3*x) + 18*(8*m*cos(x)*e^(m*x) + (m^2 - 1 5)*e^(m*x)*sin(x))*cos(2*x) - 6*(m^4 + (m^4 + 34*m^2 + 225)*cos(6*x)^2 + 9 *(m^4 + 34*m^2 + 225)*cos(4*x)^2 + 9*(m^4 + 34*m^2 + 225)*cos(2*x)^2 + (m^ 4 + 34*m^2 + 225)*sin(6*x)^2 + 9*(m^4 + 34*m^2 + 225)*sin(4*x)^2 + 18*(m^4 + 34*m^2 + 225)*sin(4*x)*sin(2*x) + 9*(m^4 + 34*m^2 + 225)*sin(2*x)^2 + 3 4*m^2 + 2*(m^4 + 34*m^2 + 3*(m^4 + 34*m^2 + 225)*cos(4*x) + 3*(m^4 + 34*m^ 2 + 225)*cos(2*x) + 225)*cos(6*x) + 6*(m^4 + 34*m^2 + 3*(m^4 + 34*m^2 + 22 5)*cos(2*x) + 225)*cos(4*x) + 6*(m^4 + 34*m^2 + 225)*cos(2*x) + 6*((m^4 + 34*m^2 + 225)*sin(4*x) + (m^4 + 34*m^2 + 225)*sin(2*x))*sin(6*x) + 225)*in tegrate(((m^2 - 15)*cos(x)*e^(m*x) - 8*m*e^(m*x)*sin(x) + ((m^2 - 15)*cos( x)*e^(m*x) - 8*m*e^(m*x)*sin(x))*cos(8*x) + 4*((m^2 - 15)*cos(x)*e^(m*x) - 8*m*e^(m*x)*sin(x))*cos(6*x) + 6*((m^2 - 15)*cos(x)*e^(m*x) - 8*m*e^(m*x) *sin(x))*cos(4*x) + 4*((m^2 - 15)*cos(x)*e^(m*x) - 8*m*e^(m*x)*sin(x))*cos (2*x) + (8*m*cos(x)*e^(m*x) + (m^2 - 15)*e^(m*x)*sin(x))*sin(8*x) + 4*(8*m *cos(x)*e^(m*x) + (m^2 - 15)*e^(m*x)*sin(x))*sin(6*x) + 6*(8*m*cos(x)*e...
\[ \int e^{m x} \sec ^3(x) \, dx=\int { \frac {e^{\left (m x\right )}}{\cos \left (x\right )^{3}} \,d x } \]
Timed out. \[ \int e^{m x} \sec ^3(x) \, dx=\int \frac {{\mathrm {e}}^{m\,x}}{{\cos \left (x\right )}^3} \,d x \]