3.551 \(\int e^{m x} \sec ^3(x) \, dx\)

Optimal. Leaf size=51 \[ \frac{8 e^{(m+3 i) x} \, _2F_1\left (3,\frac{1}{2} (3-i m);\frac{1}{2} (5-i m);-e^{2 i x}\right )}{m+3 i} \]

[Out]

(8*E^((3*I + m)*x)*Hypergeometric2F1[3, (3 - I*m)/2, (5 - I*m)/2, -E^((2*I)*x)])/(3*I + m)

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Rubi [A]  time = 0.0363634, antiderivative size = 77, normalized size of antiderivative = 1.51, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4448, 4451} \[ (-m+i) \left (-e^{(m+i) x}\right ) \text{Hypergeometric2F1}\left (1,\frac{1}{2} (1-i m),\frac{1}{2} (3-i m),-e^{2 i x}\right )-\frac{1}{2} m e^{m x} \sec (x)+\frac{1}{2} e^{m x} \tan (x) \sec (x) \]

Antiderivative was successfully verified.

[In]

Int[E^(m*x)*Sec[x]^3,x]

[Out]

-(E^((I + m)*x)*(I - m)*Hypergeometric2F1[1, (1 - I*m)/2, (3 - I*m)/2, -E^((2*I)*x)]) - (E^(m*x)*m*Sec[x])/2 +
(E^(m*x)*Sec[x]*Tan[x])/2

Rule 4448

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Sec[d + e*x]^(n - 2))/(e^2*(n - 1)*(n - 2)), x] + (Dist[(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] + Simp[(F^(c*(a + b*x))*Sec[d + e*x]^(n - 1)*Sin[d +
e*x])/(e*(n - 1)), 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] && NeQ[n, 2]

Rule 4451

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(I*n*(d + e*x))*
F^(c*(a + b*x))*Hypergeometric2F1[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))])/(I*e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin{align*} \int e^{m x} \sec ^3(x) \, dx &=-\frac{1}{2} e^{m x} m \sec (x)+\frac{1}{2} e^{m x} \sec (x) \tan (x)+\frac{1}{2} \left (1+m^2\right ) \int e^{m x} \sec (x) \, dx\\ &=-e^{(i+m) x} (i-m) \, _2F_1\left (1,\frac{1}{2} (1-i m);\frac{1}{2} (3-i m);-e^{2 i x}\right )-\frac{1}{2} e^{m x} m \sec (x)+\frac{1}{2} e^{m x} \sec (x) \tan (x)\\ \end{align*}

Mathematica [A]  time = 0.0470558, size = 66, normalized size = 1.29 \[ \frac{1}{2} e^{m x} \left (\sec (x) (\tan (x)-m)+2 (m-i) e^{i x} \, _2F_1\left (1,\frac{1}{2}-\frac{i m}{2};\frac{3}{2}-\frac{i m}{2};-e^{2 i x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(m*x)*Sec[x]^3,x]

[Out]

(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 + T
an[x])))/2

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Maple [F]  time = 0.113, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{mx}}}{ \left ( \cos \left ( x \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(m*x)/cos(x)^3,x)

[Out]

int(exp(m*x)/cos(x)^3,x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)/cos(x)^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e^{\left (m x\right )}}{\cos \left (x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)/cos(x)^3,x, algorithm="fricas")

[Out]

integral(e^(m*x)/cos(x)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{m x}}{\cos ^{3}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)/cos(x)**3,x)

[Out]

Integral(exp(m*x)/cos(x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (m x\right )}}{\cos \left (x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(m*x)/cos(x)^3,x, algorithm="giac")

[Out]

integrate(e^(m*x)/cos(x)^3, x)