\(\int \sec ^{1+m}(x) \sin (x) \, dx\) [917]

Optimal result

Integrand size = 9, antiderivative size = 8 \[ \int \sec ^{1+m}(x) \sin (x) \, dx=\frac {\sec ^m(x)}{m} \]

[Out]

sec(x)^m/m

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2702, 30} \[ \int \sec ^{1+m}(x) \sin (x) \, dx=\frac {\sec ^m(x)}{m} \]

[In]

Int[Sec[x]^(1 + m)*Sin[x],x]

[Out]

Sec[x]^m/m

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int x^{-1+m} \, dx,x,\sec (x)\right ) \\ & = \frac {\sec ^m(x)}{m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \sec ^{1+m}(x) \sin (x) \, dx=\frac {\sec ^m(x)}{m} \]

[In]

Integrate[Sec[x]^(1 + m)*Sin[x],x]

[Out]

Sec[x]^m/m

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.12

\[\frac {{\mathrm e}^{\left (1+m \right ) \ln \left (\sec \left (x \right )\right )}}{m \sec \left (x \right )}\]

[In]

int(sec(x)^(1+m)*sin(x),x)

[Out]

1/m*exp((1+m)*ln(sec(x)))/sec(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \sec ^{1+m}(x) \sin (x) \, dx=\frac {\frac {1}{\cos \left (x\right )}^{m + 1} \cos \left (x\right )}{m} \]

[In]

integrate(sec(x)^(1+m)*sin(x),x, algorithm="fricas")

[Out]

(1/cos(x))^(m + 1)*cos(x)/m

Sympy [F]

\[ \int \sec ^{1+m}(x) \sin (x) \, dx=\int \sin {\left (x \right )} \sec ^{m + 1}{\left (x \right )}\, dx \]

[In]

integrate(sec(x)**(1+m)*sin(x),x)

[Out]

Integral(sin(x)*sec(x)**(m + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.25 \[ \int \sec ^{1+m}(x) \sin (x) \, dx=\frac {\cos \left (x\right )^{-m}}{m} \]

[In]

integrate(sec(x)^(1+m)*sin(x),x, algorithm="maxima")

[Out]

cos(x)^(-m)/m

Giac [F]

\[ \int \sec ^{1+m}(x) \sin (x) \, dx=\int { \sec \left (x\right )^{m + 1} \sin \left (x\right ) \,d x } \]

[In]

integrate(sec(x)^(1+m)*sin(x),x, algorithm="giac")

[Out]

integrate(sec(x)^(m + 1)*sin(x), x)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.25 \[ \int \sec ^{1+m}(x) \sin (x) \, dx=\frac {{\left (\frac {1}{\cos \left (x\right )}\right )}^m}{m} \]

[In]

int(sin(x)*(1/cos(x))^(m + 1),x)

[Out]

(1/cos(x))^m/m