3.494 \(\int (b \sec (e+f x))^n \sin (e+f x) \, dx\)

Optimal. Leaf size=25 \[ -\frac {b (b \sec (e+f x))^{n-1}}{f (1-n)} \]

[Out]

-b*(b*sec(f*x+e))^(-1+n)/f/(1-n)

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Rubi [A]  time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2622, 30} \[ -\frac {b (b \sec (e+f x))^{n-1}}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^n*Sin[e + f*x],x]

[Out]

-((b*(b*Sec[e + f*x])^(-1 + n))/(f*(1 - n)))

Rule 30

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

Rule 2622

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*} \int (b \sec (e+f x))^n \sin (e+f x) \, dx &=\frac {b \operatorname {Subst}\left (\int x^{-2+n} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac {b (b \sec (e+f x))^{-1+n}}{f (1-n)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.88 \[ \frac {b (b \sec (e+f x))^{n-1}}{f (n-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x],x]

[Out]

(b*(b*Sec[e + f*x])^(-1 + n))/(f*(-1 + n))

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fricas [A]  time = 0.50, size = 28, normalized size = 1.12 \[ \frac {\left (\frac {b}{\cos \left (f x + e\right )}\right )^{n} \cos \left (f x + e\right )}{f n - f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e),x, algorithm="fricas")

[Out]

(b/cos(f*x + e))^n*cos(f*x + e)/(f*n - f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n*sin(f*x + e), x)

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maple [B]  time = 0.04, size = 120, normalized size = 4.80 \[ \frac {\frac {{\mathrm e}^{n \ln \left (\frac {b \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{1-\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}\right )}}{f \left (-1+n \right )}-\frac {\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) {\mathrm e}^{n \ln \left (\frac {b \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{1-\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}\right )}}{f \left (-1+n \right )}}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^n*sin(f*x+e),x)

[Out]

(1/f/(-1+n)*exp(n*ln(b*(1+tan(1/2*f*x+1/2*e)^2)/(1-tan(1/2*f*x+1/2*e)^2)))-1/f/(-1+n)*tan(1/2*f*x+1/2*e)^2*exp
(n*ln(b*(1+tan(1/2*f*x+1/2*e)^2)/(1-tan(1/2*f*x+1/2*e)^2))))/(1+tan(1/2*f*x+1/2*e)^2)

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maxima [A]  time = 0.36, size = 28, normalized size = 1.12 \[ \frac {b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )}{f {\left (n - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e),x, algorithm="maxima")

[Out]

b^n*cos(f*x + e)^(-n)*cos(f*x + e)/(f*(n - 1))

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mupad [B]  time = 0.19, size = 27, normalized size = 1.08 \[ \frac {\cos \left (e+f\,x\right )\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n}{f\,\left (n-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)*(b/cos(e + f*x))^n,x)

[Out]

(cos(e + f*x)*(b/cos(e + f*x))^n)/(f*(n - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (e + f x \right )}\right )^{n} \sin {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**n*sin(f*x+e),x)

[Out]

Integral((b*sec(e + f*x))**n*sin(e + f*x), x)

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