3.353 \(\int (b \sec (e+f x))^m \tan (e+f x) \, dx\)

Optimal. Leaf size=17 \[ \frac{(b \sec (e+f x))^m}{f m} \]

[Out]

(b*Sec[e + f*x])^m/(f*m)

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Rubi [A]  time = 0.0210696, antiderivative size = 17, normalized size of antiderivative = 1., 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 = {2606, 32} \[ \frac{(b \sec (e+f x))^m}{f m} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^m*Tan[e + f*x],x]

[Out]

(b*Sec[e + f*x])^m/(f*m)

Rule 2606

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0216875, size = 17, normalized size = 1. \[ \frac{(b \sec (e+f x))^m}{f m} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^m*Tan[e + f*x],x]

[Out]

(b*Sec[e + f*x])^m/(f*m)

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Maple [A]  time = 0.01, size = 18, normalized size = 1.1 \begin{align*}{\frac{ \left ( b\sec \left ( fx+e \right ) \right ) ^{m}}{fm}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^m*tan(f*x+e),x)

[Out]

(b*sec(f*x+e))^m/f/m

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Maxima [A]  time = 0.999073, size = 27, normalized size = 1.59 \begin{align*} \frac{b^{m} \cos \left (f x + e\right )^{-m}}{f m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^m*cos(f*x + e)^(-m)/(f*m)

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Fricas [A]  time = 1.6066, size = 35, normalized size = 2.06 \begin{align*} \frac{\left (\frac{b}{\cos \left (f x + e\right )}\right )^{m}}{f m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b/cos(f*x + e))^m/(f*m)

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Sympy [A]  time = 0.538158, size = 44, normalized size = 2.59 \begin{align*} \begin{cases} x \tan{\left (e \right )} & \text{for}\: f = 0 \wedge m = 0 \\x \left (b \sec{\left (e \right )}\right )^{m} \tan{\left (e \right )} & \text{for}\: f = 0 \\\frac{\log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} & \text{for}\: m = 0 \\\frac{b^{m} \sec ^{m}{\left (e + f x \right )}}{f m} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**m*tan(f*x+e),x)

[Out]

Piecewise((x*tan(e), Eq(f, 0) & Eq(m, 0)), (x*(b*sec(e))**m*tan(e), Eq(f, 0)), (log(tan(e + f*x)**2 + 1)/(2*f)
, Eq(m, 0)), (b**m*sec(e + f*x)**m/(f*m), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(f*x + e))^m*tan(f*x + e), x)