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

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Rubi [A]  time = 0.02, antiderivative size = 17, 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 = {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 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]

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])

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*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \[ \frac {(b \sec (e+f x))^m}{f m} \]

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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fricas [A]  time = 0.62, size = 19, normalized size = 1.12 \[ \frac {\left (\frac {b}{\cos \left (f x + e\right )}\right )^{m}}{f m} \]

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

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)

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maple [A]  time = 0.06, size = 18, normalized size = 1.06 \[ \frac {\left (b \sec \left (f x +e \right )\right )^{m}}{f m} \]

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 = 0.44, size = 20, normalized size = 1.18 \[ \frac {b^{m} \cos \left (f x + e\right )^{-m}}{f m} \]

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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mupad [B]  time = 0.12, size = 19, normalized size = 1.12 \[ \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^m}{f\,m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b/cos(e + f*x))^m/(f*m)

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sympy [A]  time = 0.42, size = 44, normalized size = 2.59 \[ \begin {cases} x \tan {\relax (e )} & \text {for}\: f = 0 \wedge m = 0 \\x \left (b \sec {\relax (e )}\right )^{m} \tan {\relax (e )} & \text {for}\: f = 0 \\\frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} & \text {for}\: m = 0 \\\frac {b^{m} \sec ^{m}{\left (e + f x \right )}}{f m} & \text {otherwise} \end {cases} \]

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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