3.4.53 \(\int (b \sec (e+f x))^m \tan (e+f x) \, dx\) [353]

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 = {2686, 32} \begin {gather*} \frac {(b \sec (e+f x))^m}{f m} \end {gather*}

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 2686

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 \text {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 = 0.03, size = 17, normalized size = 1.00 \begin {gather*} \frac {(b \sec (e+f x))^m}{f m} \end {gather*}

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 = 0.07, size = 18, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {\left (b \sec \left (f x +e \right )\right )^{m}}{f m}\)	\(18\)
default	\(\frac {\left (b \sec \left (f x +e \right )\right )^{m}}{f m}\)	\(18\)
risch	\(\frac {{\mathrm e}^{\frac {m \left (-i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{i \left (f x +e \right )}\right )+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )-i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{i \left (f x +e \right )}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )+i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right ) \mathrm {csgn}\left (\frac {i b \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right ) \mathrm {csgn}\left (\frac {i b \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right ) \mathrm {csgn}\left (i b \right )-i \pi \mathrm {csgn}\left (\frac {i b \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i b \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}\right )^{2} \mathrm {csgn}\left (i b \right )+2 \ln \left (b \right )+2 \ln \left (2\right )+2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )-2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )\right )}{2}}}{f m}\)	\(425\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 21, normalized size = 1.24 \begin {gather*} \frac {b^{m} \cos \left (f x + e\right )^{-m}}{f m} \end {gather*}

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 = 0.37, size = 20, normalized size = 1.18 \begin {gather*} \frac {\left (\frac {b}{\cos \left (f x + e\right )}\right )^{m}}{f m} \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (12) = 24\).
time = 0.14, size = 42, normalized size = 2.47 \begin {gather*} \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 {\left (b \sec {\left (e + f x \right )}\right )^{m}}{f m} & \text {otherwise} \end {cases} \end {gather*}

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*sec(e + f*x))**m/(f*m), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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