3.488 \(\int \sec ^n(e+f x) \sin ^m(e+f x) \, dx\)

Optimal. Leaf size=86 \[ -\frac{\sin ^{m-1}(e+f x) \sin ^2(e+f x)^{\frac{1-m}{2}} \sec ^{n-1}(e+f x) \, _2F_1\left (\frac{1-m}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f (1-n)} \]

[Out]

-((Hypergeometric2F1[(1 - m)/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*Sec[e + f*x]^(-1 + n)*Sin[e + f*x]^(-1 +
m)*(Sin[e + f*x]^2)^((1 - m)/2))/(f*(1 - n)))

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Rubi [A]  time = 0.0764943, antiderivative size = 86, 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 = {2587, 2576} \[ -\frac{\sin ^{m-1}(e+f x) \sin ^2(e+f x)^{\frac{1-m}{2}} \sec ^{n-1}(e+f x) \, _2F_1\left (\frac{1-m}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^n*Sin[e + f*x]^m,x]

[Out]

-((Hypergeometric2F1[(1 - m)/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*Sec[e + f*x]^(-1 + n)*Sin[e + f*x]^(-1 +
m)*(Sin[e + f*x]^2)^((1 - m)/2))/(f*(1 - n)))

Rule 2587

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

Rule 2576

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^(2*IntPar
t[(n - 1)/2] + 1)*(b*Sin[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Cos[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/
2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2])/(a*f*(m + 1)*(Sin[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a,
b, e, f, m, n}, x] && SimplerQ[n, m]

Rubi steps

\begin{align*} \int \sec ^n(e+f x) \sin ^m(e+f x) \, dx &=\left (\cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-n}(e+f x) \sin ^m(e+f x) \, dx\\ &=-\frac{\, _2F_1\left (\frac{1-m}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right ) \sec ^{-1+n}(e+f x) \sin ^{-1+m}(e+f x) \sin ^2(e+f x)^{\frac{1-m}{2}}}{f (1-n)}\\ \end{align*}

Mathematica [C]  time = 1.38711, size = 285, normalized size = 3.31 \[ \frac{4 (m+3) \sin \left (\frac{1}{2} (e+f x)\right ) \cos ^3\left (\frac{1}{2} (e+f x)\right ) \sin ^m(e+f x) \sec ^n(e+f x) F_1\left (\frac{m+1}{2};n,m-n+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{f (m+1) \left ((m+3) (\cos (e+f x)+1) F_1\left (\frac{m+1}{2};n,m-n+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-4 \sin ^2\left (\frac{1}{2} (e+f x)\right ) \left ((m-n+1) F_1\left (\frac{m+3}{2};n,m-n+2;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-n F_1\left (\frac{m+3}{2};n+1,m-n+1;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]^n*Sin[e + f*x]^m,x]

[Out]

(4*(3 + m)*AppellF1[(1 + m)/2, n, 1 + m - n, (3 + m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)
/2]^3*Sec[e + f*x]^n*Sin[(e + f*x)/2]*Sin[e + f*x]^m)/(f*(1 + m)*((3 + m)*AppellF1[(1 + m)/2, n, 1 + m - n, (3
+ m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x]) - 4*((1 + m - n)*AppellF1[(3 + m)/2, n, 2
+ m - n, (5 + m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[(3 + m)/2, 1 + n, 1 + m - n, (5 + m
)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sin[(e + f*x)/2]^2))

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Maple [F]  time = 0.516, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{n} \left ( \sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sec \left (f x + e\right )^{n} \sin \left (f x + e\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sec(f*x + e)^n*sin(f*x + e)^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin ^{m}{\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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