3.5.88 \(\int \sec ^n(e+f x) \sin ^m(e+f x) \, dx\) [488]

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

[Out]

-hypergeom([1/2-1/2*n, 1/2-1/2*m],[3/2-1/2*n],cos(f*x+e)^2)*sec(f*x+e)^(-1+n)*sin(f*x+e)^(-1+m)*(sin(f*x+e)^2)
^(1/2-1/2*m)/f/(1-n)

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Rubi [A]
time = 0.06, antiderivative size = 86, 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 = {2667, 2656} \begin {gather*} -\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)} \end {gather*}

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 2656

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)/(a*f*(m + 1)*(Sin[e + f*
x]^2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rule 2667

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]

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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 1.01, size = 285, normalized size = 3.31 \begin {gather*} \frac {4 (3+m) F_1\left (\frac {1+m}{2};n,1+m-n;\frac {3+m}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^n(e+f x) \sin \left (\frac {1}{2} (e+f x)\right ) \sin ^m(e+f x)}{f (1+m) \left ((3+m) F_1\left (\frac {1+m}{2};n,1+m-n;\frac {3+m}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))-4 \left ((1+m-n) F_1\left (\frac {3+m}{2};n,2+m-n;\frac {5+m}{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 {3+m}{2};1+n,1+m-n;\frac {5+m}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

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 = 0.10, size = 0, normalized size = 0.00 \[\int \left (\sec ^{n}\left (f x +e \right )\right ) \left (\sin ^{m}\left (f x +e \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{m}{\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx \end {gather*}

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 = 0.00, 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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (e+f\,x\right )}^m\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^m*(1/cos(e + f*x))^n,x)

[Out]

int(sin(e + f*x)^m*(1/cos(e + f*x))^n, x)

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