3.491 \(\int (b \sec (e+f x))^n (a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=92 \[ -\frac {a b \sin ^2(e+f x)^{\frac {1-m}{2}} (a \sin (e+f x))^{m-1} (b \sec (e+f x))^{n-1} \, _2F_1\left (\frac {1-m}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f (1-n)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2587, 2576} \[ -\frac {a b \sin ^2(e+f x)^{\frac {1-m}{2}} (a \sin (e+f x))^{m-1} (b \sec (e+f x))^{n-1} \, _2F_1\left (\frac {1-m}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^n*(a*Sin[e + f*x])^m,x]

[Out]

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

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]

Rubi steps

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

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Mathematica [C]  time = 0.13, size = 289, normalized size = 3.14 \[ \frac {4 (m+3) \sin \left (\frac {1}{2} (e+f x)\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) (a \sin (e+f x))^m (b \sec (e+f x))^n F_1\left (\frac {m+1}{2};n,m-n+1;\frac {m+3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{f (m+1) \left ((m+3) (\cos (e+f x)+1) F_1\left (\frac {m+1}{2};n,m-n+1;\frac {m+3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 \sin ^2\left (\frac {1}{2} (e+f x)\right ) \left ((m-n+1) F_1\left (\frac {m+3}{2};n,m-n+2;\frac {m+5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n F_1\left (\frac {m+3}{2};n+1,m-n+1;\frac {m+5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Sec[e + f*x])^n*(a*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*(b*Sec[e + f*x])^n*Sin[(e + f*x)/2]*(a*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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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \sec \left (f x + e\right )\right )^{n} \left (a \sin \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.64, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x +e \right )\right )^{n} \left (a \sin \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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