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

Optimal. Leaf size=89 \[ \frac {b \cos ^2(e+f x)^{\frac {m+1}{2}} (a \sec (e+f x))^{m+1} (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {m+1}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{a f (1-n)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 89, 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 = {2631, 2577} \[ \frac {b \cos ^2(e+f x)^{\frac {m+1}{2}} (a \sec (e+f x))^{m+1} (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {m+1}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{a f (1-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2577

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

Rule 2631

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

Rubi steps

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

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Mathematica [C]  time = 0.19, size = 283, normalized size = 3.18 \[ -\frac {b (n-3) (a \sec (e+f x))^m (b \csc (e+f x))^{n-1} F_1\left (\frac {1-n}{2};m,-m-n+1;\frac {3-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{f (n-1) \left ((n-3) F_1\left (\frac {1-n}{2};m,-m-n+1;\frac {3-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac {1}{2} (e+f x)\right ) \left ((m+n-1) F_1\left (\frac {3-n}{2};m,-m-n+2;\frac {5-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+m F_1\left (\frac {3-n}{2};m+1,-m-n+1;\frac {5-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((b*(-3 + n)*AppellF1[(1 - n)/2, m, 1 - m - n, (3 - n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(b*Csc[e +
f*x])^(-1 + n)*(a*Sec[e + f*x])^m)/(f*(-1 + n)*((-3 + n)*AppellF1[(1 - n)/2, m, 1 - m - n, (3 - n)/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((-1 + m + n)*AppellF1[(3 - n)/2, m, 2 - m - n, (5 - n)/2, Tan[(e + f*x)
/2]^2, -Tan[(e + f*x)/2]^2] + m*AppellF1[(3 - n)/2, 1 + m, 1 - m - n, (5 - n)/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2])*Tan[(e + f*x)/2]^2)))

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fricas [F]  time = 1.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \csc \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.87, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x +e \right )\right )^{n} \left (a \sec \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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