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

Optimal. Leaf size=84 \[ \frac {b \cos ^2(e+f x)^{\frac {m+1}{2}} \sec ^{m+1}(e+f x) (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {m+1}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{f (1-n)} \]

[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)*
sec(f*x+e)^(1+m)/f/(1-n)

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2631, 2577} \[ \frac {b \cos ^2(e+f x)^{\frac {m+1}{2}} \sec ^{m+1}(e+f x) (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {m+1}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Csc[e + f*x])^n*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]*Sec[e + f*x]^(1 + m))/(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 \sec ^m(e+f x) \, dx &=\left (b^2 \cos ^{1+m}(e+f x) (b \csc (e+f x))^{-1+n} \sec ^{1+m}(e+f x) (b \sin (e+f x))^{-1+n}\right ) \int \cos ^{-m}(e+f x) (b \sin (e+f x))^{-n} \, dx\\ &=\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 ) \sec ^{1+m}(e+f x)}{f (1-n)}\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 281, normalized size = 3.35 \[ -\frac {b (n-3) \sec ^m(e+f x) (b \csc (e+f x))^{n-1} F_1\left (\frac {1-n}{2};m,-m-n+1;\frac {3-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{f (n-1) \left ((n-3) F_1\left (\frac {1-n}{2};m,-m-n+1;\frac {3-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac {1}{2} (e+f x)\right ) \left ((m+n-1) F_1\left (\frac {3-n}{2};m,-m-n+2;\frac {5-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+m F_1\left (\frac {3-n}{2};m+1,-m-n+1;\frac {5-n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Csc[e + f*x])^n*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)*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 = 3.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \csc \left (f x + e\right )\right )^{n} \sec \left (f x + e\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.88, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x +e \right )\right )^{n} \left (\sec ^{m}\left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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