3.3.98 \(\int \frac {(b \csc (e+f x))^n}{\sqrt {c \sec (e+f x)}} \, dx\) [298]

Optimal. Leaf size=81 \[ \frac {b \sqrt [4]{\cos ^2(e+f x)} (b \csc (e+f x))^{-1+n} \, _2F_1\left (\frac {1}{4},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right ) \sqrt {c \sec (e+f x)}}{c f (1-n)} \]

[Out]

b*(cos(f*x+e)^2)^(1/4)*(b*csc(f*x+e))^(-1+n)*hypergeom([1/4, 1/2-1/2*n],[3/2-1/2*n],sin(f*x+e)^2)*(c*sec(f*x+e
))^(1/2)/c/f/(1-n)

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Rubi [A]
time = 0.07, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2711, 2657} \begin {gather*} \frac {b \sqrt [4]{\cos ^2(e+f x)} \sqrt {c \sec (e+f x)} (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {1}{4},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{c f (1-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Csc[e + f*x])^n/Sqrt[c*Sec[e + f*x]],x]

[Out]

(b*(Cos[e + f*x]^2)^(1/4)*(b*Csc[e + f*x])^(-1 + n)*Hypergeometric2F1[1/4, (1 - n)/2, (3 - n)/2, Sin[e + f*x]^
2]*Sqrt[c*Sec[e + f*x]])/(c*f*(1 - n))

Rule 2657

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

Rule 2711

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^2/b^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), 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 \frac {(b \csc (e+f x))^n}{\sqrt {c \sec (e+f x)}} \, dx &=\frac {\left (b^2 \sqrt {c \cos (e+f x)} (b \csc (e+f x))^{-1+n} \sqrt {c \sec (e+f x)} (b \sin (e+f x))^{-1+n}\right ) \int \sqrt {c \cos (e+f x)} (b \sin (e+f x))^{-n} \, dx}{c^2}\\ &=\frac {b \sqrt [4]{\cos ^2(e+f x)} (b \csc (e+f x))^{-1+n} \, _2F_1\left (\frac {1}{4},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right ) \sqrt {c \sec (e+f x)}}{c 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 = 32.71, size = 326, normalized size = 4.02 \begin {gather*} -\frac {4 (-3+n) F_1\left (\frac {1}{2}-\frac {n}{2};-\frac {1}{2},\frac {3}{2}-n;\frac {3}{2}-\frac {n}{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 ) (b \csc (e+f x))^n \sin \left (\frac {1}{2} (e+f x)\right )}{f (-1+n) \sqrt {c \sec (e+f x)} \left (2 (-3+n) F_1\left (\frac {1}{2}-\frac {n}{2};-\frac {1}{2},\frac {3}{2}-n;\frac {3}{2}-\frac {n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )-F_1\left (\frac {3}{2}-\frac {n}{2};\frac {1}{2},\frac {3}{2}-n;\frac {5}{2}-\frac {n}{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))+2 (3-2 n) F_1\left (\frac {3}{2}-\frac {n}{2};-\frac {1}{2},\frac {5}{2}-n;\frac {5}{2}-\frac {n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Csc[e + f*x])^n/Sqrt[c*Sec[e + f*x]],x]

[Out]

(-4*(-3 + n)*AppellF1[1/2 - n/2, -1/2, 3/2 - n, 3/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f
*x)/2]^3*(b*Csc[e + f*x])^n*Sin[(e + f*x)/2])/(f*(-1 + n)*Sqrt[c*Sec[e + f*x]]*(2*(-3 + n)*AppellF1[1/2 - n/2,
-1/2, 3/2 - n, 3/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 - AppellF1[3/2 - n/2, 1
/2, 3/2 - n, 5/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(-1 + Cos[e + f*x]) + 2*(3 - 2*n)*AppellF1[3/
2 - n/2, -1/2, 5/2 - n, 5/2 - n/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 = 0.43, size = 0, normalized size = 0.00 \[\int \frac {\left (b \csc \left (f x +e \right )\right )^{n}}{\sqrt {c \sec \left (f x +e \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x)

[Out]

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),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((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^n/sqrt(c*sec(f*x + e)), 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((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*sec(f*x + e))*(b*csc(f*x + e))^n/(c*sec(f*x + e)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \csc {\left (e + f x \right )}\right )^{n}}{\sqrt {c \sec {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))**n/(c*sec(f*x+e))**(1/2),x)

[Out]

Integral((b*csc(e + f*x))**n/sqrt(c*sec(e + f*x)), 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((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^n/sqrt(c*sec(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/sin(e + f*x))^n/(c/cos(e + f*x))^(1/2),x)

[Out]

int((b/sin(e + f*x))^n/(c/cos(e + f*x))^(1/2), x)

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