3.6.2 \(\int \csc ^4(e+f x) (b \sec (e+f x))^n \, dx\) [502]
Optimal. Leaf size=73 \[ -\frac {b \csc (e+f x) \, _2F_1\left (\frac {5}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sqrt {\sin ^2(e+f x)}}{f (1-n)} \]
[Out]
-b*csc(f*x+e)*hypergeom([5/2, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*sec(f*x+e))^(-1+n)*(sin(f*x+e)^2)^(1/2)/
f/(1-n)
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Rubi [A]
time = 0.06, antiderivative size = 73, 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 = {2712, 2656}
\begin {gather*} -\frac {b \sqrt {\sin ^2(e+f x)} \csc (e+f x) (b \sec (e+f x))^{n-1} \, _2F_1\left (\frac {5}{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[Csc[e + f*x]^4*(b*Sec[e + f*x])^n,x]
[Out]
-((b*Csc[e + f*x]*Hypergeometric2F1[5/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Sec[e + f*x])^(-1 + n)*Sqrt[
Sin[e + f*x]^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 2712
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^2/b^2)*(a*
Sec[e + f*x])^(m - 1)*(b*Csc[e + f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1), Int[1/((a*Co
s[e + f*x])^m*(b*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x]
Rubi steps
\begin {align*} \int \csc ^4(e+f x) (b \sec (e+f x))^n \, 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} \csc ^4(e+f x) \, dx\\ &=-\frac {b \csc (e+f x) \, _2F_1\left (\frac {5}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sqrt {\sin ^2(e+f x)}}{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 = 15.36, size = 3833, normalized size = 52.51 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Csc[e + f*x]^4*(b*Sec[e + f*x])^n,x]
[Out]
(Cot[(e + f*x)/2]^3*Csc[e + f*x]^4*(b*Sec[e + f*x])^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*(-(AppellF1[-3/2, n,
-n, -1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n) - 9*AppellF1[-1/2, n,
-n, 1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^2 + Ap
pellF1[3/2, n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e +
f*x)/2]^6 + (27*AppellF1[1/2, n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n*Tan[
(e + f*x)/2]^4)/(3*AppellF1[1/2, n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*n*(AppellF1[3/2, n,
1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[3/2, 1 + n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[
(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)))/(24*f*(-1/16*(Cot[(e + f*x)/2]^2*Csc[(e + f*x)/2]^2*(Cos[(e + f*x)/2]^2
*Sec[e + f*x])^n*(-(AppellF1[-3/2, n, -n, -1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e
+ f*x)/2]^2)^n) - 9*AppellF1[-1/2, n, -n, 1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e +
f*x)/2]^2)^n*Tan[(e + f*x)/2]^2 + AppellF1[3/2, n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e +
f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^6 + (27*AppellF1[1/2, n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f
*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^4)/(3*AppellF1[1/2, n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2] + 2*n*(AppellF1[3/2, n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[3/2, 1 +
n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))) + (Cot[(e + f*x)/2]^3*(Cos[(e + f*
x)/2]^2*Sec[e + f*x])^n*(-9*AppellF1[-1/2, n, -n, 1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/
2]^2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2] + 3*AppellF1[3/2, n, -n, 5/2, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^5 - (Cos[e + f*x]*S
ec[(e + f*x)/2]^2)^n*(3*n*AppellF1[-1/2, n, 1 - n, 1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)
/2]^2*Tan[(e + f*x)/2] + 3*n*AppellF1[-1/2, 1 + n, -n, 1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e +
f*x)/2]^2*Tan[(e + f*x)/2]) - 9*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^2*(-(n*AppellF1[1/2, n, 1
- n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) - n*AppellF1[1/2, 1 +
n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + (Cos[e + f*x]*Sec
[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^6*((3*n*AppellF1[5/2, n, 1 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]
^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 + (3*n*AppellF1[5/2, 1 + n, -n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5) - n*AppellF1[-3/2, n, -n, -1/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1 + n)*(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e + f*x]*Se
c[(e + f*x)/2]^2*Tan[(e + f*x)/2]) - 9*n*AppellF1[-1/2, n, -n, 1/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(
Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1 + n)*Tan[(e + f*x)/2]^2*(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e + f*x
]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + n*AppellF1[3/2, n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*
(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1 + n)*Tan[(e + f*x)/2]^6*(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e + f*
x]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + (54*AppellF1[1/2, n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^
2]*(Sec[(e + f*x)/2]^2)^(1 + n)*Tan[(e + f*x)/2]^3)/(3*AppellF1[1/2, n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2] + 2*n*(AppellF1[3/2, n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[3/2, 1 + n
, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (27*n*AppellF1[1/2, n, -n, 3/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^5)/(3*AppellF1[1/2, n, -n, 3/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*n*(AppellF1[3/2, n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x
)/2]^2] + AppellF1[3/2, 1 + n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (27*(S
ec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^4*((n*AppellF1[3/2, n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]
^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3 + (n*AppellF1[3/2, 1 + n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3))/(3*AppellF1[1/2, n, -n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f
*x)/2]^2] + 2*n*(AppellF1[3/2, n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[3/2, 1 + n,
-n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) - (27*AppellF1[1/2, n, -n, 3/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^4*(2*n*(AppellF1[3/2, n, 1 - n, 5/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[3/2, 1 + n, -n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2
]^2])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2] + 3*(...
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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \left (\csc ^{4}\left (f x +e \right )\right ) \left (b \sec \left (f x +e \right )\right )^{n}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)^4*(b*sec(f*x+e))^n,x)
[Out]
int(csc(f*x+e)^4*(b*sec(f*x+e))^n,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(csc(f*x+e)^4*(b*sec(f*x+e))^n,x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e))^n*csc(f*x + e)^4, 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(csc(f*x+e)^4*(b*sec(f*x+e))^n,x, algorithm="fricas")
[Out]
integral((b*sec(f*x + e))^n*csc(f*x + e)^4, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \sec {\left (e + f x \right )}\right )^{n} \csc ^{4}{\left (e + f x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)**4*(b*sec(f*x+e))**n,x)
[Out]
Integral((b*sec(e + f*x))**n*csc(e + f*x)**4, 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(csc(f*x+e)^4*(b*sec(f*x+e))^n,x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e))^n*csc(f*x + e)^4, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n}{{\sin \left (e+f\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b/cos(e + f*x))^n/sin(e + f*x)^4,x)
[Out]
int((b/cos(e + f*x))^n/sin(e + f*x)^4, x)
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