3.184 \(\int \csc ^5(e+f x) (b \tan (e+f x))^n \, dx\)
Optimal. Leaf size=78 \[ -\frac {\cos (e+f x) \sin ^2(e+f x)^{-n/2} (b \tan (e+f x))^n \, _2F_1\left (\frac {1-n}{2},\frac {6-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f (1-n)} \]
[Out]
-cos(f*x+e)*hypergeom([3-1/2*n, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*tan(f*x+e))^n/f/(1-n)/((sin(f*x+e)^2)^
(1/2*n))
________________________________________________________________________________________
Rubi [A] time = 0.08, antiderivative size = 78, 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 =
{2601, 2576} \[ -\frac {\cos (e+f x) \sin ^2(e+f x)^{-n/2} (b \tan (e+f x))^n \, _2F_1\left (\frac {1-n}{2},\frac {6-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f (1-n)} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^5*(b*Tan[e + f*x])^n,x]
[Out]
-((Cos[e + f*x]*Hypergeometric2F1[(1 - n)/2, (6 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Tan[e + f*x])^n)/(f*(1 -
n)*(Sin[e + f*x]^2)^(n/2)))
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 2601
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])
Rubi steps
\begin {align*} \int \csc ^5(e+f x) (b \tan (e+f x))^n \, dx &=\left (\cos ^n(e+f x) \sin ^{-n}(e+f x) (b \tan (e+f x))^n\right ) \int \cos ^{-n}(e+f x) \sin ^{-5+n}(e+f x) \, dx\\ &=-\frac {\cos (e+f x) \, _2F_1\left (\frac {1-n}{2},\frac {6-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right ) \sin ^2(e+f x)^{-n/2} (b \tan (e+f x))^n}{f (1-n)}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] time = 17.75, size = 1516, normalized size = 19.44 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Csc[e + f*x]^5*(b*Tan[e + f*x])^n,x]
[Out]
(3*Cot[(e + f*x)/2]^2*Hypergeometric2F1[-1 + n/2, n, n/2, Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2
)^n*(b*Tan[e + f*x])^n)/(16*f*(-2 + n)) + (Cot[(e + f*x)/2]^2*((-2 + n)*Cot[(e + f*x)/2]^2*Hypergeometric2F1[-
2 + n/2, n, -1 + n/2, Tan[(e + f*x)/2]^2] + (-4 + n)*Hypergeometric2F1[-1 + n/2, n, n/2, Tan[(e + f*x)/2]^2])*
(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*(b*Tan[e + f*x])^n)/(16*f*(-4 + n)*(-2 + n)) + (3*(4 + n)*AppellF1[1 + n/2
, n, 1, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[e + f*x]^2*(b*Tan[e + f*x])^n)/(16*f*(2 + n)*(2*
(AppellF1[2 + n/2, n, 2, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[2 + n/2, 1 + n, 1, 3 +
n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*(-1 + Cos[e + f*x]) + (4 + n)*AppellF1[1 + n/2, n, 1, 2 + n/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x]))) + (3*Hypergeometric2F1[1 + n/2, n, 2 + n/2, Tan
[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^2*(b*Tan[e + f*x])^n)/(16*f*(2 + n)) + (
(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^2*((4 + n)*Hypergeometric2F1[1 + n/2, n, 2 + n/2, Tan[(e
+ f*x)/2]^2] + (2 + n)*Hypergeometric2F1[2 + n/2, n, 3 + n/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*(b*Tan[e
+ f*x])^n)/(16*f*(2 + n)*(4 + n)) + (9*Cot[(e + f*x)/2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*((2 + n)*Hypergeo
metric2F1[n/2, n, 1 + n/2, Tan[(e + f*x)/2]^2] - n*AppellF1[1 + n/2, n, 1, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(
e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*Tan[e + f*x]^n*(b*Tan[e + f*x])^n)/(128*f*n*(2 + n)*((3*(Cos[e + f*x]*Sec[(
e + f*x)/2]^2)^n*Sec[e + f*x]^2*((2 + n)*Hypergeometric2F1[n/2, n, 1 + n/2, Tan[(e + f*x)/2]^2] - n*AppellF1[1
+ n/2, n, 1, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*Tan[e + f*x]^(-1 + n))/(8*
(2 + n)) + (3*(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])*((2 + n)*Hypergeometric2F1[n/2, n, 1 + n/2, Tan[(e + f*x)/2]^2] - n*AppellF
1[1 + n/2, n, 1, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*Tan[e + f*x]^n)/(8*(2 +
n)) + (3*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*(-(n*AppellF1[1 + n/2, n, 1, 2 + n/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*Tan[(e + f*x)/2]^2*(-(((1 + n/2)*AppellF1[2 + n/2, n,
2, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(2 + n/2)) + ((1 + n
/2)*n*AppellF1[2 + n/2, 1 + n, 1, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e
+ f*x)/2])/(2 + n/2)) + (n*(2 + n)*Csc[(e + f*x)/2]*Sec[(e + f*x)/2]*(-Hypergeometric2F1[n/2, n, 1 + n/2, Tan[
(e + f*x)/2]^2] + (1 - Tan[(e + f*x)/2]^2)^(-n)))/2)*Tan[e + f*x]^n)/(8*n*(2 + n))))
________________________________________________________________________________________
fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{5}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^5*(b*tan(f*x+e))^n,x, algorithm="fricas")
[Out]
integral((b*tan(f*x + e))^n*csc(f*x + e)^5, x)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^5*(b*tan(f*x+e))^n,x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e))^n*csc(f*x + e)^5, x)
________________________________________________________________________________________
maple [F] time = 0.81, size = 0, normalized size = 0.00 \[ \int \left (\csc ^{5}\left (f x +e \right )\right ) \left (b \tan \left (f x +e \right )\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)^5*(b*tan(f*x+e))^n,x)
[Out]
int(csc(f*x+e)^5*(b*tan(f*x+e))^n,x)
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^5*(b*tan(f*x+e))^n,x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e))^n*csc(f*x + e)^5, x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{{\sin \left (e+f\,x\right )}^5} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*tan(e + f*x))^n/sin(e + f*x)^5,x)
[Out]
int((b*tan(e + f*x))^n/sin(e + f*x)^5, x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan {\left (e + f x \right )}\right )^{n} \csc ^{5}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)**5*(b*tan(f*x+e))**n,x)
[Out]
Integral((b*tan(e + f*x))**n*csc(e + f*x)**5, x)
________________________________________________________________________________________