\(\int \csc ^3(e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [173]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 92 \[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=-\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \csc ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2+n p),\frac {1}{2} (1+n p),\frac {n p}{2},\sin ^2(e+f x)\right ) \sec (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 13.65 (sec) , antiderivative size = 1399, normalized size of antiderivative = 15.21 \[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx =\text {Too large to display} \] Input:

Integrate[Csc[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
 

Output:

(Cot[(e + f*x)/2]^2*Hypergeometric2F1[n*p, -1 + (n*p)/2, (n*p)/2, Tan[(e + 
 f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n*p)*(b*(c*Tan[e + f*x])^n) 
^p)/(f*(-8 + 4*n*p)) + ((4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n*p)/ 
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[e + f*x]^2*(b*(c*Tan[e + f 
*x])^n)^p)/(8*f*(2 + n*p)*((4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n* 
p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + Appell 
F1[2 + (n*p)/2, n*p, 2, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2] 
^2]*(-1 + Cos[e + f*x]) + 2*n*p*AppellF1[2 + (n*p)/2, 1 + n*p, 1, 3 + (n*p 
)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[(e + f*x)/2]^2)) + (Hype 
rgeometric2F1[n*p, 1 + (n*p)/2, 2 + (n*p)/2, Tan[(e + f*x)/2]^2]*(Cos[e + 
f*x]*Sec[(e + f*x)/2]^2)^(n*p)*Tan[(e + f*x)/2]^2*(b*(c*Tan[e + f*x])^n)^p 
)/(f*(8 + 4*n*p)) + (Cot[(e + f*x)/2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(n 
*p)*((2 + n*p)*Hypergeometric2F1[(n*p)/2, n*p, 1 + (n*p)/2, Tan[(e + f*x)/ 
2]^2] - n*p*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n*p)/2, Tan[(e + f*x)/2]^2, 
 -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*Tan[e + f*x]^(n*p)*(b*(c*Tan[e + 
 f*x])^n)^p)/(8*f*n*p*(2 + n*p)*(((Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1 + 
n*p)*(-(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])*((2 + n*p)*Hypergeometric2F1[(n*p)/2, n*p, 1 + (n*p)/2, 
 Tan[(e + f*x)/2]^2] - n*p*AppellF1[1 + (n*p)/2, n*p, 1, 2 + (n*p)/2, Tan[ 
(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*Tan[e + f*x]^(...
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4142, 3042, 3081, 3042, 3057}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b (c \tan (e+f x))^n\right )^p}{\sin (e+f x)^3}dx\)

\(\Big \downarrow \) 4142

\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \csc ^3(e+f x) (c \tan (e+f x))^{n p}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \frac {(c \tan (e+f x))^{n p}}{\sin (e+f x)^3}dx\)

\(\Big \downarrow \) 3081

\(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos ^{-n p}(e+f x) \sin ^{n p-3}(e+f x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos (e+f x)^{-n p} \sin (e+f x)^{n p-3}dx\)

\(\Big \downarrow \) 3057

\(\displaystyle -\frac {\csc ^2(e+f x) \sec (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (n p-2),\frac {1}{2} (n p+1),\frac {n p}{2},\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2-n p)}\)

Input:

Int[Csc[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p,x]
 

Output:

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

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3057
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*Frac 
Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr 
acPart[(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 3081
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[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])
 

rule 4142
Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> S 
imp[b^IntPart[p]*((b*(c*Tan[e + f*x])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*Fr 
acPart[p]))   Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{ 
b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || Ma 
tchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, 
 cos, tan, cot, sec, csc}, trig]])
 
Maple [F]

\[\int \csc \left (f x +e \right )^{3} \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]

Input:

int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
 

Output:

int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
 

Fricas [F]

\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:

integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")
 

Output:

integral(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
 

Sympy [F]

\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \csc ^{3}{\left (e + f x \right )}\, dx \] Input:

integrate(csc(f*x+e)**3*(b*(c*tan(f*x+e))**n)**p,x)
 

Output:

Integral((b*(c*tan(e + f*x))**n)**p*csc(e + f*x)**3, x)
 

Maxima [F]

\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:

integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")
 

Output:

integrate(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
 

Giac [F]

\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:

integrate(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")
 

Output:

integrate(((c*tan(f*x + e))^n*b)^p*csc(f*x + e)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \frac {{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\sin \left (e+f\,x\right )}^3} \,d x \] Input:

int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^3,x)
 

Output:

int((b*(c*tan(e + f*x))^n)^p/sin(e + f*x)^3, x)
 

Reduce [F]

\[ \int \csc ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=c^{n p} b^{p} \left (\int \tan \left (f x +e \right )^{n p} \csc \left (f x +e \right )^{3}d x \right ) \] Input:

int(csc(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p,x)
 

Output:

c**(n*p)*b**p*int(tan(e + f*x)**(n*p)*csc(e + f*x)**3,x)