\(\int \csc ^3(e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [158]

Optimal result

Integrand size = 23, antiderivative size = 96 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},2,-p,\frac {5}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a-b}\right ) \sec ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^p \left (\frac {a-b+b \sec ^2(e+f x)}{a-b}\right )^{-p}}{3 f} \] Output:

1/3*AppellF1(3/2,2,-p,5/2,sec(f*x+e)^2,-b*sec(f*x+e)^2/(a-b))*sec(f*x+e)^3 
*(a-b+b*sec(f*x+e)^2)^p/f/(((a-b+b*sec(f*x+e)^2)/(a-b))^p)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(252\) vs. \(2(96)=192\).

Time = 15.22 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.62 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {b (-3+2 p) \operatorname {AppellF1}\left (\frac {1}{2}-p,-\frac {1}{2},-p,\frac {3}{2}-p,-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right ) \cot (e+f x) \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^p}{f (-1+2 p) \left (b (-3+2 p) \operatorname {AppellF1}\left (\frac {1}{2}-p,-\frac {1}{2},-p,\frac {3}{2}-p,-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right )-\left (2 a p \operatorname {AppellF1}\left (\frac {3}{2}-p,-\frac {1}{2},1-p,\frac {5}{2}-p,-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right )+b \operatorname {AppellF1}\left (\frac {3}{2}-p,\frac {1}{2},-p,\frac {5}{2}-p,-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right )\right ) \cot ^2(e+f x)\right )} \] Input:

Integrate[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]
 

Output:

(b*(-3 + 2*p)*AppellF1[1/2 - p, -1/2, -p, 3/2 - p, -Cot[e + f*x]^2, -((a*C 
ot[e + f*x]^2)/b)]*Cot[e + f*x]*Csc[e + f*x]*(a + b*Tan[e + f*x]^2)^p)/(f* 
(-1 + 2*p)*(b*(-3 + 2*p)*AppellF1[1/2 - p, -1/2, -p, 3/2 - p, -Cot[e + f*x 
]^2, -((a*Cot[e + f*x]^2)/b)] - (2*a*p*AppellF1[3/2 - p, -1/2, 1 - p, 5/2 
- p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^2)/b)] + b*AppellF1[3/2 - p, 1/2, 
-p, 5/2 - p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^2)/b)])*Cot[e + f*x]^2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4147, 395, 394}

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.

\(\displaystyle \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \tan (e+f x)^2\right )^p}{\sin (e+f x)^3}dx\)

\(\Big \downarrow \) 4147

\(\displaystyle \frac {\int \frac {\sec ^2(e+f x) \left (b \sec ^2(e+f x)+a-b\right )^p}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {\left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac {b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \int \frac {\sec ^2(e+f x) \left (\frac {b \sec ^2(e+f x)}{a-b}+1\right )^p}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 394

\(\displaystyle \frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac {b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},2,-p,\frac {5}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a-b}\right )}{3 f}\)

Input:

Int[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]
 

Output:

(AppellF1[3/2, 2, -p, 5/2, Sec[e + f*x]^2, -((b*Sec[e + f*x]^2)/(a - b))]* 
Sec[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^p)/(3*f*(1 + (b*Sec[e + f*x]^2)/ 
(a - b))^p)
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 3042

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

rule 4147

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ 
m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 
)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( 
m - 1)/2]
 

Maple [F]

\[\int \csc \left (f x +e \right )^{3} \left (a +b \tan \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)
 

Output:

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, x)
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)**3*(a+b*tan(f*x+e)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, x)
 

Giac [F]

\[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3} \,d x } \] Input:

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p}{{\sin \left (e+f\,x\right )}^3} \,d x \] Input:

int((a + b*tan(e + f*x)^2)^p/sin(e + f*x)^3,x)
 

Output:

int((a + b*tan(e + f*x)^2)^p/sin(e + f*x)^3, x)
 

Reduce [F]

\[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int \left (\tan \left (f x +e \right )^{2} b +a \right )^{p} \csc \left (f x +e \right )^{3}d x \] Input:

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)
 

Output:

int((tan(e + f*x)**2*b + a)**p*csc(e + f*x)**3,x)