\(\int \csc (e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [157]

Optimal result

Integrand size = 21, antiderivative size = 92 \[ \int \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a-b}\right ) \sec (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^p \left (\frac {a-b+b \sec ^2(e+f x)}{a-b}\right )^{-p}}{f} \] Output:

-AppellF1(1/2,1,-p,3/2,sec(f*x+e)^2,-b*sec(f*x+e)^2/(a-b))*sec(f*x+e)*(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. \(1215\) vs. \(2(92)=184\).

Time = 14.20 (sec) , antiderivative size = 1215, normalized size of antiderivative = 13.21 \[ \int \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx =\text {Too large to display} \] Input:

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

Output:

(Csc[e + f*x]*(a + b*Tan[e + f*x]^2)^(2*p)*((2*AppellF1[-1/2 - p, -1/2, -p 
, 1/2 - p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^2)/b)]*Sqrt[Sec[e + f*x]^2]) 
/((1 + 2*p)*(1 + (a*Cot[e + f*x]^2)/b)^p*Sqrt[Csc[e + f*x]^2]) - (AppellF1 
[1, 1/2, -p, 2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]^2)/ 
(1 + (b*Tan[e + f*x]^2)/a)^p))/(2*f*(b*p*Sec[e + f*x]^2*Tan[e + f*x]*(a + 
b*Tan[e + f*x]^2)^(-1 + p)*((2*AppellF1[-1/2 - p, -1/2, -p, 1/2 - p, -Cot[ 
e + f*x]^2, -((a*Cot[e + f*x]^2)/b)]*Sqrt[Sec[e + f*x]^2])/((1 + 2*p)*(1 + 
 (a*Cot[e + f*x]^2)/b)^p*Sqrt[Csc[e + f*x]^2]) - (AppellF1[1, 1/2, -p, 2, 
-Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]^2)/(1 + (b*Tan[e + 
f*x]^2)/a)^p) + ((a + b*Tan[e + f*x]^2)^p*((2*AppellF1[-1/2 - p, -1/2, -p, 
 1/2 - p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^2)/b)]*Cot[e + f*x]*Sqrt[Sec[ 
e + f*x]^2])/((1 + 2*p)*(1 + (a*Cot[e + f*x]^2)/b)^p*Sqrt[Csc[e + f*x]^2]) 
 + (4*a*p*AppellF1[-1/2 - p, -1/2, -p, 1/2 - p, -Cot[e + f*x]^2, -((a*Cot[ 
e + f*x]^2)/b)]*Cot[e + f*x]*(1 + (a*Cot[e + f*x]^2)/b)^(-1 - p)*Sqrt[Csc[ 
e + f*x]^2]*Sqrt[Sec[e + f*x]^2])/(b*(1 + 2*p)) + (2*((-2*a*(-1/2 - p)*p*A 
ppellF1[1/2 - p, -1/2, 1 - p, 3/2 - p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^ 
2)/b)]*Cot[e + f*x]*Csc[e + f*x]^2)/(b*(1/2 - p)) - ((-1/2 - p)*AppellF1[1 
/2 - p, 1/2, -p, 3/2 - p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^2)/b)]*Cot[e 
+ f*x]*Csc[e + f*x]^2)/(1/2 - p))*Sqrt[Sec[e + f*x]^2])/((1 + 2*p)*(1 + (a 
*Cot[e + f*x]^2)/b)^p*Sqrt[Csc[e + f*x]^2]) + (2*AppellF1[-1/2 - p, -1/...
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4147, 25, 334, 333}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

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

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

Input:

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

Output:

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

Fricas [F]

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

integrate(csc(f*x+e)*(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), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(csc(f*x+e)*(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), x)
 

Giac [F]

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

integrate(csc(f*x+e)*(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), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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