\(\int \sin ^2(e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [159]

Optimal result

Integrand size = 23, antiderivative size = 84 \[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},2,-p,\frac {5}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {a+b \tan ^2(e+f x)}{a}\right )^{-p}}{3 f} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.17 (sec) , antiderivative size = 3698, normalized size of antiderivative = 44.02 \[ \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \] Input:

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

Output:

(3*a*Cos[e + f*x]^3*Sin[e + f*x]*(a + b*Tan[e + f*x]^2)^p*(AppellF1[1/2, 2 
, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]/(-3*a*AppellF1[1/2, 2 
, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*(b*p*AppellF1[3/2 
, 2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*a*AppellF1[ 
3/2, 3, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x]^2 
) + (AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*S 
ec[e + f*x]^2)/(3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Ta 
n[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a 
), -Tan[e + f*x]^2] - a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), 
 -Tan[e + f*x]^2])*Tan[e + f*x]^2))*(-1/4*(Cos[2*(e + f*x)]^3*(a + b*Tan[e 
 + f*x]^2)^p) + (I/4)*Sin[2*(e + f*x)]*(a + b*Tan[e + f*x]^2)^p + (Sin[2*( 
e + f*x)]^2*(a + b*Tan[e + f*x]^2)^p)/2 - (I/4)*Sin[2*(e + f*x)]^3*(a + b* 
Tan[e + f*x]^2)^p + Cos[2*(e + f*x)]^2*((a + b*Tan[e + f*x]^2)^p/2 - (I/4) 
*Sin[2*(e + f*x)]*(a + b*Tan[e + f*x]^2)^p) + Cos[2*(e + f*x)]*(-1/4*(a + 
b*Tan[e + f*x]^2)^p - (Sin[2*(e + f*x)]^2*(a + b*Tan[e + f*x]^2)^p)/4)))/( 
f*(6*a*b*p*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(-1 + p)*(AppellF1[1/2, 2 
, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]/(-3*a*AppellF1[1/2, 2 
, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*(b*p*AppellF1[3/2 
, 2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - 2*a*AppellF1[ 
3/2, 3, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x...
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4146, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^2 \left (a+b \tan (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

(AppellF1[3/2, 2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e 
 + f*x]^3*(a + b*Tan[e + f*x]^2)^p)/(3*f*(1 + (b*Tan[e + f*x]^2)/a)^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 4146

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(cos(f*x + e)^2 - 1)*(b*tan(f*x + e)^2 + a)^p, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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