\(\int (a+b \sin ^2(e+f x))^p (d \tan (e+f x))^m \, dx\) [474]

Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \left (a+b \sin ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1+m}{2},-p,\frac {3+m}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \cos ^2(e+f x)^{\frac {1+m}{2}} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} (d \tan (e+f x))^{1+m}}{d f (1+m)} \] Output:

AppellF1(1/2+1/2*m,1/2+1/2*m,-p,3/2+1/2*m,sin(f*x+e)^2,-b*sin(f*x+e)^2/a)* 
(cos(f*x+e)^2)^(1/2+1/2*m)*(a+b*sin(f*x+e)^2)^p*(d*tan(f*x+e))^(1+m)/d/f/( 
1+m)/((1+b*sin(f*x+e)^2/a)^p)
 

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01 \[ \int \left (a+b \sin ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1+m}{2},-p,\frac {3+m}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \cos ^2(e+f x)^{\frac {1+m}{2}} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x) (d \tan (e+f x))^m}{f (1+m)} \] Input:

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

Output:

(AppellF1[(1 + m)/2, (1 + m)/2, -p, (3 + m)/2, Sin[e + f*x]^2, -((b*Sin[e 
+ f*x]^2)/a)]*(Cos[e + f*x]^2)^((1 + m)/2)*(a + b*Sin[e + f*x]^2)^p*Tan[e 
+ f*x]*(d*Tan[e + f*x])^m)/(f*(1 + m)*(1 + (b*Sin[e + f*x]^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3676, 393, 152, 150}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3676

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

\(\Big \downarrow \) 393

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

(AppellF1[(1 + m)/2, (1 + m)/2, -p, (3 + m)/2, Sin[e + f*x]^2, -((b*Sin[e 
+ f*x]^2)/a)]*(Cos[e + f*x]^2)^((1 + m)/2)*(Sin[e + f*x]^2)^(-1 + (1 - m)/ 
2 + (1 + m)/2)*(a + b*Sin[e + f*x]^2)^p*(d*Tan[e + f*x])^(1 + m))/(d*f*(1 
+ m)*(1 + (b*Sin[e + f*x]^2)/a)^p)
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 152

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) 
Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]
 

rule 393

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

rule 3042

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

rule 3676

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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