\(\int (d \sin (e+f x))^m (a+b \tan ^2(e+f x))^p \, dx\) [153]

Optimal result

Integrand size = 25, antiderivative size = 122 \[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {a+b \tan ^2(e+f x)}{a}\right )^{-p}}{f (1+m)} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(275\) vs. \(2(122)=244\).

Time = 2.46 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.25 \[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {a (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \cos (e+f x) \sin (e+f x) (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p}{f (1+m) \left (a (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )+\left (2 b p \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {2+m}{2},1-p,\frac {5+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )-a (2+m) \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {4+m}{2},-p,\frac {5+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )\right ) \tan ^2(e+f x)\right )} \] Input:

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

Output:

(a*(3 + m)*AppellF1[(1 + m)/2, (2 + m)/2, -p, (3 + m)/2, -Tan[e + f*x]^2, 
-((b*Tan[e + f*x]^2)/a)]*Cos[e + f*x]*Sin[e + f*x]*(d*Sin[e + f*x])^m*(a + 
 b*Tan[e + f*x]^2)^p)/(f*(1 + m)*(a*(3 + m)*AppellF1[(1 + m)/2, (2 + m)/2, 
 -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] + (2*b*p*AppellF 
1[(3 + m)/2, (2 + m)/2, 1 - p, (5 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f* 
x]^2)/a)] - a*(2 + m)*AppellF1[(3 + m)/2, (4 + m)/2, -p, (5 + m)/2, -Tan[e 
 + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x]^2))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4150, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4150

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

\(\Big \downarrow \) 393

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

(AppellF1[(1 + m)/2, (2 + m)/2, -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e 
 + f*x]^2)/a)]*Cot[e + f*x]*(Sec[e + f*x]^2)^(m/2)*(d*Sin[e + f*x])^m*(Tan 
[e + f*x]^2)^((1 - m)/2 + (1 + m)/2)*(a + b*Tan[e + f*x]^2)^p)/(f*(1 + m)* 
(1 + (b*Tan[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 4150

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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