\(\int (a+b \sec ^2(e+f x))^p \sin (e+f x) \, dx\) [135]

Optimal result

Integrand size = 21, antiderivative size = 69 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin (e+f x) \, dx=-\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {a+b \sec ^2(e+f x)}{a}\right )^{-p}}{f} \] Output:

-cos(f*x+e)*hypergeom([-1/2, -p],[1/2],-b*sec(f*x+e)^2/a)*(a+b*sec(f*x+e)^ 
2)^p/f/(((a+b*sec(f*x+e)^2)/a)^p)
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin (e+f x) \, dx=-\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{f} \] Input:

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

Output:

-((Cos[e + f*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/a)]* 
(a + b*Sec[e + f*x]^2)^p)/(f*(1 + (b*Sec[e + f*x]^2)/a)^p))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4622, 279, 278}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

\(\displaystyle \frac {\int \cos ^2(e+f x) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{f}\)

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

\(\displaystyle -\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )}{f}\)

Input:

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

Output:

-((Cos[e + f*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/a)]* 
(a + b*Sec[e + f*x]^2)^p)/(f*(1 + (b*Sec[e + f*x]^2)/a)^p))
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*sec(f*x + e)^2 + a)^p*sin(f*x + e), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 14.84 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin (e+f x) \, dx=\frac {\cos \left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2}-p,-p;\ \frac {3}{2}-p;\ -\frac {a\,{\cos \left (e+f\,x\right )}^2}{b}\right )}{f\,\left (2\,p-1\right )\,{\left (\frac {a\,{\cos \left (e+f\,x\right )}^2}{b}+1\right )}^p} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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