\(\int (a+b \sec ^2(e+f x))^p \sin ^5(e+f x) \, dx\) [133]

Optimal result

Integrand size = 23, antiderivative size = 178 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (15 a^2+b (10 a+b (3-2 p)) (1-2 p)\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {a+b \sec ^2(e+f x)}{a}\right )^{-p}}{15 a^2 f} \] Output:

1/15*(10*a+b*(3-2*p))*cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(p+1)/a^2/f-1/5*cos( 
f*x+e)^5*(a+b*sec(f*x+e)^2)^(p+1)/a/f-1/15*(15*a^2+b*(10*a+b*(3-2*p))*(1-2 
*p))*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/a^2/f/(((a+b*sec(f*x+e)^2)/a)^p)
 

Mathematica [A] (warning: unable to verify)

Time = 3.36 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.42 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\frac {2 \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \left (4 \left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )+(a+2 b+a \cos (2 (e+f x))) (-17 a-6 b+4 b p+3 a \cos (2 (e+f x))) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )}{15 a^2 f \left (4 \cos (2 (e+f x)) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p-2^{-p} \left (3 \left (\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x)}{a}\right )^p+2^p \cos (4 (e+f x)) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )\right )} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^5,x]
 

Output:

(2*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^4*(4*(15*a^2 + 10*a* 
b*(1 - 2*p) + b^2*(3 - 8*p + 4*p^2))*Hypergeometric2F1[-1/2, -p, 1/2, -((b 
*Sec[e + f*x]^2)/a)] + (a + 2*b + a*Cos[2*(e + f*x)])*(-17*a - 6*b + 4*b*p 
 + 3*a*Cos[2*(e + f*x)])*((a + b + b*Tan[e + f*x]^2)/a)^p))/(15*a^2*f*(4*C 
os[2*(e + f*x)]*((a + b + b*Tan[e + f*x]^2)/a)^p - (3*(((a + 2*b + a*Cos[2 
*(e + f*x)])*Sec[e + f*x]^2)/a)^p + 2^p*Cos[4*(e + f*x)]*((a + b + b*Tan[e 
 + f*x]^2)/a)^p)/2^p))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4622, 365, 25, 359, 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.

\(\displaystyle \int \sin ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^5 \left (a+b \sec (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 4622

\(\displaystyle \frac {\int \cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{f}\)

\(\Big \downarrow \) 365

\(\displaystyle \frac {\frac {\int -\cos ^4(e+f x) \left (-5 a \sec ^2(e+f x)+10 a+b (3-2 p)\right ) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{5 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \cos ^4(e+f x) \left (-5 a \sec ^2(e+f x)+10 a+b (3-2 p)\right ) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{5 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a}}{f}\)

\(\Big \downarrow \) 359

\(\displaystyle \frac {-\frac {-\frac {\left (15 a^2+b (1-2 p) (10 a+b (3-2 p))\right ) \int \cos ^2(e+f x) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{3 a}-\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a}}{5 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a}}{f}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {-\frac {-\frac {\left (15 a^2+b (1-2 p) (10 a+b (3-2 p))\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \int \cos ^2(e+f x) \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^pd\sec (e+f x)}{3 a}-\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a}}{5 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a}}{f}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {-\frac {\frac {\left (15 a^2+b (1-2 p) (10 a+b (3-2 p))\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )}{3 a}-\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a}}{5 a}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a}}{f}\)

Input:

Int[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^5,x]
 

Output:

(-1/5*(Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^(1 + p))/a - (-1/3*((10*a + b 
*(3 - 2*p))*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(1 + p))/a + ((15*a^2 + 
b*(10*a + b*(3 - 2*p))*(1 - 2*p))*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)/(3*a*(1 + (b*Sec[ 
e + f*x]^2)/a)^p))/(5*a))/f
 

Defintions of rubi rules used

rule 25

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

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 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 365

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x 
_Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] 
- Simp[1/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p 
+ 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, 
 b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
 

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]

\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \sin \left (f x +e \right )^{5}d x\]

Input:

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x)
 

Output:

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x)
 

Fricas [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x, algorithm="fricas")
 

Output:

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(b*sec(f*x + e)^2 + a)^p* 
sin(f*x + e), x)
 

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\text {Timed out} \] Input:

integrate((a+b*sec(f*x+e)**2)**p*sin(f*x+e)**5,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^5, x)
 

Giac [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \] Input:

int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^p,x)
 

Output:

int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^p, x)
 

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int \left (\sec \left (f x +e \right )^{2} b +a \right )^{p} \sin \left (f x +e \right )^{5}d x \] Input:

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x)
 

Output:

int((sec(e + f*x)**2*b + a)**p*sin(e + f*x)**5,x)