Integrand size = 23, antiderivative size = 117 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}-\frac {(3 a+b-2 b p) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{3 a f} \]
1/3*cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(p+1)/a/f-1/3*(-2*b*p+3*a+b)*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/f/( (1+b*sec(f*x+e)^2/a)^p)
Time = 2.40 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.52 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin ^2(e+f x) \left (-2 (3 a+b-2 b p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )+(a+2 b+a \cos (2 (e+f x))) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )}{3 a f \left (-2 \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^p+\left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p+\cos (2 (e+f x)) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )} \]
-1/3*(Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^2*(-2*(3*a + b - 2*b*p)*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/a)] + (a + 2* b + a*Cos[2*(e + f*x)])*((a + b + b*Tan[e + f*x]^2)/a)^p))/(a*f*(-2*(1 + ( b*Sec[e + f*x]^2)/a)^p + ((a + b + b*Tan[e + f*x]^2)/a)^p + Cos[2*(e + f*x )]*((a + b + b*Tan[e + f*x]^2)/a)^p))
Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4622, 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.
\(\displaystyle \int \sin ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int -\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\frac {(3 a-2 b p+b) \int \cos ^2(e+f x) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{3 a}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a}}{f}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\frac {(3 a-2 b p+b) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \int \cos ^2(e+f x) \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^pd\sec (e+f x)}{3 a}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a}}{f}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a}-\frac {(3 a-2 b p+b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )}{3 a}}{f}\) |
((Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(1 + p))/(3*a) - ((3*a + b - 2*b*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))/f
3.2.34.3.1 Defintions of rubi rules used
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])
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])
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]
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])
\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \sin \left (f x +e \right )^{3}d x\]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\text {Timed out} \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]