Integrand size = 23, antiderivative size = 239 \[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{4},2,-p,\frac {5}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{f}+\frac {2 \operatorname {AppellF1}\left (\frac {3}{4},2,-p,\frac {7}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f}+\frac {\operatorname {AppellF1}\left (\frac {5}{4},2,-p,\frac {9}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{5 f} \]
AppellF1(1/4,2,-p,5/4,sin(f*x+e)^4,-b*sin(f*x+e)^4/a)*sin(f*x+e)*(a+b*sin( f*x+e)^4)^p/f/((1+b*sin(f*x+e)^4/a)^p)+2/3*AppellF1(3/4,2,-p,7/4,sin(f*x+e )^4,-b*sin(f*x+e)^4/a)*sin(f*x+e)^3*(a+b*sin(f*x+e)^4)^p/f/((1+b*sin(f*x+e )^4/a)^p)+1/5*AppellF1(5/4,2,-p,9/4,sin(f*x+e)^4,-b*sin(f*x+e)^4/a)*sin(f* x+e)^5*(a+b*sin(f*x+e)^4)^p/f/((1+b*sin(f*x+e)^4/a)^p)
\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx \]
Time = 0.41 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3702, 1569, 2009}
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.
\(\displaystyle \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^4\right )^p}{\cos (e+f x)^3}dx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \frac {\int \frac {\left (b \sin ^4(e+f x)+a\right )^p}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 1569 |
\(\displaystyle \frac {\int \left (\frac {\sin ^4(e+f x) \left (b \sin ^4(e+f x)+a\right )^p}{\left (\sin ^4(e+f x)-1\right )^2}+\frac {2 \sin ^2(e+f x) \left (b \sin ^4(e+f x)+a\right )^p}{\left (\sin ^4(e+f x)-1\right )^2}+\frac {\left (b \sin ^4(e+f x)+a\right )^p}{\left (\sin ^4(e+f x)-1\right )^2}\right )d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},2,-p,\frac {5}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )+\frac {1}{5} \sin ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},2,-p,\frac {9}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )+\frac {2}{3} \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},2,-p,\frac {7}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{f}\) |
((AppellF1[1/4, 2, -p, 5/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p)/(1 + (b*Sin[e + f*x]^4)/a)^p + (2*Appell F1[3/4, 2, -p, 7/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]^ 3*(a + b*Sin[e + f*x]^4)^p)/(3*(1 + (b*Sin[e + f*x]^4)/a)^p) + (AppellF1[5 /4, 2, -p, 9/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]^5*(a + b*Sin[e + f*x]^4)^p)/(5*(1 + (b*Sin[e + f*x]^4)/a)^p))/f
3.5.24.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int [ExpandIntegrand[(a + c*x^4)^p, (d/(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4) ))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] && ! IntegerQ[p] && ILtQ[q, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
\[\int \left (\sec ^{3}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{4}\left (f x +e \right )\right )\right )}^{p}d x\]
\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \]
\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^4+a\right )}^p}{{\cos \left (e+f\,x\right )}^3} \,d x \]