Integrand size = 31, antiderivative size = 104 \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a^2 (A (4-m)-B m) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-2+m}}{16 f (2-m)}+\frac {a^4 (A+B) (a+a \sin (e+f x))^{-2+m}}{4 f (a-a \sin (e+f x))^2} \]
-1/16*a^2*(A*(4-m)-B*m)*hypergeom([2, -2+m],[-1+m],1/2+1/2*sin(f*x+e))*(a+ a*sin(f*x+e))^(-2+m)/f/(2-m)+1/4*a^4*(A+B)*(a+a*sin(f*x+e))^(-2+m)/f/(a-a* sin(f*x+e))^2
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.73 \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {a^2 \left (-\frac {(A (-4+m)+B m) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,\frac {1}{2} (1+\sin (e+f x))\right )}{-2+m}+\frac {4 (A+B)}{(-1+\sin (e+f x))^2}\right ) (a (1+\sin (e+f x)))^{-2+m}}{16 f} \]
(a^2*(-(((A*(-4 + m) + B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (1 + Sin[ e + f*x])/2])/(-2 + m)) + (4*(A + B))/(-1 + Sin[e + f*x])^2)*(a*(1 + Sin[e + f*x]))^(-2 + m))/(16*f)
Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 87, 78}
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.
\(\displaystyle \int \sec ^5(e+f x) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{\cos (e+f x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {(\sin (e+f x) a+a)^{m-3} (a A+a B \sin (e+f x))}{a (a-a \sin (e+f x))^3}d(a \sin (e+f x))}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^4 \int \frac {(\sin (e+f x) a+a)^{m-3} (a A+a B \sin (e+f x))}{(a-a \sin (e+f x))^3}d(a \sin (e+f x))}{f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {a^4 \left (\frac {1}{4} (A (4-m)-B m) \int \frac {(\sin (e+f x) a+a)^{m-3}}{(a-a \sin (e+f x))^2}d(a \sin (e+f x))+\frac {(A+B) (a \sin (e+f x)+a)^{m-2}}{4 (a-a \sin (e+f x))^2}\right )}{f}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {a^4 \left (\frac {(A+B) (a \sin (e+f x)+a)^{m-2}}{4 (a-a \sin (e+f x))^2}-\frac {(A (4-m)-B m) (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (2,m-2,m-1,\frac {\sin (e+f x) a+a}{2 a}\right )}{16 a^2 (2-m)}\right )}{f}\) |
(a^4*(-1/16*((A*(4 - m) - B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (a + a *Sin[e + f*x])/(2*a)]*(a + a*Sin[e + f*x])^(-2 + m))/(a^2*(2 - m)) + ((A + B)*(a + a*Sin[e + f*x])^(-2 + m))/(4*(a - a*Sin[e + f*x])^2)))/f
3.11.26.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
\[\int \left (\sec ^{5}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]
\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \]
\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^5} \,d x \]