Integrand size = 23, antiderivative size = 266 \[ \int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx=-\frac {b \left (b^2 (1-n)+3 a^2 (2-n)\right ) (d \cos (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{f (2-n) n \sqrt {\sin ^2(e+f x)}}-\frac {a \left (a^2 (1-n)-3 b^2 n\right ) \cos (e+f x) (d \cos (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{f (1-n) (1+n) \sqrt {\sin ^2(e+f x)}}+\frac {a b^2 (5-2 n) (d \cos (e+f x))^n \tan (e+f x)}{f (1-n) (2-n)}+\frac {b^2 (d \cos (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2-n)} \]
-b*(b^2*(1-n)+3*a^2*(2-n))*(d*cos(f*x+e))^n*hypergeom([1/2, 1/2*n],[1+1/2* n],cos(f*x+e)^2)*sin(f*x+e)/f/(2-n)/n/(sin(f*x+e)^2)^(1/2)-a*(a^2*(1-n)-3* b^2*n)*cos(f*x+e)*(d*cos(f*x+e))^n*hypergeom([1/2, 1/2+1/2*n],[3/2+1/2*n], cos(f*x+e)^2)*sin(f*x+e)/f/(-n^2+1)/(sin(f*x+e)^2)^(1/2)+a*b^2*(5-2*n)*(d* cos(f*x+e))^n*tan(f*x+e)/f/(n^2-3*n+2)+b^2*(d*cos(f*x+e))^n*(a+b*sec(f*x+e ))*tan(f*x+e)/f/(2-n)
Time = 0.65 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.83 \[ \int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx=-\frac {(d \cos (e+f x))^n \csc (e+f x) \left (b^3 n \left (-1+n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\cos ^2(e+f x)\right )+\frac {1}{2} a (-2+n) \cos (e+f x) \left (6 b^2 n (1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(e+f x)\right )+2 a (-1+n) \cos (e+f x) \left (3 b (1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(e+f x)\right )+a n \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(e+f x)\right )\right )\right )\right ) \sec ^2(e+f x) \sqrt {\sin ^2(e+f x)}}{f (-2+n) (-1+n) n (1+n)} \]
-(((d*Cos[e + f*x])^n*Csc[e + f*x]*(b^3*n*(-1 + n^2)*Hypergeometric2F1[1/2 , (-2 + n)/2, n/2, Cos[e + f*x]^2] + (a*(-2 + n)*Cos[e + f*x]*(6*b^2*n*(1 + n)*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[e + f*x]^2] + 2*a*( -1 + n)*Cos[e + f*x]*(3*b*(1 + n)*Hypergeometric2F1[1/2, n/2, (2 + n)/2, C os[e + f*x]^2] + a*n*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n )/2, Cos[e + f*x]^2])))/2)*Sec[e + f*x]^2*Sqrt[Sin[e + f*x]^2])/(f*(-2 + n )*(-1 + n)*n*(1 + n)))
Time = 1.64 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.12, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4752, 3042, 4329, 3042, 4535, 3042, 4259, 3042, 3122, 4534, 3042, 4259, 3042, 3122}
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.
| \(\displaystyle \int (a+b \sec (e+f x))^3 (d \cos (e+f x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3 \left (d \sin \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
| \(\Big \downarrow \) 4752 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \int (d \sec (e+f x))^{-n} (a+b \sec (e+f x))^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{-n} \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3dx\) |
| \(\Big \downarrow \) 4329 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\int (d \sec (e+f x))^{-n} \left (a b^2 d (5-2 n) \sec ^2(e+f x)+b d \left (3 (2-n) a^2+b^2 (1-n)\right ) \sec (e+f x)+a d \left (a^2 (2-n)-b^2 n\right )\right )dx}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{-n} \left (a b^2 d (5-2 n) \csc \left (e+f x+\frac {\pi }{2}\right )^2+b d \left (3 (2-n) a^2+b^2 (1-n)\right ) \csc \left (e+f x+\frac {\pi }{2}\right )+a d \left (a^2 (2-n)-b^2 n\right )\right )dx}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\int (d \sec (e+f x))^{-n} \left (a b^2 d (5-2 n) \sec ^2(e+f x)+a d \left (a^2 (2-n)-b^2 n\right )\right )dx+b \left (3 a^2 (2-n)+b^2 (1-n)\right ) \int (d \sec (e+f x))^{1-n}dx}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {b \left (3 a^2 (2-n)+b^2 (1-n)\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{1-n}dx+\int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{-n} \left (a b^2 d (5-2 n) \csc \left (e+f x+\frac {\pi }{2}\right )^2+a d \left (a^2 (2-n)-b^2 n\right )\right )dx}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{-n} \left (a b^2 d (5-2 n) \csc \left (e+f x+\frac {\pi }{2}\right )^2+a d \left (a^2 (2-n)-b^2 n\right )\right )dx+b \left (3 a^2 (2-n)+b^2 (1-n)\right ) \left (\frac {\cos (e+f x)}{d}\right )^{-n} (d \sec (e+f x))^{-n} \int \left (\frac {\cos (e+f x)}{d}\right )^{n-1}dx}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{-n} \left (a b^2 d (5-2 n) \csc \left (e+f x+\frac {\pi }{2}\right )^2+a d \left (a^2 (2-n)-b^2 n\right )\right )dx+b \left (3 a^2 (2-n)+b^2 (1-n)\right ) \left (\frac {\cos (e+f x)}{d}\right )^{-n} (d \sec (e+f x))^{-n} \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}\right )^{n-1}dx}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{-n} \left (a b^2 d (5-2 n) \csc \left (e+f x+\frac {\pi }{2}\right )^2+a d \left (a^2 (2-n)-b^2 n\right )\right )dx-\frac {b d \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \sec (e+f x))^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\frac {a d (2-n) \left (a^2 (1-n)-3 b^2 n\right ) \int (d \sec (e+f x))^{-n}dx}{1-n}-\frac {b d \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \sec (e+f x))^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {a b^2 d (5-2 n) \tan (e+f x) (d \sec (e+f x))^{-n}}{f (1-n)}}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\frac {a d (2-n) \left (a^2 (1-n)-3 b^2 n\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{-n}dx}{1-n}-\frac {b d \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \sec (e+f x))^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {a b^2 d (5-2 n) \tan (e+f x) (d \sec (e+f x))^{-n}}{f (1-n)}}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\frac {a d (2-n) \left (a^2 (1-n)-3 b^2 n\right ) \left (\frac {\cos (e+f x)}{d}\right )^{-n} (d \sec (e+f x))^{-n} \int \left (\frac {\cos (e+f x)}{d}\right )^ndx}{1-n}-\frac {b d \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \sec (e+f x))^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {a b^2 d (5-2 n) \tan (e+f x) (d \sec (e+f x))^{-n}}{f (1-n)}}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {\frac {a d (2-n) \left (a^2 (1-n)-3 b^2 n\right ) \left (\frac {\cos (e+f x)}{d}\right )^{-n} (d \sec (e+f x))^{-n} \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}\right )^ndx}{1-n}-\frac {b d \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \sec (e+f x))^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {a b^2 d (5-2 n) \tan (e+f x) (d \sec (e+f x))^{-n}}{f (1-n)}}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle (d \cos (e+f x))^n (d \sec (e+f x))^n \left (\frac {-\frac {a d^2 (2-n) \left (a^2 (1-n)-3 b^2 n\right ) \sin (e+f x) (d \sec (e+f x))^{-n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\cos ^2(e+f x)\right )}{f (1-n) (n+1) \sqrt {\sin ^2(e+f x)}}-\frac {b d \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \sec (e+f x))^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {n+2}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {a b^2 d (5-2 n) \tan (e+f x) (d \sec (e+f x))^{-n}}{f (1-n)}}{d (2-n)}+\frac {b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^{-n}}{f (2-n)}\right )\) |
(d*Cos[e + f*x])^n*(d*Sec[e + f*x])^n*((b^2*(a + b*Sec[e + f*x])*Tan[e + f *x])/(f*(2 - n)*(d*Sec[e + f*x])^n) + (-((a*d^2*(2 - n)*(a^2*(1 - n) - 3*b ^2*n)*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[e + f*x]^2]*(d*Sec[ e + f*x])^(-1 - n)*Sin[e + f*x])/(f*(1 - n)*(1 + n)*Sqrt[Sin[e + f*x]^2])) - (b*d*(b^2*(1 - n) + 3*a^2*(2 - n))*Hypergeometric2F1[1/2, n/2, (2 + n)/ 2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*n*(d*Sec[e + f*x])^n*Sqrt[Sin[e + f*x] ^2]) + (a*b^2*d*(5 - 2*n)*Tan[e + f*x])/(f*(1 - n)*(d*Sec[e + f*x])^n))/(d *(2 - n)))
3.9.75.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[ (a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a* b^2*d*n + b*(b^2*d*(m + n - 2) + 3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2* d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, n}, x ] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && !IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
\[\int \left (d \cos \left (f x +e \right )\right )^{n} \left (a +b \sec \left (f x +e \right )\right )^{3}d x\]
\[ \int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \cos \left (f x + e\right )\right )^{n} \,d x } \]
integral((b^3*sec(f*x + e)^3 + 3*a*b^2*sec(f*x + e)^2 + 3*a^2*b*sec(f*x + e) + a^3)*(d*cos(f*x + e))^n, x)
\[ \int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx=\int \left (d \cos {\left (e + f x \right )}\right )^{n} \left (a + b \sec {\left (e + f x \right )}\right )^{3}\, dx \]
\[ \int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \cos \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \cos \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx=\int {\left (d\,\cos \left (e+f\,x\right )\right )}^n\,{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^3 \,d x \]