Integrand size = 29, antiderivative size = 295 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=-\frac {\left (6 a^2 b^2 \left (1-n^2\right )-a^4 \left (3-4 n+n^2\right )-b^4 \left (3+4 n+n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{8 d (1+n)}-\frac {a b n \left (a^2 (2-n)-b^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{2 d (2+n)}+\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d} \]
-1/8*(6*a^2*b^2*(-n^2+1)-a^4*(n^2-4*n+3)-b^4*(n^2+4*n+3))*hypergeom([1, 1/ 2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(1+n)/d/(1+n)-1/2*a*b*n*(a^2 *(2-n)-b^2*(2+n))*hypergeom([1, 1+1/2*n],[1/2*n+2],sin(d*x+c)^2)*sin(d*x+c )^(2+n)/d/(2+n)+1/4*sec(d*x+c)^4*sin(d*x+c)^(1+n)*(a^4+6*a^2*b^2+b^4+4*a*b *(a^2+b^2)*sin(d*x+c))/d+1/8*sec(d*x+c)^2*sin(d*x+c)^(1+n)*(a^4*(3-n)-6*a^ 2*b^2*(1+n)-b^4*(5+n)+4*a*b*(a^2*(2-n)-b^2*(2+n))*sin(d*x+c))/d
Time = 0.15 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.56 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\frac {\left (6 \left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+(a-b)^3 (3 a+5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x))+(3 a-5 b) (a+b)^3 \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x))+2 (a-b)^4 \operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x))+2 (a+b)^4 \operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x))\right ) \sin ^{1+n}(c+d x)}{16 d (1+n)} \]
((6*(a^2 - b^2)^2*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^ 2] + (a - b)^3*(3*a + 5*b)*Hypergeometric2F1[2, 1 + n, 2 + n, -Sin[c + d*x ]] + (3*a - 5*b)*(a + b)^3*Hypergeometric2F1[2, 1 + n, 2 + n, Sin[c + d*x] ] + 2*(a - b)^4*Hypergeometric2F1[3, 1 + n, 2 + n, -Sin[c + d*x]] + 2*(a + b)^4*Hypergeometric2F1[3, 1 + n, 2 + n, Sin[c + d*x]])*Sin[c + d*x]^(1 + n))/(16*d*(1 + n))
Time = 0.73 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3316, 558, 25, 2337, 557, 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.
| \(\displaystyle \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^n (a+b \sin (c+d x))^4}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^n(c+d x) (a+b \sin (c+d x))^4}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 558 |
| \(\displaystyle \frac {b^5 \left (\frac {\sin ^{n+1}(c+d x) \left (a^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+b^4\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {\sin ^n(c+d x) \left ((3-n) a^4-6 b^2 (n+1) a^2+4 b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x) a-4 b^4 \sin ^2(c+d x)-b^4 (n+1)\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {\int \frac {\sin ^n(c+d x) \left ((3-n) a^4-6 b^2 (n+1) a^2+4 b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x) a-4 b^4 \sin ^2(c+d x)-b^4 (n+1)\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {\sin ^{n+1}(c+d x) \left (a^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+b^4\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2337 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\sin ^{n+1}(c+d x) \left (a^4 (3-n)+4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)-b^4 (n+5)\right )}{2 b \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {\sin ^n(c+d x) \left (-\left (\left (n^2-4 n+3\right ) a^4\right )+6 b^2 \left (1-n^2\right ) a^2+4 b n \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x) a-b^4 \left (n^2+4 n+3\right )\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}+\frac {\sin ^{n+1}(c+d x) \left (a^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+b^4\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 557 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\sin ^{n+1}(c+d x) \left (a^4 (3-n)+4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)-b^4 (n+5)\right )}{2 b \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {4 a b n \left (a^2 (2-n)-b^2 (n+2)\right ) \int \frac {\sin ^{n+1}(c+d x)}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))+\left (-\left (a^4 \left (n^2-4 n+3\right )\right )+6 a^2 b^2 \left (1-n^2\right )-b^4 \left (n^2+4 n+3\right )\right ) \int \frac {\sin ^n(c+d x)}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}+\frac {\sin ^{n+1}(c+d x) \left (a^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+b^4\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\sin ^{n+1}(c+d x) \left (a^4 (3-n)+4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)-b^4 (n+5)\right )}{2 b \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\frac {4 a n \left (a^2 (2-n)-b^2 (n+2)\right ) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{n+2}+\frac {\left (-\left (a^4 \left (n^2-4 n+3\right )\right )+6 a^2 b^2 \left (1-n^2\right )-b^4 \left (n^2+4 n+3\right )\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{b (n+1)}}{2 b^2}}{4 b^2}+\frac {\sin ^{n+1}(c+d x) \left (a^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+b^4\right )}{4 b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
(b^5*((Sin[c + d*x]^(1 + n)*(a^4 + 6*a^2*b^2 + b^4 + 4*a*b*(a^2 + b^2)*Sin [c + d*x]))/(4*b*(b^2 - b^2*Sin[c + d*x]^2)^2) + ((Sin[c + d*x]^(1 + n)*(a ^4*(3 - n) - 6*a^2*b^2*(1 + n) - b^4*(5 + n) + 4*a*b*(a^2*(2 - n) - b^2*(2 + n))*Sin[c + d*x]))/(2*b*(b^2 - b^2*Sin[c + d*x]^2)) - (((6*a^2*b^2*(1 - n^2) - a^4*(3 - 4*n + n^2) - b^4*(3 + 4*n + n^2))*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + n))/(b*(1 + n)) + (4 *a*n*(a^2*(2 - n) - b^2*(2 + n))*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2 , Sin[c + d*x]^2]*Sin[c + d*x]^(2 + n))/(2 + n))/(2*b^2))/(4*b^2)))/d
3.16.9.3.1 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[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* (f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt Q[n, 1] && !IntegerQ[m] && LtQ[p, -1]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq , a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(-(c*x)^(m + 1))*(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))) , x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2 *a*(p + 1)*Q + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a , b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] && !GtQ[m, 0]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
\[\int \left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{4}d x\]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
integral(-(4*(a*b^3*cos(d*x + c)^2 - a^3*b - a*b^3)*sec(d*x + c)^5*sin(d*x + c) - (b^4*cos(d*x + c)^4 + a^4 + 6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)* cos(d*x + c)^2)*sec(d*x + c)^5)*sin(d*x + c)^n, x)
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\text {Timed out} \]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4}{{\cos \left (c+d\,x\right )}^5} \,d x \]