Integrand size = 21, antiderivative size = 215 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {(4+n) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n)}-\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}-\frac {2^{\frac {1}{2}+n} n \left (5+3 n+n^2\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-n} (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n)}-\frac {n \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d \left (6+5 n+n^2\right )} \]
-(4+n)*cos(d*x+c)*(a+a*sin(d*x+c))^n/d/(n^3+6*n^2+11*n+6)-cos(d*x+c)*sin(d *x+c)^2*(a+a*sin(d*x+c))^n/d/(3+n)-2^(1/2+n)*n*(n^2+3*n+5)*cos(d*x+c)*hype rgeom([1/2, 1/2-n],[3/2],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(-1/2-n)*(a+a* sin(d*x+c))^n/d/(n^3+6*n^2+11*n+6)-n*cos(d*x+c)*(a+a*sin(d*x+c))^(1+n)/a/d /(n^2+5*n+6)
Leaf count is larger than twice the leaf count of optimal. \(2131\) vs. \(2(215)=430\).
Time = 23.74 (sec) , antiderivative size = 2131, normalized size of antiderivative = 9.91 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\text {Result too large to show} \]
(2*(Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Ta n[c + d*x])^2] - 6*Hypergeometric2F1[1/2 + n, 2 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + 12*Hypergeometric2F1[1/2 + n, 3 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*Hypergeometric2F1[1/2 + n, 4 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2])*Sin[c + d*x] ^3*(a + a*Sin[c + d*x])^n*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)^2*(1 + (Sqrt[Se c[c + d*x]^2] + Tan[c + d*x])^2)^n)/(d*(1 + 2*n)*(Sqrt[Sec[c + d*x]^2] + T an[c + d*x])^3*((4*n*(Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Se c[c + d*x]^2] + Tan[c + d*x])^2] - 6*Hypergeometric2F1[1/2 + n, 2 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + 12*Hypergeometric2F1[1/2 + n, 3 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*Hyperg eometric2F1[1/2 + n, 4 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x] )^2])*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt [Sec[c + d*x]^2]*Tan[c + d*x])*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)^2*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-1 + n) )/((1 + 2*n)*(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2) - (6*(Hypergeometric 2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 6 *Hypergeometric2F1[1/2 + n, 2 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + 12*Hypergeometric2F1[1/2 + n, 3 + n, 3/2 + n, -(Sqrt[Sec[...
Time = 0.89 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3262, 3042, 3447, 3042, 3502, 3042, 3230, 3042, 3131, 3042, 3130}
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.
| \(\displaystyle \int \sin ^3(c+d x) (a \sin (c+d x)+a)^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^3 (a \sin (c+d x)+a)^ndx\) |
| \(\Big \downarrow \) 3262 |
| \(\displaystyle \frac {\int \sin (c+d x) (\sin (c+d x) a+a)^n (n \sin (c+d x) a+2 a)dx}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sin (c+d x) (\sin (c+d x) a+a)^n (n \sin (c+d x) a+2 a)dx}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int (\sin (c+d x) a+a)^n \left (a n \sin ^2(c+d x)+2 a \sin (c+d x)\right )dx}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (c+d x) a+a)^n \left (a n \sin (c+d x)^2+2 a \sin (c+d x)\right )dx}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\int (\sin (c+d x) a+a)^n \left (n (n+1) a^2+(n+4) \sin (c+d x) a^2\right )dx}{a (n+2)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int (\sin (c+d x) a+a)^n \left (n (n+1) a^2+(n+4) \sin (c+d x) a^2\right )dx}{a (n+2)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {\frac {a^2 n \left (n^2+3 n+5\right ) \int (\sin (c+d x) a+a)^ndx}{n+1}-\frac {a^2 (n+4) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^2 n \left (n^2+3 n+5\right ) \int (\sin (c+d x) a+a)^ndx}{n+1}-\frac {a^2 (n+4) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3131 |
| \(\displaystyle \frac {\frac {\frac {a^2 n \left (n^2+3 n+5\right ) (\sin (c+d x)+1)^{-n} (a \sin (c+d x)+a)^n \int (\sin (c+d x)+1)^ndx}{n+1}-\frac {a^2 (n+4) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^2 n \left (n^2+3 n+5\right ) (\sin (c+d x)+1)^{-n} (a \sin (c+d x)+a)^n \int (\sin (c+d x)+1)^ndx}{n+1}-\frac {a^2 (n+4) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
| \(\Big \downarrow \) 3130 |
| \(\displaystyle \frac {\frac {-\frac {a^2 2^{n+\frac {1}{2}} n \left (n^2+3 n+5\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1)}-\frac {a^2 (n+4) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}\) |
-((Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^n)/(d*(3 + n))) + (-(( n*Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(d*(2 + n))) + (-((a^2*(4 + n )*Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(1 + n))) - (2^(1/2 + n)*a^2*n*( 5 + 3*n + n^2)*Cos[c + d*x]*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1 - Sin[ c + d*x])/2]*(1 + Sin[c + d*x])^(-1/2 - n)*(a + a*Sin[c + d*x])^n)/(d*(1 + n)))/(a*(2 + n)))/(a*(3 + n))
3.2.41.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] ) Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*(a + b*Sin[e + f*x]) ^m*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[1/(b*(m + n)) In t[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 2)*Simp[d*(a*c*m + b*d*( n - 1)) + b*c^2*(m + n) + d*(a*d*m + b*c*(m + 2*n - 1))*Sin[e + f*x], x], x ], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
\[\int \left (\sin ^{3}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{n}d x\]
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\text {Timed out} \]
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3} \,d x } \]
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int {\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^n \,d x \]