Integrand size = 23, antiderivative size = 130 \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=-\frac {2 \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {1+\sin (e+f x)}}+\frac {(5+4 n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+n,2+n,\sin (e+f x)\right ) (d \sin (e+f x))^{1+n}}{d f (1+n) (3+2 n) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \]
-2*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(3+2*n)/(1+sin(f*x+e))^(1/2)+(5+4*n )*cos(f*x+e)*hypergeom([1/2, 1+n],[2+n],sin(f*x+e))*(d*sin(f*x+e))^(1+n)/d /f/(1+n)/(3+2*n)/(1-sin(f*x+e))^(1/2)/(1+sin(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 6.35 (sec) , antiderivative size = 5129, normalized size of antiderivative = 39.45 \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\text {Result too large to show} \]
Time = 0.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3242, 27, 2011, 3042, 3255, 74}
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.
\(\displaystyle \int (\sin (e+f x)+1)^{3/2} (d \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (\sin (e+f x)+1)^{3/2} (d \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3242 |
\(\displaystyle \frac {2 \int \frac {(d \sin (e+f x))^n (d (4 n+5)+d \sin (e+f x) (4 n+5))}{2 \sqrt {\sin (e+f x)+1}}dx}{d (2 n+3)}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(d \sin (e+f x))^n (d (4 n+5)+d \sin (e+f x) (4 n+5))}{\sqrt {\sin (e+f x)+1}}dx}{d (2 n+3)}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle \frac {(4 n+5) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x)+1}dx}{2 n+3}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(4 n+5) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x)+1}dx}{2 n+3}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 3255 |
\(\displaystyle \frac {(4 n+5) \cos (e+f x) \int \frac {(d \sin (e+f x))^n}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (2 n+3) \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {(4 n+5) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n+1,n+2,\sin (e+f x)\right )}{d f (n+1) (2 n+3) \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {2 \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {\sin (e+f x)+1}}\) |
(-2*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(3 + 2*n)*Sqrt[1 + Sin[e + f*x]]) + ((5 + 4*n)*Cos[e + f*x]*Hypergeometric2F1[1/2, 1 + n, 2 + n, Sin [e + f*x]]*(d*Sin[e + f*x])^(1 + n))/(d*f*(1 + n)*(3 + 2*n)*Sqrt[1 - Sin[e + f*x]]*Sqrt[1 + Sin[e + f*x]])
3.2.24.3.1 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[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(c + d*x)^n/Sqrt[a - b*x], x] , x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !IntegerQ[2*n]
\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (\sin \left (f x +e \right )+1\right )^{\frac {3}{2}}d x\]
\[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{n} {\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\int \left (d \sin {\left (e + f x \right )}\right )^{n} \left (\sin {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}\, dx \]
\[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{n} {\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{n} {\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (d \sin (e+f x))^n (1+\sin (e+f x))^{3/2} \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,{\left (\sin \left (e+f\,x\right )+1\right )}^{3/2} \,d x \]