Integrand size = 27, antiderivative size = 248 \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx=\frac {54 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {3+3 \sin (e+f x)}}-\frac {18 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}-\frac {54 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{d^2 f (3+2 n) (5+2 n) \sqrt {3+3 \sin (e+f x)}} \]
2*a^3*(3*c-d*(11+4*n))*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)/d^2/f/(3+2*n)/(5+ 2*n)/(a+a*sin(f*x+e))^(1/2)-2*a^3*(3*c^2-2*c*d*(7+4*n)+d^2*(16*n^2+56*n+43 ))*cos(f*x+e)*hypergeom([1/2, -n],[3/2],d*(1-sin(f*x+e))/(c+d))*(c+d*sin(f *x+e))^n/d^2/f/(3+2*n)/(5+2*n)/(((c+d*sin(f*x+e))/(c+d))^n)/(a+a*sin(f*x+e ))^(1/2)-2*a^2*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(1/2)/d/ f/(5+2*n)
Time = 10.33 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.75 \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx=-\frac {9 \sqrt {3} \cos (e+f x) (c+d \sin (e+f x))^n \left (-((3 c-d (11+4 n)) (c+d \sin (e+f x)))+d (3+2 n) (1+\sin (e+f x)) (c+d \sin (e+f x))+\left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d (-1+\sin (e+f x))}{c+d}\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}\right )}{2 d^2 f \left (\frac {3}{2}+n\right ) \left (\frac {5}{2}+n\right ) \sqrt {1+\sin (e+f x)}} \]
(-9*Sqrt[3]*Cos[e + f*x]*(c + d*Sin[e + f*x])^n*(-((3*c - d*(11 + 4*n))*(c + d*Sin[e + f*x])) + d*(3 + 2*n)*(1 + Sin[e + f*x])*(c + d*Sin[e + f*x]) + ((3*c^2 - 2*c*d*(7 + 4*n) + d^2*(43 + 56*n + 16*n^2))*Hypergeometric2F1[ 1/2, -n, 3/2, -((d*(-1 + Sin[e + f*x]))/(c + d))])/((c + d*Sin[e + f*x])/( c + d))^n))/(2*d^2*f*(3/2 + n)*(5/2 + n)*Sqrt[1 + Sin[e + f*x]])
Time = 0.90 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3242, 27, 3042, 3460, 3042, 3255, 80, 79}
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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x))^ndx\) |
| \(\Big \downarrow \) 3242 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^n \left (a^2 (c+d (4 n+7))-a^2 (3 c-11 d-4 d n) \sin (e+f x)\right )dx}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^n \left (a^2 (c+d (4 n+7))-a^2 (3 c-11 d-4 d n) \sin (e+f x)\right )dx}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^n \left (a^2 (c+d (4 n+7))-a^2 (3 c-11 d-4 d n) \sin (e+f x)\right )dx}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^ndx}{d (2 n+3)}+\frac {2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^ndx}{d (2 n+3)}+\frac {2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
| \(\Big \downarrow \) 3255 |
| \(\displaystyle \frac {\frac {a^4 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{d f (2 n+3) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {\frac {a^4 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \int \frac {\left (\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}\right )^n}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{d f (2 n+3) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {\frac {2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^3 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right )}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}}{d (2 n+5)}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)}\) |
(-2*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(1 + n) )/(d*f*(5 + 2*n)) + ((2*a^3*(3*c - d*(11 + 4*n))*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(3 + 2*n)*Sqrt[a + a*Sin[e + f*x]]) - (2*a^3*(3*c^2 - 2*c*d*(7 + 4*n) + d^2*(43 + 56*n + 16*n^2))*Cos[e + f*x]*Hypergeometric 2F1[1/2, -n, 3/2, (d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^n)/ (d*f*(3 + 2*n)*Sqrt[a + a*Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)) /(d*(5 + 2*n))
3.7.62.3.1 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[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
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[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]
\[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
integral(-(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n, x)
Timed out. \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx=\text {Timed out} \]
\[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
\[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
Timed out. \[ \int (3+3 \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]