Integrand size = 29, antiderivative size = 273 \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx=\frac {15 \sqrt {3} (c-15 d) (c+d)^3 \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{64 d^{3/2} f}+\frac {45 (c-15 d) (c+d)^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{64 d f \sqrt {3+3 \sin (e+f x)}}+\frac {15 (c-15 d) (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{32 d f \sqrt {3+3 \sin (e+f x)}}+\frac {3 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{8 d f \sqrt {3+3 \sin (e+f x)}}-\frac {9 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {3+3 \sin (e+f x)}} \]
5/64*a^(3/2)*(c-15*d)*(c+d)^3*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f *x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/d^(3/2)/f+5/96*a^2*(c-15*d)*(c+d)*cos (f*x+e)*(c+d*sin(f*x+e))^(3/2)/d/f/(a+a*sin(f*x+e))^(1/2)+1/24*a^2*(c-15*d )*cos(f*x+e)*(c+d*sin(f*x+e))^(5/2)/d/f/(a+a*sin(f*x+e))^(1/2)-1/4*a^2*cos (f*x+e)*(c+d*sin(f*x+e))^(7/2)/d/f/(a+a*sin(f*x+e))^(1/2)+5/64*a^2*(c-15*d )*(c+d)^2*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d/f/(a+a*sin(f*x+e))^(1/2)
Time = 3.70 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.18 \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx=\frac {3 \sqrt {3} (1+\sin (e+f x))^{3/2} \left (-\frac {5 (c-15 d) (c+d)^3 \left (2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )\right )}{d^{3/2}}-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)} \left (15 c^3+455 c^2 d+653 c d^2+285 d^3-4 d^2 (17 c+15 d) \cos (2 (e+f x))+2 d \left (59 c^2+190 c d+93 d^2\right ) \sin (e+f x)-12 d^3 \sin (3 (e+f x))\right )}{3 d}\right )}{128 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]
(3*Sqrt[3]*(1 + Sin[e + f*x])^(3/2)*((-5*(c - 15*d)*(c + d)^3*(2*ArcTan[(S qrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + ArcT anh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] - Log[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]] ]))/d^(3/2) - (2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*(15*c^3 + 455*c^2*d + 653*c*d^2 + 285*d^3 - 4*d^2*(17*c + 15*d)*Cos[ 2*(e + f*x)] + 2*d*(59*c^2 + 190*c*d + 93*d^2)*Sin[e + f*x] - 12*d^3*Sin[3 *(e + f*x)]))/(3*d)))/(128*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)
Time = 1.14 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.97, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3242, 27, 2011, 3042, 3249, 3042, 3249, 3042, 3249, 3042, 3254, 218}
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.
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{5/2}dx\) |
\(\Big \downarrow \) 3242 |
\(\displaystyle \frac {\int -\frac {\left ((c-15 d) a^2+(c-15 d) \sin (e+f x) a^2\right ) (c+d \sin (e+f x))^{5/2}}{2 \sqrt {\sin (e+f x) a+a}}dx}{4 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\left ((c-15 d) a^2+(c-15 d) \sin (e+f x) a^2\right ) (c+d \sin (e+f x))^{5/2}}{\sqrt {\sin (e+f x) a+a}}dx}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle -\frac {a (c-15 d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{5/2}dx}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (c-15 d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{5/2}dx}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3249 |
\(\displaystyle -\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3249 |
\(\displaystyle -\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \left (\frac {3}{4} (c+d) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \left (\frac {3}{4} (c+d) \int \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}dx-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3249 |
\(\displaystyle -\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \left (\frac {3}{4} (c+d) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \left (\frac {3}{4} (c+d) \left (\frac {1}{2} (c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3254 |
\(\displaystyle -\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \left (\frac {3}{4} (c+d) \left (-\frac {a (c+d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}-\frac {a (c-15 d) \left (\frac {5}{6} (c+d) \left (\frac {3}{4} (c+d) \left (-\frac {\sqrt {a} (c+d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {d} f}-\frac {a \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\right )}{8 d}\) |
-1/4*(a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(7/2))/(d*f*Sqrt[a + a*Sin[e + f*x]]) - (a*(c - 15*d)*(-1/3*(a*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/ (f*Sqrt[a + a*Sin[e + f*x]]) + (5*(c + d)*(-1/2*(a*Cos[e + f*x]*(c + d*Sin [e + f*x])^(3/2))/(f*Sqrt[a + a*Sin[e + f*x]]) + (3*(c + d)*(-((Sqrt[a]*(c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt [c + d*Sin[e + f*x]])])/(Sqrt[d]*f)) - (a*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])))/4))/6))/(8*d)
3.6.71.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 2*n + 1))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))], x ] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Timed out.
\[\int \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 560 vs. \(2 (247) = 494\).
Time = 0.99 (sec) , antiderivative size = 1579, normalized size of antiderivative = 5.78 \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx=\text {Too large to display} \]
[-1/1536*(15*(a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 - 44*a*c*d^3 - 15*a*d^4 + (a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 - 44*a*c*d^3 - 15*a*d^4)*cos(f*x + e) + (a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 - 44*a*c*d^3 - 15*a*d^4)*sin(f*x + e)) *sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^ 2 + 4*a*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*cos(f*x + e)^4 - 32*(5*a*c ^2*d^2 - 14*a*c*d^3 + 13*a*d^4)*cos(f*x + e)^3 - 32*(a*c^3*d - 2*a*c^2*d^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^4*cos(f*x + e)^4 - c^3*d + 17*c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 - d^4)*cos(f*x + e)^3 - 2*(5* c^2*d^2 - 26*c*d^3 + 33*d^4)*cos(f*x + e)^2 - (c^3*d - 7*c^2*d^2 + 31*c*d^ 3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 + c^3*d - 17*c^2*d^2 + 5 9*c*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f*x + e)^2 - 2*(5*c^2*d^2 - 14* c*d^3 + 13*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt( d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*a*c^3*d + 230*a*c^2*d^2 - 476 *a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*cos(f*x + e)^4 + a*c^4 + 4 *a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - 256*(a*c*d^3 - a*d^4)*cos(f*x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a*d^4)*cos(f*x + e)^2 + 32*(a*c^ 3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4)*cos(f*x + e))*sin(f*x + e))/(cos (f*x + e) + sin(f*x + e) + 1)) - 8*(48*a*d^3*cos(f*x + e)^4 - 15*a*c^3 - 3 37*a*c^2*d - 341*a*c*d^2 - 147*a*d^3 + 8*(17*a*c*d^2 + 15*a*d^3)*cos(f*x + e)^3 - 2*(59*a*c^2*d + 122*a*c*d^2 + 63*a*d^3)*cos(f*x + e)^2 - (15*a*...
Timed out. \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
\[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]