Integrand size = 29, antiderivative size = 114 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=-\frac {6 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{d^{3/2} f}+\frac {18 (c-d) \cos (e+f x)}{d (c+d) f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]
-2*a^(3/2)*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*s in(f*x+e))^(1/2))/d^(3/2)/f+2*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(a+a*sin(f*x+ e))^(1/2)/(c+d*sin(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(380\) vs. \(2(114)=228\).
Time = 5.37 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.33 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=-\frac {3 \sqrt {3} (1+\sin (e+f x))^{3/2} \left (-2 c \sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )+2 d^{3/2} \cos \left (\frac {1}{2} (e+f x)\right )+2 c \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )-2 d^{3/2} \sin \left (\frac {1}{2} (e+f x)\right )-2 (c+d) \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right ) \sqrt {c+d \sin (e+f x)}-(c+d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right ) \sqrt {c+d \sin (e+f x)}+c \log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right ) \sqrt {c+d \sin (e+f x)}+d \log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right ) \sqrt {c+d \sin (e+f x)}\right )}{d^{3/2} (c+d) f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c+d \sin (e+f x)}} \]
(-3*Sqrt[3]*(1 + Sin[e + f*x])^(3/2)*(-2*c*Sqrt[d]*Cos[(e + f*x)/2] + 2*d^ (3/2)*Cos[(e + f*x)/2] + 2*c*Sqrt[d]*Sin[(e + f*x)/2] - 2*d^(3/2)*Sin[(e + f*x)/2] - 2*(c + d)*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sq rt[c + d*Sin[e + f*x]]]*Sqrt[c + d*Sin[e + f*x]] - (c + d)*ArcTanh[(Sqrt[2 ]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]]*Sqrt[c + d* Sin[e + f*x]] + c*Log[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]*Sqrt[c + d*Sin[e + f*x]] + d*Log[Sqrt[2]*Sqrt[d]*Cos[(2* e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]*Sqrt[c + d*Sin[e + f*x]]))/ (d^(3/2)*(c + d)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c + d*Sin[ e + f*x]])
Time = 0.50 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3241, 27, 2011, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c+d \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3241 |
| \(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {2 a \int -\frac {a (c+d)+a \sin (e+f x) (c+d)}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{d (c+d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {a (c+d)+a \sin (e+f x) (c+d)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle \frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{d}+\frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{d}+\frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{d f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{d f (c+d) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{d^{3/2} f}\) |
(-2*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x] ]*Sqrt[c + d*Sin[e + f*x]])])/(d^(3/2)*f) + (2*a^2*(c - d)*Cos[e + f*x])/( d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
3.6.75.3.1 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)*(b*c - a*d)*Cos[e + f*x]*(a + b *Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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[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_)]]/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]
Leaf count of result is larger than twice the leaf count of optimal. \(4781\) vs. \(2(101)=202\).
Time = 1.09 (sec) , antiderivative size = 4782, normalized size of antiderivative = 41.95
\[\text {output too large to display}\]
-1/f*sec(f*x+e)*a*(-arctan(((c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin(f*x+ e)+d))^(1/2)/(-(d^2/c^2)^(1/2)*c)^(1/2))*sin(f*x+e)^2*((c+d*sin(f*x+e))*d/ ((d^2/c^2)^(1/2)*c*sin(f*x+e)+d))^(1/2)*d^6+2*cos(f*x+e)^2*(-(d^2/c^2)^(1/ 2)*c)^(1/2)*c^3*d^3-6*cos(f*x+e)^2*(-(d^2/c^2)^(1/2)*c)^(1/2)*c^2*d^4+6*co s(f*x+e)^2*(-(d^2/c^2)^(1/2)*c)^(1/2)*c*d^5-4*c^3*(-(d^2/c^2)^(1/2)*c)^(1/ 2)*sin(f*x+e)*d^3+4*c*(-(d^2/c^2)^(1/2)*c)^(1/2)*sin(f*x+e)*d^5-c^4*arctan (((c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin(f*x+e)+d))^(1/2)/(-(d^2/c^2)^( 1/2)*c)^(1/2))*((c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin(f*x+e)+d))^(1/2) *d^2+2*c^2*arctan(((c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin(f*x+e)+d))^(1 /2)/(-(d^2/c^2)^(1/2)*c)^(1/2))*((c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin (f*x+e)+d))^(1/2)*d^4-2*(-(d^2/c^2)^(1/2)*c)^(1/2)*sin(f*x+e)*d^6+arctan(( (c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin(f*x+e)+d))^(1/2)/(-(d^2/c^2)^(1/ 2)*c)^(1/2))*sin(f*x+e)*((c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin(f*x+e)+ d))^(1/2)*d^6+2*c^4*(-(d^2/c^2)^(1/2)*c)^(1/2)*sin(f*x+e)*d^2+cos(f*x+e)*c *(((d^2/c^2)^(1/2)*c^4+6*(d^2/c^2)^(1/2)*c^2*d^2+d^4*(d^2/c^2)^(1/2)-4*c^2 *d^2-4*d^4)*c)^(1/2)*(-(d^2/c^2)^(1/2)*c)^(1/2)*arctan(((d^2/c^2)^(1/2)*c* sin(f*x+e)+d*cos(f*x+e)-d)/((c+d*sin(f*x+e))*d/((d^2/c^2)^(1/2)*c*sin(f*x+ e)+d))^(1/2)/(-(d^2/c^2)^(1/2)*c*sin(f*x+e)+d*cos(f*x+e)-d)*((d^2/c^2)^(1/ 2)*c^2-d^2)*c*((d^2/c^2)^(1/2)-1)/(((d^2/c^2)^(1/2)*c^4+6*(d^2/c^2)^(1/2)* c^2*d^2+d^4*(d^2/c^2)^(1/2)-4*c^2*d^2-4*d^4)*c)^(1/2))*((c+d*sin(f*x+e)...
Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (101) = 202\).
Time = 0.61 (sec) , antiderivative size = 1297, normalized size of antiderivative = 11.38 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
[-1/4*((a*c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d^2 + (a*c*d + a*d^2)*cos(f*x + e))*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)*co s(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 + 59*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*(a*c - a*d + (a*c - a*d)*cos(f*x + e) - (a*c - a*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sq rt(d*sin(f*x + e) + c))/((c*d^2 + d^3)*f*cos(f*x + e)^2 - (c^2*d + c*d^2)* f*cos(f*x + e) - (c^2*d + 2*c*d^2 + d^3)*f - ((c*d^2 + d^3)*f*cos(f*x +...
\[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]