Integrand size = 29, antiderivative size = 180 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^{5/2}} \, dx=-\frac {(3 c-5 d) (c+d) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{144 \sqrt {6} (c-d)^{3/2} f}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (3+3 \sin (e+f x))^{5/2}}-\frac {(3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{48 (c-d) f (3+3 \sin (e+f x))^{3/2}} \]
-1/32*(3*c-5*d)*(c+d)*arctanh(1/2*cos(f*x+e)*a^(1/2)*(c-d)^(1/2)*2^(1/2)/( a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/a^(5/2)/(c-d)^(3/2)/f*2^(1/2 )-1/4*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^(5/2)-1/16*(3*c -d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a/(c-d)/f/(a+a*sin(f*x+e))^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(415\) vs. \(2(180)=360\).
Time = 3.44 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.31 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (7 c-5 d+(3 c-d) \sin (e+f x)) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {\left (3 c^2-2 c d-5 d^2\right ) \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}\right )}{288 \sqrt {3} (c-d) f (1+\sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*((-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(7*c - 5*d + (3*c - d)*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(Cos[ (e + f*x)/2] + Sin[(e + f*x)/2])^3 + ((3*c^2 - 2*c*d - 5*d^2)*(Log[1 + Tan [(e + f*x)/2]] - Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*S qrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]]))/(Sec[(e + f*x)/2]^2 /(2 + 2*Tan[(e + f*x)/2]) - (-1/2*((c - d)*Sec[(e + f*x)/2]^2) + (Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*(d + d*Cos[e + f*x] + c*Sin[e + f*x])) /Sqrt[c + d*Sin[e + f*x]])/(c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^ (-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]))))/(288*Sqrt[3 ]*(c - d)*f*(1 + Sin[e + f*x])^(5/2)*Sqrt[c + d*Sin[e + f*x]])
Time = 0.77 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3243, 27, 3042, 3457, 27, 3042, 3261, 221}
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.
| \(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{(a \sin (e+f x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3243 |
| \(\displaystyle \frac {\int \frac {a (3 c+d)+2 a d \sin (e+f x)}{2 (\sin (e+f x) a+a)^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (3 c+d)+2 a d \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{8 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 c+d)+2 a d \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{8 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {-\frac {\int -\frac {a^2 (3 c-5 d) (c+d)}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a^2 (c-d)}-\frac {a (3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {(3 c-5 d) (c+d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 (c-d)}-\frac {a (3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(3 c-5 d) (c+d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 (c-d)}-\frac {a (3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {-\frac {a (3 c-5 d) (c+d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{2 f (c-d)}-\frac {a (3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {(3 c-5 d) (c+d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} \sqrt {a} f (c-d)^{3/2}}-\frac {a (3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
-1/4*(Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*(a + a*Sin[e + f*x])^(5/2) ) + (-1/2*((3*c - 5*d)*(c + d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/ (Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt[2]*Sqr t[a]*(c - d)^(3/2)*f) - (a*(3*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]] )/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2)))/(8*a^2)
3.7.2.3.1 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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m* ((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c *(m + 1) - b*d*(m + 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] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c , 0]))
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], 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[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(1726\) vs. \(2(162)=324\).
Time = 4.80 (sec) , antiderivative size = 1727, normalized size of antiderivative = 9.59
-1/32/f*(3*cos(f*x+e)^2*2^(1/2)*(2*c-2*d)^(1/2)*ln(-2*((2*c-2*d)^(1/2)*2^( 1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin (f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*c^2-2*co s(f*x+e)^2*2^(1/2)*(2*c-2*d)^(1/2)*ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*si n(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos (f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*c*d-5*cos(f*x+e)^2*2^ (1/2)*(2*c-2*d)^(1/2)*ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(co s(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos (f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*d^2-6*sin(f*x+e)*2^(1/2)*(2*c-2*d) ^(1/2)*ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1 /2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(c os(f*x+e)-1-sin(f*x+e)))*c^2+4*sin(f*x+e)*2^(1/2)*(2*c-2*d)^(1/2)*ln(-2*(( 2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+ c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin (f*x+e)))*c*d+10*sin(f*x+e)*2^(1/2)*(2*c-2*d)^(1/2)*ln(-2*((2*c-2*d)^(1/2) *2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d *sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*d^2- 6*2^(1/2)*(2*c-2*d)^(1/2)*ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e)) /(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d *cos(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*c^2+4*2^(1/2)*(2*c-2*d)^(1/...
Leaf count of result is larger than twice the leaf count of optimal. 620 vs. \(2 (162) = 324\).
Time = 0.59 (sec) , antiderivative size = 1474, normalized size of antiderivative = 8.19 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
[1/128*(((3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 2*c*d - 5*d^2 )*cos(f*x + e)^2 - 12*c^2 + 8*c*d + 20*d^2 - 2*(3*c^2 - 2*c*d - 5*d^2)*cos (f*x + e) + ((3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e)^2 - 12*c^2 + 8*c*d + 20* d^2 - 2*(3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(2*a*c - 2 *a*d)*log(((a*c^2 - 14*a*c*d + 17*a*d^2)*cos(f*x + e)^3 - 4*a*c^2 - 8*a*c* d - 4*a*d^2 - (13*a*c^2 - 22*a*c*d - 3*a*d^2)*cos(f*x + e)^2 - 4*((c - 3*d )*cos(f*x + e)^2 - (3*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x + e) + 4*c - 4*d)*sin(f*x + e) - 4*c + 4*d)*sqrt(2*a*c - 2*a*d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c) - 2*(9*a*c^2 - 14*a*c*d + 9*a*d^2)*cos(f*x + e) - (4*a*c^2 + 8*a*c*d + 4*a*d^2 - (a*c^2 - 14*a*c*d + 17*a*d^2)*cos(f*x + e)^2 - 2*(7*a*c^2 - 18*a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x + e))/(cos (f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin (f*x + e) - 2*cos(f*x + e) - 4)) + 8*((3*c^2 - 4*c*d + d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 12*c*d + 5*d^2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - (3*c^2 - 4*c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a *sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a^3*c^2 - 2*a^3*c*d + a^3*d ^2)*f*cos(f*x + e)^3 + 3*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e) - 4*(a^3*c^2 - 2*a^3*c* d + a^3*d^2)*f + ((a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^ 3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e) - 4*(a^3*c^2 - 2*a^3*c*d + ...
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {c + d \sin {\left (e + f x \right )}}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]