Integrand size = 29, antiderivative size = 190 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx=-\frac {\left (3 c^2-10 c d+19 d^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{144 \sqrt {6} (c-d)^{5/2} f}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 (c-d) f (3+3 \sin (e+f x))^{5/2}}-\frac {(c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 (c-d)^2 f (3+3 \sin (e+f x))^{3/2}} \]
-1/32*(3*c^2-10*c*d+19*d^2)*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)^(5/2)/f* 2^(1/2)-1/4*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/(c-d)/f/(a+a*sin(f*x+e))^(5/ 2)-3/16*(c-3*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a/(c-d)^2/f/(a+a*sin(f*x +e))^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(414\) vs. \(2(190)=380\).
Time = 6.55 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.18 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (7 c-13 d+3 (c-3 d) \sin (e+f x)) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {\left (3 c^2-10 c d+19 d^2\right ) \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}\right )}{288 \sqrt {3} (c-d)^2 f (1+\sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*((-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(7*c - 13*d + 3*(c - 3*d)*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(C os[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + ((3*c^2 - 10*c*d + 19*d^2)*(Log[1 + Tan[(e + f*x)/2]] - Log[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]]))/(Sec[(e + f*x) /2]^2/(2 + 2*Tan[(e + f*x)/2]) - (-1/2*((c - d)*Sec[(e + f*x)/2]^2) + (Sqr t[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*S qrt[3]*(c - d)^2*f*(1 + Sin[e + f*x])^(5/2)*Sqrt[c + d*Sin[e + f*x]])
Time = 0.81 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3245, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} \sqrt {c+d \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle -\frac {\int -\frac {a (3 c-7 d)+2 a d \sin (e+f x)}{2 (\sin (e+f x) a+a)^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (3 c-7 d)+2 a d \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{8 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 c-7 d)+2 a d \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2} \sqrt {c+d \sin (e+f x)}}dx}{8 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {-\frac {\int -\frac {a^2 \left (3 c^2-10 d c+19 d^2\right )}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a^2 (c-d)}-\frac {3 a (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (3 c^2-10 c d+19 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 (c-d)}-\frac {3 a (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (3 c^2-10 c d+19 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 (c-d)}-\frac {3 a (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {-\frac {a \left (3 c^2-10 c d+19 d^2\right ) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{2 f (c-d)}-\frac {3 a (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {\left (3 c^2-10 c d+19 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} \sqrt {a} f (c-d)^{3/2}}-\frac {3 a (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\) |
-1/4*(Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/((c - d)*f*(a + a*Sin[e + f*x ])^(5/2)) + (-1/2*((3*c^2 - 10*c*d + 19*d^2)*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]*Sqrt[a]*(c - d)^(3/2)*f) - (3*a*(c - 3*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*(c - d) )
3.7.3.3.1 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^2*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*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + 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] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[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. \(1557\) vs. \(2(172)=344\).
Time = 4.69 (sec) , antiderivative size = 1558, normalized size of antiderivative = 8.20
1/16/f*(-3*2^(1/2)*cos(f*x+e)^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+10*2^(1/2)*cos(f*x+e )^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-19*2^(1/2)*cos(f*x+e)^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*si n(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+3* (2*c-2*d)^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c*cos(f*x+e)^2+3*( 2*c-2*d)^(1/2)*sin(f*x+e)*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/ 2)*c-9*(2*c-2*d)^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^ 2*d-9*(2*c-2*d)^(1/2)*sin(f*x+e)*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+ 1))^(1/2)*d+6*2^(1/2)*sin(f*x+e)*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^2-20*2^(1/2)*sin(f*x+ e)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*s in(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+38*2^(1/2)*sin(f*x+e)*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+7*...
Leaf count of result is larger than twice the leaf count of optimal. 705 vs. \(2 (172) = 344\).
Time = 0.68 (sec) , antiderivative size = 1644, normalized size of antiderivative = 8.65 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx=\text {Too large to display} \]
[1/128*(((3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 10*c*d + 19 *d^2)*cos(f*x + e)^2 - 12*c^2 + 40*c*d - 76*d^2 - 2*(3*c^2 - 10*c*d + 19*d ^2)*cos(f*x + e) + ((3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^2 + 40 *c*d - 76*d^2 - 2*(3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e))*sin(f*x + e))*sq rt(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*si n(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 + 3*d^2)* cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 20*c*d + 13*d^2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - 3*(c^2 - 4*c*d + 3*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^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3 *c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2* d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a ^3*c*d^2 - a^3*d^3)*f + ((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3...
\[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {1}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \sqrt {c + d \sin {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\sqrt {c+d\,\sin \left (e+f\,x\right )}} \,d x \]