Integrand size = 29, antiderivative size = 186 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {(c-7 d) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{6 \sqrt {6} (c-d)^{5/2} f}-\frac {\cos (e+f x)}{2 (c-d) f (3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}}-\frac {d (c+5 d) \cos (e+f x)}{6 (c-d)^2 (c+d) f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]
-1/4*(c-7*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^(3/2)/(c-d)^(5/2)/f*2^(1/2)-1/2*cos (f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(1/2)-1/2*d*(c+5*d )*cos(f*x+e)/a/(c-d)^2/(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. \(404\) vs. \(2(186)=372\).
Time = 4.80 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.17 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (c^2+c d+4 d^2+d (c+5 d) \sin (e+f x)\right )}{(c+d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {(c-7 d) \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}\right )}{12 \sqrt {3} (c-d)^2 f (1+\sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*((-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(c^2 + c*d + 4*d^2 + d*(c + 5*d)*Sin[e + f*x]))/((c + d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + ((c - 7*d)*(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) + (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]))))/(12*Sqrt[3]*(c - d)^2*f*(1 + Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]])
Time = 0.83 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.18, 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, 3463, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle -\frac {\int -\frac {a (c-5 d)+2 a d \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{2 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (c-5 d)+2 a d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (c-5 d)+2 a d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^{3/2}}dx}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {a^2 (c-7 d) (c+d)}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a \left (c^2-d^2\right )}-\frac {2 a d (c+5 d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a (c-7 d) (c+d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}-\frac {2 a d (c+5 d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a (c-7 d) (c+d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}-\frac {2 a d (c+5 d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {-\frac {2 a^2 (c-7 d) (c+d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f \left (c^2-d^2\right )}-\frac {2 a d (c+5 d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {\sqrt {2} \sqrt {a} (c-7 d) (c+d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f \sqrt {c-d} \left (c^2-d^2\right )}-\frac {2 a d (c+5 d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{4 a^2 (c-d)}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}\) |
-1/2*Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]) + (-((Sqrt[2]*Sqrt[a]*(c - 7*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[c - d]*(c^2 - d^2)*f)) - (2*a*d*(c + 5*d)*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))/(4*a^2*(c - d) )
3.6.98.3.1 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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(1983\) vs. \(2(168)=336\).
Time = 5.03 (sec) , antiderivative size = 1984, normalized size of antiderivative = 10.67
-1/2/f*(2^(1/2)*cos(f*x+e)*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*((c+d*sin(f*x+ e))/(cos(f*x+e)+1))^(1/2)-6*cos(f*x+e)*sin(f*x+e)*2^(1/2)*((c+d*sin(f*x+e) )/(cos(f*x+e)+1))^(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*c os(f*x+e)-c+d)/(cos(f*x+e)-1-sin(f*x+e)))*c*d-7*2^(1/2)*cos(f*x+e)*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*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)+2^(1/2)* cos(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*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)+l n(-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^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f *x+e)-6*2^(1/2)*cos(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*d*((c+d*sin(f*x+e))/(cos(f* x+e)+1))^(1/2)-6*ln(-2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(...
Leaf count of result is larger than twice the leaf count of optimal. 860 vs. \(2 (168) = 336\).
Time = 0.69 (sec) , antiderivative size = 1954, normalized size of antiderivative = 10.51 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
[-1/16*(((c^2*d - 6*c*d^2 - 7*d^3)*cos(f*x + e)^3 - 2*c^3 + 10*c^2*d + 26* c*d^2 + 14*d^3 + (c^3 - 4*c^2*d - 19*c*d^2 - 14*d^3)*cos(f*x + e)^2 - (c^3 - 5*c^2*d - 13*c*d^2 - 7*d^3)*cos(f*x + e) - (2*c^3 - 10*c^2*d - 26*c*d^2 - 14*d^3 - (c^2*d - 6*c*d^2 - 7*d^3)*cos(f*x + e)^2 + (c^3 - 5*c^2*d - 13 *c*d^2 - 7*d^3)*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 - (1 3*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*co s(f*x + e) - 4)) - 8*(c^3 - c^2*d - c*d^2 + d^3 + (c^2*d + 4*c*d^2 - 5*d^3 )*cos(f*x + e)^2 + (c^3 + 3*c*d^2 - 4*d^3)*cos(f*x + e) - (c^3 - c^2*d - c *d^2 + d^3 - (c^2*d + 4*c*d^2 - 5*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a* sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e)^3 + (a^2*c^5 - 4*a^2*c^3*d^2 + 2*a^2 *c^2*d^3 + 3*a^2*c*d^4 - 2*a^2*d^5)*f*cos(f*x + e)^2 - (a^2*c^5 - a^2*c^4* d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e)...
\[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]