Integrand size = 29, antiderivative size = 127 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx=-\frac {(c-3 d) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{6 \sqrt {6} (c-d)^{3/2} f}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 (c-d) f (3+3 \sin (e+f x))^{3/2}} \]
-1/4*(c-3*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)^(3/2)/f*2^(1/2)-1/2*cos (f*x+e)*(c+d*sin(f*x+e))^(1/2)/(c-d)/f/(a+a*sin(f*x+e))^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(384\) vs. \(2(127)=254\).
Time = 3.51 (sec) , antiderivative size = 384, normalized size of antiderivative = 3.02 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {(c-3 d) \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}\right )}{12 \sqrt {3} (c-d) f (1+\sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*((-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (( c - 3*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]*Sqr t[(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)*f*(1 + Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]])
Time = 0.47 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 3245, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{3/2} \sqrt {c+d \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle -\frac {\int -\frac {a (c-3 d)}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(c-3 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(c-3 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle -\frac {(c-3 d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{2 f (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {(c-3 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f (c-d)^{3/2}}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\) |
-1/2*((c - 3*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]*a^(3/2)*(c - d)^(3 /2)*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2))
3.6.97.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(642\) vs. \(2(112)=224\).
Time = 4.91 (sec) , antiderivative size = 643, normalized size of antiderivative = 5.06
| method | result | size |
| default | \(-\frac {\left (\sin \left (f x +e \right ) \ln \left (\frac {2 \sqrt {2 c -2 d}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right ) \sqrt {2}\, c -3 \sin \left (f x +e \right ) \ln \left (\frac {2 \sqrt {2 c -2 d}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right ) \sqrt {2}\, d +\ln \left (\frac {2 \sqrt {2 c -2 d}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right ) \sqrt {2}\, c -3 \ln \left (\frac {2 \sqrt {2 c -2 d}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right ) \sqrt {2}\, d +\sqrt {2 c -2 d}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )-\sin \left (f x +e \right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {2 c -2 d}+\sqrt {2 c -2 d}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\right ) \sqrt {c +d \sin \left (f x +e \right )}}{2 f \left (1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) a \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2 c -2 d}\, \left (c -d \right )}\) | \(643\) |
-1/2/f*(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)))*2^(1/2)*c-3*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 )))*2^(1/2)*d+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)))*2^(1/2)*c-3*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)))*2^(1/2)*d+(2 *c-2*d)^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)-sin(f*x+e )*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*(2*c-2*d)^(1/2)+(2*c-2*d)^(1/2)* ((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2))*(c+d*sin(f*x+e))^(1/2)/(1+cos(f*x +e)+sin(f*x+e))/a/((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)/(a*(sin(f*x+e)+1 ))^(1/2)/(2*c-2*d)^(1/2)/(c-d)
Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (112) = 224\).
Time = 0.52 (sec) , antiderivative size = 1008, normalized size of antiderivative = 7.94 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx=\text {Too large to display} \]
[1/16*(((c - 3*d)*cos(f*x + e)^2 - (c - 3*d)*cos(f*x + e) - ((c - 3*d)*cos (f*x + e) + 2*c - 6*d)*sin(f*x + e) - 2*c + 6*d)*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*((c - d)*cos(f*x + e) - (c - d)*sin(f*x + e) + c - d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a^2*c^2 - 2*a^2 *c*d + a^2*d^2)*f*cos(f*x + e)^2 - (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f *x + e) - 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f - ((a^2*c^2 - 2*a^2*c*d + a^ 2*d^2)*f*cos(f*x + e) + 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f)*sin(f*x + e)) , -1/8*(((c - 3*d)*cos(f*x + e)^2 - (c - 3*d)*cos(f*x + e) - ((c - 3*d)*co s(f*x + e) + 2*c - 6*d)*sin(f*x + e) - 2*c + 6*d)*sqrt(-2*a*c + 2*a*d)*arc tan(1/4*sqrt(-2*a*c + 2*a*d)*sqrt(a*sin(f*x + e) + a)*((c - 3*d)*sin(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c)/((a*c*d - a*d^2)*cos(f*x + e)*sin( f*x + e) + (a*c^2 - a*c*d)*cos(f*x + e))) - 4*((c - d)*cos(f*x + e) - (...
\[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {1}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {c + d \sin {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {d \sin \left (f x + e\right ) + c}} \,d x } \]
\[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {d \sin \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} \sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\sqrt {c+d\,\sin \left (e+f\,x\right )}} \,d x \]