\(\int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx\) [41]

Optimal result

Integrand size = 12, antiderivative size = 42 \[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=-4 \text {arctanh}\left (\frac {\cos (x)}{\sqrt {1+\sin (x)}}\right )+\frac {4 \cos (x)}{\sqrt {1+\sin (x)}}-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {1+\sin (x)}} \] Output:

-4*arctanh(cos(x)/(1+sin(x))^(1/2))+4*cos(x)/(1+sin(x))^(1/2)-2*cos(x)*ln( 
sin(x))/(1+sin(x))^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(42)=84\).

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.07 \[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=\frac {2 \left (-\log \left (1+\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (1-\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-\cos \left (\frac {x}{2}\right ) (-2+\log (\sin (x)))+(-2+\log (\sin (x))) \sin \left (\frac {x}{2}\right )\right ) \sqrt {1+\sin (x)}}{\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )} \] Input:

Integrate[Log[Sin[x]]*Sqrt[1 + Sin[x]],x]
 

Output:

(2*(-Log[1 + Cos[x/2] - Sin[x/2]] + Log[1 - Cos[x/2] + Sin[x/2]] - Cos[x/2 
]*(-2 + Log[Sin[x]]) + (-2 + Log[Sin[x]])*Sin[x/2])*Sqrt[1 + Sin[x]])/(Cos 
[x/2] + Sin[x/2])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3034, 27, 3042, 3353, 3042, 3460, 3042, 3252, 219}

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.

\(\displaystyle \int \sqrt {\sin (x)+1} \log (\sin (x)) \, dx\)

\(\Big \downarrow \) 3034

\(\displaystyle -\int -\frac {2 \cos (x) \cot (x)}{\sqrt {\sin (x)+1}}dx-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int \frac {\cos (x) \cot (x)}{\sqrt {\sin (x)+1}}dx-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {\cos (x)^2}{\sin (x) \sqrt {\sin (x)+1}}dx-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 3353

\(\displaystyle 2 \int \csc (x) (1-\sin (x)) \sqrt {\sin (x)+1}dx-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {(1-\sin (x)) \sqrt {\sin (x)+1}}{\sin (x)}dx-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 3460

\(\displaystyle 2 \left (\int \csc (x) \sqrt {\sin (x)+1}dx+\frac {2 \cos (x)}{\sqrt {\sin (x)+1}}\right )-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \left (\int \frac {\sqrt {\sin (x)+1}}{\sin (x)}dx+\frac {2 \cos (x)}{\sqrt {\sin (x)+1}}\right )-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 3252

\(\displaystyle 2 \left (\frac {2 \cos (x)}{\sqrt {\sin (x)+1}}-2 \int \frac {1}{1-\frac {\cos ^2(x)}{\sin (x)+1}}d\frac {\cos (x)}{\sqrt {\sin (x)+1}}\right )-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (\frac {2 \cos (x)}{\sqrt {\sin (x)+1}}-2 \text {arctanh}\left (\frac {\cos (x)}{\sqrt {\sin (x)+1}}\right )\right )-\frac {2 \cos (x) \log (\sin (x))}{\sqrt {\sin (x)+1}}\)

Input:

Int[Log[Sin[x]]*Sqrt[1 + Sin[x]],x]
 

Output:

(-2*Cos[x]*Log[Sin[x]])/Sqrt[1 + Sin[x]] + 2*(-2*ArcTanh[Cos[x]/Sqrt[1 + S 
in[x]]] + (2*Cos[x])/Sqrt[1 + Sin[x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3034

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[Log[u]   w, x 
] - Int[SimplifyIntegrand[w*(D[u, x]/u), x], x] /; InverseFunctionFreeQ[w, 
x]] /; InverseFunctionFreeQ[u, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*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]
 

rule 3353

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/b^2   Int[(d*Sin[e 
 + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /; Fre 
eQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[ 
n, 0])
 

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]
 

Maple [F]

\[\int \ln \left (\sin \left (x \right )\right ) \sqrt {\sin \left (x \right )+1}d x\]

Input:

int(ln(sin(x))*(sin(x)+1)^(1/2),x)
 

Output:

int(ln(sin(x))*(sin(x)+1)^(1/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (36) = 72\).

Time = 0.10 (sec) , antiderivative size = 146, normalized size of antiderivative = 3.48 \[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=-\frac {{\left (\cos \left (x\right ) + \sin \left (x\right ) + 1\right )} \log \left (\frac {\cos \left (x\right )^{2} - {\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right ) + 2 \, {\left (\cos \left (x\right ) - \sin \left (x\right ) + 1\right )} \sqrt {\sin \left (x\right ) + 1} + 2 \, \cos \left (x\right ) + 1}{2 \, {\left (\cos \left (x\right ) + \sin \left (x\right ) + 1\right )}}\right ) - {\left (\cos \left (x\right ) + \sin \left (x\right ) + 1\right )} \log \left (\frac {\cos \left (x\right )^{2} - {\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right ) - 2 \, {\left (\cos \left (x\right ) - \sin \left (x\right ) + 1\right )} \sqrt {\sin \left (x\right ) + 1} + 2 \, \cos \left (x\right ) + 1}{2 \, {\left (\cos \left (x\right ) + \sin \left (x\right ) + 1\right )}}\right ) + 2 \, {\left ({\left (\cos \left (x\right ) - \sin \left (x\right ) + 1\right )} \log \left (\sin \left (x\right )\right ) - 2 \, \cos \left (x\right ) + 2 \, \sin \left (x\right ) - 2\right )} \sqrt {\sin \left (x\right ) + 1}}{\cos \left (x\right ) + \sin \left (x\right ) + 1} \] Input:

integrate(log(sin(x))*(1+sin(x))^(1/2),x, algorithm="fricas")
 

Output:

-((cos(x) + sin(x) + 1)*log(1/2*(cos(x)^2 - (cos(x) - 1)*sin(x) + 2*(cos(x 
) - sin(x) + 1)*sqrt(sin(x) + 1) + 2*cos(x) + 1)/(cos(x) + sin(x) + 1)) - 
(cos(x) + sin(x) + 1)*log(1/2*(cos(x)^2 - (cos(x) - 1)*sin(x) - 2*(cos(x) 
- sin(x) + 1)*sqrt(sin(x) + 1) + 2*cos(x) + 1)/(cos(x) + sin(x) + 1)) + 2* 
((cos(x) - sin(x) + 1)*log(sin(x)) - 2*cos(x) + 2*sin(x) - 2)*sqrt(sin(x) 
+ 1))/(cos(x) + sin(x) + 1)
 

Sympy [F]

\[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=\int \sqrt {\sin {\left (x \right )} + 1} \log {\left (\sin {\left (x \right )} \right )}\, dx \] Input:

integrate(ln(sin(x))*(1+sin(x))**(1/2),x)
 

Output:

Integral(sqrt(sin(x) + 1)*log(sin(x)), x)
 

Maxima [F]

\[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=\int { \sqrt {\sin \left (x\right ) + 1} \log \left (\sin \left (x\right )\right ) \,d x } \] Input:

integrate(log(sin(x))*(1+sin(x))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(sin(x) + 1)*log(sin(x)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (36) = 72\).

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.14 \[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=\sqrt {2} {\left (2 \, \log \left (\sin \left (x\right )\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + {\left (\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{4} \, \pi - \frac {1}{2} \, x\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{4} \, \pi - \frac {1}{2} \, x\right ) \right |}}\right ) + 4 \, \sin \left (\frac {1}{4} \, \pi - \frac {1}{2} \, x\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )\right )\right )} \] Input:

integrate(log(sin(x))*(1+sin(x))^(1/2),x, algorithm="giac")
 

Output:

sqrt(2)*(2*log(sin(x))*sgn(cos(-1/4*pi + 1/2*x))*sin(-1/4*pi + 1/2*x) + (s 
qrt(2)*log(abs(-2*sqrt(2) + 4*sin(1/4*pi - 1/2*x))/abs(2*sqrt(2) + 4*sin(1 
/4*pi - 1/2*x))) + 4*sin(1/4*pi - 1/2*x))*sgn(cos(-1/4*pi + 1/2*x)))
 

Mupad [F(-1)]

Timed out. \[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=\int \ln \left (\sin \left (x\right )\right )\,\sqrt {\sin \left (x\right )+1} \,d x \] Input:

int(log(sin(x))*(sin(x) + 1)^(1/2),x)
 

Output:

int(log(sin(x))*(sin(x) + 1)^(1/2), x)
 

Reduce [F]

\[ \int \log (\sin (x)) \sqrt {1+\sin (x)} \, dx=\int \sqrt {\sin \left (x \right )+1}\, \mathrm {log}\left (\sin \left (x \right )\right )d x \] Input:

int(log(sin(x))*(1+sin(x))^(1/2),x)
 

Output:

int(sqrt(sin(x) + 1)*log(sin(x)),x)