\(\int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx\) [203]

Optimal result

Integrand size = 21, antiderivative size = 172 \[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}+\frac {2 a E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (e+f x)}}{3 b f \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}{3 b f \sqrt {a+b \sin (e+f x)}} \] Output:

-2/3*cos(f*x+e)*(a+b*sin(f*x+e))^(1/2)/f-2/3*a*EllipticE(cos(1/2*e+1/4*Pi+ 
1/2*f*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(f*x+e))^(1/2)/b/f/((a+b*sin(f*x 
+e))/(a+b))^(1/2)-2/3*(a^2-b^2)*InverseJacobiAM(1/2*e-1/4*Pi+1/2*f*x,2^(1/ 
2)*(b/(a+b))^(1/2))*((a+b*sin(f*x+e))/(a+b))^(1/2)/b/f/(a+b*sin(f*x+e))^(1 
/2)
 

Mathematica [A] (verified)

Time = 4.50 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.83 \[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=-\frac {2 \left (b \cos (e+f x) (a+b \sin (e+f x))+a (a+b) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (e+f x)}{a+b}}-\left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (e+f x)}{a+b}}\right )}{3 b f \sqrt {a+b \sin (e+f x)}} \] Input:

Integrate[Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]],x]
 

Output:

(-2*(b*Cos[e + f*x]*(a + b*Sin[e + f*x]) + a*(a + b)*EllipticE[(-2*e + Pi 
- 2*f*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[e + f*x])/(a + b)] - (a^2 - b^2 
)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[e + f*x] 
)/(a + b)]))/(3*b*f*Sqrt[a + b*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3232, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

\(\displaystyle \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x) \sqrt {a+b \sin (e+f x)}dx\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {2}{3} \int \frac {b+a \sin (e+f x)}{2 \sqrt {a+b \sin (e+f x)}}dx-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {b+a \sin (e+f x)}{\sqrt {a+b \sin (e+f x)}}dx-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {b+a \sin (e+f x)}{\sqrt {a+b \sin (e+f x)}}dx-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {1}{3} \left (\frac {a \int \sqrt {a+b \sin (e+f x)}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)}}dx}{b}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {a \int \sqrt {a+b \sin (e+f x)}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)}}dx}{b}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{3} \left (\frac {a \sqrt {a+b \sin (e+f x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)}}dx}{b}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {a \sqrt {a+b \sin (e+f x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)}}dx}{b}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{3} \left (\frac {2 a \sqrt {a+b \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b f \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (e+f x)}}dx}{b}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{3} \left (\frac {2 a \sqrt {a+b \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b f \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (e+f x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}}}dx}{b \sqrt {a+b \sin (e+f x)}}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {2 a \sqrt {a+b \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b f \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (e+f x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}}}dx}{b \sqrt {a+b \sin (e+f x)}}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{3} \left (\frac {2 a \sqrt {a+b \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b f \sqrt {\frac {a+b \sin (e+f x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (e+f x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{b f \sqrt {a+b \sin (e+f x)}}\right )-\frac {2 \cos (e+f x) \sqrt {a+b \sin (e+f x)}}{3 f}\)

Input:

Int[Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]],x]
 

Output:

(-2*Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]])/(3*f) + ((2*a*EllipticE[(e - Pi 
/2 + f*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[e + f*x]])/(b*f*Sqrt[(a + b*Sin 
[e + f*x])/(a + b)]) - (2*(a^2 - b^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*b)/ 
(a + b)]*Sqrt[(a + b*Sin[e + f*x])/(a + b)])/(b*f*Sqrt[a + b*Sin[e + f*x]] 
))/3
 

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 3042

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

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 3232

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[1/(m + 1)   Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* 
d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ 
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 
 0] && IntegerQ[2*m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(459\) vs. \(2(161)=322\).

Time = 0.90 (sec) , antiderivative size = 460, normalized size of antiderivative = 2.67

method	result	size
default	\(-\frac {2 \left (\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}\, \sqrt {-\frac {b \left (-1+\sin \left (f x +e \right )\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (f x +e \right )\right )}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}-\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}\, \sqrt {-\frac {b \left (-1+\sin \left (f x +e \right )\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (f x +e \right )\right )}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}-\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}\, \sqrt {-\frac {b \left (-1+\sin \left (f x +e \right )\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (f x +e \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b +\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}\, \sqrt {-\frac {b \left (-1+\sin \left (f x +e \right )\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (f x +e \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (f x +e \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}-\sin \left (f x +e \right )^{3} b^{3}-\sin \left (f x +e \right )^{2} a \,b^{2}+b^{3} \sin \left (f x +e \right )+a \,b^{2}\right )}{3 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )}\, f}\)	\(460\)

Input:

int(sin(f*x+e)*(a+b*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3*(((a+b*sin(f*x+e))/(a-b))^(1/2)*(-b*(-1+sin(f*x+e))/(a+b))^(1/2)*(-b* 
(1+sin(f*x+e))/(a-b))^(1/2)*EllipticE(((a+b*sin(f*x+e))/(a-b))^(1/2),((a-b 
)/(a+b))^(1/2))*a^3-((a+b*sin(f*x+e))/(a-b))^(1/2)*(-b*(-1+sin(f*x+e))/(a+ 
b))^(1/2)*(-b*(1+sin(f*x+e))/(a-b))^(1/2)*EllipticE(((a+b*sin(f*x+e))/(a-b 
))^(1/2),((a-b)/(a+b))^(1/2))*a*b^2-((a+b*sin(f*x+e))/(a-b))^(1/2)*(-b*(-1 
+sin(f*x+e))/(a+b))^(1/2)*(-b*(1+sin(f*x+e))/(a-b))^(1/2)*EllipticF(((a+b* 
sin(f*x+e))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b+((a+b*sin(f*x+e))/(a-b 
))^(1/2)*(-b*(-1+sin(f*x+e))/(a+b))^(1/2)*(-b*(1+sin(f*x+e))/(a-b))^(1/2)* 
EllipticF(((a+b*sin(f*x+e))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^3-sin(f*x+ 
e)^3*b^3-sin(f*x+e)^2*a*b^2+b^3*sin(f*x+e)+a*b^2)/b^2/cos(f*x+e)/(a+b*sin( 
f*x+e))^(1/2)/f
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.30 \[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=-\frac {2 \, {\left (3 \, \sqrt {b \sin \left (f x + e\right ) + a} b^{2} \cos \left (f x + e\right ) + 3 i \, a \sqrt {\frac {1}{2} i \, b} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (f x + e\right ) - 3 i \, b \sin \left (f x + e\right ) - 2 i \, a}{3 \, b}\right )\right ) - 3 i \, a \sqrt {-\frac {1}{2} i \, b} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (f x + e\right ) + 3 i \, b \sin \left (f x + e\right ) + 2 i \, a}{3 \, b}\right )\right ) + {\left (2 \, a^{2} - 3 \, b^{2}\right )} \sqrt {\frac {1}{2} i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (f x + e\right ) - 3 i \, b \sin \left (f x + e\right ) - 2 i \, a}{3 \, b}\right ) + {\left (2 \, a^{2} - 3 \, b^{2}\right )} \sqrt {-\frac {1}{2} i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (f x + e\right ) + 3 i \, b \sin \left (f x + e\right ) + 2 i \, a}{3 \, b}\right )\right )}}{9 \, b^{2} f} \] Input:

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

-2/9*(3*sqrt(b*sin(f*x + e) + a)*b^2*cos(f*x + e) + 3*I*a*sqrt(1/2*I*b)*b* 
weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 
 weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2) 
/b^3, 1/3*(3*b*cos(f*x + e) - 3*I*b*sin(f*x + e) - 2*I*a)/b)) - 3*I*a*sqrt 
(-1/2*I*b)*b*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9 
*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a 
^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(f*x + e) + 3*I*b*sin(f*x + e) + 2*I*a)/b 
)) + (2*a^2 - 3*b^2)*sqrt(1/2*I*b)*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2 
)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(f*x + e) - 3*I*b*sin( 
f*x + e) - 2*I*a)/b) + (2*a^2 - 3*b^2)*sqrt(-1/2*I*b)*weierstrassPInverse( 
-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(f 
*x + e) + 3*I*b*sin(f*x + e) + 2*I*a)/b))/(b^2*f)
 

Sympy [F]

\[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=\int \sqrt {a + b \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx \] Input:

integrate(sin(f*x+e)*(a+b*sin(f*x+e))**(1/2),x)
 

Output:

Integral(sqrt(a + b*sin(e + f*x))*sin(e + f*x), x)
 

Maxima [F]

\[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right ) + a} \sin \left (f x + e\right ) \,d x } \] Input:

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sin(f*x + e) + a)*sin(f*x + e), x)
 

Giac [F]

\[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right ) + a} \sin \left (f x + e\right ) \,d x } \] Input:

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*sin(f*x + e) + a)*sin(f*x + e), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=\int \sin \left (e+f\,x\right )\,\sqrt {a+b\,\sin \left (e+f\,x\right )} \,d x \] Input:

int(sin(e + f*x)*(a + b*sin(e + f*x))^(1/2),x)
 

Output:

int(sin(e + f*x)*(a + b*sin(e + f*x))^(1/2), x)
 

Reduce [F]

\[ \int \sin (e+f x) \sqrt {a+b \sin (e+f x)} \, dx=\int \sqrt {\sin \left (f x +e \right ) b +a}\, \sin \left (f x +e \right )d x \] Input:

int(sin(f*x+e)*(a+b*sin(f*x+e))^(1/2),x)
 

Output:

int(sqrt(sin(e + f*x)*b + a)*sin(e + f*x),x)