\(\int \sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)} \, dx\) [211]

Optimal result

Integrand size = 25, antiderivative size = 371 \[ \int \sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)} \, dx=-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}+\frac {(a-b) \sqrt {a+b} \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (1+\csc (c+d x))}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \tan (c+d x)}{a d}-\frac {\sqrt {a+b} \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (1+\csc (c+d x))}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \tan (c+d x)}{d}+\frac {a \sqrt {a+b} \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (1+\csc (c+d x))}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \tan (c+d x)}{b d} \] Output:

-cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/d/sin(d*x+c)^(1/2)+(a-b)*(a+b)^(1/2)*(a 
*(1-csc(d*x+c))/(a+b))^(1/2)*(a*(1+csc(d*x+c))/(a-b))^(1/2)*EllipticE((a+b 
*sin(d*x+c))^(1/2)/(a+b)^(1/2)/sin(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*tan( 
d*x+c)/a/d-(a+b)^(1/2)*(a*(1-csc(d*x+c))/(a+b))^(1/2)*(a*(1+csc(d*x+c))/(a 
-b))^(1/2)*EllipticF((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2)/sin(d*x+c)^(1/2),( 
-(a+b)/(a-b))^(1/2))*tan(d*x+c)/d+a*(a+b)^(1/2)*(a*(1-csc(d*x+c))/(a+b))^( 
1/2)*(a*(1+csc(d*x+c))/(a-b))^(1/2)*EllipticPi((a+b*sin(d*x+c))^(1/2)/(a+b 
)^(1/2)/sin(d*x+c)^(1/2),(a+b)/b,(-(a+b)/(a-b))^(1/2))*tan(d*x+c)/b/d
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 29.28 (sec) , antiderivative size = 10847, normalized size of antiderivative = 29.24 \[ \int \sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)} \, dx=\text {Result too large to show} \] Input:

Integrate[Sqrt[Sin[c + d*x]]*Sqrt[a + b*Sin[c + d*x]],x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3300, 27, 3042, 3533, 27, 3042, 3280, 3042, 3288, 3295, 3473}

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.

\(\displaystyle \int \sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)}dx\)

\(\Big \downarrow \) 3300

\(\displaystyle \frac {\int -\frac {a b-a b \sin ^2(c+d x)}{2 \sin ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sin (c+d x)}}dx}{b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {a b-a b \sin ^2(c+d x)}{\sin ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sin (c+d x)}}dx}{2 b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {a b-a b \sin (c+d x)^2}{\sin (c+d x)^{3/2} \sqrt {a+b \sin (c+d x)}}dx}{2 b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 3533

\(\displaystyle -\frac {\int \frac {a b}{\sin ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sin (c+d x)}}dx-a b \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {a+b \sin (c+d x)}}dx}{2 b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a b \int \frac {1}{\sin (c+d x)^{3/2} \sqrt {a+b \sin (c+d x)}}dx-a b \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {a+b \sin (c+d x)}}dx}{2 b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 3280

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

\(\displaystyle -\frac {a b \left (\int \frac {\sin (c+d x)+1}{\sin (c+d x)^{3/2} \sqrt {a+b \sin (c+d x)}}dx-\int \frac {1}{\sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 a \sqrt {a+b} \tan (c+d x) \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (\csc (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 3295

\(\displaystyle -\frac {a b \left (\int \frac {\sin (c+d x)+1}{\sin (c+d x)^{3/2} \sqrt {a+b \sin (c+d x)}}dx+\frac {2 \sqrt {a+b} \tan (c+d x) \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (\csc (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )-\frac {2 a \sqrt {a+b} \tan (c+d x) \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (\csc (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

\(\Big \downarrow \) 3473

\(\displaystyle -\frac {a b \left (\frac {2 \sqrt {a+b} \tan (c+d x) \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (\csc (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} \tan (c+d x) \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (\csc (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}\right )-\frac {2 a \sqrt {a+b} \tan (c+d x) \sqrt {\frac {a (1-\csc (c+d x))}{a+b}} \sqrt {\frac {a (\csc (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b} \sqrt {\sin (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\)

Input:

Int[Sqrt[Sin[c + d*x]]*Sqrt[a + b*Sin[c + d*x]],x]
 

Output:

-((Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(d*Sqrt[Sin[c + d*x]])) - ((-2*a 
*Sqrt[a + b]*Sqrt[(a*(1 - Csc[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Csc[c + d*x 
]))/(a - b)]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Sin[c + d*x]]/(Sqrt[a 
 + b]*Sqrt[Sin[c + d*x]])], -((a + b)/(a - b))]*Tan[c + d*x])/d + a*b*((-2 
*(a - b)*Sqrt[a + b]*Sqrt[(a*(1 - Csc[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Csc 
[c + d*x]))/(a - b)]*EllipticE[ArcSin[Sqrt[a + b*Sin[c + d*x]]/(Sqrt[a + b 
]*Sqrt[Sin[c + d*x]])], -((a + b)/(a - b))]*Tan[c + d*x])/(a^2*d) + (2*Sqr 
t[a + b]*Sqrt[(a*(1 - Csc[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Csc[c + d*x]))/ 
(a - b)]*EllipticF[ArcSin[Sqrt[a + b*Sin[c + d*x]]/(Sqrt[a + b]*Sqrt[Sin[c 
 + d*x]])], -((a + b)/(a - b))]*Tan[c + d*x])/(a*d)))/(2*b)
 

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 3280

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin 
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[1/(a - b)   Int[1/(Sqrt[a + b*Sin 
[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Simp[b/(a - b)   Int[(1 + Si 
n[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*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] && 
 NeQ[c^2 - d^2, 0]
 

rule 3288

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

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3300

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

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3533

Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] : 
> Simp[C/b^2   Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x 
] + Simp[1/b^2   Int[(A*b^2 - a^2*C - 2*a*b*C*Sin[e + f*x])/((a + b*Sin[e + 
 f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, 
A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.52 (sec) , antiderivative size = 4217, normalized size of antiderivative = 11.37

method	result	size
default	\(\text {Expression too large to display}\)	\(4217\)

Input:

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

Output:

-1/2/d*(I*(cos(d*x+c)+1)*(-a^2+b^2)^(1/2)*((cot(d*x+c)*a+(-a^2+b^2)^(1/2)- 
b-csc(d*x+c)*a)/(-a^2+b^2)^(1/2))^(1/2)*(-(csc(d*x+c)-cot(d*x+c))/(b+(-a^2 
+b^2)^(1/2))*a)^(1/2)*((-cot(d*x+c)*a+(-a^2+b^2)^(1/2)+b+csc(d*x+c)*a)/(b+ 
(-a^2+b^2)^(1/2)))^(1/2)*a^3*EllipticPi(((-cot(d*x+c)*a+(-a^2+b^2)^(1/2)+b 
+csc(d*x+c)*a)/(b+(-a^2+b^2)^(1/2)))^(1/2),(b+(-a^2+b^2)^(1/2))/(I*a+(-a^2 
+b^2)^(1/2)+b),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))+ 
I*(cos(d*x+c)+1)*((cot(d*x+c)*a+(-a^2+b^2)^(1/2)-b-csc(d*x+c)*a)/(-a^2+b^2 
)^(1/2))^(1/2)*(-(csc(d*x+c)-cot(d*x+c))/(b+(-a^2+b^2)^(1/2))*a)^(1/2)*((- 
cot(d*x+c)*a+(-a^2+b^2)^(1/2)+b+csc(d*x+c)*a)/(b+(-a^2+b^2)^(1/2)))^(1/2)* 
a^3*b*EllipticPi(((-cot(d*x+c)*a+(-a^2+b^2)^(1/2)+b+csc(d*x+c)*a)/(b+(-a^2 
+b^2)^(1/2)))^(1/2),(b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)+b),1/2*2^(1 
/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))+(-2*cos(d*x+c)-2)*(-a^2 
+b^2)^(1/2)*((cot(d*x+c)*a+(-a^2+b^2)^(1/2)-b-csc(d*x+c)*a)/(-a^2+b^2)^(1/ 
2))^(1/2)*(-(csc(d*x+c)-cot(d*x+c))/(b+(-a^2+b^2)^(1/2))*a)^(1/2)*((-cot(d 
*x+c)*a+(-a^2+b^2)^(1/2)+b+csc(d*x+c)*a)/(b+(-a^2+b^2)^(1/2)))^(1/2)*a^2*b 
*EllipticPi(((-cot(d*x+c)*a+(-a^2+b^2)^(1/2)+b+csc(d*x+c)*a)/(b+(-a^2+b^2) 
^(1/2)))^(1/2),-(b+(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)-b),1/2*2^(1/2)* 
((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))+(cos(d*x+c)+1)*((cot(d*x+c) 
*a+(-a^2+b^2)^(1/2)-b-csc(d*x+c)*a)/(-a^2+b^2)^(1/2))^(1/2)*(-(csc(d*x+c)- 
cot(d*x+c))/(b+(-a^2+b^2)^(1/2))*a)^(1/2)*((-cot(d*x+c)*a+(-a^2+b^2)^(1...
 

Fricas [F(-1)]

Timed out. \[ \int \sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

integrate(sin(d*x+c)**(1/2)*(a+b*sin(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(a + b*sin(c + d*x))*sqrt(sin(c + d*x)), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(b*sin(d*x + c) + a)*sqrt(sin(d*x + c)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \sqrt {\sin (c+d x)} \sqrt {a+b \sin (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

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

int(sin(c + d*x)^(1/2)*(a + b*sin(c + d*x))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(sin(c + d*x)^(1/2)*(a + b*sin(c + d*x))^(1/2), x)
 

Reduce [F]

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

int(sin(d*x+c)^(1/2)*(a+b*sin(d*x+c))^(1/2),x)
 

Output:

int(sqrt(sin(c + d*x))*sqrt(sin(c + d*x)*b + a),x)