3.3.74 \(\int \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)} \, dx\) [274]

3.3.74.1 Optimal result
3.3.74.2 Mathematica [C] (verified)
3.3.74.3 Rubi [A] (verified)
3.3.74.4 Maple [A] (verified)
3.3.74.5 Fricas [C] (verification not implemented)
3.3.74.6 Sympy [F]
3.3.74.7 Maxima [F]
3.3.74.8 Giac [F]
3.3.74.9 Mupad [F(-1)]

3.3.74.1 Optimal result

Integrand size = 27, antiderivative size = 194 \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)} \, dx=-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a+a \sin (c+d x)}}+\frac {\sqrt {e} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d (1+\cos (c+d x)+\sin (c+d x))}+\frac {\sqrt {e} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d (1+\cos (c+d x)+\sin (c+d x))} \]

output
-a*(e*cos(d*x+c))^(3/2)/d/e/(a+a*sin(d*x+c))^(1/2)+arcsinh((e*cos(d*x+c))^ 
(1/2)/e^(1/2))*e^(1/2)*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/(1+co 
s(d*x+c)+sin(d*x+c))+arctan(sin(d*x+c)*e^(1/2)/(e*cos(d*x+c))^(1/2)/(1+cos 
(d*x+c))^(1/2))*e^(1/2)*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/(1+c 
os(d*x+c)+sin(d*x+c))
 
3.3.74.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.40 \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)} \, dx=-\frac {4 \sqrt [4]{2} (e \cos (c+d x))^{3/2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {a (1+\sin (c+d x))}}{3 d e (1+\sin (c+d x))^{5/4}} \]

input
Integrate[Sqrt[e*Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]],x]
 
output
(-4*2^(1/4)*(e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[-1/4, 3/4, 7/4, (1 - 
Sin[c + d*x])/2]*Sqrt[a*(1 + Sin[c + d*x])])/(3*d*e*(1 + Sin[c + d*x])^(5/ 
4))
 
3.3.74.3 Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3157, 3042, 3163, 3042, 25, 3254, 216, 3312, 63, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3157

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3163

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 3312

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

\(\Big \downarrow \) 63

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

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{2} a \left (\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\)

input
Int[Sqrt[e*Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]],x]
 
output
-((a*(e*Cos[c + d*x])^(3/2))/(d*e*Sqrt[a + a*Sin[c + d*x]])) + (a*((2*Sqrt 
[e]*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + 
a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x])) + (2*Sqrt[e]*Ar 
cTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])] 
*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + 
 a*Sin[c + d*x]))))/2
 

3.3.74.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 63
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[c, 0]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 222
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3157
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + p, 0] && Integers 
Q[2*m, 2*p]
 

rule 3163
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e + f*x 
]]/(a + a*Cos[e + f*x] + b*Sin[e + f*x]))   Int[Sqrt[1 + Cos[e + f*x]]/Sqrt 
[g*Cos[e + f*x]], x], x] - Simp[g*Sqrt[1 + Cos[e + f*x]]*(Sqrt[a + b*Sin[e 
+ f*x]]/(b + b*Cos[e + f*x] + a*Sin[e + f*x]))   Int[Sin[e + f*x]/(Sqrt[g*C 
os[e + f*x]]*Sqrt[1 + Cos[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, g}, x] & 
& EqQ[a^2 - b^2, 0]
 

rule 3254
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b + d*x^2), x], 
x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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 3312
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( 
c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f)   Su 
bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x]
 
3.3.74.4 Maple [A] (verified)

Time = 6.70 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.42

method result size
default \(-\frac {\sqrt {e \cos \left (d x +c \right )}\, \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+\cos \left (d x +c \right )-\sin \left (d x +c \right )+\sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-\sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+1\right )}{d \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right )}\) \(276\)

input
int((e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/d*(e*cos(d*x+c))^(1/2)*(a*(1+sin(d*x+c)))^(1/2)/(1+cos(d*x+c)+sin(d*x+c 
))*((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan((-cos(d*x+c)/(1+cos(d*x+c))) 
^(1/2))-(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(sin(d*x+c)/(1+cos(d*x+c 
))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+cos(d*x+c)-sin(d*x+c)+sec(d*x+c)*(- 
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2) 
)-sec(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(sin(d*x+c)/(1+cos( 
d*x+c))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+1)
 
3.3.74.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 966, normalized size of antiderivative = 4.98 \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

input
integrate((e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas 
")
 
output
1/4*((d*cos(d*x + c) + d*sin(d*x + c) + d)*(-a^2*e^2/d^4)^(1/4)*log(1/2*(2 
*(a*e*sin(d*x + c) + (d^2*cos(d*x + c) + d^2)*sqrt(-a^2*e^2/d^4))*sqrt(e*c 
os(d*x + c))*sqrt(a*sin(d*x + c) + a) + (2*d^3*cos(d*x + c)^2 + d^3*cos(d* 
x + c) - d^3*sin(d*x + c) - d^3)*(-a^2*e^2/d^4)^(3/4) + (a*d*e*cos(d*x + c 
) + a*d*e + (2*a*d*e*cos(d*x + c) + a*d*e)*sin(d*x + c))*(-a^2*e^2/d^4)^(1 
/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - (d*cos(d*x + c) + d*sin(d*x + c) 
 + d)*(-a^2*e^2/d^4)^(1/4)*log(1/2*(2*(a*e*sin(d*x + c) + (d^2*cos(d*x + c 
) + d^2)*sqrt(-a^2*e^2/d^4))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) 
 - (2*d^3*cos(d*x + c)^2 + d^3*cos(d*x + c) - d^3*sin(d*x + c) - d^3)*(-a^ 
2*e^2/d^4)^(3/4) - (a*d*e*cos(d*x + c) + a*d*e + (2*a*d*e*cos(d*x + c) + a 
*d*e)*sin(d*x + c))*(-a^2*e^2/d^4)^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1 
)) + (I*d*cos(d*x + c) + I*d*sin(d*x + c) + I*d)*(-a^2*e^2/d^4)^(1/4)*log( 
1/2*(2*(a*e*sin(d*x + c) - (d^2*cos(d*x + c) + d^2)*sqrt(-a^2*e^2/d^4))*sq 
rt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) - (2*I*d^3*cos(d*x + c)^2 + I* 
d^3*cos(d*x + c) - I*d^3*sin(d*x + c) - I*d^3)*(-a^2*e^2/d^4)^(3/4) + (I*a 
*d*e*cos(d*x + c) + I*a*d*e + (2*I*a*d*e*cos(d*x + c) + I*a*d*e)*sin(d*x + 
 c))*(-a^2*e^2/d^4)^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) + (-I*d*cos( 
d*x + c) - I*d*sin(d*x + c) - I*d)*(-a^2*e^2/d^4)^(1/4)*log(1/2*(2*(a*e*si 
n(d*x + c) - (d^2*cos(d*x + c) + d^2)*sqrt(-a^2*e^2/d^4))*sqrt(e*cos(d*x + 
 c))*sqrt(a*sin(d*x + c) + a) - (-2*I*d^3*cos(d*x + c)^2 - I*d^3*cos(d*...
 
3.3.74.6 Sympy [F]

\[ \int \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sqrt {e \cos {\left (c + d x \right )}}\, dx \]

input
integrate((e*cos(d*x+c))**(1/2)*(a+a*sin(d*x+c))**(1/2),x)
 
output
Integral(sqrt(a*(sin(c + d*x) + 1))*sqrt(e*cos(c + d*x)), x)
 
3.3.74.7 Maxima [F]

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

input
integrate((e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima 
")
 
output
integrate(sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a), x)
 
3.3.74.8 Giac [F]

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

input
integrate((e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 
output
integrate(sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a), x)
 
3.3.74.9 Mupad [F(-1)]

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

input
int((e*cos(c + d*x))^(1/2)*(a + a*sin(c + d*x))^(1/2),x)
 
output
int((e*cos(c + d*x))^(1/2)*(a + a*sin(c + d*x))^(1/2), x)