\(\int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx\) [499]

Optimal result

Integrand size = 27, antiderivative size = 239 \[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=\frac {4 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}-\frac {4 a^2 \left (c^2-5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c-5 d) \left (c^2-d^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^2 f \sqrt {c+d \sin (e+f x)}} \] Output:

4/15*a^2*(c-5*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d/f-2/5*a^2*cos(f*x+e)* 
(c+d*sin(f*x+e))^(3/2)/d/f+4/15*a^2*(c^2-5*c*d-12*d^2)*EllipticE(cos(1/2*e 
+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/d^2/f/((c 
+d*sin(f*x+e))/(c+d))^(1/2)+4/15*a^2*(c-5*d)*(c^2-d^2)*InverseJacobiAM(1/2 
*e-1/4*Pi+1/2*f*x,2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/ 
d^2/f/(c+d*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.02 \[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=-\frac {a^2 (1+\sin (e+f x))^2 \left (-4 \left (c^3-4 c^2 d-17 c d^2-12 d^3\right ) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+4 \left (c^3-5 c^2 d-c d^2+5 d^3\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-d \cos (e+f x) \left (-2 c^2-20 c d-3 d^2+3 d^2 \cos (2 (e+f x))-4 d (2 c+5 d) \sin (e+f x)\right )\right )}{15 d^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {c+d \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]],x]
 

Output:

-1/15*(a^2*(1 + Sin[e + f*x])^2*(-4*(c^3 - 4*c^2*d - 17*c*d^2 - 12*d^3)*El 
lipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c 
 + d)] + 4*(c^3 - 5*c^2*d - c*d^2 + 5*d^3)*EllipticF[(-2*e + Pi - 2*f*x)/4 
, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - d*Cos[e + f*x]*(-2*c 
^2 - 20*c*d - 3*d^2 + 3*d^2*Cos[2*(e + f*x)] - 4*d*(2*c + 5*d)*Sin[e + f*x 
])))/(d^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sqrt[c + d*Sin[e + f*x 
]])
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 3242, 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.

\(\displaystyle \int (a \sin (e+f x)+a)^2 \sqrt {c+d \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^2 \sqrt {c+d \sin (e+f x)}dx\)

\(\Big \downarrow \) 3242

\(\displaystyle \frac {2 \int \left (4 a^2 d-a^2 (c-5 d) \sin (e+f x)\right ) \sqrt {c+d \sin (e+f x)}dx}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \left (4 a^2 d-a^2 (c-5 d) \sin (e+f x)\right ) \sqrt {c+d \sin (e+f x)}dx}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {2 \left (\frac {2}{3} \int \frac {a^2 d (11 c+5 d)-a^2 \left (c^2-5 d c-12 d^2\right ) \sin (e+f x)}{2 \sqrt {c+d \sin (e+f x)}}dx+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {1}{3} \int \frac {a^2 d (11 c+5 d)-a^2 \left (c^2-5 d c-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {1}{3} \int \frac {a^2 d (11 c+5 d)-a^2 \left (c^2-5 d c-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {a^2 (c-5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a^2 \left (c^2-5 c d-12 d^2\right ) \int \sqrt {c+d \sin (e+f x)}dx}{d}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {a^2 (c-5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a^2 \left (c^2-5 c d-12 d^2\right ) \int \sqrt {c+d \sin (e+f x)}dx}{d}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {a^2 (c-5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {a^2 (c-5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {a^2 (c-5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {2 a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {a^2 (c-5 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {a^2 (c-5 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {2 \left (\frac {1}{3} \left (\frac {2 a^2 (c-5 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{d f \sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 \left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {2 a^2 (c-5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}\right )}{5 d}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

Input:

Int[(a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]],x]
 

Output:

(-2*a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(5*d*f) + (2*((2*a^2*(c - 
 5*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*f) + ((-2*a^2*(c^2 - 5*c*d 
 - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + 
 f*x]])/(d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*a^2*(c - 5*d)*(c^2 - 
 d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x 
])/(c + d)])/(d*f*Sqrt[c + d*Sin[e + f*x]]))/3))/(5*d)
 

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]
 

rule 3242

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 - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*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] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1034\) vs. \(2(224)=448\).

Time = 3.07 (sec) , antiderivative size = 1035, normalized size of antiderivative = 4.33

method	result	size
default	\(\text {Expression too large to display}\)	\(1035\)
parts	\(\text {Expression too large to display}\)	\(1499\)

Input:

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

Output:

-2/15*a^2*(2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-d*(-1+sin(f*x+e))/(c+d))^(1/ 
2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2 
),((c-d)/(c+d))^(1/2))*c^3*d-34*c^2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-d*(-1 
+sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d* 
sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^2-2*c*((c+d*sin(f*x+e))/(c 
-d))^(1/2)*(-d*(-1+sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2 
)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^3+34*((c 
+d*sin(f*x+e))/(c-d))^(1/2)*(-d*(-1+sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f* 
x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^ 
(1/2))*d^4-2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-d*(-1+sin(f*x+e))/(c+d))^(1/ 
2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2 
),((c-d)/(c+d))^(1/2))*c^4+10*(-d*(-1+sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin( 
f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d) 
)^(1/2))*((c+d*sin(f*x+e))/(c-d))^(1/2)*c^3*d+26*((c+d*sin(f*x+e))/(c-d))^ 
(1/2)*(-d*(-1+sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*Ell 
ipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^2*d^2-10*(-d* 
(-1+sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c 
+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c-d))^ 
(1/2)*c*d^3-24*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-d*(-1+sin(f*x+e))/(c+d))^( 
1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.23 \[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

2/45*(2*(2*a^2*c^3 - 10*a^2*c^2*d + 9*a^2*c*d^2 + 15*a^2*d^3)*sqrt(1/2*I*d 
)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2 
)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d) + 2*(2*a^2*c 
^3 - 10*a^2*c^2*d + 9*a^2*c*d^2 + 15*a^2*d^3)*sqrt(-1/2*I*d)*weierstrassPI 
nverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3* 
d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) - 6*(-I*a^2*c^2*d + 5*I*a^ 
2*c*d^2 + 12*I*a^2*d^3)*sqrt(1/2*I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2) 
/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3 
*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d* 
sin(f*x + e) - 2*I*c)/d)) - 6*(I*a^2*c^2*d - 5*I*a^2*c*d^2 - 12*I*a^2*d^3) 
*sqrt(-1/2*I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 
+ 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8* 
I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c 
)/d)) - 3*(3*a^2*d^3*cos(f*x + e)*sin(f*x + e) + (a^2*c*d^2 + 10*a^2*d^3)* 
cos(f*x + e))*sqrt(d*sin(f*x + e) + c))/(d^3*f)
 

Sympy [F]

\[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=a^{2} \left (\int 2 \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int \sqrt {c + d \sin {\left (e + f x \right )}}\, dx\right ) \] Input:

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

Output:

a**2*(Integral(2*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integral(sqrt 
(c + d*sin(e + f*x))*sin(e + f*x)**2, x) + Integral(sqrt(c + d*sin(e + f*x 
)), x))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \sqrt {d \sin \left (f x + e\right ) + c} \,d x } \] Input:

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

Output:

integrate((a*sin(f*x + e) + a)^2*sqrt(d*sin(f*x + e) + c), x)
 

Giac [F]

\[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \sqrt {d \sin \left (f x + e\right ) + c} \,d x } \] Input:

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

Output:

integrate((a*sin(f*x + e) + a)^2*sqrt(d*sin(f*x + e) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,\sqrt {c+d\,\sin \left (e+f\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=a^{2} \left (\int \sqrt {\sin \left (f x +e \right ) d +c}d x +\int \sqrt {\sin \left (f x +e \right ) d +c}\, \sin \left (f x +e \right )^{2}d x +2 \left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \sin \left (f x +e \right )d x \right )\right ) \] Input:

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

Output:

a**2*(int(sqrt(sin(e + f*x)*d + c),x) + int(sqrt(sin(e + f*x)*d + c)*sin(e 
 + f*x)**2,x) + 2*int(sqrt(sin(e + f*x)*d + c)*sin(e + f*x),x))