\(\int (a+b \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx\) [739]

Optimal result

Integrand size = 27, antiderivative size = 254 \[ \int (a+b \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d f}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac {2 \left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 b (b c-5 a d) \left (c^2-d^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^2 f \sqrt {c+d \sin (e+f x)}} \] Output:

4/15*b*(-5*a*d+b*c)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d/f-2/5*b^2*cos(f*x+ 
e)*(c+d*sin(f*x+e))^(3/2)/d/f-2/15*(3*(5*a^2+3*b^2)*d^2-2*b*c*(-5*a*d+b*c) 
)*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*b*(-5*a*d+b*c)*(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*s 
in(f*x+e))/(c+d))^(1/2)/d^2/f/(c+d*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.84 \[ \int (a+b \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx=\frac {2 \left (-d^2 \left (15 a^2 c+7 b^2 c+10 a b d\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )+\left (-10 a b c d-15 a^2 d^2+b^2 \left (2 c^2-9 d^2\right )\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-2 b d \cos (e+f x) (c+d \sin (e+f x)) (b c+10 a d+3 b d \sin (e+f x))}{15 d^2 f \sqrt {c+d \sin (e+f x)}} \] Input:

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

Output:

(2*(-(d^2*(15*a^2*c + 7*b^2*c + 10*a*b*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, 
 (2*d)/(c + d)]) + (-10*a*b*c*d - 15*a^2*d^2 + b^2*(2*c^2 - 9*d^2))*((c + 
d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + P 
i - 2*f*x)/4, (2*d)/(c + d)]))*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - 2*b*d* 
Cos[e + f*x]*(c + d*Sin[e + f*x])*(b*c + 10*a*d + 3*b*d*Sin[e + f*x]))/(15 
*d^2*f*Sqrt[c + d*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 3270, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3270

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {2}{3} \int \frac {d \left (15 c a^2+10 b d a+7 b^2 c\right )+\left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) \sin (e+f x)}{2 \sqrt {c+d \sin (e+f x)}}dx+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {d \left (15 c a^2+10 b d a+7 b^2 c\right )+\left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {d \left (15 c a^2+10 b d a+7 b^2 c\right )+\left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \int \sqrt {c+d \sin (e+f x)}dx}{d}+\frac {2 b \left (c^2-d^2\right ) (b c-5 a d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \int \sqrt {c+d \sin (e+f x)}dx}{d}+\frac {2 b \left (c^2-d^2\right ) (b c-5 a d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 b \left (c^2-d^2\right ) (b c-5 a d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 b \left (c^2-d^2\right ) (b c-5 a d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 b \left (c^2-d^2\right ) (b c-5 a d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}+\frac {2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 b \left (c^2-d^2\right ) (b c-5 a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 b \left (c^2-d^2\right ) (b c-5 a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 b \left (c^2-d^2\right ) (b c-5 a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{d f \sqrt {c+d \sin (e+f x)}}\right )+\frac {4 b (b c-5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f}}{5 d}-\frac {2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}\)

Input:

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

Output:

(-2*b^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(5*d*f) + ((4*b*(b*c - 5* 
a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(3*f) + ((2*(3*(5*a^2 + 3*b^2) 
*d^2 - 2*b*c*(b*c - 5*a*d))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*S 
qrt[c + d*Sin[e + f*x]])/(d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*b*( 
b*c - 5*a*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 3270

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1089\) vs. \(2(239)=478\).

Time = 4.23 (sec) , antiderivative size = 1090, normalized size of antiderivative = 4.29

method	result	size
default	\(\text {Expression too large to display}\)	\(1090\)
parts	\(\text {Expression too large to display}\)	\(1498\)

Input:

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

Output:

(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*(2*a^2*c*(c/d-1)*((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)/(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c- 
d))^(1/2),((c-d)/(c+d))^(1/2))+b*(2*a*d+b*c)*(-2/3/d*(-(-c-d*sin(f*x+e))*c 
os(f*x+e)^2)^(1/2)+2/3*(c/d-1)*((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)/(-(-c-d*sin(f*x+e))*cos 
(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1 
/2))-4/3*c/d*(c/d-1)*((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)/(-(-c-d*sin(f*x+e))*cos(f*x+e)^2) 
^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1 
/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+2*a*( 
a*d+2*b*c)*(c/d-1)*((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)/(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^( 
1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2 
))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))+b^2*d*(- 
2/5/d*sin(f*x+e)*(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)+8/15*c/d^2*(-(-c- 
d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)+4/15*c/d*(c/d-1)*((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)/(-( 
-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1 
/2),((c-d)/(c+d))^(1/2))+2*(3/5+8/15*c^2/d^2)*(c/d-1)*((c+d*sin(f*x+e))...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 551, normalized size of antiderivative = 2.17 \[ \int (a+b \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

2/45*((4*b^2*c^3 - 20*a*b*c^2*d + 30*a*b*d^3 + 3*(5*a^2 + b^2)*c*d^2)*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) + (4 
*b^2*c^3 - 20*a*b*c^2*d + 30*a*b*d^3 + 3*(5*a^2 + b^2)*c*d^2)*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) - 3*(-2*I* 
b^2*c^2*d + 10*I*a*b*c*d^2 + 3*I*(5*a^2 + 3*b^2)*d^3)*sqrt(1/2*I*d)*weiers 
trassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, weier 
strassPInverse(-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*(2*I*b^2*c^2*d 
 - 10*I*a*b*c*d^2 - 3*I*(5*a^2 + 3*b^2)*d^3)*sqrt(-1/2*I*d)*weierstrassZet 
a(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrassP 
Inverse(-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*b^2*d^3*cos(f*x + 
 e)*sin(f*x + e) + (b^2*c*d^2 + 10*a*b*d^3)*cos(f*x + e))*sqrt(d*sin(f*x + 
 e) + c))/(d^3*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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