\(\int \cos ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) [480]

Optimal result

Integrand size = 23, antiderivative size = 215 \[ \int \cos ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {4 \left (a^2+3 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{15 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{15 b^2 d \sqrt {a+b \sin (c+d x)}} \]

[Out]

2/5*cos(d*x+c)*(a+b*sin(d*x+c))^(3/2)/b/d-4/15*a*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/b/d-4/15*(a^2+3*b^2)*(sin(1
/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^
(1/2))*(a+b*sin(d*x+c))^(1/2)/b^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)+4/15*a*(a^2-b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)
^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x
+c))/(a+b))^(1/2)/b^2/d/(a+b*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2774, 2832, 2831, 2742, 2740, 2734, 2732} \[ \int \cos ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {4 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {4 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d} \]

[In]

Int[Cos[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-4*a*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(15*b*d) + (2*Cos[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(5*b*d) +
(4*(a^2 + 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*b^2*d*Sqrt[(a + b*
Sin[c + d*x])/(a + b)]) - (4*a*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*
x])/(a + b)])/(15*b^2*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2732

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 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[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 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/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 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[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 2774

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

Rule 2831

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[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 2832

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] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {2 \int (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)} \, dx}{5 b} \\ & = -\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {4 \int \frac {2 a b+\frac {1}{2} \left (a^2+3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 b} \\ & = -\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac {\left (2 a \left (a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 b^2}+\frac {\left (2 \left (a^2+3 b^2\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{15 b^2} \\ & = -\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {\left (2 \left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{15 b^2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (2 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{15 b^2 \sqrt {a+b \sin (c+d x)}} \\ & = -\frac {4 a \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 b d}+\frac {2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}+\frac {4 \left (a^2+3 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{15 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{15 b^2 d \sqrt {a+b \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {-4 \left (a^3+a^2 b+3 a b^2+3 b^3\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+4 a \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+b \cos (c+d x) \left (2 a^2+3 b^2-3 b^2 \cos (2 (c+d x))+8 a b \sin (c+d x)\right )}{15 b^2 d \sqrt {a+b \sin (c+d x)}} \]

[In]

Integrate[Cos[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-4*(a^3 + a^2*b + 3*a*b^2 + 3*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/
(a + b)] + 4*a*(a^2 - b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)]
+ b*Cos[c + d*x]*(2*a^2 + 3*b^2 - 3*b^2*Cos[2*(c + d*x)] + 8*a*b*Sin[c + d*x]))/(15*b^2*d*Sqrt[a + b*Sin[c + d
*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(791\) vs. \(2(261)=522\).

Time = 2.11 (sec) , antiderivative size = 792, normalized size of antiderivative = 3.68

method	result	size
default	\(\frac {\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b}{15}+\frac {4 a^{2} \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}}{5}-\frac {4 a \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}}{15}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{4}}{5}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, E\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4}}{15}-\frac {8 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, E\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{2}}{15}+\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, E\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{4}}{5}-\frac {2 b^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{5}-\frac {8 a \,b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{15}-\frac {2 a^{2} b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{15}+\frac {2 b^{4} \left (\sin ^{2}\left (d x +c \right )\right )}{5}+\frac {8 a \,b^{3} \sin \left (d x +c \right )}{15}+\frac {2 a^{2} b^{2}}{15}}{b^{3} \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\)	\(792\)

[In]

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

[Out]

2/15*(2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b+6*a^2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c
)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/
2))*b^2-2*a*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ell
ipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^3-6*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-
1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2)
)*b^4-2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4-4*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b
/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^
2*b^2+6*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^4-3*b^4*sin(d*x+c)^4-4*a*b^3*sin(d*x+c)^3-a^2*b^2*sin
(d*x+c)^2+3*b^4*sin(d*x+c)^2+4*a*b^3*sin(d*x+c)+a^2*b^2)/b^3/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.13 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.08 \[ \int \cos ^2(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {2 \, {\left (2 \, \sqrt {2} {\left (a^{3} - 3 \, a b^{2}\right )} \sqrt {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 (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right ) + 2 \, \sqrt {2} {\left (a^{3} - 3 \, a b^{2}\right )} \sqrt {-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 (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right ) + 3 \, \sqrt {2} {\left (i \, a^{2} b + 3 i \, b^{3}\right )} \sqrt {i \, 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 (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right )\right ) + 3 \, \sqrt {2} {\left (-i \, a^{2} b - 3 i \, b^{3}\right )} \sqrt {-i \, 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 (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right ) - 3 \, {\left (3 \, b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a b^{2} \cos \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}\right )}}{45 \, b^{3} d} \]

[In]

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

[Out]

-2/45*(2*sqrt(2)*(a^3 - 3*a*b^2)*sqrt(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(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b) + 2*sqrt(2)*(a^3 - 3*a*b^2)*sqrt(-I*b)*weie
rstrassPInverse(-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(d*x + c) + 3*I*b*sin(
d*x + c) + 2*I*a)/b) + 3*sqrt(2)*(I*a^2*b + 3*I*b^3)*sqrt(I*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(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b)) + 3*sqrt(2)*(-I*a^2*b - 3*I*b^3)*sqrt(-I*b)*weierstrassZe
ta(-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(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) - 3*(3*b^3*cos(d*x +
c)*sin(d*x + c) + a*b^2*cos(d*x + c))*sqrt(b*sin(d*x + c) + a))/(b^3*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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