\(\int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx\) [517]

Optimal result

Integrand size = 20, antiderivative size = 212 \[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}+\frac {2 \sqrt {2} a E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {\left (4 a^2-b^2\right ) \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}{6 \sqrt {2} d \sqrt {2 a+b \sin (2 c+2 d x)}} \] Output:

-1/12*b*cos(2*d*x+2*c)*(2*a+b*sin(2*d*x+2*c))^(1/2)*2^(1/2)/d-2/3*2^(1/2)* 
a*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2)*(b/(2*a+b))^(1/2))*(2*a+b*sin(2*d*x+ 
2*c))^(1/2)/d/((2*a+b*sin(2*d*x+2*c))/(2*a+b))^(1/2)-1/12*(4*a^2-b^2)*Inve 
rseJacobiAM(c-1/4*Pi+d*x,2^(1/2)*(b/(2*a+b))^(1/2))*((2*a+b*sin(2*d*x+2*c) 
)/(2*a+b))^(1/2)*2^(1/2)/d/(2*a+b*sin(2*d*x+2*c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.79 \[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=\frac {-b \cos (2 (c+d x)) (2 a+b \sin (2 (c+d x)))+8 a (2 a+b) E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}-\left (4 a^2-b^2\right ) \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}}{6 d \sqrt {4 a+2 b \sin (2 (c+d x))}} \] Input:

Integrate[(a + b*Cos[c + d*x]*Sin[c + d*x])^(3/2),x]
 

Output:

(-(b*Cos[2*(c + d*x)]*(2*a + b*Sin[2*(c + d*x)])) + 8*a*(2*a + b)*Elliptic 
E[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*(c + d*x)])/(2*a + 
b)] - (4*a^2 - b^2)*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + 
 b*Sin[2*(c + d*x)])/(2*a + b)])/(6*d*Sqrt[4*a + 2*b*Sin[2*(c + d*x)]])
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3042, 3145, 3042, 3135, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3145

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3135

\(\displaystyle \frac {2}{3} \int \frac {12 a^2+8 b \sin (2 c+2 d x) a+b^2}{4 \sqrt {2} \sqrt {2 a+b \sin (2 c+2 d x)}}dx-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {12 a^2+8 b \sin (2 c+2 d x) a+b^2}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {12 a^2+8 b \sin (2 c+2 d x) a+b^2}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {8 a \int \sqrt {2 a+b \sin (2 c+2 d x)}dx-\left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 a \int \sqrt {2 a+b \sin (2 c+2 d x)}dx-\left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\frac {8 a \sqrt {2 a+b \sin (2 c+2 d x)} \int \sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}dx}{\sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {8 a \sqrt {2 a+b \sin (2 c+2 d x)} \int \sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}dx}{\sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\frac {8 a \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\frac {8 a \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {\left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \int \frac {1}{\sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}}dx}{\sqrt {2 a+b \sin (2 c+2 d x)}}}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {8 a \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {\left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \int \frac {1}{\sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}}dx}{\sqrt {2 a+b \sin (2 c+2 d x)}}}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\frac {8 a \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {\left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},\frac {2 b}{2 a+b}\right )}{d \sqrt {2 a+b \sin (2 c+2 d x)}}}{6 \sqrt {2}}-\frac {b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{6 \sqrt {2} d}\)

Input:

Int[(a + b*Cos[c + d*x]*Sin[c + d*x])^(3/2),x]
 

Output:

-1/6*(b*Cos[2*c + 2*d*x]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(Sqrt[2]*d) + ((8 
*a*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[2*a + b*Sin[2*c + 2*d*x 
]])/(d*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)]) - ((4*a^2 - b^2)*Ellipt 
icF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a 
+ b)])/(d*Sqrt[2*a + b*Sin[2*c + 2*d*x]]))/(6*Sqrt[2])
 

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 3135

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[1/n   Int[(a + b* 
Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*x] 
, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && 
 IntegerQ[2*n]
 

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 3145

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ 
Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, 
 x]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(843\) vs. \(2(197)=394\).

Time = 0.62 (sec) , antiderivative size = 844, normalized size of antiderivative = 3.98

method	result	size
default	\(\frac {24 a^{3} \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right )+4 \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, a^{2} b -6 \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) b^{2} a -\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) b^{3}-32 \operatorname {EllipticE}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, a^{3}+8 \operatorname {EllipticE}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, a \,b^{2}+\sin \left (2 d x +2 c \right )^{3} b^{3}+2 \sin \left (2 d x +2 c \right )^{2} a \,b^{2}-b^{3} \sin \left (2 d x +2 c \right )-2 a \,b^{2}}{6 b \cos \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sin \left (2 d x +2 c \right )}\, d}\)	\(844\)

Input:

int((a+b*cos(d*x+c)*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(24*a^3*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-b*(sin(2*d*x+2*c)-1)/ 
(2*a+b))^(1/2)*(-b*(1+sin(2*d*x+2*c))/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin 
(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))+4*EllipticF(((2*a+b*s 
in(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*((2*a+b*sin(2*d*x+2 
*c))/(2*a-b))^(1/2)*(-b*(sin(2*d*x+2*c)-1)/(2*a+b))^(1/2)*(-b*(1+sin(2*d*x 
+2*c))/(2*a-b))^(1/2)*a^2*b-6*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-b*( 
sin(2*d*x+2*c)-1)/(2*a+b))^(1/2)*(-b*(1+sin(2*d*x+2*c))/(2*a-b))^(1/2)*Ell 
ipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b^2 
*a-((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-b*(sin(2*d*x+2*c)-1)/(2*a+b))^ 
(1/2)*(-b*(1+sin(2*d*x+2*c))/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2* 
c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b^3-32*EllipticE(((2*a+b*sin(2 
*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*((2*a+b*sin(2*d*x+2*c)) 
/(2*a-b))^(1/2)*(-b*(sin(2*d*x+2*c)-1)/(2*a+b))^(1/2)*(-b*(1+sin(2*d*x+2*c 
))/(2*a-b))^(1/2)*a^3+8*EllipticE(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),( 
(2*a-b)/(2*a+b))^(1/2))*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-b*(sin(2* 
d*x+2*c)-1)/(2*a+b))^(1/2)*(-b*(1+sin(2*d*x+2*c))/(2*a-b))^(1/2)*a*b^2+sin 
(2*d*x+2*c)^3*b^3+2*sin(2*d*x+2*c)^2*a*b^2-b^3*sin(2*d*x+2*c)-2*a*b^2)/b/c 
os(2*d*x+2*c)/(4*a+2*b*sin(2*d*x+2*c))^(1/2)/d
 

Fricas [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

integral((b*cos(d*x + c)*sin(d*x + c) + a)^(3/2), x)
 

Sympy [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=\int \left (a + b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))**(3/2),x)
 

Output:

Integral((a + b*sin(c + d*x)*cos(c + d*x))**(3/2), x)
 

Maxima [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=\int {\left (a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:

int((a + b*cos(c + d*x)*sin(c + d*x))^(3/2),x)
 

Output:

int((a + b*cos(c + d*x)*sin(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx=\left (\int \sqrt {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )d x \right ) b \] Input:

int((a+b*cos(d*x+c)*sin(d*x+c))^(3/2),x)
 

Output:

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