\(\int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx\) [516]

Optimal result

Integrand size = 20, antiderivative size = 265 \[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=-\frac {2 \sqrt {2} a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{15 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {\left (92 a^2+9 b^2\right ) E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}{60 \sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {2 \sqrt {2} a \left (4 a^2-b^2\right ) \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}{15 d \sqrt {2 a+b \sin (2 c+2 d x)}} \] Output:

-2/15*2^(1/2)*a*b*cos(2*d*x+2*c)*(2*a+b*sin(2*d*x+2*c))^(1/2)/d-1/40*b*cos 
(2*d*x+2*c)*(2*a+b*sin(2*d*x+2*c))^(3/2)*2^(1/2)/d-1/120*(92*a^2+9*b^2)*El 
lipticE(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)*2^(1/2)/d/((2*a+b*sin(2*d*x+2*c))/(2*a+b))^(1/2)-2/15*2^(1/2)*a*(4 
*a^2-b^2)*InverseJacobiAM(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)/d/(2*a+b*sin(2*d*x+2*c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.82 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.76 \[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=\frac {2 \left (184 a^3+92 a^2 b+18 a b^2+9 b^3\right ) E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}-32 a \left (4 a^2-b^2\right ) \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}-b \left (88 a^2 \cos (2 (c+d x))+b (28 a+3 b \sin (2 (c+d x))) \sin (4 (c+d x))\right )}{120 d \sqrt {4 a+2 b \sin (2 (c+d x))}} \] Input:

Integrate[(a + b*Cos[c + d*x]*Sin[c + d*x])^(5/2),x]
 

Output:

(2*(184*a^3 + 92*a^2*b + 18*a*b^2 + 9*b^3)*EllipticE[c - Pi/4 + d*x, (2*b) 
/(2*a + b)]*Sqrt[(2*a + b*Sin[2*(c + d*x)])/(2*a + b)] - 32*a*(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)] - b*(88*a^2*Cos[2*(c + d*x)] + b*(28*a + 3*b*Sin[2*(c + d*x) 
])*Sin[4*(c + d*x)]))/(120*d*Sqrt[4*a + 2*b*Sin[2*(c + d*x)]])
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.850, Rules used = {3042, 3145, 3042, 3135, 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.

\(\displaystyle \int (a+b \sin (c+d x) \cos (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \sin (c+d x) \cos (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 3145

\(\displaystyle \int \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^{5/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^{5/2}dx\)

\(\Big \downarrow \) 3135

\(\displaystyle \frac {2}{5} \int \frac {\sqrt {2 a+b \sin (2 c+2 d x)} \left (20 a^2+16 b \sin (2 c+2 d x) a+3 b^2\right )}{8 \sqrt {2}}dx-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt {2 a+b \sin (2 c+2 d x)} \left (20 a^2+16 b \sin (2 c+2 d x) a+3 b^2\right )dx}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt {2 a+b \sin (2 c+2 d x)} \left (20 a^2+16 b \sin (2 c+2 d x) a+3 b^2\right )dx}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {2}{3} \int \frac {2 a \left (60 a^2+17 b^2\right )+b \left (92 a^2+9 b^2\right ) \sin (2 c+2 d x)}{2 \sqrt {2 a+b \sin (2 c+2 d x)}}dx-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {2 a \left (60 a^2+17 b^2\right )+b \left (92 a^2+9 b^2\right ) \sin (2 c+2 d x)}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {2 a \left (60 a^2+17 b^2\right )+b \left (92 a^2+9 b^2\right ) \sin (2 c+2 d x)}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {\frac {1}{3} \left (\left (92 a^2+9 b^2\right ) \int \sqrt {2 a+b \sin (2 c+2 d x)}dx-16 a \left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\left (92 a^2+9 b^2\right ) \int \sqrt {2 a+b \sin (2 c+2 d x)}dx-16 a \left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} \int \sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}dx}{\sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-16 a \left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} \int \sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}dx}{\sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-16 a \left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-16 a \left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {2 a+b \sin (2 c+2 d x)}}dx\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {16 a \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \int \frac {1}{\sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}}dx}{\sqrt {2 a+b \sin (2 c+2 d x)}}\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {16 a \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \int \frac {1}{\sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}}dx}{\sqrt {2 a+b \sin (2 c+2 d x)}}\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\frac {1}{3} \left (\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {16 a \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},\frac {2 b}{2 a+b}\right )}{d \sqrt {2 a+b \sin (2 c+2 d x)}}\right )-\frac {16 a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{3 d}}{20 \sqrt {2}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}\)

Input:

Int[(a + b*Cos[c + d*x]*Sin[c + d*x])^(5/2),x]
 

Output:

-1/20*(b*Cos[2*c + 2*d*x]*(2*a + b*Sin[2*c + 2*d*x])^(3/2))/(Sqrt[2]*d) + 
((-16*a*b*Cos[2*c + 2*d*x]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(3*d) + (((92*a 
^2 + 9*b^2)*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)]) - (16*a*(4*a^2 
 - b^2)*EllipticF[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]]))/3)/(20*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]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1137\) vs. \(2(244)=488\).

Time = 1.47 (sec) , antiderivative size = 1138, normalized size of antiderivative = 4.29

method	result	size
default	\(\text {Expression too large to display}\)	\(1138\)

Input:

int((a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/60*(240*a^4*((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*s 
in(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))+64*EllipticF(((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*b-24*a^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)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1 
/2))*b^2-16*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*s 
in(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b^3-9*EllipticF(((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)*b^4-368*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^4+56*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^2*b^2+9*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*si...
 

Fricas [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

integral(-(b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*si 
n(d*x + c) - a^2)*sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^(5/2), x)
 

Giac [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=\int {\left (a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int((a + b*cos(c + d*x)*sin(c + d*x))^(5/2),x)
 

Output:

int((a + b*cos(c + d*x)*sin(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx=\left (\int \sqrt {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}d x \right ) a^{2}+2 \left (\int \sqrt {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )d x \right ) a b +\left (\int \sqrt {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}d x \right ) b^{2} \] Input:

int((a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x)
 

Output:

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