3.3.15 \(\int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx\) [215]

3.3.15.1 Optimal result
3.3.15.2 Mathematica [C] (verified)
3.3.15.3 Rubi [A] (verified)
3.3.15.4 Maple [A] (verified)
3.3.15.5 Fricas [C] (verification not implemented)
3.3.15.6 Sympy [F(-1)]
3.3.15.7 Maxima [F]
3.3.15.8 Giac [F]
3.3.15.9 Mupad [F(-1)]

3.3.15.1 Optimal result

Integrand size = 25, antiderivative size = 170 \[ \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx=-\frac {10 a^3 (e \cos (c+d x))^{7/2}}{21 d e}+\frac {2 a^3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^3 e (e \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}-\frac {10 (e \cos (c+d x))^{7/2} \left (a^3+a^3 \sin (c+d x)\right )}{33 d e} \]

output
-10/21*a^3*(e*cos(d*x+c))^(7/2)/d/e+2/3*a^3*e*(e*cos(d*x+c))^(3/2)*sin(d*x 
+c)/d-2/11*a*(e*cos(d*x+c))^(7/2)*(a+a*sin(d*x+c))^2/d/e-10/33*(e*cos(d*x+ 
c))^(7/2)*(a^3+a^3*sin(d*x+c))/d/e+2*a^3*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/ 
cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1 
/2)/d/cos(d*x+c)^(1/2)
 
3.3.15.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.39 \[ \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx=-\frac {32\ 2^{3/4} a^3 (e \cos (c+d x))^{7/2} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{7 d e (1+\sin (c+d x))^{7/4}} \]

input
Integrate[(e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^3,x]
 
output
(-32*2^(3/4)*a^3*(e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[-15/4, 7/4, 11/4 
, (1 - Sin[c + d*x])/2])/(7*d*e*(1 + Sin[c + d*x])^(7/4))
 
3.3.15.3 Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3157, 3042, 3157, 3042, 3148, 3042, 3115, 3042, 3121, 3042, 3119}

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.

\(\displaystyle \int (a \sin (c+d x)+a)^3 (e \cos (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (c+d x)+a)^3 (e \cos (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 3157

\(\displaystyle \frac {15}{11} a \int (e \cos (c+d x))^{5/2} (\sin (c+d x) a+a)^2dx-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {15}{11} a \int (e \cos (c+d x))^{5/2} (\sin (c+d x) a+a)^2dx-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3157

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \int (e \cos (c+d x))^{5/2} (\sin (c+d x) a+a)dx-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \int (e \cos (c+d x))^{5/2} (\sin (c+d x) a+a)dx-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3148

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \left (a \int (e \cos (c+d x))^{5/2}dx-\frac {2 a (e \cos (c+d x))^{7/2}}{7 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \left (a \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx-\frac {2 a (e \cos (c+d x))^{7/2}}{7 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \left (a \left (\frac {3}{5} e^2 \int \sqrt {e \cos (c+d x)}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )-\frac {2 a (e \cos (c+d x))^{7/2}}{7 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \left (a \left (\frac {3}{5} e^2 \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )-\frac {2 a (e \cos (c+d x))^{7/2}}{7 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \left (a \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )-\frac {2 a (e \cos (c+d x))^{7/2}}{7 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \left (a \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )-\frac {2 a (e \cos (c+d x))^{7/2}}{7 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {15}{11} a \left (\frac {11}{9} a \left (a \left (\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )-\frac {2 a (e \cos (c+d x))^{7/2}}{7 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{7/2}}{9 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e}\)

input
Int[(e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^3,x]
 
output
(-2*a*(e*Cos[c + d*x])^(7/2)*(a + a*Sin[c + d*x])^2)/(11*d*e) + (15*a*((-2 
*(e*Cos[c + d*x])^(7/2)*(a^2 + a^2*Sin[c + d*x]))/(9*d*e) + (11*a*((-2*a*( 
e*Cos[c + d*x])^(7/2))/(7*d*e) + a*((6*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[ 
(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*e*(e*Cos[c + d*x])^(3/2)*Si 
n[c + d*x])/(5*d))))/9))/11
 

3.3.15.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3148
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 

rule 3157
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + p, 0] && Integers 
Q[2*m, 2*p]
 
3.3.15.4 Maple [A] (verified)

Time = 44.11 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.55

method result size
default \(-\frac {2 a^{3} e^{3} \left (-1344 \left (\sin ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2464 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4032 \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4928 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2928 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3080 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-864 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-616 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1908 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-231 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-804 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+111 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{231 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) \(264\)
parts \(-\frac {2 a^{3} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, e^{3} \left (-8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}+\frac {2 a^{3} \left (\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {e^{2} \left (e \cos \left (d x +c \right )\right )^{\frac {7}{2}}}{7}\right )}{d \,e^{3}}-\frac {6 a^{3} \left (e \cos \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d e}+\frac {4 a^{3} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, e^{3} \left (80 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-240 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+272 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-144 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\) \(502\)

input
int((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
-2/231/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^3*e^3*(-13 
44*sin(1/2*d*x+1/2*c)^13+2464*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+403 
2*sin(1/2*d*x+1/2*c)^11-4928*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-2928* 
sin(1/2*d*x+1/2*c)^9+3080*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-864*sin( 
1/2*d*x+1/2*c)^7-616*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+1908*sin(1/2* 
d*x+1/2*c)^5-231*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^( 
1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-804*sin(1/2*d*x+1/2*c)^3+111*si 
n(1/2*d*x+1/2*c))/d
 
3.3.15.5 Fricas [C] (verification not implemented)

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

Time = 0.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.90 \[ \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx=\frac {231 i \, \sqrt {2} a^{3} e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 231 i \, \sqrt {2} a^{3} e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (21 \, a^{3} e^{2} \cos \left (d x + c\right )^{5} - 132 \, a^{3} e^{2} \cos \left (d x + c\right )^{3} - 77 \, {\left (a^{3} e^{2} \cos \left (d x + c\right )^{3} - a^{3} e^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{231 \, d} \]

input
integrate((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")
 
output
1/231*(231*I*sqrt(2)*a^3*e^(5/2)*weierstrassZeta(-4, 0, weierstrassPInvers 
e(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 231*I*sqrt(2)*a^3*e^(5/2)*weier 
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) 
)) + 2*(21*a^3*e^2*cos(d*x + c)^5 - 132*a^3*e^2*cos(d*x + c)^3 - 77*(a^3*e 
^2*cos(d*x + c)^3 - a^3*e^2*cos(d*x + c))*sin(d*x + c))*sqrt(e*cos(d*x + c 
)))/d
 
3.3.15.6 Sympy [F(-1)]

Timed out. \[ \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

input
integrate((e*cos(d*x+c))**(5/2)*(a+a*sin(d*x+c))**3,x)
 
output
Timed out
 
3.3.15.7 Maxima [F]

\[ \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{3} \,d x } \]

input
integrate((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")
 
output
integrate((e*cos(d*x + c))^(5/2)*(a*sin(d*x + c) + a)^3, x)
 
3.3.15.8 Giac [F]

\[ \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{3} \,d x } \]

input
integrate((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^3,x, algorithm="giac")
 
output
integrate((e*cos(d*x + c))^(5/2)*(a*sin(d*x + c) + a)^3, x)
 
3.3.15.9 Mupad [F(-1)]

Timed out. \[ \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3 \,d x \]

input
int((e*cos(c + d*x))^(5/2)*(a + a*sin(c + d*x))^3,x)
 
output
int((e*cos(c + d*x))^(5/2)*(a + a*sin(c + d*x))^3, x)