\(\int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx\) [268]

Optimal result

Integrand size = 25, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {2 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^4 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:

2/15*e^2*(e*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d/ 
cos(d*x+c)^(1/2)-4/9*e*(e*cos(d*x+c))^(3/2)/a/d/(a+a*sin(d*x+c))^3+2/15*e* 
(e*cos(d*x+c))^(3/2)/d/(a^2+a^2*sin(d*x+c))^2+2/15*e*(e*cos(d*x+c))^(3/2)/ 
d/(a^4+a^4*sin(d*x+c))
 

Mathematica [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {(e \cos (c+d x))^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {13}{4},\frac {11}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{14 \sqrt [4]{2} a^4 d e (1+\sin (c+d x))^{7/4}} \] Input:

Integrate[(e*Cos[c + d*x])^(5/2)/(a + a*Sin[c + d*x])^4,x]
 

Output:

-1/14*((e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[7/4, 13/4, 11/4, (1 - Sin[ 
c + d*x])/2])/(2^(1/4)*a^4*d*e*(1 + Sin[c + d*x])^(7/4))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3159, 3042, 3160, 3042, 3162, 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.

\(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a \sin (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a \sin (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle -\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)}}{(\sin (c+d x) a+a)^2}dx}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)}}{(\sin (c+d x) a+a)^2}dx}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3160

\(\displaystyle -\frac {e^2 \left (\frac {\int \frac {\sqrt {e \cos (c+d x)}}{\sin (c+d x) a+a}dx}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (\frac {\int \frac {\sqrt {e \cos (c+d x)}}{\sin (c+d x) a+a}dx}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3162

\(\displaystyle -\frac {e^2 \left (\frac {-\frac {\int \sqrt {e \cos (c+d x)}dx}{a}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (\frac {-\frac {\int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {e^2 \left (\frac {-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (\frac {-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {e^2 \left (\frac {-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a d \sqrt {\cos (c+d x)}}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3}\)

Input:

Int[(e*Cos[c + d*x])^(5/2)/(a + a*Sin[c + d*x])^4,x]
 

Output:

(-4*e*(e*Cos[c + d*x])^(3/2))/(9*a*d*(a + a*Sin[c + d*x])^3) - (e^2*((-2*( 
e*Cos[c + d*x])^(3/2))/(5*d*e*(a + a*Sin[c + d*x])^2) + ((-2*Sqrt[e*Cos[c 
+ d*x]]*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]) - (2*(e*Cos[c 
+ d*x])^(3/2))/(d*e*(a + a*Sin[c + d*x])))/(5*a)))/(3*a^2)
 

Defintions of rubi rules used

rule 3042

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

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 3159

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

rule 3160

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/(a*f*g*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1))   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] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & 
& IntegersQ[2*m, 2*p]
 

rule 3162

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[b*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p - 1)*(a + b*S 
in[e + f*x]))), x] + Simp[p/(a*(p - 1))   Int[(g*Cos[e + f*x])^p, x], x] /; 
 FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] &&  !GeQ[p, 1] && Intege 
rQ[2*p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(513\) vs. \(2(138)=276\).

Time = 3.63 (sec) , antiderivative size = 514, normalized size of antiderivative = 3.34

\[-\frac {2 \left (96 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-192 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+96 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+272 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-72 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-176 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+24 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-144 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-42 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \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 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+144 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{3}}{45 \left (16 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-32 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\]

Input:

int((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^4,x)
 

Output:

-2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2* 
c)^4-8*sin(1/2*d*x+1/2*c)^2+1)/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2* 
c)^2*e+e)^(1/2)*(96*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-48*(2*sin(1/2 
*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+ 
1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^8-192*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/ 
2*c)^8+96*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*El 
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^6+272*sin(1/2*d*x+1 
/2*c)^6*cos(1/2*d*x+1/2*c)-72*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c) 
^4-176*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+24*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2 
)*sin(1/2*d*x+1/2*c)^2-144*sin(1/2*d*x+1/2*c)^5-42*sin(1/2*d*x+1/2*c)^2*co 
s(1/2*d*x+1/2*c)-3*(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))+144*sin(1/2*d*x+1/2*c)^3+4*si 
n(1/2*d*x+1/2*c))*e^3/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.60 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} {\left (-i \, e^{2} \cos \left (d x + c\right )^{3} - 3 i \, e^{2} \cos \left (d x + c\right )^{2} + 2 i \, e^{2} \cos \left (d x + c\right ) + 4 i \, e^{2} + {\left (-i \, e^{2} \cos \left (d x + c\right )^{2} + 2 i \, e^{2} \cos \left (d x + c\right ) + 4 i \, e^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (i \, e^{2} \cos \left (d x + c\right )^{3} + 3 i \, e^{2} \cos \left (d x + c\right )^{2} - 2 i \, e^{2} \cos \left (d x + c\right ) - 4 i \, e^{2} + {\left (i \, e^{2} \cos \left (d x + c\right )^{2} - 2 i \, e^{2} \cos \left (d x + c\right ) - 4 i \, e^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, e^{2} \cos \left (d x + c\right )^{3} - 6 \, e^{2} \cos \left (d x + c\right )^{2} + e^{2} \cos \left (d x + c\right ) + 10 \, e^{2} - {\left (3 \, e^{2} \cos \left (d x + c\right )^{2} + 9 \, e^{2} \cos \left (d x + c\right ) + 10 \, e^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{45 \, {\left (a^{4} d \cos \left (d x + c\right )^{3} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \cos \left (d x + c\right ) - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \cos \left (d x + c\right ) - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:

integrate((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
 

Output:

-2/45*(3*sqrt(1/2)*(-I*e^2*cos(d*x + c)^3 - 3*I*e^2*cos(d*x + c)^2 + 2*I*e 
^2*cos(d*x + c) + 4*I*e^2 + (-I*e^2*cos(d*x + c)^2 + 2*I*e^2*cos(d*x + c) 
+ 4*I*e^2)*sin(d*x + c))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInvers 
e(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(1/2)*(I*e^2*cos(d*x + c) 
^3 + 3*I*e^2*cos(d*x + c)^2 - 2*I*e^2*cos(d*x + c) - 4*I*e^2 + (I*e^2*cos( 
d*x + c)^2 - 2*I*e^2*cos(d*x + c) - 4*I*e^2)*sin(d*x + c))*sqrt(e)*weierst 
rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) 
 - (3*e^2*cos(d*x + c)^3 - 6*e^2*cos(d*x + c)^2 + e^2*cos(d*x + c) + 10*e^ 
2 - (3*e^2*cos(d*x + c)^2 + 9*e^2*cos(d*x + c) + 10*e^2)*sin(d*x + c))*sqr 
t(e*cos(d*x + c)))/(a^4*d*cos(d*x + c)^3 + 3*a^4*d*cos(d*x + c)^2 - 2*a^4* 
d*cos(d*x + c) - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 2*a^4*d*cos(d*x + c) - 
4*a^4*d)*sin(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate((e*cos(d*x+c))**(5/2)/(a+a*sin(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
 

Output:

integrate((e*cos(d*x + c))^(5/2)/(a*sin(d*x + c) + a)^4, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")
 

Output:

integrate((e*cos(d*x + c))^(5/2)/(a*sin(d*x + c) + a)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \] Input:

int((e*cos(c + d*x))^(5/2)/(a + a*sin(c + d*x))^4,x)
 

Output:

int((e*cos(c + d*x))^(5/2)/(a + a*sin(c + d*x))^4, x)
 

Reduce [F]

\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{4}+4 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )+1}d x \right ) e^{2}}{a^{4}} \] Input:

int((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^4,x)
 

Output:

(sqrt(e)*int((sqrt(cos(c + d*x))*cos(c + d*x)**2)/(sin(c + d*x)**4 + 4*sin 
(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1),x)*e**2)/a**4