3.6.60 \(\int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{3/2}} \, dx\) [560]

3.6.60.1 Optimal result
3.6.60.2 Mathematica [A] (verified)
3.6.60.3 Rubi [A] (verified)
3.6.60.4 Maple [A] (verified)
3.6.60.5 Fricas [C] (verification not implemented)
3.6.60.6 Sympy [F(-1)]
3.6.60.7 Maxima [F]
3.6.60.8 Giac [F]
3.6.60.9 Mupad [F(-1)]

3.6.60.1 Optimal result

Integrand size = 25, antiderivative size = 160 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{3/2}} \, dx=\frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac {2 a \left (a^2+6 b^2\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {\cos (c+d x)}}+\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}} \]

output
2/3*b*(3*a^2+4*b^2)*(e*cos(d*x+c))^(3/2)/d/e^3+2*a*b*(e*cos(d*x+c))^(3/2)* 
(a+b*sin(d*x+c))/d/e^3+2*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d/e/(e*cos(d* 
x+c))^(1/2)-2*a*(a^2+6*b^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/e^2/cos(d*x 
+c)^(1/2)
 
3.6.60.2 Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.61 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{3/2}} \, dx=\frac {2 b^3 \cos ^2(c+d x)-6 a \left (a^2+6 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 \left (3 a^2 b+b^3+a \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{3 d e \sqrt {e \cos (c+d x)}} \]

input
Integrate[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(3/2),x]
 
output
(2*b^3*Cos[c + d*x]^2 - 6*a*(a^2 + 6*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c 
+ d*x)/2, 2] + 6*(3*a^2*b + b^3 + a*(a^2 + 3*b^2)*Sin[c + d*x]))/(3*d*e*Sq 
rt[e*Cos[c + d*x]])
 
3.6.60.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3170, 27, 3042, 3341, 27, 3042, 3148, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3170

\(\displaystyle \frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {2 \int \frac {1}{2} \sqrt {e \cos (c+d x)} (a+b \sin (c+d x)) \left (a^2+5 b \sin (c+d x) a+4 b^2\right )dx}{e^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3341

\(\displaystyle \frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {\frac {2}{5} \int \frac {5}{2} \sqrt {e \cos (c+d x)} \left (a \left (a^2+6 b^2\right )+b \left (3 a^2+4 b^2\right ) \sin (c+d x)\right )dx-\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2+6 b^2\right ) \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}-\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}\)

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {\frac {a \left (a^2+6 b^2\right ) \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)}}-\frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}-\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {-\frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e}+\frac {2 a \left (a^2+6 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e}}{e^2}\)

input
Int[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(3/2),x]
 
output
(2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(d*e*Sqrt[e*Cos[c + d*x]]) 
 - ((-2*b*(3*a^2 + 4*b^2)*(e*Cos[c + d*x])^(3/2))/(3*d*e) + (2*a*(a^2 + 6* 
b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]] 
) - (2*a*b*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x]))/(d*e))/e^2
 

3.6.60.3.1 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 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 3170
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x 
])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) 
  Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + 
a^2*(p + 2) + a*b*(m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g 
}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2* 
p] || IntegerQ[m])
 

rule 3341
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* 
(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S 
imp[1/(m + p + 1)   Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim 
p[a*c*(m + p + 1) + b*d*m + (a*d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] 
 &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S 
implerQ[c + d*x, a + b*x])
 
3.6.60.4 Maple [A] (verified)

Time = 4.63 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.55

method result size
default \(\frac {\frac {8 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{3}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}+12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}-12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-\frac {8 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{3}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b +\frac {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}}{3}}{e \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\) \(248\)
parts \(-\frac {2 a^{3} \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\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}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{e \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 b^{3} \left (\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {e^{2}}{\sqrt {e \cos \left (d x +c \right )}}\right )}{d \,e^{3}}-\frac {12 a \,b^{2} \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\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}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{e \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 {6 a^{2} b}{\sqrt {e \cos \left (d x +c \right )}\, e d}\) \(463\)

input
int((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 
output
2/3/e/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)*(4*sin(1/2*d* 
x+1/2*c)^5*b^3+6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3+18*cos(1/2*d* 
x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2-3*(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))*a^3-18*(si 
n(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))*a*b^2-4*sin(1/2*d*x+1/2*c)^3*b^3+9*sin(1/2*d*x+1/2* 
c)*a^2*b+4*sin(1/2*d*x+1/2*c)*b^3)/d
 
3.6.60.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{2} b + 3 \, b^{3} + 3 \, {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, d e^{2} \cos \left (d x + c\right )} \]

input
integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(3/2),x, algorithm="fricas")
 
output
-1/3*(3*sqrt(2)*(I*a^3 + 6*I*a*b^2)*sqrt(e)*cos(d*x + c)*weierstrassZeta(- 
4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt( 
2)*(-I*a^3 - 6*I*a*b^2)*sqrt(e)*cos(d*x + c)*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(b^3*cos(d*x + c) 
^2 + 9*a^2*b + 3*b^3 + 3*(a^3 + 3*a*b^2)*sin(d*x + c))*sqrt(e*cos(d*x + c) 
))/(d*e^2*cos(d*x + c))
 
3.6.60.6 Sympy [F(-1)]

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

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

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

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

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

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

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

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