3.6.61 \(\int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{5/2}} \, dx\) [561]

3.6.61.1 Optimal result
3.6.61.2 Mathematica [A] (verified)
3.6.61.3 Rubi [A] (verified)
3.6.61.4 Maple [B] (verified)
3.6.61.5 Fricas [C] (verification not implemented)
3.6.61.6 Sympy [F(-1)]
3.6.61.7 Maxima [F]
3.6.61.8 Giac [F]
3.6.61.9 Mupad [F(-1)]

3.6.61.1 Optimal result

Integrand size = 25, antiderivative size = 164 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 b \left (a^2+4 b^2\right ) \sqrt {e \cos (c+d x)}}{3 d e^3}+\frac {2 a \left (a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d e (e \cos (c+d x))^{3/2}} \]

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

Time = 1.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.63 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{5/2}} \, dx=\frac {6 a^2 b+5 b^3+3 b^3 \cos (2 (c+d x))+2 a \left (a^2-6 b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 a^3 \sin (c+d x)+6 a b^2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}} \]

input
Integrate[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(5/2),x]
 
output
(6*a^2*b + 5*b^3 + 3*b^3*Cos[2*(c + d*x)] + 2*a*(a^2 - 6*b^2)*Cos[c + d*x] 
^(3/2)*EllipticF[(c + d*x)/2, 2] + 2*a^3*Sin[c + d*x] + 6*a*b^2*Sin[c + d* 
x])/(3*d*e*(e*Cos[c + d*x])^(3/2))
 
3.6.61.3 Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.99, 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, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3341

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

input
Int[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(5/2),x]
 
output
(2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(3*d*e*(e*Cos[c + d*x])^(3 
/2)) + ((2*b*(a^2 + 4*b^2)*Sqrt[e*Cos[c + d*x]])/(d*e) + (2*a*(a^2 - 6*b^2 
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d*x]]) + 
 (2*a*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x]))/(d*e))/(3*e^2)
 

3.6.61.3.1 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 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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])
 
3.6.61.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(383\) vs. \(2(172)=344\).

Time = 4.97 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.34

method result size
default \(-\frac {2 \left (2 F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}-12 F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+12 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}+6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-12 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b +4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}\right )}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\) \(384\)
parts \(-\frac {2 a^{3} \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )+\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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \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 )}}{3 e^{2} \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 )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\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 (\sqrt {e \cos \left (d x +c \right )}+\frac {e^{2}}{3 \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,e^{3}}-\frac {4 a \,b^{2} \left (2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \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^{2} \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \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^{2} b}{\left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} e d}\) \(549\)

input
int((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 
output
-2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^ 
2*e+e)^(1/2)/e^2*(2*EllipticF(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*a^3-12* 
EllipticF(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*a*b^2+12*sin(1/2*d*x+1/2*c) 
^5*b^3+2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3+6*cos(1/2*d*x+1/2*c)* 
sin(1/2*d*x+1/2*c)^2*a*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+6*(sin(1/2*d*x+1/ 
2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c 
),2^(1/2))*a*b^2-12*sin(1/2*d*x+1/2*c)^3*b^3+3*sin(1/2*d*x+1/2*c)*a^2*b+4* 
sin(1/2*d*x+1/2*c)*b^3)/d
 
3.6.61.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, b^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b + b^{3} + {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, d e^{3} \cos \left (d x + c\right )^{2}} \]

input
integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(5/2),x, algorithm="fricas")
 
output
1/3*(sqrt(2)*(-I*a^3 + 6*I*a*b^2)*sqrt(e)*cos(d*x + c)^2*weierstrassPInver 
se(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*a^3 - 6*I*a*b^2)*sqr 
t(e)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + 
c)) + 2*(3*b^3*cos(d*x + c)^2 + 3*a^2*b + b^3 + (a^3 + 3*a*b^2)*sin(d*x + 
c))*sqrt(e*cos(d*x + c)))/(d*e^3*cos(d*x + c)^2)
 
3.6.61.6 Sympy [F(-1)]

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

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

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

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

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

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

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

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