\(\int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx\) [559]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 187 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac {6 a \left (a^2-2 b^2\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac {2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}} \] Output:

2/5*b*(3*a^2-4*b^2)*(e*cos(d*x+c))^(3/2)/d/e^5-6/5*a*(a^2-2*b^2)*(e*cos(d* 
x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^4/cos(d*x+c)^(1/2)+2 
/5*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d/e/(e*cos(d*x+c))^(5/2)-2/5*(a+b*s 
in(d*x+c))*(a*b-(3*a^2-4*b^2)*sin(d*x+c))/d/e^3/(e*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.96 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 \left (-5 b^3-3 a \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+b \left (3 a^2+b^2\right ) \sec ^2(c+d x)+3 a^3 \sin (c+d x)-6 a b^2 \sin (c+d x)+a \left (a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)\right )}{5 d e^3 \sqrt {e \cos (c+d x)}} \] Input:

Integrate[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]
 

Output:

(2*(-5*b^3 - 3*a*(a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2 
] + b*(3*a^2 + b^2)*Sec[c + d*x]^2 + 3*a^3*Sin[c + d*x] - 6*a*b^2*Sin[c + 
d*x] + a*(a^2 + 3*b^2)*Sec[c + d*x]*Tan[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c + 
 d*x]])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.04, 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, 3340, 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.

\(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

Int[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]
 

Output:

(2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(5*d*e*(e*Cos[c + d*x])^(5 
/2)) + ((-3*((-2*b*(3*a^2 - 4*b^2)*(e*Cos[c + d*x])^(3/2))/(3*d*e) + (2*a* 
(a^2 - 2*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[ 
c + d*x]])))/e^2 - (2*(a + b*Sin[c + d*x])*(a*b - (3*a^2 - 4*b^2)*Sin[c + 
d*x]))/(d*e*Sqrt[e*Cos[c + d*x]]))/(5*e^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 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 3340
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(g* 
Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*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*Si 
n[e + f*x])^(m - 1)*Simp[a*c*(p + 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x] 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, 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] 
&& SimplerQ[c + d*x, a + b*x])
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(617\) vs. \(2(171)=342\).

Time = 4.15 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.30

method result size
default \(\frac {\frac {48 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a^{3}}{5}-\frac {96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a \,b^{2}}{5}-\frac {24 \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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{3}}{5}+\frac {48 \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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a \,b^{2}}{5}-\frac {48 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{3}}{5}+\frac {96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a \,b^{2}}{5}+\frac {24 \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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}}{5}-\frac {48 \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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{2}}{5}-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b^{3}+\frac {16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}}{5}-\frac {12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{2}}{5}-\frac {6 \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 ) a^{3}}{5}+\frac {12 \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 ) a \,b^{2}}{5}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{3}+\frac {6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b}{5}-\frac {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}}{5}}{\left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \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}\, e^{3} d}\) \(618\)
parts \(\text {Expression too large to display}\) \(799\)

Input:

int((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2/5/(4*sin(1/2*d*x+1/2*c)^4-4*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^3*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+ 
1/2*c)^6*a^3-48*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a*b^2-12*(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))*sin(1/2*d*x+1/2*c)^4*a^3+24*(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))*sin 
(1/2*d*x+1/2*c)^4*a*b^2-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^3+48* 
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+12*(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)) 
*sin(1/2*d*x+1/2*c)^2*a^3-24*(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))*sin(1/2*d*x+1/2*c)^ 
2*a*b^2-20*sin(1/2*d*x+1/2*c)^5*b^3+8*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^2-3*(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+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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+20*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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (5 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} - {\left (a^{3} + 3 \, a b^{2} + 3 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{5 \, d e^{4} \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-2/5*(3*sqrt(1/2)*(I*a^3 - 2*I*a*b^2)*sqrt(e)*cos(d*x + c)^3*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*s 
qrt(1/2)*(-I*a^3 + 2*I*a*b^2)*sqrt(e)*cos(d*x + c)^3*weierstrassZeta(-4, 0 
, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + (5*b^3*cos( 
d*x + c)^2 - 3*a^2*b - b^3 - (a^3 + 3*a*b^2 + 3*(a^3 - 2*a*b^2)*cos(d*x + 
c)^2)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(d*e^4*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*sin(d*x+c))**3/(e*cos(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

integrate((b*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

integrate((b*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:

int((a + b*sin(c + d*x))^3/(e*cos(c + d*x))^(7/2),x)
                                                                                    
                                                                                    
 

Output:

int((a + b*sin(c + d*x))^3/(e*cos(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, \left (5 \cos \left (d x +c \right )^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a^{3} d +5 \cos \left (d x +c \right )^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{4}}d x \right ) b^{3} d +15 \cos \left (d x +c \right )^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{4}}d x \right ) a \,b^{2} d +6 \sqrt {\cos \left (d x +c \right )}\, a^{2} b \right )}{5 \cos \left (d x +c \right )^{3} d \,e^{4}} \] Input:

int((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x)
 

Output:

(sqrt(e)*(5*cos(c + d*x)**3*int(sqrt(cos(c + d*x))/cos(c + d*x)**4,x)*a**3 
*d + 5*cos(c + d*x)**3*int((sqrt(cos(c + d*x))*sin(c + d*x)**3)/cos(c + d* 
x)**4,x)*b**3*d + 15*cos(c + d*x)**3*int((sqrt(cos(c + d*x))*sin(c + d*x)* 
*2)/cos(c + d*x)**4,x)*a*b**2*d + 6*sqrt(cos(c + d*x))*a**2*b))/(5*cos(c + 
 d*x)**3*d*e**4)