\(\int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx\) [515]

Optimal result

Integrand size = 18, antiderivative size = 149 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\frac {4 \left (8 a^2+b^2\right ) \arctan \left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\left (4 a^2-b^2\right )^{5/2} d}+\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^2}+\frac {12 a b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right )^2 d (2 a+b \sin (2 c+2 d x))} \] Output:

4*(8*a^2+b^2)*arctan((b+2*a*tan(d*x+c))/(4*a^2-b^2)^(1/2))/(4*a^2-b^2)^(5/ 
2)/d+2*b*cos(2*d*x+2*c)/(4*a^2-b^2)/d/(2*a+b*sin(2*d*x+2*c))^2+12*a*b*cos( 
2*d*x+2*c)/(4*a^2-b^2)^2/d/(2*a+b*sin(2*d*x+2*c))
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\frac {2 \left (\frac {2 \left (8 a^2+b^2\right ) \arctan \left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\left (4 a^2-b^2\right )^{5/2}}+\frac {b \cos (2 (c+d x)) \left (16 a^2-b^2+6 a b \sin (2 (c+d x))\right )}{\left (-4 a^2+b^2\right )^2 (2 a+b \sin (2 (c+d x)))^2}\right )}{d} \] Input:

Integrate[(a + b*Cos[c + d*x]*Sin[c + d*x])^(-3),x]
 

Output:

(2*((2*(8*a^2 + b^2)*ArcTan[(b + 2*a*Tan[c + d*x])/Sqrt[4*a^2 - b^2]])/(4* 
a^2 - b^2)^(5/2) + (b*Cos[2*(c + d*x)]*(16*a^2 - b^2 + 6*a*b*Sin[2*(c + d* 
x)]))/((-4*a^2 + b^2)^2*(2*a + b*Sin[2*(c + d*x)])^2)))/d
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {3042, 3145, 3042, 3143, 27, 3042, 3233, 25, 27, 3042, 3139, 1083, 217}

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.

\(\displaystyle \int \frac {1}{(a+b \sin (c+d x) \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+b \sin (c+d x) \cos (c+d x))^3}dx\)

\(\Big \downarrow \) 3145

\(\displaystyle \int \frac {1}{\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^3}dx\)

\(\Big \downarrow \) 3143

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

\(\displaystyle \frac {4 \left (\frac {3 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))}-\frac {\int -\frac {8 a^2+b^2}{2 a+b \sin (2 c+2 d x)}dx}{4 a^2-b^2}\right )}{4 a^2-b^2}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 \left (\frac {\int \frac {8 a^2+b^2}{2 a+b \sin (2 c+2 d x)}dx}{4 a^2-b^2}+\frac {3 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))}\right )}{4 a^2-b^2}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 \left (\frac {\left (8 a^2+b^2\right ) \int \frac {1}{2 a+b \sin (2 c+2 d x)}dx}{4 a^2-b^2}+\frac {3 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))}\right )}{4 a^2-b^2}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \left (\frac {\left (8 a^2+b^2\right ) \int \frac {1}{2 a+b \sin (2 c+2 d x)}dx}{4 a^2-b^2}+\frac {3 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))}\right )}{4 a^2-b^2}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {4 \left (\frac {\left (8 a^2+b^2\right ) \int \frac {1}{2 a \tan ^2(c+d x)+2 b \tan (c+d x)+2 a}d\tan (c+d x)}{d \left (4 a^2-b^2\right )}+\frac {3 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))}\right )}{4 a^2-b^2}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {4 \left (\frac {3 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))}-\frac {2 \left (8 a^2+b^2\right ) \int \frac {1}{-(2 b+4 a \tan (c+d x))^2-4 \left (4 a^2-b^2\right )}d(2 b+4 a \tan (c+d x))}{d \left (4 a^2-b^2\right )}\right )}{4 a^2-b^2}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {4 \left (\frac {\left (8 a^2+b^2\right ) \arctan \left (\frac {4 a \tan (c+d x)+2 b}{2 \sqrt {4 a^2-b^2}}\right )}{d \left (4 a^2-b^2\right )^{3/2}}+\frac {3 a b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))}\right )}{4 a^2-b^2}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^2}\)

Input:

Int[(a + b*Cos[c + d*x]*Sin[c + d*x])^(-3),x]
 

Output:

(2*b*Cos[2*c + 2*d*x])/((4*a^2 - b^2)*d*(2*a + b*Sin[2*c + 2*d*x])^2) + (4 
*(((8*a^2 + b^2)*ArcTan[(2*b + 4*a*Tan[c + d*x])/(2*Sqrt[4*a^2 - b^2])])/( 
(4*a^2 - b^2)^(3/2)*d) + (3*a*b*Cos[2*c + 2*d*x])/((4*a^2 - b^2)*d*(2*a + 
b*Sin[2*c + 2*d*x]))))/(4*a^2 - b^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3143

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp 
[1/((n + 1)*(a^2 - b^2))   Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) 
- b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3145

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ 
Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, 
 x]
 

rule 3233

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.82

method	result	size
derivativedivides	\(\frac {\frac {\frac {b^{2} \left (10 a^{2}-b^{2}\right ) \tan \left (d x +c \right )^{3}}{\left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (32 a^{4}+14 a^{2} b^{2}-b^{4}\right ) \tan \left (d x +c \right )^{2}}{2 \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (22 a^{2}-b^{2}\right ) \tan \left (d x +c \right )}{a \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (16 a^{2}-b^{2}\right )}{16 a^{4}-8 a^{2} b^{2}+b^{4}}}{\left (\tan \left (d x +c \right )^{2} a +\tan \left (d x +c \right ) b +a \right )^{2}}+\frac {4 \left (8 a^{2}+b^{2}\right ) \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right )}{\left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) \sqrt {4 a^{2}-b^{2}}}}{d}\)	\(271\)
default	\(\frac {\frac {\frac {b^{2} \left (10 a^{2}-b^{2}\right ) \tan \left (d x +c \right )^{3}}{\left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (32 a^{4}+14 a^{2} b^{2}-b^{4}\right ) \tan \left (d x +c \right )^{2}}{2 \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (22 a^{2}-b^{2}\right ) \tan \left (d x +c \right )}{a \left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (16 a^{2}-b^{2}\right )}{16 a^{4}-8 a^{2} b^{2}+b^{4}}}{\left (\tan \left (d x +c \right )^{2} a +\tan \left (d x +c \right ) b +a \right )^{2}}+\frac {4 \left (8 a^{2}+b^{2}\right ) \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right )}{\left (16 a^{4}-8 a^{2} b^{2}+b^{4}\right ) \sqrt {4 a^{2}-b^{2}}}}{d}\)	\(271\)
risch	\(-\frac {4 i \left (-8 i a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-i {\mathrm e}^{6 i \left (d x +c \right )} b^{3}+48 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{2} a \,{\mathrm e}^{4 i \left (d x +c \right )}+40 i a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-6 a \,b^{2}\right )}{\left (i b +4 a \,{\mathrm e}^{2 i \left (d x +c \right )}-i {\mathrm e}^{4 i \left (d x +c \right )} b \right )^{2} \left (4 a^{2}-b^{2}\right )^{2} d}-\frac {16 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}-4 a^{2}+b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-4 a^{2}+b^{2}}\, \left (2 a +b \right )^{2} \left (2 a -b \right )^{2} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}-4 a^{2}+b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-4 a^{2}+b^{2}}\, \left (2 a +b \right )^{2} \left (2 a -b \right )^{2} d}+\frac {16 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}+4 a^{2}-b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-4 a^{2}+b^{2}}\, \left (2 a +b \right )^{2} \left (2 a -b \right )^{2} d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}+4 a^{2}-b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-4 a^{2}+b^{2}}\, \left (2 a +b \right )^{2} \left (2 a -b \right )^{2} d}\)	\(495\)

Input:

int(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*((b^2*(10*a^2-b^2)/(16*a^4-8*a^2*b^2+b^4)/a*tan(d*x+c)^3+1/2*b*(32*a^4 
+14*a^2*b^2-b^4)/(16*a^4-8*a^2*b^2+b^4)/a^2*tan(d*x+c)^2+b^2*(22*a^2-b^2)/ 
a/(16*a^4-8*a^2*b^2+b^4)*tan(d*x+c)+b*(16*a^2-b^2)/(16*a^4-8*a^2*b^2+b^4)) 
/(tan(d*x+c)^2*a+tan(d*x+c)*b+a)^2+4*(8*a^2+b^2)/(16*a^4-8*a^2*b^2+b^4)/(4 
*a^2-b^2)^(1/2)*arctan((b+2*a*tan(d*x+c))/(4*a^2-b^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (145) = 290\).

Time = 0.12 (sec) , antiderivative size = 969, normalized size of antiderivative = 6.50 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

[1/2*(64*a^4*b - 20*a^2*b^3 + b^5 - 2*(64*a^4*b - 20*a^2*b^3 + b^5)*cos(d* 
x + c)^2 - 2*((8*a^2*b^2 + b^4)*cos(d*x + c)^4 - 8*a^4 - a^2*b^2 - (8*a^2* 
b^2 + b^4)*cos(d*x + c)^2 - 2*(8*a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c)) 
*sqrt(-4*a^2 + b^2)*log(-(2*(8*a^2 - b^2)*cos(d*x + c)^4 - 4*a*b*cos(d*x + 
 c)*sin(d*x + c) - 2*(8*a^2 - b^2)*cos(d*x + c)^2 + 2*a^2 - b^2 + (2*b*cos 
(d*x + c)^2 + 4*(2*a*cos(d*x + c)^3 - a*cos(d*x + c))*sin(d*x + c) - b)*sq 
rt(-4*a^2 + b^2))/(b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x 
 + c)*sin(d*x + c) - a^2)) - 12*(2*(4*a^3*b^2 - a*b^4)*cos(d*x + c)^3 - (4 
*a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((64*a^6*b^2 - 48*a^4*b^4 + 
12*a^2*b^6 - b^8)*d*cos(d*x + c)^4 - (64*a^6*b^2 - 48*a^4*b^4 + 12*a^2*b^6 
 - b^8)*d*cos(d*x + c)^2 - 2*(64*a^7*b - 48*a^5*b^3 + 12*a^3*b^5 - a*b^7)* 
d*cos(d*x + c)*sin(d*x + c) - (64*a^8 - 48*a^6*b^2 + 12*a^4*b^4 - a^2*b^6) 
*d), 1/2*(64*a^4*b - 20*a^2*b^3 + b^5 - 2*(64*a^4*b - 20*a^2*b^3 + b^5)*co 
s(d*x + c)^2 - 4*((8*a^2*b^2 + b^4)*cos(d*x + c)^4 - 8*a^4 - a^2*b^2 - (8* 
a^2*b^2 + b^4)*cos(d*x + c)^2 - 2*(8*a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + 
 c))*sqrt(4*a^2 - b^2)*arctan(-(4*a*cos(d*x + c)*sin(d*x + c) + b)*sqrt(4* 
a^2 - b^2)/(2*(4*a^2 - b^2)*cos(d*x + c)^2 - 4*a^2 + b^2)) - 12*(2*(4*a^3* 
b^2 - a*b^4)*cos(d*x + c)^3 - (4*a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + 
c))/((64*a^6*b^2 - 48*a^4*b^4 + 12*a^2*b^6 - b^8)*d*cos(d*x + c)^4 - (64*a 
^6*b^2 - 48*a^4*b^4 + 12*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(64*a^7*b ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b^2-4*a^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.69 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {2 \, a \tan \left (d x + c\right ) + b}{\sqrt {4 \, a^{2} - b^{2}}}\right )\right )} {\left (8 \, a^{2} + b^{2}\right )}}{{\left (16 \, a^{4} - 8 \, a^{2} b^{2} + b^{4}\right )} \sqrt {4 \, a^{2} - b^{2}}} + \frac {20 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} - 2 \, a b^{4} \tan \left (d x + c\right )^{3} + 32 \, a^{4} b \tan \left (d x + c\right )^{2} + 14 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - b^{5} \tan \left (d x + c\right )^{2} + 44 \, a^{3} b^{2} \tan \left (d x + c\right ) - 2 \, a b^{4} \tan \left (d x + c\right ) + 32 \, a^{4} b - 2 \, a^{2} b^{3}}{{\left (16 \, a^{6} - 8 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \] Input:

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

1/2*(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(a) + arctan((2*a*tan(d*x + c) + b 
)/sqrt(4*a^2 - b^2)))*(8*a^2 + b^2)/((16*a^4 - 8*a^2*b^2 + b^4)*sqrt(4*a^2 
 - b^2)) + (20*a^3*b^2*tan(d*x + c)^3 - 2*a*b^4*tan(d*x + c)^3 + 32*a^4*b* 
tan(d*x + c)^2 + 14*a^2*b^3*tan(d*x + c)^2 - b^5*tan(d*x + c)^2 + 44*a^3*b 
^2*tan(d*x + c) - 2*a*b^4*tan(d*x + c) + 32*a^4*b - 2*a^2*b^3)/((16*a^6 - 
8*a^4*b^2 + a^2*b^4)*(a*tan(d*x + c)^2 + b*tan(d*x + c) + a)^2))/d
 

Mupad [B] (verification not implemented)

Time = 16.54 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.66 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\frac {\frac {16\,a^2\,b-b^3}{16\,a^4-8\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (c+d\,x\right )\,\left (22\,a^2\,b-b^3\right )}{a\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (16\,a^2\,b-b^3\right )\,\left (2\,a^2+b^2\right )}{2\,a^2\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (10\,a^2\,b-b^3\right )}{a\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+b^2\right )+a^2+a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {4\,\mathrm {atan}\left (\frac {\left (\frac {4\,a\,\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^2+b^2\right )}{{\left (2\,a+b\right )}^{5/2}\,{\left (2\,a-b\right )}^{5/2}}+\frac {2\,\left (8\,a^2+b^2\right )\,\left (16\,a^4\,b-8\,a^2\,b^3+b^5\right )}{{\left (2\,a+b\right )}^{5/2}\,{\left (2\,a-b\right )}^{5/2}\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}\right )\,\left (16\,a^4-8\,a^2\,b^2+b^4\right )}{16\,a^2+2\,b^2}\right )\,\left (8\,a^2+b^2\right )}{d\,{\left (2\,a+b\right )}^{5/2}\,{\left (2\,a-b\right )}^{5/2}} \] Input:

int(1/(a + b*cos(c + d*x)*sin(c + d*x))^3,x)
 

Output:

((16*a^2*b - b^3)/(16*a^4 + b^4 - 8*a^2*b^2) + (b*tan(c + d*x)*(22*a^2*b - 
 b^3))/(a*(16*a^4 + b^4 - 8*a^2*b^2)) + (tan(c + d*x)^2*(16*a^2*b - b^3)*( 
2*a^2 + b^2))/(2*a^2*(16*a^4 + b^4 - 8*a^2*b^2)) + (b*tan(c + d*x)^3*(10*a 
^2*b - b^3))/(a*(16*a^4 + b^4 - 8*a^2*b^2)))/(d*(tan(c + d*x)^2*(2*a^2 + b 
^2) + a^2 + a^2*tan(c + d*x)^4 + 2*a*b*tan(c + d*x) + 2*a*b*tan(c + d*x)^3 
)) + (4*atan((((4*a*tan(c + d*x)*(8*a^2 + b^2))/((2*a + b)^(5/2)*(2*a - b) 
^(5/2)) + (2*(8*a^2 + b^2)*(16*a^4*b + b^5 - 8*a^2*b^3))/((2*a + b)^(5/2)* 
(2*a - b)^(5/2)*(16*a^4 + b^4 - 8*a^2*b^2)))*(16*a^4 + b^4 - 8*a^2*b^2))/( 
16*a^2 + 2*b^2))*(8*a^2 + b^2))/(d*(2*a + b)^(5/2)*(2*a - b)^(5/2))
 

Reduce [F]

\[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^3} \, dx=\int \frac {1}{\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:

int(1/(a+b*cos(d*x+c)*sin(d*x+c))^3,x)
 

Output:

int(1/(cos(c + d*x)**3*sin(c + d*x)**3*b**3 + 3*cos(c + d*x)**2*sin(c + d* 
x)**2*a*b**2 + 3*cos(c + d*x)*sin(c + d*x)*a**2*b + a**3),x)