\(\int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx\) [133]

Optimal result

Integrand size = 28, antiderivative size = 119 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}-\frac {b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^2} \] Output:

(2*a^2-b^2)*arctanh((-b+a*tan(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/(a^2+b^2)^( 
5/2)/d-1/2*b*((4*a^2+b^2)*cos(d*x+c)+3*a*b*sin(d*x+c))/(a^2+b^2)^2/d/(a*co 
s(d*x+c)+b*sin(d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=\frac {\frac {2 \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{\left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}}{2 d} \] Input:

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

Output:

((2*(2*a^2 - b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 
 + b^2)^(5/2) - (b*((4*a^2 + b^2)*Cos[c + d*x] + 3*a*b*Sin[c + d*x]))/((a^ 
2 + b^2)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2))/(2*d)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4902, 2191, 27, 2191, 27, 1083, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4902

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

\(\Big \downarrow \) 2191

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2191

\(\displaystyle \frac {2 \left (\frac {-\frac {\int -\frac {2 \left (2 a^2-b^2\right )}{-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{4 \left (a^2+b^2\right )}-\frac {b \left (\frac {2 b^4}{a^3}+b \left (\frac {2 b^2}{a^2}+5\right ) \tan \left (\frac {1}{2} (c+d x)\right )+\frac {3 b^2}{a}+4 a\right )}{2 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}}{a^2+b^2}+\frac {b^2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a b\right )}{a^3 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {\frac {\left (2 a^2-b^2\right ) \int \frac {1}{-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{2 \left (a^2+b^2\right )}-\frac {b \left (\frac {2 b^4}{a^3}+b \left (\frac {2 b^2}{a^2}+5\right ) \tan \left (\frac {1}{2} (c+d x)\right )+\frac {3 b^2}{a}+4 a\right )}{2 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}}{a^2+b^2}+\frac {b^2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a b\right )}{a^3 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{d}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {2 \left (\frac {-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}d\left (2 b-2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2+b^2}-\frac {b \left (\frac {2 b^4}{a^3}+b \left (\frac {2 b^2}{a^2}+5\right ) \tan \left (\frac {1}{2} (c+d x)\right )+\frac {3 b^2}{a}+4 a\right )}{2 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}}{a^2+b^2}+\frac {b^2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a b\right )}{a^3 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (\frac {-\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {2 b-2 a \tan \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{2 \left (a^2+b^2\right )^{3/2}}-\frac {b \left (\frac {2 b^4}{a^3}+b \left (\frac {2 b^2}{a^2}+5\right ) \tan \left (\frac {1}{2} (c+d x)\right )+\frac {3 b^2}{a}+4 a\right )}{2 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )}}{a^2+b^2}+\frac {b^2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a b\right )}{a^3 \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{d}\)

Input:

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

Output:

(2*((b^2*(a*b + (a^2 + 2*b^2)*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)*(a + 2*b 
*Tan[(c + d*x)/2] - a*Tan[(c + d*x)/2]^2)^2) + (-1/2*((2*a^2 - b^2)*ArcTan 
h[(2*b - 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 + b^2])])/(a^2 + b^2)^(3/2) - ( 
b*(4*a + (3*b^2)/a + (2*b^4)/a^3 + b*(5 + (2*b^2)/a^2)*Tan[(c + d*x)/2]))/ 
(2*(a^2 + b^2)*(a + 2*b*Tan[(c + d*x)/2] - a*Tan[(c + d*x)/2]^2)))/(a^2 + 
b^2)))/d
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[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 2191

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = 
PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P 
q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + 
c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ 
(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c))   Int 
[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* 
(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 
2 - 4*a*c, 0] && LtQ[p, -1]
 

rule 3042

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

rule 4902

Int[u_, x_Symbol] :> With[{w = Block[{$ShowSteps = False, $StepCounter = Nu 
ll}, Int[SubstFor[1/(1 + FreeFactors[Tan[FunctionOfTrig[u, x]/2], x]^2*x^2) 
, Tan[FunctionOfTrig[u, x]/2]/FreeFactors[Tan[FunctionOfTrig[u, x]/2], x], 
u, x], x]]}, Module[{v = FunctionOfTrig[u, x], d}, Simp[d = FreeFactors[Tan 
[v/2], x]; 2*(d/Coefficient[v, x, 1])   Subst[Int[SubstFor[1/(1 + d^2*x^2), 
 Tan[v/2]/d, u, x], x], x, Tan[v/2]/d], x]] /; CalculusFreeQ[w, x]] /; Inve 
rseFunctionFreeQ[u, x] &&  !FalseQ[FunctionOfTrig[u, x]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(279\) vs. \(2(112)=224\).

Time = 0.60 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.35

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(280\)
default	\(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(280\)
risch	\(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (-3 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a b +4 a^{2}+b^{2}\right )}{\left (-i a +b \right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} d \left (i a +b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}\)	\(452\)

Input:

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

Output:

1/d*(-2*(-1/2*b^2*(5*a^2+2*b^2)/a/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^3 
-1/2*b*(4*a^4-7*a^2*b^2-2*b^4)/(a^4+2*a^2*b^2+b^4)/a^2*tan(1/2*d*x+1/2*c)^ 
2+1/2*b^2*(11*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4)/a*tan(1/2*d*x+1/2*c)+1/2*b*(4 
*a^2+b^2)/(a^4+2*a^2*b^2+b^4))/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2 
*c)-a)^2+(2*a^2-b^2)/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a* 
tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (112) = 224\).

Time = 0.09 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.96 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=-\frac {{\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{4} - 3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d\right )}} \] Input:

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

Output:

-1/4*((2*a^2*b^2 - b^4 + (2*a^4 - 3*a^2*b^2 + b^4)*cos(d*x + c)^2 + 2*(2*a 
^3*b - a*b^3)*cos(d*x + c)*sin(d*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d* 
x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^ 
2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + 
c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) + 2*(4*a^4*b + 5*a^2*b^3 + b^5)*co 
s(d*x + c) + 6*(a^3*b^2 + a*b^4)*sin(d*x + c))/((a^8 + 2*a^6*b^2 - 2*a^2*b 
^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*c 
os(d*x + c)*sin(d*x + c) + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (112) = 224\).

Time = 0.12 (sec) , antiderivative size = 412, normalized size of antiderivative = 3.46 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (4 \, a^{4} b + a^{2} b^{3} + \frac {{\left (11 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (4 \, a^{4} b - 7 \, a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {{\left (5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4} + \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, {\left (a^{8} - 3 \, a^{4} b^{4} - 2 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {{\left (a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{2 \, d} \] Input:

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

Output:

-1/2*((2*a^2 - b^2)*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 
+ b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/((a^4 + 
 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + 2*(4*a^4*b + a^2*b^3 + (11*a^3*b^2 + 
2*a*b^4)*sin(d*x + c)/(cos(d*x + c) + 1) - (4*a^4*b - 7*a^2*b^3 - 2*b^5)*s 
in(d*x + c)^2/(cos(d*x + c) + 1)^2 - (5*a^3*b^2 + 2*a*b^4)*sin(d*x + c)^3/ 
(cos(d*x + c) + 1)^3)/(a^8 + 2*a^6*b^2 + a^4*b^4 + 4*(a^7*b + 2*a^5*b^3 + 
a^3*b^5)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*(a^8 - 3*a^4*b^4 - 2*a^2*b^6) 
*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*(a^7*b + 2*a^5*b^3 + a^3*b^5)*sin 
(d*x + c)^3/(cos(d*x + c) + 1)^3 + (a^8 + 2*a^6*b^2 + a^4*b^4)*sin(d*x + c 
)^4/(cos(d*x + c) + 1)^4))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (112) = 224\).

Time = 0.20 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{4} b - a^{2} b^{3}\right )}}{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \] Input:

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

Output:

-1/2*((2*a^2 - b^2)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + 
b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a 
^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*a 
*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 7*a^2*b^3*t 
an(1/2*d*x + 1/2*c)^2 - 2*b^5*tan(1/2*d*x + 1/2*c)^2 - 11*a^3*b^2*tan(1/2* 
d*x + 1/2*c) - 2*a*b^4*tan(1/2*d*x + 1/2*c) - 4*a^4*b - a^2*b^3)/((a^6 + 2 
*a^4*b^2 + a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - 
 a)^2))/d
 

Mupad [B] (verification not implemented)

Time = 18.08 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.72 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=\frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}-a^4\,b-b^5-2\,a^2\,b^3+a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-\frac {b^2}{2}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}+a^4\,b+b^5+2\,a^2\,b^3-a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{2\,d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\frac {4\,a^2\,b+b^3}{a^4+2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-2\,b^2\right )\,\left (4\,a^2\,b+b^3\right )}{a^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-4\,b^2\right )+a^2-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \] Input:

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

Output:

(log((a^2 + b^2)^(5/2) - a^4*b - b^5 - 2*a^2*b^3 + a^5*tan(c/2 + (d*x)/2) 
+ a*b^4*tan(c/2 + (d*x)/2) + 2*a^3*b^2*tan(c/2 + (d*x)/2))*(a^2 - b^2/2))/ 
(d*(a^2 + b^2)^(5/2)) - (log((a^2 + b^2)^(5/2) + a^4*b + b^5 + 2*a^2*b^3 - 
 a^5*tan(c/2 + (d*x)/2) - a*b^4*tan(c/2 + (d*x)/2) - 2*a^3*b^2*tan(c/2 + ( 
d*x)/2))*(2*a^2 - b^2))/(2*d*(a^2 + b^2)^(5/2)) - ((4*a^2*b + b^3)/(a^4 + 
b^4 + 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^2*(a^2 - 2*b^2)*(4*a^2*b + b^3))/(a 
^2*(a^4 + b^4 + 2*a^2*b^2)) + (b*tan(c/2 + (d*x)/2)*(11*a^2*b + 2*b^3))/(a 
*(a^4 + b^4 + 2*a^2*b^2)) - (b*tan(c/2 + (d*x)/2)^3*(5*a^2*b + 2*b^3))/(a* 
(a^4 + b^4 + 2*a^2*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2 
)^2*(2*a^2 - 4*b^2) + a^2 - 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*tan(c/2 + ( 
d*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 3126, normalized size of antiderivative = 26.27 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 16*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b** 
2))*cos(c + d*x)**3*sin(c + d*x)*a**7*b**2*i + 40*sqrt(a**2 + b**2)*atan(( 
tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)**3*sin(c + d*x 
)*a**5*b**4*i - 16*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqr 
t(a**2 + b**2))*cos(c + d*x)**3*sin(c + d*x)*a**3*b**6*i + 8*sqrt(a**2 + b 
**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)**2* 
sin(c + d*x)**2*a**8*b*i - 60*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i 
 - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)**2*sin(c + d*x)**2*a**6*b**3*i + 1 
08*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))* 
cos(c + d*x)**2*sin(c + d*x)**2*a**4*b**5*i - 40*sqrt(a**2 + b**2)*atan((t 
an((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)**2*sin(c + d*x) 
**2*a**2*b**7*i - 8*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sq 
rt(a**2 + b**2))*cos(c + d*x)**2*a**8*b*i + 20*sqrt(a**2 + b**2)*atan((tan 
((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)**2*a**6*b**3*i - 
8*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*c 
os(c + d*x)**2*a**4*b**5*i + 16*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a 
*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**3*a**7*b**2*i - 72 
*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*co 
s(c + d*x)*sin(c + d*x)**3*a**5*b**4*i + 96*sqrt(a**2 + b**2)*atan((tan((c 
 + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**3*a...