\(\int \frac {e+f x}{(a+b \cos (c+d x))^2} \, dx\) [189]

Optimal result

Integrand size = 18, antiderivative size = 296 \[ \int \frac {e+f x}{(a+b \cos (c+d x))^2} \, dx=-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {f \log (a+b \cos (c+d x))}{\left (a^2-b^2\right ) d^2}-\frac {a f \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}+\frac {a f \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {b (e+f x) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-I*a*(f*x+e)*ln(1+b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d+ 
I*a*(f*x+e)*ln(1+b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d-f 
*ln(a+b*cos(d*x+c))/(a^2-b^2)/d^2-a*f*polylog(2,-b*exp(I*(d*x+c))/(a-(a^2- 
b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^2+a*f*polylog(2,-b*exp(I*(d*x+c))/(a+(a^2-b 
^2)^(1/2)))/(a^2-b^2)^(3/2)/d^2-b*(f*x+e)*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos( 
d*x+c))
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(933\) vs. \(2(296)=592\).

Time = 10.93 (sec) , antiderivative size = 933, normalized size of antiderivative = 3.15 \[ \int \frac {e+f x}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(-(b*d*e*Sin[c + d*x]) + b*c*f*Sin[c + d*x] - b*f*(c + d*x)*Sin[c + d*x])/ 
((a - b)*(a + b)*d^2*(a + b*Cos[c + d*x])) + (Cos[(c + d*x)/2]^2*((2*a*(d* 
e - c*f)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]* 
Sqrt[a + b]) + f*Log[Sec[(c + d*x)/2]^2] - f*Log[(a + b*Cos[c + d*x])*Sec[ 
(c + d*x)/2]^2] - (I*a*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(Sqrt[a + b] - S 
qrt[-a + b]*Tan[(c + d*x)/2])/(I*Sqrt[-a + b] + Sqrt[a + b])] + PolyLog[2, 
 (Sqrt[-a + b]*(1 - I*Tan[(c + d*x)/2]))/(Sqrt[-a + b] - I*Sqrt[a + b])])) 
/(Sqrt[-a + b]*Sqrt[a + b]) + (I*a*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(I*( 
Sqrt[a + b] + Sqrt[-a + b]*Tan[(c + d*x)/2]))/(Sqrt[-a + b] + I*Sqrt[a + b 
])] + PolyLog[2, (Sqrt[-a + b]*(1 - I*Tan[(c + d*x)/2]))/(Sqrt[-a + b] + I 
*Sqrt[a + b])]))/(Sqrt[-a + b]*Sqrt[a + b]) - (I*a*f*(Log[1 + I*Tan[(c + d 
*x)/2]]*Log[(Sqrt[a + b] + Sqrt[-a + b]*Tan[(c + d*x)/2])/(I*Sqrt[-a + b] 
+ Sqrt[a + b])] + PolyLog[2, (Sqrt[-a + b]*(1 + I*Tan[(c + d*x)/2]))/(Sqrt 
[-a + b] - I*Sqrt[a + b])]))/(Sqrt[-a + b]*Sqrt[a + b]) + (I*a*f*(Log[1 + 
I*Tan[(c + d*x)/2]]*Log[(I*(Sqrt[a + b] - Sqrt[-a + b]*Tan[(c + d*x)/2]))/ 
(Sqrt[-a + b] + I*Sqrt[a + b])] + PolyLog[2, (Sqrt[-a + b]*(1 + I*Tan[(c + 
 d*x)/2]))/(Sqrt[-a + b] + I*Sqrt[a + b])]))/(Sqrt[-a + b]*Sqrt[a + b]))*( 
a*d*e + a*d*f*x + b*f*Sin[c + d*x])*(Sqrt[a + b] - Sqrt[-a + b]*Tan[(c + d 
*x)/2])*(Sqrt[a + b] + Sqrt[-a + b]*Tan[(c + d*x)/2]))/((a^2 - b^2)*d^2*(a 
 + b*Cos[c + d*x])*(a*(d*e - c*f + I*f*Log[1 - I*Tan[(c + d*x)/2]] - I*...
 

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3805, 25, 3042, 3147, 16, 3802, 2694, 27, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3805

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 3802

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

\(\displaystyle \frac {2 a \left (\frac {b \left (\frac {f \int e^{-i (c+d x)} \log \left (\frac {e^{i (c+d x)} b}{a-\sqrt {a^2-b^2}}+1\right )de^{i (c+d x)}}{b d^2}-\frac {i (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (\frac {f \int e^{-i (c+d x)} \log \left (\frac {e^{i (c+d x)} b}{a+\sqrt {a^2-b^2}}+1\right )de^{i (c+d x)}}{b d^2}-\frac {i (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}-\frac {f \log (a+b \cos (c+d x))}{d^2 \left (a^2-b^2\right )}-\frac {b (e+f x) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {2 a \left (\frac {b \left (-\frac {f \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {i (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (-\frac {f \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {i (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}-\frac {f \log (a+b \cos (c+d x))}{d^2 \left (a^2-b^2\right )}-\frac {b (e+f x) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

Input:

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

Output:

-((f*Log[a + b*Cos[c + d*x]])/((a^2 - b^2)*d^2)) + (2*a*((b*(((-I)*(e + f* 
x)*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) - (f*PolyLog[ 
2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))])/(b*d^2)))/(2*Sqrt[a^2 - 
b^2]) - (b*(((-I)*(e + f*x)*Log[1 + (b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^ 
2])])/(b*d) - (f*PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))]) 
/(b*d^2)))/(2*Sqrt[a^2 - b^2])))/(a^2 - b^2) - (b*(e + f*x)*Sin[c + d*x])/ 
((a^2 - b^2)*d*(a + b*Cos[c + d*x]))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

rule 3802

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

rule 3805

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (270 ) = 540\).

Time = 3.73 (sec) , antiderivative size = 674, normalized size of antiderivative = 2.28

method	result	size
risch	\(\frac {2 i \left (f x +e \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{\left (-a^{2}+b^{2}\right ) d^{2}}+\frac {f \ln \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{\left (-a^{2}+b^{2}\right ) d^{2}}+\frac {2 i a e \arctan \left (\frac {2 \,{\mathrm e}^{i \left (d x +c \right )} b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d}+\frac {i f a \ln \left (\frac {-{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right ) d \sqrt {a^{2}-b^{2}}}-\frac {i f a \ln \left (\frac {{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right ) d \sqrt {a^{2}-b^{2}}}+\frac {i f a \ln \left (\frac {-{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{\left (-a^{2}+b^{2}\right ) d^{2} \sqrt {a^{2}-b^{2}}}-\frac {i f a \ln \left (\frac {{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{\left (-a^{2}+b^{2}\right ) d^{2} \sqrt {a^{2}-b^{2}}}+\frac {f a \operatorname {dilog}\left (\frac {-{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{\left (-a^{2}+b^{2}\right ) d^{2} \sqrt {a^{2}-b^{2}}}-\frac {f a \operatorname {dilog}\left (\frac {{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{\left (-a^{2}+b^{2}\right ) d^{2} \sqrt {a^{2}-b^{2}}}-\frac {2 i a f c \arctan \left (\frac {2 \,{\mathrm e}^{i \left (d x +c \right )} b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2}}\)	\(674\)

Input:

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

Output:

2*I*(f*x+e)*(a*exp(I*(d*x+c))+b)/d/(-a^2+b^2)/(b*exp(2*I*(d*x+c))+2*a*exp( 
I*(d*x+c))+b)-2/(-a^2+b^2)/d^2*f*ln(exp(I*(d*x+c)))+1/(-a^2+b^2)/d^2*f*ln( 
b*exp(2*I*(d*x+c))+2*a*exp(I*(d*x+c))+b)+2*I/(-a^2+b^2)^(3/2)/d*a*e*arctan 
(1/2*(2*exp(I*(d*x+c))*b+2*a)/(-a^2+b^2)^(1/2))+I/(-a^2+b^2)/d*f*a/(a^2-b^ 
2)^(1/2)*ln((-exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*x- 
I/(-a^2+b^2)/d*f*a/(a^2-b^2)^(1/2)*ln((exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)+a) 
/(a+(a^2-b^2)^(1/2)))*x+I/(-a^2+b^2)/d^2*f*a/(a^2-b^2)^(1/2)*ln((-exp(I*(d 
*x+c))*b+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*c-I/(-a^2+b^2)/d^2*f*a/( 
a^2-b^2)^(1/2)*ln((exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)) 
)*c+1/(-a^2+b^2)/d^2*f*a/(a^2-b^2)^(1/2)*dilog((-exp(I*(d*x+c))*b+(a^2-b^2 
)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))-1/(-a^2+b^2)/d^2*f*a/(a^2-b^2)^(1/2)*dilo 
g((exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))-2*I/(-a^2+b^2) 
^(3/2)/d^2*a*f*c*arctan(1/2*(2*exp(I*(d*x+c))*b+2*a)/(-a^2+b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1482 vs. \(2 (266) = 532\).

Time = 0.28 (sec) , antiderivative size = 1482, normalized size of antiderivative = 5.01 \[ \int \frac {e+f x}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*((a*b^2*f*cos(d*x + c) + a^2*b*f)*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos 
(d*x + c) + I*a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a 
^2 - b^2)/b^2) + b)/b + 1) - (a*b^2*f*cos(d*x + c) + a^2*b*f)*sqrt((a^2 - 
b^2)/b^2)*dilog(-(a*cos(d*x + c) + I*a*sin(d*x + c) - (b*cos(d*x + c) + I* 
b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) + (a*b^2*f*cos(d*x + c) 
+ a^2*b*f)*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) - I*a*sin(d*x + c) 
 + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - 
 (a*b^2*f*cos(d*x + c) + a^2*b*f)*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x 
+ c) - I*a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - 
b^2)/b^2) + b)/b + 1) - (-I*a^2*b*d*f*x - I*a^2*b*c*f + (-I*a*b^2*d*f*x - 
I*a*b^2*c*f)*cos(d*x + c))*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) + I*a 
*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) 
+ b)/b) - (I*a^2*b*d*f*x + I*a^2*b*c*f + (I*a*b^2*d*f*x + I*a*b^2*c*f)*cos 
(d*x + c))*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) + I*a*sin(d*x + c) - 
(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) - (I*a^2 
*b*d*f*x + I*a^2*b*c*f + (I*a*b^2*d*f*x + I*a*b^2*c*f)*cos(d*x + c))*sqrt( 
(a^2 - b^2)/b^2)*log((a*cos(d*x + c) - I*a*sin(d*x + c) + (b*cos(d*x + c) 
- I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) - (-I*a^2*b*d*f*x - I*a^ 
2*b*c*f + (-I*a*b^2*d*f*x - I*a*b^2*c*f)*cos(d*x + c))*sqrt((a^2 - b^2)/b^ 
2)*log((a*cos(d*x + c) - I*a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d...
 

Sympy [F(-1)]

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

integrate((f*x+e)/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e+f x}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((f*x+e)/(a+b*cos(d*x+c))^2,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(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {e+f x}{(a+b \cos (c+d x))^2} \, dx=\text {too large to display} \] Input:

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

Output:

(4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a 
**2 - b**2))*tan((c + d*x)/2)**2*a**3*d*e + 4*sqrt(a**2 - b**2)*atan((tan( 
(c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*tan((c + d*x)/2)** 
2*a**2*b*d*e - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x 
)/2)*b)/sqrt(a**2 - b**2))*tan((c + d*x)/2)**2*a*b**2*d*e + 4*sqrt(a**2 - 
b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a* 
*3*d*e + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)* 
b)/sqrt(a**2 - b**2))*a**2*b*d*e + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x) 
/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a*b**2*d*e - 8*int(x/(tan(( 
c + d*x)/2)**4*a**5 - 2*tan((c + d*x)/2)**4*a**4*b - 2*tan((c + d*x)/2)**4 
*a**3*b**2 + 8*tan((c + d*x)/2)**4*a**2*b**3 - 7*tan((c + d*x)/2)**4*a*b** 
4 + 2*tan((c + d*x)/2)**4*b**5 + 2*tan((c + d*x)/2)**2*a**5 - 8*tan((c + d 
*x)/2)**2*a**3*b**2 + 4*tan((c + d*x)/2)**2*a**2*b**3 + 6*tan((c + d*x)/2) 
**2*a*b**4 - 4*tan((c + d*x)/2)**2*b**5 + a**5 + 2*a**4*b - 2*a**3*b**2 - 
4*a**2*b**3 + a*b**4 + 2*b**5),x)*tan((c + d*x)/2)**2*a**7*b**2*d**2*f - 8 
*int(x/(tan((c + d*x)/2)**4*a**5 - 2*tan((c + d*x)/2)**4*a**4*b - 2*tan((c 
 + d*x)/2)**4*a**3*b**2 + 8*tan((c + d*x)/2)**4*a**2*b**3 - 7*tan((c + d*x 
)/2)**4*a*b**4 + 2*tan((c + d*x)/2)**4*b**5 + 2*tan((c + d*x)/2)**2*a**5 - 
 8*tan((c + d*x)/2)**2*a**3*b**2 + 4*tan((c + d*x)/2)**2*a**2*b**3 + 6*tan 
((c + d*x)/2)**2*a*b**4 - 4*tan((c + d*x)/2)**2*b**5 + a**5 + 2*a**4*b ...