3.1.41 \(\int \frac {c+d x}{(a+b \sec (e+f x))^2} \, dx\) [41]

3.1.41.1 Optimal result

Integrand size = 18, antiderivative size = 582 \[ \int \frac {c+d x}{(a+b \sec (e+f x))^2} \, dx=\frac {(c+d x)^2}{2 a^2 d}-\frac {i b^3 (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} f}+\frac {2 i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} f}+\frac {i b^3 (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} f}-\frac {2 i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} f}+\frac {b^2 d \log (b+a \cos (e+f x))}{a^2 \left (a^2-b^2\right ) f^2}-\frac {b^3 d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} f^2}+\frac {2 b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} f^2}+\frac {b^3 d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} f^2}-\frac {2 b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} f^2}+\frac {b^2 (c+d x) \sin (e+f x)}{a \left (a^2-b^2\right ) f (b+a \cos (e+f x))} \]

output

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

3.1.41.2 Mathematica [A] (warning: unable to verify)

Time = 11.64 (sec) , antiderivative size = 1037, normalized size of antiderivative = 1.78 \[ \int \frac {c+d x}{(a+b \sec (e+f x))^2} \, dx=\frac {(e+f x) (-2 d e+2 c f+d (e+f x)) (b+a \cos (e+f x))^2 \sec ^2(e+f x)}{2 a^2 f^2 (a+b \sec (e+f x))^2}+\frac {(b+a \cos (e+f x)) \sec ^2(e+f x) \left (b^2 d e \sin (e+f x)-b^2 c f \sin (e+f x)-b^2 d (e+f x) \sin (e+f x)\right )}{a (-a+b) (a+b) f^2 (a+b \sec (e+f x))^2}+\frac {b \cos ^2\left (\frac {1}{2} (e+f x)\right ) (b+a \cos (e+f x)) \left (-\frac {2 \left (2 a^2-b^2\right ) (d e-c f) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-a-b}}\right )}{\sqrt {-a-b} \sqrt {a-b}}-b d \log \left (\sec ^2\left (\frac {1}{2} (e+f x)\right )\right )+b d \log \left (-\left ((b+a \cos (e+f x)) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )-\frac {i \left (2 a^2-b^2\right ) d \left (\log \left (1+i \tan \left (\frac {1}{2} (e+f x)\right )\right ) \log \left (\frac {i \left (\sqrt {a+b}-\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {a-b}+i \sqrt {a+b}}\right )-\log \left (1-i \tan \left (\frac {1}{2} (e+f x)\right )\right ) \log \left (\frac {\sqrt {a+b}-\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{i \sqrt {a-b}+\sqrt {a+b}}\right )+\log \left (1-i \tan \left (\frac {1}{2} (e+f x)\right )\right ) \log \left (\frac {i \left (\sqrt {a+b}+\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {a-b}+i \sqrt {a+b}}\right )-\log \left (1+i \tan \left (\frac {1}{2} (e+f x)\right )\right ) \log \left (\frac {\sqrt {a+b}+\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{i \sqrt {a-b}+\sqrt {a+b}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {a-b} \left (1-i \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {a-b}-i \sqrt {a+b}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {a-b} \left (1-i \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {a-b}+i \sqrt {a+b}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {a-b} \left (1+i \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {a-b}-i \sqrt {a+b}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {a-b} \left (1+i \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {a-b}+i \sqrt {a+b}}\right )\right )}{\sqrt {a-b} \sqrt {a+b}}\right ) \sec ^2(e+f x) \left (\left (2 a^2-b^2\right ) (c f+d f x)+a b d \sin (e+f x)\right ) \left (\sqrt {a+b}-\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sqrt {a+b}+\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )\right )}{a^2 \left (a^2-b^2\right ) f^2 (a+b \sec (e+f x))^2 \left (-\left (\left (2 a^2-b^2\right ) \left (d e-c f-i d \log \left (1-i \tan \left (\frac {1}{2} (e+f x)\right )\right )+i d \log \left (1+i \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )+a b d \sin (e+f x)\right )} \]

input

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

output

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

3.1.41.3 Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 582, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4679, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a^2 f \sqrt {b^2-a^2}}-\frac {2 i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a^2 f \sqrt {b^2-a^2}}+\frac {b^2 (c+d x) \sin (e+f x)}{a f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac {2 b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a^2 f^2 \sqrt {b^2-a^2}}-\frac {2 b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a^2 f^2 \sqrt {b^2-a^2}}+\frac {b^2 d \log (a \cos (e+f x)+b)}{a^2 f^2 \left (a^2-b^2\right )}-\frac {i b^3 (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a^2 f \left (b^2-a^2\right )^{3/2}}+\frac {i b^3 (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a^2 f \left (b^2-a^2\right )^{3/2}}-\frac {b^3 d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a^2 f^2 \left (b^2-a^2\right )^{3/2}}+\frac {b^3 d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a^2 f^2 \left (b^2-a^2\right )^{3/2}}+\frac {(c+d x)^2}{2 a^2 d}\)

input

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

output

(c + d*x)^2/(2*a^2*d) - (I*b^3*(c + d*x)*Log[1 + (a*E^(I*(e + f*x)))/(b - 
Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*f) + ((2*I)*b*(c + d*x)*Log[1 
+ (a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*f) + 
(I*b^3*(c + d*x)*Log[1 + (a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2])])/(a^2 
*(-a^2 + b^2)^(3/2)*f) - ((2*I)*b*(c + d*x)*Log[1 + (a*E^(I*(e + f*x)))/(b 
 + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*f) + (b^2*d*Log[b + a*Cos[e + 
 f*x]])/(a^2*(a^2 - b^2)*f^2) - (b^3*d*PolyLog[2, -((a*E^(I*(e + f*x)))/(b 
 - Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*f^2) + (2*b*d*PolyLog[2, - 
((a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*f^2) 
+ (b^3*d*PolyLog[2, -((a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2]))])/(a^2*( 
-a^2 + b^2)^(3/2)*f^2) - (2*b*d*PolyLog[2, -((a*E^(I*(e + f*x)))/(b + Sqrt 
[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*f^2) + (b^2*(c + d*x)*Sin[e + f*x]) 
/(a*(a^2 - b^2)*f*(b + a*Cos[e + f*x]))
 

3.1.41.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

3.1.41.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1288 vs. \(2 (528 ) = 1056\).

Time = 0.64 (sec) , antiderivative size = 1289, normalized size of antiderivative = 2.21

method	result	size
risch	\(\text {Expression too large to display}\)	\(1289\)

input

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

output

1/2/a^2*d*x^2+1/a^2*x*c+2*I/f/(a^2-b^2)*b*d/(-a^2+b^2)^(1/2)*ln((-a*exp(I* 
(f*x+e))+(-a^2+b^2)^(1/2)-b)/(-b+(-a^2+b^2)^(1/2)))*x-4*I/f^2/(a^2-b^2)^(3 
/2)*b*d*e*arctan(1/2*(2*a*exp(I*(f*x+e))+2*b)/(a^2-b^2)^(1/2))+1/f^2/(a^2- 
b^2)/a^2*b^3*d/(-a^2+b^2)^(1/2)*dilog((a*exp(I*(f*x+e))+(-a^2+b^2)^(1/2)+b 
)/(b+(-a^2+b^2)^(1/2)))-2/f^2/(a^2-b^2)*b*d/(-a^2+b^2)^(1/2)*dilog((a*exp( 
I*(f*x+e))+(-a^2+b^2)^(1/2)+b)/(b+(-a^2+b^2)^(1/2)))+2/f^2/(a^2-b^2)*b*d/( 
-a^2+b^2)^(1/2)*dilog((-a*exp(I*(f*x+e))+(-a^2+b^2)^(1/2)-b)/(-b+(-a^2+b^2 
)^(1/2)))-1/f^2/(a^2-b^2)/a^2*b^3*d/(-a^2+b^2)^(1/2)*dilog((-a*exp(I*(f*x+ 
e))+(-a^2+b^2)^(1/2)-b)/(-b+(-a^2+b^2)^(1/2)))-2*I/f^2/(a^2-b^2)*b*d/(-a^2 
+b^2)^(1/2)*ln((a*exp(I*(f*x+e))+(-a^2+b^2)^(1/2)+b)/(b+(-a^2+b^2)^(1/2))) 
*e-I/f/(a^2-b^2)/a^2*b^3*d/(-a^2+b^2)^(1/2)*ln((-a*exp(I*(f*x+e))+(-a^2+b^ 
2)^(1/2)-b)/(-b+(-a^2+b^2)^(1/2)))*x+2*I*b^2*(d*x+c)*(exp(I*(f*x+e))*b+a)/ 
a^2/(a^2-b^2)/f/(exp(2*I*(f*x+e))*a+2*exp(I*(f*x+e))*b+a)-I/f^2/(a^2-b^2)/ 
a^2*b^3*d/(-a^2+b^2)^(1/2)*ln((-a*exp(I*(f*x+e))+(-a^2+b^2)^(1/2)-b)/(-b+( 
-a^2+b^2)^(1/2)))*e+2*I/f^2/(a^2-b^2)^(3/2)/a^2*b^3*d*e*arctan(1/2*(2*a*ex 
p(I*(f*x+e))+2*b)/(a^2-b^2)^(1/2))+2*I/f^2/(a^2-b^2)*b*d/(-a^2+b^2)^(1/2)* 
ln((-a*exp(I*(f*x+e))+(-a^2+b^2)^(1/2)-b)/(-b+(-a^2+b^2)^(1/2)))*e+4*I/f/( 
a^2-b^2)^(3/2)*b*c*arctan(1/2*(2*a*exp(I*(f*x+e))+2*b)/(a^2-b^2)^(1/2))-2/ 
f^2/(a^2-b^2)/a^2*b^2*d*ln(exp(I*(f*x+e)))+1/f^2/(a^2-b^2)/a^2*b^2*d*ln(ex 
p(2*I*(f*x+e))*a+2*exp(I*(f*x+e))*b+a)+I/f/(a^2-b^2)/a^2*b^3*d/(-a^2+b^...
 

3.1.41.5 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. 2080 vs. \(2 (520) = 1040\).

Time = 0.49 (sec) , antiderivative size = 2080, normalized size of antiderivative = 3.57 \[ \int \frac {c+d x}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \]

input

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

output

1/2*((a^4*b - 2*a^2*b^3 + b^5)*d*f^2*x^2 + 2*(a^4*b - 2*a^2*b^3 + b^5)*c*f 
^2*x - ((2*a^4*b - a^2*b^3)*d*cos(f*x + e) + (2*a^3*b^2 - a*b^4)*d)*sqrt(- 
(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) + I*b*sin(f*x + e) + (a*cos(f*x + 
e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + ((2*a^4*b - a^ 
2*b^3)*d*cos(f*x + e) + (2*a^3*b^2 - a*b^4)*d)*sqrt(-(a^2 - b^2)/a^2)*dilo 
g(-(b*cos(f*x + e) + I*b*sin(f*x + e) - (a*cos(f*x + e) + I*a*sin(f*x + e) 
)*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) - ((2*a^4*b - a^2*b^3)*d*cos(f*x + e) 
 + (2*a^3*b^2 - a*b^4)*d)*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) - 
I*b*sin(f*x + e) + (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a 
^2) + a)/a + 1) + ((2*a^4*b - a^2*b^3)*d*cos(f*x + e) + (2*a^3*b^2 - a*b^4 
)*d)*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) - I*b*sin(f*x + e) - (a 
*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + (-I 
*(2*a^3*b^2 - a*b^4)*d*f*x - I*(2*a^3*b^2 - a*b^4)*d*e + (-I*(2*a^4*b - a^ 
2*b^3)*d*f*x - I*(2*a^4*b - a^2*b^3)*d*e)*cos(f*x + e))*sqrt(-(a^2 - b^2)/ 
a^2)*log((b*cos(f*x + e) + I*b*sin(f*x + e) + (a*cos(f*x + e) + I*a*sin(f* 
x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a) + (I*(2*a^3*b^2 - a*b^4)*d*f*x + I* 
(2*a^3*b^2 - a*b^4)*d*e + (I*(2*a^4*b - a^2*b^3)*d*f*x + I*(2*a^4*b - a^2* 
b^3)*d*e)*cos(f*x + e))*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(f*x + e) + I*b*s 
in(f*x + e) - (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 
 a)/a) + (I*(2*a^3*b^2 - a*b^4)*d*f*x + I*(2*a^3*b^2 - a*b^4)*d*e + (I*...
 

3.1.41.6 Sympy [F]

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

input

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

output

Integral((c + d*x)/(a + b*sec(e + f*x))**2, x)
 

3.1.41.7 Maxima [F(-2)]

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

input

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

3.1.41.8 Giac [F]

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

input

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

output

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

3.1.41.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}