\(\int \frac {(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [308]

Optimal result

Integrand size = 24, antiderivative size = 413 \[ \int \frac {(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 i a (e+f x) \arctan \left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac {i a f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i a f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i b f \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^2} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1185\) vs. \(2(413)=826\).

Time = 10.97 (sec) , antiderivative size = 1185, normalized size of antiderivative = 2.87 \[ \int \frac {(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-((b*e*Log[1 + (b*Sin[c + d*x])/a])/((a^2 - b^2)*d)) + (b*c*f*Log[1 + (b*S 
in[c + d*x])/a])/((a^2 - b^2)*d^2) - (b^2*f*(((c + d*x)*Log[a + b*Sin[c + 
d*x]])/b - ((-1/2*I)*(-c + Pi/2 - d*x)^2 + (4*I)*ArcSin[Sqrt[(a + b)/b]/Sq 
rt[2]]*ArcTan[((a - b)*Tan[(-c + Pi/2 - d*x)/2])/Sqrt[a^2 - b^2]] + (-c + 
Pi/2 - d*x + 2*ArcSin[Sqrt[(a + b)/b]/Sqrt[2]])*Log[1 + ((a - Sqrt[a^2 - b 
^2])*E^(I*(-c + Pi/2 - d*x)))/b] + (-c + Pi/2 - d*x - 2*ArcSin[Sqrt[(a + b 
)/b]/Sqrt[2]])*Log[1 + ((a + Sqrt[a^2 - b^2])*E^(I*(-c + Pi/2 - d*x)))/b] 
- (-c + Pi/2 - d*x)*Log[a + b*Sin[c + d*x]] - I*(PolyLog[2, ((-a - Sqrt[a^ 
2 - b^2])*E^(I*(-c + Pi/2 - d*x)))/b] + PolyLog[2, ((-a + Sqrt[a^2 - b^2]) 
*E^(I*(-c + Pi/2 - d*x)))/b]))/b))/((a^2 - b^2)*d^2) + ((d*e + d*f*x)*(((- 
I)*b*(d*e + d*f*x)^2)/f + 2*(a - b)*(d*e - c*f)*Log[1 - Tan[(c + d*x)/2]] 
- 4*b*(d*e + d*f*x)*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - 2*(a + b)*(d*e - 
 c*f)*Log[1 + Tan[(c + d*x)/2]] - (4*I)*b*f*PolyLog[2, -Cos[c + d*x] + I*S 
in[c + d*x]] + (2*I)*(a + b)*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(1/2 - I/2 
)*(1 + Tan[(c + d*x)/2])] + PolyLog[2, ((1 + I) - (1 - I)*Tan[(c + d*x)/2] 
)/2]) - (2*I)*(a + b)*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(1/2 + I/2)*(1 + 
Tan[(c + d*x)/2])] + PolyLog[2, (-1/2 - I/2)*(I + Tan[(c + d*x)/2])]) + (2 
*I)*(a - b)*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(-1/2 + I/2)*(-1 + Tan[(c + 
 d*x)/2])] + PolyLog[2, ((1 + I) + (1 - I)*Tan[(c + d*x)/2])/2]) - (2*I)*( 
a - b)*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(-1/2 - I/2)*(-1 + Tan[(c + d...
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 386, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {5044, 5030, 2620, 2715, 2838, 7293, 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.

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

\(\Big \downarrow \) 5044

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

\(\Big \downarrow \) 5030

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

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 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 5030

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

rule 5044

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

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 845 vs. \(2 (373 ) = 746\).

Time = 1.40 (sec) , antiderivative size = 846, normalized size of antiderivative = 2.05

method	result	size
risch	\(-\frac {e b \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{d \left (a -b \right ) \left (a +b \right )}-\frac {4 e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (4 a +4 b \right )}+\frac {4 e \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (4 a -4 b \right )}-\frac {f b \ln \left (\frac {-i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (a -b \right ) \left (a +b \right )}-\frac {f b \ln \left (\frac {-i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (a -b \right ) \left (a +b \right )}-\frac {f b \ln \left (\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (a -b \right ) \left (a +b \right )}-\frac {f b \ln \left (\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (a -b \right ) \left (a +b \right )}-\frac {4 i f \ln \left (-i {\mathrm e}^{i \left (d x +c \right )}\right ) \ln \left (-i \left (-{\mathrm e}^{i \left (d x +c \right )}+i\right )\right )}{d^{2} \left (4 a +4 b \right )}+\frac {i f b \operatorname {dilog}\left (\frac {-i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (a -b \right ) \left (a +b \right )}-\frac {4 f \ln \left (-i \left (-{\mathrm e}^{i \left (d x +c \right )}+i\right )\right ) x}{d \left (4 a +4 b \right )}-\frac {4 f \ln \left (-i \left (-{\mathrm e}^{i \left (d x +c \right )}+i\right )\right ) c}{d^{2} \left (4 a +4 b \right )}+\frac {i f b \operatorname {dilog}\left (\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (a -b \right ) \left (a +b \right )}-\frac {4 i f \operatorname {dilog}\left (-i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left (4 a +4 b \right )}+\frac {4 f \ln \left (-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )\right ) x}{d \left (4 a -4 b \right )}+\frac {4 f \ln \left (-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )\right ) c}{d^{2} \left (4 a -4 b \right )}-\frac {4 i f \operatorname {dilog}\left (-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )\right )}{d^{2} \left (4 a -4 b \right )}+\frac {c f b \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{d^{2} \left (a -b \right ) \left (a +b \right )}+\frac {4 c f \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d^{2} \left (4 a +4 b \right )}-\frac {4 c f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{2} \left (4 a -4 b \right )}\)	\(846\)

Input:

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

Output:

-1/d*e*b/(a-b)/(a+b)*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-4/d*e 
/(4*a+4*b)*ln(exp(I*(d*x+c))-I)+4/d*e/(4*a-4*b)*ln(exp(I*(d*x+c))+I)-1/d*f 
*b/(a-b)/(a+b)*ln((-I*exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/ 
2)))*x-1/d^2*f*b/(a-b)/(a+b)*ln((-I*exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)+a)/(a 
+(a^2-b^2)^(1/2)))*c-1/d*f*b/(a-b)/(a+b)*ln((I*b*exp(I*(d*x+c))+(a^2-b^2)^ 
(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*x-1/d^2*f*b/(a-b)/(a+b)*ln((I*b*exp(I*(d*x+ 
c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*c-4*I/d^2*f/(4*a+4*b)*ln(-I*e 
xp(I*(d*x+c)))*ln(-I*(-exp(I*(d*x+c))+I))+I/d^2*f*b/(a-b)/(a+b)*dilog((-I* 
exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))-4/d*f/(4*a+4*b)*l 
n(-I*(-exp(I*(d*x+c))+I))*x-4/d^2*f/(4*a+4*b)*ln(-I*(-exp(I*(d*x+c))+I))*c 
+I/d^2*f*b/(a-b)/(a+b)*dilog((I*b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a 
^2-b^2)^(1/2)))-4*I/d^2*f/(4*a+4*b)*dilog(-I*exp(I*(d*x+c)))+4/d*f/(4*a-4* 
b)*ln(-I*(exp(I*(d*x+c))+I))*x+4/d^2*f/(4*a-4*b)*ln(-I*(exp(I*(d*x+c))+I)) 
*c-4*I/d^2*f/(4*a-4*b)*dilog(-I*(exp(I*(d*x+c))+I))+1/d^2*c*f*b/(a-b)/(a+b 
)*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+4/d^2*c*f/(4*a+4*b)*ln(e 
xp(I*(d*x+c))-I)-4/d^2*c*f/(4*a-4*b)*ln(exp(I*(d*x+c))+I)
 

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. 1181 vs. \(2 (356) = 712\).

Time = 0.29 (sec) , antiderivative size = 1181, normalized size of antiderivative = 2.86 \[ \int \frac {(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(-I*b*f*dilog((I*a*cos(d*x + c) - 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*b*f*dilog((I*a*co 
s(d*x + c) - 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*b*f*dilog((-I*a*cos(d*x + c) - 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*b*f*dilog((-I*a*cos(d*x + c) - 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 - b)*f*dilog(I*c 
os(d*x + c) + sin(d*x + c)) + I*(a + b)*f*dilog(I*cos(d*x + c) - sin(d*x + 
 c)) - I*(a - b)*f*dilog(-I*cos(d*x + c) + sin(d*x + c)) - I*(a + b)*f*dil 
og(-I*cos(d*x + c) - sin(d*x + c)) + (b*d*e - b*c*f)*log(2*b*cos(d*x + c) 
+ 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (b*d*e - b*c* 
f)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) 
- 2*I*a) + (b*d*e - b*c*f)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2* 
b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (b*d*e - b*c*f)*log(-2*b*cos(d*x + c) 
- 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (b*d*f*x + b* 
c*f)*log(-(I*a*cos(d*x + c) - 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) + (b*d*f*x + b*c*f)*log(-(I*a*cos( 
d*x + c) - 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) + (b*d*f*x + b*c*f)*log(-(-I*a*cos(d*x + c) - a*sin(d 
*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (\int \frac {\sec \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) a^{2} d f -\left (\int \frac {\sec \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) b^{2} d f -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a e +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b e +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a e +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b e -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b e}{d \left (a^{2}-b^{2}\right )} \] Input:

int((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x)
 

Output:

(int((sec(c + d*x)*x)/(sin(c + d*x)*b + a),x)*a**2*d*f - int((sec(c + d*x) 
*x)/(sin(c + d*x)*b + a),x)*b**2*d*f - log(tan((c + d*x)/2) - 1)*a*e + log 
(tan((c + d*x)/2) - 1)*b*e + log(tan((c + d*x)/2) + 1)*a*e + log(tan((c + 
d*x)/2) + 1)*b*e - log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b 
*e)/(d*(a**2 - b**2))