\(\int (e+f x) \log (a+b \tan (c+d x)) \, dx\) [23]

Optimal result

Integrand size = 17, antiderivative size = 235 \[ \int (e+f x) \log (a+b \tan (c+d x)) \, dx=\frac {(e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 f}-\frac {(e+f x)^2 \log \left (1+\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{2 f}+\frac {(e+f x)^2 \log (a+b \tan (c+d x))}{2 f}-\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}+\frac {i (e+f x) \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{2 d}+\frac {f \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{4 d^2}-\frac {f \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{4 d^2} \] Output:

1/2*(f*x+e)^2*ln(1+exp(2*I*(d*x+c)))/f-1/2*(f*x+e)^2*ln(1+(a-I*b)*exp(2*I* 
(d*x+c))/(a+I*b))/f+1/2*(f*x+e)^2*ln(a+b*tan(d*x+c))/f-1/2*I*(f*x+e)*polyl 
og(2,-exp(2*I*(d*x+c)))/d+1/2*I*(f*x+e)*polylog(2,-(a-I*b)*exp(2*I*(d*x+c) 
)/(a+I*b))/d+1/4*f*polylog(3,-exp(2*I*(d*x+c)))/d^2-1/4*f*polylog(3,-(a-I* 
b)*exp(2*I*(d*x+c))/(a+I*b))/d^2
 

Mathematica [A] (verified)

Time = 2.24 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.45 \[ \int (e+f x) \log (a+b \tan (c+d x)) \, dx=\frac {2 d^2 f x^2 \log \left (1+e^{-2 i (c+d x)}\right )-2 d^2 f x^2 \log \left (1+\frac {(a+i b) e^{-2 i (c+d x)}}{a-i b}\right )+2 d^2 f x^2 \log (a+b \tan (c+d x))-2 i d e \log \left (-\frac {b (-i+\tan (c+d x))}{a+i b}\right ) \log (a+b \tan (c+d x))+2 i d e \log \left (-\frac {b (i+\tan (c+d x))}{a-i b}\right ) \log (a+b \tan (c+d x))+2 i d f x \operatorname {PolyLog}\left (2,-e^{-2 i (c+d x)}\right )-2 i d f x \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i (c+d x)}}{a-i b}\right )+2 i d e \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a-i b}\right )-2 i d e \operatorname {PolyLog}\left (2,\frac {a+b \tan (c+d x)}{a+i b}\right )+f \operatorname {PolyLog}\left (3,-e^{-2 i (c+d x)}\right )-f \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i (c+d x)}}{a-i b}\right )}{4 d^2} \] Input:

Integrate[(e + f*x)*Log[a + b*Tan[c + d*x]],x]
 

Output:

(2*d^2*f*x^2*Log[1 + E^((-2*I)*(c + d*x))] - 2*d^2*f*x^2*Log[1 + (a + I*b) 
/((a - I*b)*E^((2*I)*(c + d*x)))] + 2*d^2*f*x^2*Log[a + b*Tan[c + d*x]] - 
(2*I)*d*e*Log[-((b*(-I + Tan[c + d*x]))/(a + I*b))]*Log[a + b*Tan[c + d*x] 
] + (2*I)*d*e*Log[-((b*(I + Tan[c + d*x]))/(a - I*b))]*Log[a + b*Tan[c + d 
*x]] + (2*I)*d*f*x*PolyLog[2, -E^((-2*I)*(c + d*x))] - (2*I)*d*f*x*PolyLog 
[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x)))] + (2*I)*d*e*PolyLog[2, (a 
+ b*Tan[c + d*x])/(a - I*b)] - (2*I)*d*e*PolyLog[2, (a + b*Tan[c + d*x])/( 
a + I*b)] + f*PolyLog[3, -E^((-2*I)*(c + d*x))] - f*PolyLog[3, (-a - I*b)/ 
((a - I*b)*E^((2*I)*(c + d*x)))])/(4*d^2)
 

Rubi [F]

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.

\(\displaystyle \int (e+f x) \log (a+b \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3031

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(e + f*x)*Log[a + b*Tan[c + d*x]],x]
 

Output:

$Aborted
 

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 2009

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

rule 3031

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1) 
*(Log[u]/(b*(m + 1))), x] - Simp[1/(b*(m + 1))   Int[SimplifyIntegrand[(a + 
 b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && InverseFunc 
tionFreeQ[u, x] && NeQ[m, -1]
 

rule 7293

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 8.98 (sec) , antiderivative size = 3464, normalized size of antiderivative = 14.74

method	result	size
risch	\(\text {Expression too large to display}\)	\(3464\)

Input:

int((f*x+e)*ln(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

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

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. 995 vs. \(2 (197) = 394\).

Time = 0.12 (sec) , antiderivative size = 995, normalized size of antiderivative = 4.23 \[ \int (e+f x) \log (a+b \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)*log(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

-1/8*(2*(I*d*f*x + I*d*e)*dilog(2*((I*a*b - b^2)*tan(d*x + c)^2 - a^2 - I* 
a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x + c))/((a^2 + b^2)*tan(d*x + c)^2 + 
a^2 + b^2) + 1) + 2*(-I*d*f*x - I*d*e)*dilog(2*((-I*a*b - b^2)*tan(d*x + c 
)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*x + c))/((a^2 + b^2)*ta 
n(d*x + c)^2 + a^2 + b^2) + 1) + 2*(-I*d*f*x - I*d*e)*dilog(2*(I*tan(d*x + 
 c) - 1)/(tan(d*x + c)^2 + 1) + 1) + 2*(I*d*f*x + I*d*e)*dilog(2*(-I*tan(d 
*x + c) - 1)/(tan(d*x + c)^2 + 1) + 1) - 4*(d^2*f*x^2 + 2*d^2*e*x)*log(b*t 
an(d*x + c) + a) + 2*(d^2*f*x^2 + 2*d^2*e*x + 2*c*d*e - c^2*f)*log(-2*((I* 
a*b - b^2)*tan(d*x + c)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x 
+ c))/((a^2 + b^2)*tan(d*x + c)^2 + a^2 + b^2)) + 2*(d^2*f*x^2 + 2*d^2*e*x 
 + 2*c*d*e - c^2*f)*log(-2*((-I*a*b - b^2)*tan(d*x + c)^2 - a^2 + I*a*b + 
(-I*a^2 - 2*a*b + I*b^2)*tan(d*x + c))/((a^2 + b^2)*tan(d*x + c)^2 + a^2 + 
 b^2)) - 2*(2*c*d*e - c^2*f)*log(((I*a*b + b^2)*tan(d*x + c)^2 - a^2 + I*a 
*b + (I*a^2 + I*b^2)*tan(d*x + c))/(tan(d*x + c)^2 + 1)) - 2*(2*c*d*e - c^ 
2*f)*log(((I*a*b - b^2)*tan(d*x + c)^2 + a^2 + I*a*b + (I*a^2 + I*b^2)*tan 
(d*x + c))/(tan(d*x + c)^2 + 1)) - 2*(d^2*f*x^2 + 2*d^2*e*x)*log(-2*(I*tan 
(d*x + c) - 1)/(tan(d*x + c)^2 + 1)) - 2*(d^2*f*x^2 + 2*d^2*e*x)*log(-2*(- 
I*tan(d*x + c) - 1)/(tan(d*x + c)^2 + 1)) + f*polylog(3, ((a^2 + 2*I*a*b - 
 b^2)*tan(d*x + c)^2 - a^2 - 2*I*a*b + b^2 - 2*(-I*a^2 + 2*a*b + I*b^2)*ta 
n(d*x + c))/((a^2 + b^2)*tan(d*x + c)^2 + a^2 + b^2)) + f*polylog(3, ((...
 

Sympy [F]

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

integrate((f*x+e)*ln(a+b*tan(d*x+c)),x)
 

Output:

Integral((e + f*x)*log(a + b*tan(c + d*x)), x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (197) = 394\).

Time = 0.18 (sec) , antiderivative size = 520, normalized size of antiderivative = 2.21 \[ \int (e+f x) \log (a+b \tan (c+d x)) \, dx=\frac {2 \, {\left (2 \, {\left (d x + c\right )} e + \frac {{\left ({\left (d x + c\right )}^{2} - 2 \, {\left (d x + c\right )} c\right )} f}{d}\right )} \log \left (b \tan \left (d x + c\right ) + a\right ) + \frac {2 \, {\left (i \, {\left (d x + c\right )}^{2} f + 2 \, {\left (i \, d e - i \, c f\right )} {\left (d x + c\right )}\right )} \arctan \left (\frac {2 \, a b \cos \left (2 \, d x + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d x + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{a^{2} + b^{2}}\right ) + 2 \, {\left (i \, {\left (d x + c\right )}^{2} f + 2 \, {\left (i \, d e - i \, c f\right )} {\left (d x + c\right )}\right )} \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) + 2 \, {\left (i \, d e + i \, {\left (d x + c\right )} f - i \, c f\right )} {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x + 2 i \, c\right )}}{-i \, a + b}\right ) + 2 \, {\left (-i \, d e - i \, {\left (d x + c\right )} f + i \, c f\right )} {\rm Li}_2\left (-e^{\left (2 i \, d x + 2 i \, c\right )}\right ) + {\left ({\left (d x + c\right )}^{2} f + 2 \, {\left (d e - c f\right )} {\left (d x + c\right )}\right )} \log \left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right ) - {\left ({\left (d x + c\right )}^{2} f + 2 \, {\left (d e - c f\right )} {\left (d x + c\right )}\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{a^{2} + b^{2}}\right ) - f {\rm Li}_{3}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x + 2 i \, c\right )}}{-i \, a + b}) + f {\rm Li}_{3}(-e^{\left (2 i \, d x + 2 i \, c\right )})}{d}}{4 \, d} \] Input:

integrate((f*x+e)*log(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

1/4*(2*(2*(d*x + c)*e + ((d*x + c)^2 - 2*(d*x + c)*c)*f/d)*log(b*tan(d*x + 
 c) + a) + (2*(I*(d*x + c)^2*f + 2*(I*d*e - I*c*f)*(d*x + c))*arctan2((2*a 
*b*cos(2*d*x + 2*c) - (a^2 - b^2)*sin(2*d*x + 2*c))/(a^2 + b^2), (2*a*b*si 
n(2*d*x + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*x + 2*c))/(a^2 + b^2)) + 
2*(I*(d*x + c)^2*f + 2*(I*d*e - I*c*f)*(d*x + c))*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c) + 1) + 2*(I*d*e + I*(d*x + c)*f - I*c*f)*dilog((I*a + b 
)*e^(2*I*d*x + 2*I*c)/(-I*a + b)) + 2*(-I*d*e - I*(d*x + c)*f + I*c*f)*dil 
og(-e^(2*I*d*x + 2*I*c)) + ((d*x + c)^2*f + 2*(d*e - c*f)*(d*x + c))*log(c 
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) - ((d*x + 
 c)^2*f + 2*(d*e - c*f)*(d*x + c))*log(((a^2 + b^2)*cos(2*d*x + 2*c)^2 + 4 
*a*b*sin(2*d*x + 2*c) + (a^2 + b^2)*sin(2*d*x + 2*c)^2 + a^2 + b^2 + 2*(a^ 
2 - b^2)*cos(2*d*x + 2*c))/(a^2 + b^2)) - f*polylog(3, (I*a + b)*e^(2*I*d* 
x + 2*I*c)/(-I*a + b)) + f*polylog(3, -e^(2*I*d*x + 2*I*c)))/d)/d
 

Giac [F]

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

integrate((f*x+e)*log(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)*log(b*tan(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x) \log (a+b \tan (c+d x)) \, dx=\int \ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (e+f\,x\right ) \,d x \] Input:

int(log(a + b*tan(c + d*x))*(e + f*x),x)
 

Output:

int(log(a + b*tan(c + d*x))*(e + f*x), x)
 

Reduce [F]

\[ \int (e+f x) \log (a+b \tan (c+d x)) \, dx=\left (\int \mathrm {log}\left (\tan \left (d x +c \right ) b +a \right )d x \right ) e +\left (\int \mathrm {log}\left (\tan \left (d x +c \right ) b +a \right ) x d x \right ) f \] Input:

int((f*x+e)*log(a+b*tan(d*x+c)),x)
 

Output:

int(log(tan(c + d*x)*b + a),x)*e + int(log(tan(c + d*x)*b + a)*x,x)*f