\(\int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx\) [21]

Optimal result

Integrand size = 19, antiderivative size = 399 \[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx=\frac {(e+f x)^4 \log \left (1+e^{2 i (c+d x)}\right )}{4 f}-\frac {(e+f x)^4 \log \left (1+\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{4 f}+\frac {(e+f x)^4 \log (a+b \tan (c+d x))}{4 f}-\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}+\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{2 d}+\frac {3 f (e+f x)^2 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{4 d^2}-\frac {3 f (e+f x)^2 \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{4 d^2}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (4,-e^{2 i (c+d x)}\right )}{4 d^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (4,-\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{4 d^3}-\frac {3 f^3 \operatorname {PolyLog}\left (5,-e^{2 i (c+d x)}\right )}{8 d^4}+\frac {3 f^3 \operatorname {PolyLog}\left (5,-\frac {(a-i b) e^{2 i (c+d x)}}{a+i b}\right )}{8 d^4} \] Output:

1/4*(f*x+e)^4*ln(1+exp(2*I*(d*x+c)))/f-1/4*(f*x+e)^4*ln(1+(a-I*b)*exp(2*I* 
(d*x+c))/(a+I*b))/f+1/4*(f*x+e)^4*ln(a+b*tan(d*x+c))/f-1/2*I*(f*x+e)^3*pol 
ylog(2,-exp(2*I*(d*x+c)))/d+1/2*I*(f*x+e)^3*polylog(2,-(a-I*b)*exp(2*I*(d* 
x+c))/(a+I*b))/d+3/4*f*(f*x+e)^2*polylog(3,-exp(2*I*(d*x+c)))/d^2-3/4*f*(f 
*x+e)^2*polylog(3,-(a-I*b)*exp(2*I*(d*x+c))/(a+I*b))/d^2+3/4*I*f^2*(f*x+e) 
*polylog(4,-exp(2*I*(d*x+c)))/d^3-3/4*I*f^2*(f*x+e)*polylog(4,-(a-I*b)*exp 
(2*I*(d*x+c))/(a+I*b))/d^3-3/8*f^3*polylog(5,-exp(2*I*(d*x+c)))/d^4+3/8*f^ 
3*polylog(5,-(a-I*b)*exp(2*I*(d*x+c))/(a+I*b))/d^4
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1710\) vs. \(2(399)=798\).

Time = 13.18 (sec) , antiderivative size = 1710, normalized size of antiderivative = 4.29 \[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx =\text {Too large to display} \] Input:

Integrate[(e + f*x)^3*Log[a + b*Tan[c + d*x]],x]
 

Output:

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*Log[a + b*Tan[c + d*x]])/4 
+ ((a^2 + b^2)*d*((40*e^3*x^2)/(I*a + b) + (40*e^2*f*x^3)/(I*a + b) + (20* 
e*f^2*x^4)/(I*a + b) + (4*f^3*x^5)/(I*a + b) - (40*e^3*((-I)*b*(-1 + E^((2 
*I)*c)) + a*(1 + E^((2*I)*c)))*x*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c 
+ d*x)))])/((a^2 + b^2)*d) - (60*e^2*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E 
^((2*I)*c)))*f*x^2*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x)))])/((a 
^2 + b^2)*d) - (40*e*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*f^2 
*x^3*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x)))])/((a^2 + b^2)*d) - 
 (10*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*f^3*x^4*Log[1 + (a 
+ I*b)/((a - I*b)*E^((2*I)*(c + d*x)))])/((a^2 + b^2)*d) + (20*e^3*((-I)*b 
*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*PolyLog[2, (-a - I*b)/((a - I*b 
)*E^((2*I)*(c + d*x)))])/((a + I*b)*(I*a + b)*d^2) + (30*e^2*(b - b*E^((2* 
I)*c) - I*a*(1 + E^((2*I)*c)))*f*(2*d*x*PolyLog[2, (-a - I*b)/((a - I*b)*E 
^((2*I)*(c + d*x)))] - I*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d* 
x)))]))/((a^2 + b^2)*d^3) + (30*e*(b - b*E^((2*I)*c) - I*a*(1 + E^((2*I)*c 
)))*f^2*(2*d^2*x^2*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x)))] 
- (2*I)*d*x*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x)))] - PolyL 
og[4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x)))]))/((a^2 + b^2)*d^4) + (5 
*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*f^3*((-4*I)*d^3*x^3*Pol 
yLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x)))] - 6*d^2*x^2*PolyLog...
 

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.

\(\displaystyle \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3031

\(\displaystyle \frac {(e+f x)^4 \log (a+b \tan (c+d x))}{4 f}-\frac {\int \frac {b d (e+f x)^4 \sec ^2(c+d x)}{a+b \tan (c+d x)}dx}{4 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(e+f x)^4 \log (a+b \tan (c+d x))}{4 f}-\frac {b d \int \frac {(e+f x)^4 \sec ^2(c+d x)}{a+b \tan (c+d x)}dx}{4 f}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {(e+f x)^4 \log (a+b \tan (c+d x))}{4 f}-\frac {b d \int \left (\frac {\sec ^2(c+d x) e^4}{a+b \tan (c+d x)}+\frac {4 f x \sec ^2(c+d x) e^3}{a+b \tan (c+d x)}+\frac {6 f^2 x^2 \sec ^2(c+d x) e^2}{a+b \tan (c+d x)}+\frac {4 f^3 x^3 \sec ^2(c+d x) e}{a+b \tan (c+d x)}+\frac {f^4 x^4 \sec ^2(c+d x)}{a+b \tan (c+d x)}\right )dx}{4 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(e+f x)^4 \log (a+b \tan (c+d x))}{4 f}-\frac {b d \left (4 e^3 f \int \frac {x \sec ^2(c+d x)}{a+b \tan (c+d x)}dx+6 e^2 f^2 \int \frac {x^2 \sec ^2(c+d x)}{a+b \tan (c+d x)}dx+4 e f^3 \int \frac {x^3 \sec ^2(c+d x)}{a+b \tan (c+d x)}dx+f^4 \int \frac {x^4 \sec ^2(c+d x)}{a+b \tan (c+d x)}dx+\frac {e^4 \log (a+b \tan (c+d x))}{b d}\right )}{4 f}\)

Input:

Int[(e + f*x)^3*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 = 10.88 (sec) , antiderivative size = 8233, normalized size of antiderivative = 20.63

method	result	size
risch	\(\text {Expression too large to display}\)	\(8233\)

Input:

int((f*x+e)^3*ln(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

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. 2053 vs. \(2 (327) = 654\).

Time = 0.15 (sec) , antiderivative size = 2053, normalized size of antiderivative = 5.15 \[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*log(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

1/16*(3*f^3*polylog(5, ((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)*tan(d*x + c))/((a^2 + b^2)*tan(d*x + 
 c)^2 + a^2 + b^2)) + 3*f^3*polylog(5, ((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)*tan(d*x + c))/((a^2 + 
 b^2)*tan(d*x + c)^2 + a^2 + b^2)) - 3*f^3*polylog(5, (tan(d*x + c)^2 + 2* 
I*tan(d*x + c) - 1)/(tan(d*x + c)^2 + 1)) - 3*f^3*polylog(5, (tan(d*x + c) 
^2 - 2*I*tan(d*x + c) - 1)/(tan(d*x + c)^2 + 1)) - 4*(I*d^3*f^3*x^3 + 3*I* 
d^3*e*f^2*x^2 + 3*I*d^3*e^2*f*x + I*d^3*e^3)*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) - 4*(-I*d^3*f^3*x^3 - 3*I*d^3*e*f^2*x^2 
 - 3*I*d^3*e^2*f*x - I*d^3*e^3)*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) - 4*(-I*d^3*f^3*x^3 - 3*I*d^3*e*f^2*x^2 - 3*I*d^3* 
e^2*f*x - I*d^3*e^3)*dilog(2*(I*tan(d*x + c) - 1)/(tan(d*x + c)^2 + 1) + 1 
) - 4*(I*d^3*f^3*x^3 + 3*I*d^3*e*f^2*x^2 + 3*I*d^3*e^2*f*x + I*d^3*e^3)*di 
log(2*(-I*tan(d*x + c) - 1)/(tan(d*x + c)^2 + 1) + 1) + 4*(d^4*f^3*x^4 + 4 
*d^4*e*f^2*x^3 + 6*d^4*e^2*f*x^2 + 4*d^4*e^3*x)*log(b*tan(d*x + c) + a) - 
2*(d^4*f^3*x^4 + 4*d^4*e*f^2*x^3 + 6*d^4*e^2*f*x^2 + 4*d^4*e^3*x + 4*c*d^3 
*e^3 - 6*c^2*d^2*e^2*f + 4*c^3*d*e*f^2 - c^4*f^3)*log(-2*((I*a*b - b^2)*ta 
n(d*x + c)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x + c))/((a^...
 

Sympy [F]

\[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx=\int \left (e + f x\right )^{3} \log {\left (a + b \tan {\left (c + d x \right )} \right )}\, dx \] Input:

integrate((f*x+e)**3*ln(a+b*tan(d*x+c)),x)
 

Output:

Integral((e + f*x)**3*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. 1385 vs. \(2 (327) = 654\).

Time = 0.32 (sec) , antiderivative size = 1385, normalized size of antiderivative = 3.47 \[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*log(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

1/12*(3*(4*(d*x + c)*e^3 + 6*((d*x + c)^2 - 2*(d*x + c)*c)*e^2*f/d + 4*((d 
*x + c)^3 - 3*(d*x + c)^2*c + 3*(d*x + c)*c^2)*e*f^2/d^2 + ((d*x + c)^4 - 
4*(d*x + c)^3*c + 6*(d*x + c)^2*c^2 - 4*(d*x + c)*c^3)*f^3/d^3)*log(b*tan( 
d*x + c) + a) + (9*f^3*polylog(5, (I*a + b)*e^(2*I*d*x + 2*I*c)/(-I*a + b) 
) - 9*f^3*polylog(5, -e^(2*I*d*x + 2*I*c)) + 2*(3*I*(d*x + c)^4*f^3 + 8*(I 
*d*e*f^2 - I*c*f^3)*(d*x + c)^3 + 9*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*c^2*f 
^3)*(d*x + c)^2 + 6*(I*d^3*e^3 - 3*I*c*d^2*e^2*f + 3*I*c^2*d*e*f^2 - I*c^3 
*f^3)*(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*sin(2*d*x + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos( 
2*d*x + 2*c))/(a^2 + b^2)) + 2*(3*I*(d*x + c)^4*f^3 + 8*(I*d*e*f^2 - I*c*f 
^3)*(d*x + c)^3 + 9*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*c^2*f^3)*(d*x + c)^2 
+ 6*(I*d^3*e^3 - 3*I*c*d^2*e^2*f + 3*I*c^2*d*e*f^2 - I*c^3*f^3)*(d*x + c)) 
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) + 6*(I*d^3*e^3 - 3*I*c*d^ 
2*e^2*f + 3*I*c^2*d*e*f^2 + 2*I*(d*x + c)^3*f^3 - I*c^3*f^3 + 4*(I*d*e*f^2 
 - I*c*f^3)*(d*x + c)^2 + 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*c^2*f^3)*(d*x 
 + c))*dilog((I*a + b)*e^(2*I*d*x + 2*I*c)/(-I*a + b)) + 6*(-I*d^3*e^3 + 3 
*I*c*d^2*e^2*f - 3*I*c^2*d*e*f^2 - 2*I*(d*x + c)^3*f^3 + I*c^3*f^3 + 4*(-I 
*d*e*f^2 + I*c*f^3)*(d*x + c)^2 + 3*(-I*d^2*e^2*f + 2*I*c*d*e*f^2 - I*c^2* 
f^3)*(d*x + c))*dilog(-e^(2*I*d*x + 2*I*c)) + (3*(d*x + c)^4*f^3 + 8*(d*e* 
f^2 - c*f^3)*(d*x + c)^3 + 9*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x +...
 

Giac [F]

\[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx=\int { {\left (f x + e\right )}^{3} \log \left (b \tan \left (d x + c\right ) + a\right ) \,d x } \] Input:

integrate((f*x+e)^3*log(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)^3*log(b*tan(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx=\int \ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (e+f\,x\right )}^3 \,d x \] Input:

int(log(a + b*tan(c + d*x))*(e + f*x)^3,x)
 

Output:

int(log(a + b*tan(c + d*x))*(e + f*x)^3, x)
 

Reduce [F]

\[ \int (e+f x)^3 \log (a+b \tan (c+d x)) \, dx=\left (\int \mathrm {log}\left (\tan \left (d x +c \right ) b +a \right )d x \right ) e^{3}+\left (\int \mathrm {log}\left (\tan \left (d x +c \right ) b +a \right ) x^{3}d x \right ) f^{3}+3 \left (\int \mathrm {log}\left (\tan \left (d x +c \right ) b +a \right ) x^{2}d x \right ) e \,f^{2}+3 \left (\int \mathrm {log}\left (\tan \left (d x +c \right ) b +a \right ) x d x \right ) e^{2} f \] Input:

int((f*x+e)^3*log(a+b*tan(d*x+c)),x)
 

Output:

int(log(tan(c + d*x)*b + a),x)*e**3 + int(log(tan(c + d*x)*b + a)*x**3,x)* 
f**3 + 3*int(log(tan(c + d*x)*b + a)*x**2,x)*e*f**2 + 3*int(log(tan(c + d* 
x)*b + a)*x,x)*e**2*f