\(\int (c+d x)^3 (a+b \tan (e+f x)) \, dx\) [39]

Optimal result

Integrand size = 18, antiderivative size = 152 \[ \int (c+d x)^3 (a+b \tan (e+f x)) \, dx=\frac {a (c+d x)^4}{4 d}+\frac {i b (c+d x)^4}{4 d}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}-\frac {3 i b d^3 \operatorname {PolyLog}\left (4,-e^{2 i (e+f x)}\right )}{4 f^4} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(342\) vs. \(2(152)=304\).

Time = 0.12 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.25 \[ \int (c+d x)^3 (a+b \tan (e+f x)) \, dx=a c^3 x+\frac {3}{2} a c^2 d x^2+\frac {3}{2} i b c^2 d x^2+a c d^2 x^3+i b c d^2 x^3+\frac {1}{4} a d^3 x^4+\frac {1}{4} i b d^3 x^4-\frac {3 b c^2 d x \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {3 b c d^2 x^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {b d^3 x^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {b c^3 \log (\cos (e+f x))}{f}+\frac {3 i b c^2 d \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2}+\frac {3 i b c d^2 x \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}+\frac {3 i b d^3 x^2 \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {3 b c d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}-\frac {3 b d^3 x \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}-\frac {3 i b d^3 \operatorname {PolyLog}\left (4,-e^{2 i (e+f x)}\right )}{4 f^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 152, 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, 4205, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4205

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a (c+d x)^4}{4 d}-\frac {3 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {3 i b d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^4}{4 d}-\frac {3 i b d^3 \operatorname {PolyLog}\left (4,-e^{2 i (e+f x)}\right )}{4 f^4}\)

Input:

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

Output:

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

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 4205

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (131 ) = 262\).

Time = 0.53 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.29

method	result	size
risch	\(\frac {i d^{3} b \,x^{4}}{4}+\frac {2 b \,c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}-\frac {b \,c^{3} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f}-\frac {3 b d \,c^{2} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f}-\frac {3 b \,d^{2} c \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) x^{2}}{f}+\frac {6 b \,e^{2} d^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {6 b e d \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+d^{2} a c \,x^{3}+\frac {3 d a \,c^{2} x^{2}}{2}+a \,c^{3} x -\frac {3 i b \,d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{4 f^{4}}-i b \,c^{3} x -\frac {i b \,c^{4}}{4 d}+\frac {d^{3} a \,x^{4}}{4}+\frac {a \,c^{4}}{4 d}-\frac {2 b \,e^{3} d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}-\frac {b \,d^{3} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) x^{3}}{f}-\frac {3 b \,d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{2 f^{3}}-\frac {3 b \,d^{2} c \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{3}}+\frac {3 i b \,d^{3} e^{4}}{2 f^{4}}+\frac {3 i d b \,c^{2} x^{2}}{2}+i d^{2} b c \,x^{3}+\frac {3 i b \,d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f^{2}}-\frac {6 i b \,d^{2} c \,e^{2} x}{f^{2}}+\frac {6 i b d \,c^{2} e x}{f}+\frac {2 i b \,d^{3} e^{3} x}{f^{3}}+\frac {3 i b \,d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x^{2}}{2 f^{2}}-\frac {4 i b \,d^{2} c \,e^{3}}{f^{3}}+\frac {3 i b d \,c^{2} e^{2}}{f^{2}}+\frac {3 i b d \,c^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{2}}\)	\(500\)

Input:

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

Output:

d^2*a*c*x^3+3/2*d*a*c^2*x^2+a*c^3*x-I*b*c^3*x-1/4*I/d*b*c^4-2/f^4*b*e^3*d^ 
3*ln(exp(I*(f*x+e)))-1/f*b*d^3*ln(1+exp(2*I*(f*x+e)))*x^3-3/2/f^3*b*d^3*po 
lylog(3,-exp(2*I*(f*x+e)))*x-3/2/f^3*b*d^2*c*polylog(3,-exp(2*I*(f*x+e)))+ 
3/2*I/f^4*b*d^3*e^4+3/2*I*d*b*c^2*x^2+1/4*I*d^3*b*x^4+2/f*b*c^3*ln(exp(I*( 
f*x+e)))-1/f*b*c^3*ln(1+exp(2*I*(f*x+e)))+1/4*d^3*a*x^4+1/4/d*a*c^4+I*d^2* 
b*c*x^3-3/f*b*d*c^2*ln(1+exp(2*I*(f*x+e)))*x-3/f*b*d^2*c*ln(1+exp(2*I*(f*x 
+e)))*x^2+6/f^3*b*e^2*d^2*c*ln(exp(I*(f*x+e)))-6/f^2*b*e*d*c^2*ln(exp(I*(f 
*x+e)))+2*I/f^3*b*d^3*e^3*x+3/2*I/f^2*b*d^3*polylog(2,-exp(2*I*(f*x+e)))*x 
^2-4*I/f^3*b*d^2*c*e^3+3*I/f^2*b*d*c^2*e^2+3/2*I/f^2*b*d*c^2*polylog(2,-ex 
p(2*I*(f*x+e)))+3*I/f^2*b*d^2*c*polylog(2,-exp(2*I*(f*x+e)))*x-6*I/f^2*b*d 
^2*c*e^2*x-3/4*I*b*d^3*polylog(4,-exp(2*I*(f*x+e)))/f^4+6*I/f*b*d*c^2*e*x
 

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. 500 vs. \(2 (127) = 254\).

Time = 0.09 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.29 \[ \int (c+d x)^3 (a+b \tan (e+f x)) \, dx=\frac {2 \, a d^{3} f^{4} x^{4} + 8 \, a c d^{2} f^{4} x^{3} + 12 \, a c^{2} d f^{4} x^{2} + 8 \, a c^{3} f^{4} x + 3 i \, b d^{3} {\rm polylog}\left (4, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 i \, b d^{3} {\rm polylog}\left (4, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (i \, b d^{3} f^{2} x^{2} + 2 i \, b c d^{2} f^{2} x + i \, b c^{2} d f^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \, {\left (-i \, b d^{3} f^{2} x^{2} - 2 i \, b c d^{2} f^{2} x - i \, b c^{2} d f^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b c^{3} f^{3}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b c^{3} f^{3}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right )}{8 \, f^{4}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a + b*tan(e + f*x))*(c + d*x)**3, 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. 672 vs. \(2 (127) = 254\).

Time = 0.22 (sec) , antiderivative size = 672, normalized size of antiderivative = 4.42 \[ \int (c+d x)^3 (a+b \tan (e+f x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/12*(12*(f*x + e)*a*c^3 + 3*(f*x + e)^4*a*d^3/f^3 - 12*(f*x + e)^3*a*d^3* 
e/f^3 + 18*(f*x + e)^2*a*d^3*e^2/f^3 - 12*(f*x + e)*a*d^3*e^3/f^3 + 12*(f* 
x + e)^3*a*c*d^2/f^2 - 36*(f*x + e)^2*a*c*d^2*e/f^2 + 36*(f*x + e)*a*c*d^2 
*e^2/f^2 + 18*(f*x + e)^2*a*c^2*d/f - 36*(f*x + e)*a*c^2*d*e/f + 12*b*c^3* 
log(sec(f*x + e)) - 12*b*d^3*e^3*log(sec(f*x + e))/f^3 + 36*b*c*d^2*e^2*lo 
g(sec(f*x + e))/f^2 - 36*b*c^2*d*e*log(sec(f*x + e))/f - (-3*I*(f*x + e)^4 
*b*d^3 + 12*I*b*d^3*polylog(4, -e^(2*I*f*x + 2*I*e)) - 12*(-I*b*d^3*e + I* 
b*c*d^2*f)*(f*x + e)^3 - 18*(I*b*d^3*e^2 - 2*I*b*c*d^2*e*f + I*b*c^2*d*f^2 
)*(f*x + e)^2 - 4*(-4*I*(f*x + e)^3*b*d^3 + 9*(I*b*d^3*e - I*b*c*d^2*f)*(f 
*x + e)^2 + 9*(-I*b*d^3*e^2 + 2*I*b*c*d^2*e*f - I*b*c^2*d*f^2)*(f*x + e))* 
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 6*(4*I*(f*x + e)^2*b*d^3 
 + 3*I*b*d^3*e^2 - 6*I*b*c*d^2*e*f + 3*I*b*c^2*d*f^2 + 6*(-I*b*d^3*e + I*b 
*c*d^2*f)*(f*x + e))*dilog(-e^(2*I*f*x + 2*I*e)) + 2*(4*(f*x + e)^3*b*d^3 
- 9*(b*d^3*e - b*c*d^2*f)*(f*x + e)^2 + 9*(b*d^3*e^2 - 2*b*c*d^2*e*f + b*c 
^2*d*f^2)*(f*x + e))*log(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2 
*f*x + 2*e) + 1) + 6*(4*(f*x + e)*b*d^3 - 3*b*d^3*e + 3*b*c*d^2*f)*polylog 
(3, -e^(2*I*f*x + 2*I*e)))/f^3)/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 (a+b \tan (e+f x)) \, dx=\frac {4 \left (\int \tan \left (f x +e \right ) x^{3}d x \right ) b \,d^{3} f +12 \left (\int \tan \left (f x +e \right ) x^{2}d x \right ) b c \,d^{2} f +12 \left (\int \tan \left (f x +e \right ) x d x \right ) b \,c^{2} d f +2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) b \,c^{3}+4 a \,c^{3} f x +6 a \,c^{2} d f \,x^{2}+4 a c \,d^{2} f \,x^{3}+a \,d^{3} f \,x^{4}}{4 f} \] Input:

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

Output:

(4*int(tan(e + f*x)*x**3,x)*b*d**3*f + 12*int(tan(e + f*x)*x**2,x)*b*c*d** 
2*f + 12*int(tan(e + f*x)*x,x)*b*c**2*d*f + 2*log(tan(e + f*x)**2 + 1)*b*c 
**3 + 4*a*c**3*f*x + 6*a*c**2*d*f*x**2 + 4*a*c*d**2*f*x**3 + a*d**3*f*x**4 
)/(4*f)