\(\int (c+d x)^2 (a+b \coth (e+f x)) \, dx\) [38]

Optimal result

Integrand size = 18, antiderivative size = 101 \[ \int (c+d x)^2 (a+b \coth (e+f x)) \, dx=\frac {a (c+d x)^3}{3 d}-\frac {b (c+d x)^3}{3 d}+\frac {b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{f^2}-\frac {b d^2 \operatorname {PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.70 \[ \int (c+d x)^2 (a+b \coth (e+f x)) \, dx=a c^2 x+a c d x^2-b c d x^2+\frac {1}{3} a d^2 x^3-\frac {1}{3} b d^2 x^3+\frac {2 b c d x \log \left (1-e^{2 e+2 f x}\right )}{f}+\frac {b d^2 x^2 \log \left (1-e^{2 e+2 f x}\right )}{f}+\frac {b c^2 \log (\sinh (e+f x))}{f}+\frac {b c d \operatorname {PolyLog}\left (2,e^{2 e+2 f x}\right )}{f^2}+\frac {b d^2 x \operatorname {PolyLog}\left (2,e^{2 e+2 f x}\right )}{f^2}-\frac {b d^2 \operatorname {PolyLog}\left (3,e^{2 e+2 f x}\right )}{2 f^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 101, 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.

\(\displaystyle \int (c+d x)^2 (a+b \coth (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4205

\(\displaystyle \int \left (a (c+d x)^2+b (c+d x)^2 \coth (e+f x)\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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)

Leaf count of result is larger than twice the leaf count of optimal. \(464\) vs. \(2(95)=190\).

Time = 0.12 (sec) , antiderivative size = 465, normalized size of antiderivative = 4.60

method	result	size
risch	\(\frac {b \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}-1\right )}{f^{3}}-\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {b \,d^{2} \ln \left (1-{\mathrm e}^{f x +e}\right ) x^{2}}{f}-\frac {2 b d c \,e^{2}}{f^{2}}-\frac {b \,d^{2} \ln \left (1-{\mathrm e}^{f x +e}\right ) e^{2}}{f^{3}}+\frac {2 b \,d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{f x +e}\right ) x}{f^{2}}+\frac {b \,d^{2} \ln \left ({\mathrm e}^{f x +e}+1\right ) x^{2}}{f}+\frac {2 b \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{f x +e}\right ) x}{f^{2}}+\frac {2 b d c \operatorname {polylog}\left (2, {\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {2 b d c \operatorname {polylog}\left (2, -{\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {2 b \,d^{2} e^{2} x}{f^{2}}+\frac {d^{2} a \,x^{3}}{3}-\frac {d^{2} b \,x^{3}}{3}+\frac {a \,c^{3}}{3 d}+\frac {b \,c^{3}}{3 d}-\frac {4 b d c e x}{f}-\frac {2 b c d e \ln \left ({\mathrm e}^{f x +e}-1\right )}{f^{2}}+\frac {4 b c d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {2 b d c \ln \left (1-{\mathrm e}^{f x +e}\right ) x}{f}+\frac {2 b d c \ln \left (1-{\mathrm e}^{f x +e}\right ) e}{f^{2}}+\frac {2 b d c \ln \left ({\mathrm e}^{f x +e}+1\right ) x}{f}+\frac {4 b \,d^{2} e^{3}}{3 f^{3}}-\frac {2 b \,d^{2} \operatorname {polylog}\left (3, {\mathrm e}^{f x +e}\right )}{f^{3}}-\frac {2 b \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {b \,c^{2} \ln \left ({\mathrm e}^{f x +e}-1\right )}{f}-\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {b \,c^{2} \ln \left ({\mathrm e}^{f x +e}+1\right )}{f}+d a c \,x^{2}-d b c \,x^{2}+a \,c^{2} x +b \,c^{2} x\)	\(465\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (94) = 188\).

Time = 0.10 (sec) , antiderivative size = 303, normalized size of antiderivative = 3.00 \[ \int (c+d x)^2 (a+b \coth (e+f x)) \, dx=\frac {{\left (a - b\right )} d^{2} f^{3} x^{3} + 3 \, {\left (a - b\right )} c d f^{3} x^{2} + 3 \, {\left (a - b\right )} c^{2} f^{3} x - 6 \, b d^{2} {\rm polylog}\left (3, \cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) - 6 \, b d^{2} {\rm polylog}\left (3, -\cosh \left (f x + e\right ) - \sinh \left (f x + e\right )\right ) + 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) + 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (-\cosh \left (f x + e\right ) - \sinh \left (f x + e\right )\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + 1\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - 1\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\cosh \left (f x + e\right ) - \sinh \left (f x + e\right ) + 1\right )}{3 \, f^{3}} \] Input:

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

Output:

1/3*((a - b)*d^2*f^3*x^3 + 3*(a - b)*c*d*f^3*x^2 + 3*(a - b)*c^2*f^3*x - 6 
*b*d^2*polylog(3, cosh(f*x + e) + sinh(f*x + e)) - 6*b*d^2*polylog(3, -cos 
h(f*x + e) - sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(cosh(f*x + e) 
+ sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(-cosh(f*x + e) - sinh(f*x 
 + e)) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x + b*c^2*f^2)*log(cosh(f*x + e) + 
 sinh(f*x + e) + 1) + 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(cosh(f*x 
 + e) + sinh(f*x + e) - 1) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 
+ 2*b*c*d*e*f)*log(-cosh(f*x + e) - sinh(f*x + e) + 1))/f^3
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (94) = 188\).

Time = 0.27 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.38 \[ \int (c+d x)^2 (a+b \coth (e+f x)) \, dx=\frac {1}{3} \, a d^{2} x^{3} + \frac {1}{3} \, b d^{2} x^{3} + a c d x^{2} + b c d x^{2} + a c^{2} x + \frac {b c^{2} \log \left (\sinh \left (f x + e\right )\right )}{f} + \frac {2 \, {\left (f x \log \left (e^{\left (f x + e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (f x + e\right )}\right )\right )} b c d}{f^{2}} + \frac {2 \, {\left (f x \log \left (-e^{\left (f x + e\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (f x + e\right )}\right )\right )} b c d}{f^{2}} + \frac {{\left (f^{2} x^{2} \log \left (e^{\left (f x + e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (f x + e\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (f x + e\right )})\right )} b d^{2}}{f^{3}} + \frac {{\left (f^{2} x^{2} \log \left (-e^{\left (f x + e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (e^{\left (f x + e\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (f x + e\right )})\right )} b d^{2}}{f^{3}} - \frac {2 \, {\left (b d^{2} f^{3} x^{3} + 3 \, b c d f^{3} x^{2}\right )}}{3 \, f^{3}} \] Input:

integrate((d*x+c)^2*(a+b*coth(f*x+e)),x, algorithm="maxima")
 

Output:

1/3*a*d^2*x^3 + 1/3*b*d^2*x^3 + a*c*d*x^2 + b*c*d*x^2 + a*c^2*x + b*c^2*lo 
g(sinh(f*x + e))/f + 2*(f*x*log(e^(f*x + e) + 1) + dilog(-e^(f*x + e)))*b* 
c*d/f^2 + 2*(f*x*log(-e^(f*x + e) + 1) + dilog(e^(f*x + e)))*b*c*d/f^2 + ( 
f^2*x^2*log(e^(f*x + e) + 1) + 2*f*x*dilog(-e^(f*x + e)) - 2*polylog(3, -e 
^(f*x + e)))*b*d^2/f^3 + (f^2*x^2*log(-e^(f*x + e) + 1) + 2*f*x*dilog(e^(f 
*x + e)) - 2*polylog(3, e^(f*x + e)))*b*d^2/f^3 - 2/3*(b*d^2*f^3*x^3 + 3*b 
*c*d*f^3*x^2)/f^3
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 (a+b \coth (e+f x)) \, dx=\int \left (a+b\,\mathrm {coth}\left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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