3.1.31 \(\int (c+d x)^3 \text {sech}^2(a+b x) \, dx\) [31]

3.1.31.1 Optimal result
3.1.31.2 Mathematica [A] (verified)
3.1.31.3 Rubi [C] (verified)
3.1.31.4 Maple [B] (verified)
3.1.31.5 Fricas [C] (verification not implemented)
3.1.31.6 Sympy [F]
3.1.31.7 Maxima [B] (verification not implemented)
3.1.31.8 Giac [F]
3.1.31.9 Mupad [F(-1)]

3.1.31.1 Optimal result

Integrand size = 16, antiderivative size = 103 \[ \int (c+d x)^3 \text {sech}^2(a+b x) \, dx=\frac {(c+d x)^3}{b}-\frac {3 d (c+d x)^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}+\frac {3 d^3 \operatorname {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}+\frac {(c+d x)^3 \tanh (a+b x)}{b} \]

output
(d*x+c)^3/b-3*d*(d*x+c)^2*ln(1+exp(2*b*x+2*a))/b^2-3*d^2*(d*x+c)*polylog(2 
,-exp(2*b*x+2*a))/b^3+3/2*d^3*polylog(3,-exp(2*b*x+2*a))/b^4+(d*x+c)^3*tan 
h(b*x+a)/b
 
3.1.31.2 Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.41 \[ \int (c+d x)^3 \text {sech}^2(a+b x) \, dx=\frac {-\frac {d e^{2 a} \left (\frac {4 e^{-2 a} (c+d x)^3}{d}+\frac {6 \left (1+e^{-2 a}\right ) (c+d x)^2 \log \left (1+e^{-2 (a+b x)}\right )}{b}-\frac {3 d \left (1+e^{-2 a}\right ) \left (2 b (c+d x) \operatorname {PolyLog}\left (2,-e^{-2 (a+b x)}\right )+d \operatorname {PolyLog}\left (3,-e^{-2 (a+b x)}\right )\right )}{b^3}\right )}{1+e^{2 a}}+2 (c+d x)^3 \text {sech}(a) \text {sech}(a+b x) \sinh (b x)}{2 b} \]

input
Integrate[(c + d*x)^3*Sech[a + b*x]^2,x]
 
output
(-((d*E^(2*a)*((4*(c + d*x)^3)/(d*E^(2*a)) + (6*(1 + E^(-2*a))*(c + d*x)^2 
*Log[1 + E^(-2*(a + b*x))])/b - (3*d*(1 + E^(-2*a))*(2*b*(c + d*x)*PolyLog 
[2, -E^(-2*(a + b*x))] + d*PolyLog[3, -E^(-2*(a + b*x))]))/b^3))/(1 + E^(2 
*a))) + 2*(c + d*x)^3*Sech[a]*Sech[a + b*x]*Sinh[b*x])/(2*b)
 
3.1.31.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.23, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 4672, 26, 3042, 26, 4201, 2620, 3011, 2720, 7143}

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.

\(\displaystyle \int (c+d x)^3 \text {sech}^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^3 \csc \left (i a+i b x+\frac {\pi }{2}\right )^2dx\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {(c+d x)^3 \tanh (a+b x)}{b}-\frac {3 i d \int -i (c+d x)^2 \tanh (a+b x)dx}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {(c+d x)^3 \tanh (a+b x)}{b}-\frac {3 d \int (c+d x)^2 \tanh (a+b x)dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c+d x)^3 \tanh (a+b x)}{b}-\frac {3 d \int -i (c+d x)^2 \tan (i a+i b x)dx}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {3 i d \int (c+d x)^2 \tan (i a+i b x)dx}{b}\)

\(\Big \downarrow \) 4201

\(\displaystyle \frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {3 i d \left (2 i \int \frac {e^{2 (a+b x)} (c+d x)^2}{1+e^{2 (a+b x)}}dx-\frac {i (c+d x)^3}{3 d}\right )}{b}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {3 i d \left (2 i \left (\frac {(c+d x)^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {d \int (c+d x) \log \left (1+e^{2 (a+b x)}\right )dx}{b}\right )-\frac {i (c+d x)^3}{3 d}\right )}{b}\)

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input
Int[(c + d*x)^3*Sech[a + b*x]^2,x]
 
output
((3*I)*d*(((-1/3*I)*(c + d*x)^3)/d + (2*I)*(((c + d*x)^2*Log[1 + E^(2*(a + 
 b*x))])/(2*b) - (d*(-1/2*((c + d*x)*PolyLog[2, -E^(2*(a + b*x))])/b + (d* 
PolyLog[3, -E^(2*(a + b*x))])/(4*b^2)))/b)))/b + ((c + d*x)^3*Tanh[a + b*x 
])/b
 

3.1.31.3.1 Defintions of rubi rules used

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

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 2720
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4201
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 4672
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 7143
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 
3.1.31.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(101)=202\).

Time = 0.31 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.89

method result size
risch \(-\frac {2 \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {12 d^{2} c a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {12 d^{2} c a x}{b^{2}}+\frac {6 d^{2} c \,x^{2}}{b}+\frac {6 d^{2} c \,a^{2}}{b^{3}}-\frac {6 d^{2} c \ln \left (1+{\mathrm e}^{2 b x +2 a}\right ) x}{b^{2}}-\frac {3 d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}+\frac {2 d^{3} x^{3}}{b}-\frac {4 d^{3} a^{3}}{b^{4}}-\frac {3 d^{3} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right ) x^{2}}{b^{2}}+\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{4}}-\frac {6 d^{3} a^{2} x}{b^{3}}-\frac {3 d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right ) x}{b^{3}}+\frac {6 d^{3} a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 d \,c^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}}\) \(298\)

input
int((d*x+c)^3*sech(b*x+a)^2,x,method=_RETURNVERBOSE)
 
output
-2*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/b/(1+exp(2*b*x+2*a))-12/b^3*d^2*c*a 
*ln(exp(b*x+a))+12/b^2*d^2*c*a*x+6/b*d^2*c*x^2+6/b^3*d^2*c*a^2-6/b^2*d^2*c 
*ln(1+exp(2*b*x+2*a))*x-3/b^3*d^2*c*polylog(2,-exp(2*b*x+2*a))+2/b*d^3*x^3 
-4/b^4*d^3*a^3-3/b^2*d^3*ln(1+exp(2*b*x+2*a))*x^2+3/2*d^3*polylog(3,-exp(2 
*b*x+2*a))/b^4-6/b^3*d^3*a^2*x-3/b^3*d^3*polylog(2,-exp(2*b*x+2*a))*x+6/b^ 
4*d^3*a^2*ln(exp(b*x+a))-3/b^2*d*c^2*ln(1+exp(2*b*x+2*a))+6/b^2*d*c^2*ln(e 
xp(b*x+a))
 
3.1.31.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 1332, normalized size of antiderivative = 12.93 \[ \int (c+d x)^3 \text {sech}^2(a+b x) \, dx=\text {Too large to display} \]

input
integrate((d*x+c)^3*sech(b*x+a)^2,x, algorithm="fricas")
 
output
-(2*b^3*c^3 - 6*a*b^2*c^2*d + 6*a^2*b*c*d^2 - 2*a^3*d^3 - 2*(b^3*d^3*x^3 + 
 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3 
)*cosh(b*x + a)^2 - 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a 
*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*cosh(b*x + a)*sinh(b*x + a) - 2*(b^3 
*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 
 + a^3*d^3)*sinh(b*x + a)^2 + 6*(b*d^3*x + b*c*d^2 + (b*d^3*x + b*c*d^2)*c 
osh(b*x + a)^2 + 2*(b*d^3*x + b*c*d^2)*cosh(b*x + a)*sinh(b*x + a) + (b*d^ 
3*x + b*c*d^2)*sinh(b*x + a)^2)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + 
 6*(b*d^3*x + b*c*d^2 + (b*d^3*x + b*c*d^2)*cosh(b*x + a)^2 + 2*(b*d^3*x + 
 b*c*d^2)*cosh(b*x + a)*sinh(b*x + a) + (b*d^3*x + b*c*d^2)*sinh(b*x + a)^ 
2)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 3*(b^2*c^2*d - 2*a*b*c*d^2 
+ a^2*d^3 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cosh(b*x + a)^2 + 2*(b^2*c 
^2*d - 2*a*b*c*d^2 + a^2*d^3)*cosh(b*x + a)*sinh(b*x + a) + (b^2*c^2*d - 2 
*a*b*c*d^2 + a^2*d^3)*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) + 
 I) + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3 + (b^2*c^2*d - 2*a*b*c*d^2 + a^ 
2*d^3)*cosh(b*x + a)^2 + 2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cosh(b*x + 
a)*sinh(b*x + a) + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*sinh(b*x + a)^2)*lo 
g(cosh(b*x + a) + sinh(b*x + a) - I) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2* 
a*b*c*d^2 - a^2*d^3 + (b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3 
)*cosh(b*x + a)^2 + 2*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*...
 
3.1.31.6 Sympy [F]

\[ \int (c+d x)^3 \text {sech}^2(a+b x) \, dx=\int \left (c + d x\right )^{3} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

input
integrate((d*x+c)**3*sech(b*x+a)**2,x)
 
output
Integral((c + d*x)**3*sech(a + b*x)**2, x)
 
3.1.31.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (100) = 200\).

Time = 0.30 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.31 \[ \int (c+d x)^3 \text {sech}^2(a+b x) \, dx=3 \, c^{2} d {\left (\frac {2 \, x e^{\left (2 \, b x + 2 \, a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {\log \left ({\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, a\right )}\right )}{b^{2}}\right )} - \frac {3 \, {\left (2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )\right )} c d^{2}}{b^{3}} + \frac {2 \, c^{3}}{b {\left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}} - \frac {2 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2}\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})\right )} d^{3}}{2 \, b^{4}} + \frac {2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2}\right )}}{b^{4}} \]

input
integrate((d*x+c)^3*sech(b*x+a)^2,x, algorithm="maxima")
 
output
3*c^2*d*(2*x*e^(2*b*x + 2*a)/(b*e^(2*b*x + 2*a) + b) - log((e^(2*b*x + 2*a 
) + 1)*e^(-2*a))/b^2) - 3*(2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b* 
x + 2*a)))*c*d^2/b^3 + 2*c^3/(b*(e^(-2*b*x - 2*a) + 1)) - 2*(d^3*x^3 + 3*c 
*d^2*x^2)/(b*e^(2*b*x + 2*a) + b) - 3/2*(2*b^2*x^2*log(e^(2*b*x + 2*a) + 1 
) + 2*b*x*dilog(-e^(2*b*x + 2*a)) - polylog(3, -e^(2*b*x + 2*a)))*d^3/b^4 
+ 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2)/b^4
 
3.1.31.8 Giac [F]

\[ \int (c+d x)^3 \text {sech}^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \operatorname {sech}\left (b x + a\right )^{2} \,d x } \]

input
integrate((d*x+c)^3*sech(b*x+a)^2,x, algorithm="giac")
 
output
integrate((d*x + c)^3*sech(b*x + a)^2, x)
 
3.1.31.9 Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \text {sech}^2(a+b x) \, dx=\int \frac {{\left (c+d\,x\right )}^3}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]

input
int((c + d*x)^3/cosh(a + b*x)^2,x)
 
output
int((c + d*x)^3/cosh(a + b*x)^2, x)