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

Optimal. Leaf size=103 \[ \frac {3 d^3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}-\frac {3 d^2 (c+d x) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {3 d (c+d x)^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {(c+d x)^3}{b} \]

[Out]

(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*polylo
g(3,-exp(2*b*x+2*a))/b^4+(d*x+c)^3*tanh(b*x+a)/b

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Rubi [A]  time = 0.22, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4184, 3718, 2190, 2531, 2282, 6589} \[ -\frac {3 d^2 (c+d x) \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}+\frac {3 d^3 \text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}-\frac {3 d (c+d x)^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {(c+d x)^3}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Sech[a + b*x]^2,x]

[Out]

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2531

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] + Dist[(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 3718

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] + Dist[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 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

Rubi steps

\begin {align*} \int (c+d x)^3 \text {sech}^2(a+b x) \, dx &=\frac {(c+d x)^3 \tanh (a+b x)}{b}-\frac {(3 d) \int (c+d x)^2 \tanh (a+b x) \, dx}{b}\\ &=\frac {(c+d x)^3}{b}+\frac {(c+d x)^3 \tanh (a+b x)}{b}-\frac {(6 d) \int \frac {e^{2 (a+b x)} (c+d x)^2}{1+e^{2 (a+b x)}} \, dx}{b}\\ &=\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 {(c+d x)^3 \tanh (a+b x)}{b}+\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=\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) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {\left (3 d^3\right ) \int \text {Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=\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) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {(c+d x)^3 \tanh (a+b x)}{b}+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=\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) \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {3 d^3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}+\frac {(c+d x)^3 \tanh (a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 145, normalized size = 1.41 \[ \frac {2 \text {sech}(a) \sinh (b x) (c+d x)^3 \text {sech}(a+b x)-\frac {e^{2 a} d \left (-\frac {3 \left (e^{-2 a}+1\right ) d \left (2 b (c+d x) \text {Li}_2\left (-e^{-2 (a+b x)}\right )+d \text {Li}_3\left (-e^{-2 (a+b x)}\right )\right )}{b^3}+\frac {6 \left (e^{-2 a}+1\right ) (c+d x)^2 \log \left (e^{-2 (a+b x)}+1\right )}{b}+\frac {4 e^{-2 a} (c+d x)^3}{d}\right )}{e^{2 a}+1}}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Sech[a + b*x]^2,x]

[Out]

(-((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)

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fricas [C]  time = 0.50, size = 1332, normalized size = 12.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sech(b*x+a)^2,x, algorithm="fricas")

[Out]

-(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)*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)) + 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)*log(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*d^3)*cosh(b*x + a)*sinh(b*x + a) + (b^2*d^3*x^2 + 2*b^2*c*d^2*x
+ 2*a*b*c*d^2 - a^2*d^3)*sinh(b*x + a)^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + 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*d^3)*cosh(b*x + a)*sinh(b*x + a) + (b^2*d^3*x^2 + 2*b^2*c*d^2*
x + 2*a*b*c*d^2 - a^2*d^3)*sinh(b*x + a)^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - 6*(d^3*cosh(b*x + a)
^2 + 2*d^3*cosh(b*x + a)*sinh(b*x + a) + d^3*sinh(b*x + a)^2 + d^3)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x +
a)) - 6*(d^3*cosh(b*x + a)^2 + 2*d^3*cosh(b*x + a)*sinh(b*x + a) + d^3*sinh(b*x + a)^2 + d^3)*polylog(3, -I*co
sh(b*x + a) - I*sinh(b*x + a)))/(b^4*cosh(b*x + a)^2 + 2*b^4*cosh(b*x + a)*sinh(b*x + a) + b^4*sinh(b*x + a)^2
+ b^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{3} \operatorname {sech}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sech(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^3*sech(b*x + a)^2, x)

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maple [B]  time = 0.32, size = 298, normalized size = 2.89 \[ -\frac {2 \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {3 d \,c^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {6 d^{3} a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {2 d^{3} x^{3}}{b}-\frac {6 d^{3} a^{2} x}{b^{3}}-\frac {4 d^{3} a^{3}}{b^{4}}-\frac {3 d^{3} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right ) x^{2}}{b^{2}}-\frac {3 d^{3} \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right ) x}{b^{3}}+\frac {3 d^{3} \polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{4}}-\frac {12 d^{2} c a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {6 d^{2} c \,x^{2}}{b}+\frac {12 d^{2} c a x}{b^{2}}+\frac {6 d^{2} c \,a^{2}}{b^{3}}-\frac {6 d^{2} c \ln \left (1+{\mathrm e}^{2 b x +2 a}\right ) x}{b^{2}}-\frac {3 d^{2} c \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*sech(b*x+a)^2,x)

[Out]

-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))-3*d/b^2*c^2*ln(1+exp(2*b*x+2*a))+6*d/b^2*c^2*ln(ex
p(b*x+a))+6*d^3/b^4*a^2*ln(exp(b*x+a))+2*d^3/b*x^3-6*d^3/b^3*a^2*x-4*d^3/b^4*a^3-3*d^3/b^2*ln(1+exp(2*b*x+2*a)
)*x^2-3*d^3/b^3*polylog(2,-exp(2*b*x+2*a))*x+3/2*d^3*polylog(3,-exp(2*b*x+2*a))/b^4-12*d^2/b^3*c*a*ln(exp(b*x+
a))+6*d^2/b*c*x^2+12*d^2/b^2*c*a*x+6*d^2/b^3*c*a^2-6*d^2/b^2*c*ln(1+exp(2*b*x+2*a))*x-3*d^2/b^3*c*polylog(2,-e
xp(2*b*x+2*a))

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maxima [B]  time = 0.50, size = 238, normalized size = 2.31 \[ 3 \, c^{2} d {\left (\frac {2 \, x e^{\left (2 \, b x + 2 \, a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {\log \left ({\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, a\right )}\right )}{b^{2}}\right )} - \frac {3 \, {\left (2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )\right )} c d^{2}}{b^{3}} + \frac {2 \, c^{3}}{b {\left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}} - \frac {2 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2}\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})\right )} d^{3}}{2 \, b^{4}} + \frac {2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2}\right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*sech(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^3}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/cosh(a + b*x)^2,x)

[Out]

int((c + d*x)^3/cosh(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{3} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*sech(b*x+a)**2,x)

[Out]

Integral((c + d*x)**3*sech(a + b*x)**2, x)

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