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

Optimal. Leaf size=102 \[ -\frac {i d \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {i d \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac {(c+d x) \tanh (a+b x) \text {sech}(a+b x)}{2 b} \]

[Out]

(d*x+c)*arctan(exp(b*x+a))/b-1/2*I*d*polylog(2,-I*exp(b*x+a))/b^2+1/2*I*d*polylog(2,I*exp(b*x+a))/b^2+1/2*d*se
ch(b*x+a)/b^2+1/2*(d*x+c)*sech(b*x+a)*tanh(b*x+a)/b

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Rubi [A]  time = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4185, 4180, 2279, 2391} \[ -\frac {i d \text {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac {i d \text {PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac {(c+d x) \tanh (a+b x) \text {sech}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sech[a + b*x]^3,x]

[Out]

((c + d*x)*ArcTan[E^(a + b*x)])/b - ((I/2)*d*PolyLog[2, (-I)*E^(a + b*x)])/b^2 + ((I/2)*d*PolyLog[2, I*E^(a +
b*x)])/b^2 + (d*Sech[a + b*x])/(2*b^2) + ((c + d*x)*Sech[a + b*x]*Tanh[a + b*x])/(2*b)

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rubi steps

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

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Mathematica [A]  time = 2.96, size = 178, normalized size = 1.75 \[ \frac {b c \tan ^{-1}(\sinh (a+b x))+b c \tanh (a+b x) \text {sech}(a+b x)+\frac {1}{2} d \left (-2 i \left (\text {Li}_2\left (-i e^{a+b x}\right )-\text {Li}_2\left (i e^{a+b x}\right )\right )-\left ((-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )\right )+(\pi -2 i a) \log \left (\cot \left (\frac {1}{4} (2 i a+2 i b x+\pi )\right )\right )\right )+b d x \text {sech}(a) \sinh (b x) \text {sech}^2(a+b x)+d (b x \tanh (a)+1) \text {sech}(a+b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*ArcTan[Sinh[a + b*x]] + (d*(-(((-2*I)*a + Pi - (2*I)*b*x)*(Log[1 - I*E^(a + b*x)] - Log[1 + I*E^(a + b*x)
])) + ((-2*I)*a + Pi)*Log[Cot[((2*I)*a + Pi + (2*I)*b*x)/4]] - (2*I)*(PolyLog[2, (-I)*E^(a + b*x)] - PolyLog[2
, I*E^(a + b*x)])))/2 + b*d*x*Sech[a]*Sech[a + b*x]^2*Sinh[b*x] + d*Sech[a + b*x]*(1 + b*x*Tanh[a]) + b*c*Sech
[a + b*x]*Tanh[a + b*x])/(2*b^2)

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fricas [B]  time = 0.45, size = 1243, normalized size = 12.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b*d*x + b*c + d)*cosh(b*x + a)^3 + 6*(b*d*x + b*c + d)*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(b*d*x + b*c
+ d)*sinh(b*x + a)^3 - 2*(b*d*x + b*c - d)*cosh(b*x + a) + (I*d*cosh(b*x + a)^4 + 4*I*d*cosh(b*x + a)*sinh(b*x
+ a)^3 + I*d*sinh(b*x + a)^4 + 2*I*d*cosh(b*x + a)^2 + (6*I*d*cosh(b*x + a)^2 + 2*I*d)*sinh(b*x + a)^2 + (4*I
*d*cosh(b*x + a)^3 + 4*I*d*cosh(b*x + a))*sinh(b*x + a) + I*d)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + (-I*
d*cosh(b*x + a)^4 - 4*I*d*cosh(b*x + a)*sinh(b*x + a)^3 - I*d*sinh(b*x + a)^4 - 2*I*d*cosh(b*x + a)^2 + (-6*I*
d*cosh(b*x + a)^2 - 2*I*d)*sinh(b*x + a)^2 + (-4*I*d*cosh(b*x + a)^3 - 4*I*d*cosh(b*x + a))*sinh(b*x + a) - I*
d)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + ((I*b*c - I*a*d)*cosh(b*x + a)^4 + (4*I*b*c - 4*I*a*d)*cosh(b*x
+ a)*sinh(b*x + a)^3 + (I*b*c - I*a*d)*sinh(b*x + a)^4 + (2*I*b*c - 2*I*a*d)*cosh(b*x + a)^2 + ((6*I*b*c - 6*
I*a*d)*cosh(b*x + a)^2 + 2*I*b*c - 2*I*a*d)*sinh(b*x + a)^2 + I*b*c - I*a*d + ((4*I*b*c - 4*I*a*d)*cosh(b*x +
a)^3 + (4*I*b*c - 4*I*a*d)*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + I) + ((-I*b*c + I
*a*d)*cosh(b*x + a)^4 + (-4*I*b*c + 4*I*a*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (-I*b*c + I*a*d)*sinh(b*x + a)^4
+ (-2*I*b*c + 2*I*a*d)*cosh(b*x + a)^2 + ((-6*I*b*c + 6*I*a*d)*cosh(b*x + a)^2 - 2*I*b*c + 2*I*a*d)*sinh(b*x +
a)^2 - I*b*c + I*a*d + ((-4*I*b*c + 4*I*a*d)*cosh(b*x + a)^3 + (-4*I*b*c + 4*I*a*d)*cosh(b*x + a))*sinh(b*x +
a))*log(cosh(b*x + a) + sinh(b*x + a) - I) + ((-I*b*d*x - I*a*d)*cosh(b*x + a)^4 + (-4*I*b*d*x - 4*I*a*d)*cos
h(b*x + a)*sinh(b*x + a)^3 + (-I*b*d*x - I*a*d)*sinh(b*x + a)^4 - I*b*d*x + (-2*I*b*d*x - 2*I*a*d)*cosh(b*x +
a)^2 + (-2*I*b*d*x + (-6*I*b*d*x - 6*I*a*d)*cosh(b*x + a)^2 - 2*I*a*d)*sinh(b*x + a)^2 - I*a*d + ((-4*I*b*d*x
- 4*I*a*d)*cosh(b*x + a)^3 + (-4*I*b*d*x - 4*I*a*d)*cosh(b*x + a))*sinh(b*x + a))*log(I*cosh(b*x + a) + I*sinh
(b*x + a) + 1) + ((I*b*d*x + I*a*d)*cosh(b*x + a)^4 + (4*I*b*d*x + 4*I*a*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (I
*b*d*x + I*a*d)*sinh(b*x + a)^4 + I*b*d*x + (2*I*b*d*x + 2*I*a*d)*cosh(b*x + a)^2 + (2*I*b*d*x + (6*I*b*d*x +
6*I*a*d)*cosh(b*x + a)^2 + 2*I*a*d)*sinh(b*x + a)^2 + I*a*d + ((4*I*b*d*x + 4*I*a*d)*cosh(b*x + a)^3 + (4*I*b*
d*x + 4*I*a*d)*cosh(b*x + a))*sinh(b*x + a))*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - 2*(b*d*x - 3*(b*d*x
+ b*c + d)*cosh(b*x + a)^2 + b*c - d)*sinh(b*x + a))/(b^2*cosh(b*x + a)^4 + 4*b^2*cosh(b*x + a)*sinh(b*x + a)
^3 + b^2*sinh(b*x + a)^4 + 2*b^2*cosh(b*x + a)^2 + 2*(3*b^2*cosh(b*x + a)^2 + b^2)*sinh(b*x + a)^2 + b^2 + 4*(
b^2*cosh(b*x + a)^3 + b^2*cosh(b*x + a))*sinh(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.33, size = 216, normalized size = 2.12 \[ \frac {{\mathrm e}^{b x +a} \left (b d x \,{\mathrm e}^{2 b x +2 a}+b c \,{\mathrm e}^{2 b x +2 a}-b d x +d \,{\mathrm e}^{2 b x +2 a}-c b +d \right )}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}+\frac {c \arctan \left ({\mathrm e}^{b x +a}\right )}{b}-\frac {i d \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {i d \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {i d \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{2 b}+\frac {i d \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i d \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i d \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {d a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(b*x+a)*(b*d*x*exp(2*b*x+2*a)+b*c*exp(2*b*x+2*a)-b*d*x+d*exp(2*b*x+2*a)-c*b+d)/b^2/(1+exp(2*b*x+2*a))^2+1/b
*c*arctan(exp(b*x+a))-1/2*I/b*d*ln(1+I*exp(b*x+a))*x-1/2*I/b^2*d*ln(1+I*exp(b*x+a))*a+1/2*I/b*d*ln(1-I*exp(b*x
+a))*x+1/2*I/b^2*d*ln(1-I*exp(b*x+a))*a-1/2*I/b^2*d*dilog(1+I*exp(b*x+a))+1/2*I/b^2*d*dilog(1-I*exp(b*x+a))-1/
b^2*d*a*arctan(exp(b*x+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d {\left (\frac {{\left (b x e^{\left (3 \, a\right )} + e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} - {\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + 8 \, \int \frac {x e^{\left (b x + a\right )}}{8 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x}\right )} - c {\left (\frac {\arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac {e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(((b*x*e^(3*a) + e^(3*a))*e^(3*b*x) - (b*x*e^a - e^a)*e^(b*x))/(b^2*e^(4*b*x + 4*a) + 2*b^2*e^(2*b*x + 2*a)
+ b^2) + 8*integrate(1/8*x*e^(b*x + a)/(e^(2*b*x + 2*a) + 1), x)) - c*(arctan(e^(-b*x - a))/b - (e^(-b*x - a)
- e^(-3*b*x - 3*a))/(b*(2*e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a) + 1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/cosh(a + b*x)^3,x)

[Out]

int((c + d*x)/cosh(a + b*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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