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

Optimal. Leaf size=92 \[ \frac {d \text {Li}_2\left (-e^{a+b x}\right )}{2 b^2}-\frac {d \text {Li}_2\left (e^{a+b x}\right )}{2 b^2}-\frac {d \text {csch}(a+b x)}{2 b^2}+\frac {(c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(c+d x) \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

[Out]

(d*x+c)*arctanh(exp(b*x+a))/b-1/2*d*csch(b*x+a)/b^2-1/2*(d*x+c)*coth(b*x+a)*csch(b*x+a)/b+1/2*d*polylog(2,-exp
(b*x+a))/b^2-1/2*d*polylog(2,exp(b*x+a))/b^2

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Rubi [A]  time = 0.08, antiderivative size = 92, 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, 4182, 2279, 2391} \[ \frac {d \text {PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}-\frac {d \text {PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}-\frac {d \text {csch}(a+b x)}{2 b^2}+\frac {(c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(c+d x) \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Csch[a + b*x]^3,x]

[Out]

((c + d*x)*ArcTanh[E^(a + b*x)])/b - (d*Csch[a + b*x])/(2*b^2) - ((c + d*x)*Coth[a + b*x]*Csch[a + b*x])/(2*b)
+ (d*PolyLog[2, -E^(a + b*x)])/(2*b^2) - (d*PolyLog[2, E^(a + b*x)])/(2*b^2)

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 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && 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 {csch}^3(a+b x) \, dx &=-\frac {d \text {csch}(a+b x)}{2 b^2}-\frac {(c+d x) \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {1}{2} \int (c+d x) \text {csch}(a+b x) \, dx\\ &=\frac {(c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {d \text {csch}(a+b x)}{2 b^2}-\frac {(c+d x) \coth (a+b x) \text {csch}(a+b x)}{2 b}+\frac {d \int \log \left (1-e^{a+b x}\right ) \, dx}{2 b}-\frac {d \int \log \left (1+e^{a+b x}\right ) \, dx}{2 b}\\ &=\frac {(c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {d \text {csch}(a+b x)}{2 b^2}-\frac {(c+d x) \coth (a+b x) \text {csch}(a+b x)}{2 b}+\frac {d \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}-\frac {d \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}\\ &=\frac {(c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {d \text {csch}(a+b x)}{2 b^2}-\frac {(c+d x) \coth (a+b x) \text {csch}(a+b x)}{2 b}+\frac {d \text {Li}_2\left (-e^{a+b x}\right )}{2 b^2}-\frac {d \text {Li}_2\left (e^{a+b x}\right )}{2 b^2}\\ \end {align*}

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Mathematica [C]  time = 2.31, size = 313, normalized size = 3.40 \[ -\frac {d \left (-a \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )-i \left (i \left (\text {Li}_2\left (-e^{i (i a+i b x)}\right )-\text {Li}_2\left (e^{i (i a+i b x)}\right )\right )+(i a+i b x) \left (\log \left (1-e^{i (i a+i b x)}\right )-\log \left (1+e^{i (i a+i b x)}\right )\right )\right )\right )}{2 b^2}+\frac {d \text {csch}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {csch}\left (\frac {a}{2}+\frac {b x}{2}\right )}{4 b^2}+\frac {d \text {sech}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {sech}\left (\frac {a}{2}+\frac {b x}{2}\right )}{4 b^2}-\frac {c \text {csch}^2\left (\frac {1}{2} (a+b x)\right )}{8 b}-\frac {c \text {sech}^2\left (\frac {1}{2} (a+b x)\right )}{8 b}-\frac {c \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {d x \text {csch}^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}-\frac {d x \text {sech}^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(d*x*Csch[a/2 + (b*x)/2]^2)/b - (c*Csch[(a + b*x)/2]^2)/(8*b) - (c*Log[Tanh[(a + b*x)/2]])/(2*b) - (d*(-(
a*Log[Tanh[(a + b*x)/2]]) - I*((I*a + I*b*x)*(Log[1 - E^(I*(I*a + I*b*x))] - Log[1 + E^(I*(I*a + I*b*x))]) + I
*(PolyLog[2, -E^(I*(I*a + I*b*x))] - PolyLog[2, E^(I*(I*a + I*b*x))]))))/(2*b^2) - (d*x*Sech[a/2 + (b*x)/2]^2)
/(8*b) - (c*Sech[(a + b*x)/2]^2)/(8*b) + (d*Csch[a/2]*Csch[a/2 + (b*x)/2]*Sinh[(b*x)/2])/(4*b^2) + (d*Sech[a/2
]*Sech[a/2 + (b*x)/2]*Sinh[(b*x)/2])/(4*b^2)

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fricas [B]  time = 0.44, size = 1026, normalized size = 11.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*csch(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) + (d*cosh(b*x + a)^4 + 4*d*cosh(b*x + a)*sinh(b*x +
a)^3 + d*sinh(b*x + a)^4 - 2*d*cosh(b*x + a)^2 + 2*(3*d*cosh(b*x + a)^2 - d)*sinh(b*x + a)^2 + 4*(d*cosh(b*x +
a)^3 - d*cosh(b*x + a))*sinh(b*x + a) + d)*dilog(cosh(b*x + a) + sinh(b*x + a)) - (d*cosh(b*x + a)^4 + 4*d*co
sh(b*x + a)*sinh(b*x + a)^3 + d*sinh(b*x + a)^4 - 2*d*cosh(b*x + a)^2 + 2*(3*d*cosh(b*x + a)^2 - d)*sinh(b*x +
a)^2 + 4*(d*cosh(b*x + a)^3 - d*cosh(b*x + a))*sinh(b*x + a) + d)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - ((b
*d*x + b*c)*cosh(b*x + a)^4 + 4*(b*d*x + b*c)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*d*x + b*c)*sinh(b*x + a)^4 +
b*d*x - 2*(b*d*x + b*c)*cosh(b*x + a)^2 - 2*(b*d*x - 3*(b*d*x + b*c)*cosh(b*x + a)^2 + b*c)*sinh(b*x + a)^2 +
b*c + 4*((b*d*x + b*c)*cosh(b*x + a)^3 - (b*d*x + b*c)*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh(
b*x + a) + 1) + ((b*c - a*d)*cosh(b*x + a)^4 + 4*(b*c - a*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*c - a*d)*sinh(
b*x + a)^4 - 2*(b*c - a*d)*cosh(b*x + a)^2 + 2*(3*(b*c - a*d)*cosh(b*x + a)^2 - b*c + a*d)*sinh(b*x + a)^2 + b
*c - a*d + 4*((b*c - a*d)*cosh(b*x + a)^3 - (b*c - a*d)*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh
(b*x + a) - 1) + ((b*d*x + a*d)*cosh(b*x + a)^4 + 4*(b*d*x + a*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*d*x + a*d
)*sinh(b*x + a)^4 + b*d*x - 2*(b*d*x + a*d)*cosh(b*x + a)^2 - 2*(b*d*x - 3*(b*d*x + a*d)*cosh(b*x + a)^2 + a*d
)*sinh(b*x + a)^2 + a*d + 4*((b*d*x + a*d)*cosh(b*x + a)^3 - (b*d*x + a*d)*cosh(b*x + a))*sinh(b*x + a))*log(-
cosh(b*x + a) - 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 \[ \int {\left (d x + c\right )} \operatorname {csch}\left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 197, normalized size = 2.14 \[ -\frac {{\mathrm e}^{b x +a} \left (b d x \,{\mathrm e}^{2 b x +2 a}+b c \,{\mathrm e}^{2 b x +2 a}+b d x +d \,{\mathrm e}^{2 b x +2 a}+c b -d \right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {c \arctanh \left ({\mathrm e}^{b x +a}\right )}{b}+\frac {d \ln \left (1+{\mathrm e}^{b x +a}\right ) x}{2 b}+\frac {d \ln \left (1+{\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {d \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {d \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {d \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {d \polylog \left (2, {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {d a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*csch(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/(exp(2*b*x+2*a)-1)^2+1/
b*c*arctanh(exp(b*x+a))+1/2/b*d*ln(1+exp(b*x+a))*x+1/2/b^2*d*ln(1+exp(b*x+a))*a+1/2*d*polylog(2,-exp(b*x+a))/b
^2-1/2/b*d*ln(1-exp(b*x+a))*x-1/2/b^2*d*ln(1-exp(b*x+a))*a-1/2*d*polylog(2,exp(b*x+a))/b^2-1/b^2*d*a*arctanh(e
xp(b*x+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -d {\left (\frac {{\left (b x e^{\left (3 \, a\right )} + e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} + {\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + 8 \, \int \frac {x}{16 \, {\left (e^{\left (b x + a\right )} + 1\right )}}\,{d x} + 8 \, \int \frac {x}{16 \, {\left (e^{\left (b x + a\right )} - 1\right )}}\,{d x}\right )} + \frac {1}{2} \, c {\left (\frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {2 \, {\left (e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*csch(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(1/16*x/(e^(b*x + a) + 1), x) + 8*integrate(1/16*x/(e^(b*x + a) - 1), x)) + 1/2*c*(log(e^
(-b*x - a) + 1)/b - log(e^(-b*x - a) - 1)/b + 2*(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 \[ \int \frac {c+d\,x}{{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/sinh(a + b*x)^3,x)

[Out]

int((c + d*x)/sinh(a + b*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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