3.236 \(\int \frac {x \sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx\)

Optimal. Leaf size=288 \[ \frac {\left (a^2-b^2\right ) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {\left (a^2-b^2\right ) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {x \left (a^2-b^2\right ) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b^3 d}+\frac {x \left (a^2-b^2\right ) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b^3 d}-\frac {x^2 \left (a^2-b^2\right )}{2 b^3}+\frac {a \sinh (c+d x)}{b^2 d^2}-\frac {a x \cosh (c+d x)}{b^2 d}-\frac {\sinh (c+d x) \cosh (c+d x)}{4 b d^2}+\frac {x \sinh ^2(c+d x)}{2 b d}+\frac {x}{4 b d} \]

[Out]

1/4*x/b/d-1/2*(a^2-b^2)*x^2/b^3-a*x*cosh(d*x+c)/b^2/d+(a^2-b^2)*x*ln(1+b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)))/b^3/d
+(a^2-b^2)*x*ln(1+b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b^3/d+(a^2-b^2)*polylog(2,-b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)
))/b^3/d^2+(a^2-b^2)*polylog(2,-b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b^3/d^2+a*sinh(d*x+c)/b^2/d^2-1/4*cosh(d*x+c
)*sinh(d*x+c)/b/d^2+1/2*x*sinh(d*x+c)^2/b/d

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Rubi [A]  time = 0.34, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {5566, 3296, 2637, 5372, 2635, 8, 5562, 2190, 2279, 2391} \[ \frac {\left (a^2-b^2\right ) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {\left (a^2-b^2\right ) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^2}+\frac {x \left (a^2-b^2\right ) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b^3 d}+\frac {x \left (a^2-b^2\right ) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b^3 d}-\frac {x^2 \left (a^2-b^2\right )}{2 b^3}+\frac {a \sinh (c+d x)}{b^2 d^2}-\frac {a x \cosh (c+d x)}{b^2 d}-\frac {\sinh (c+d x) \cosh (c+d x)}{4 b d^2}+\frac {x \sinh ^2(c+d x)}{2 b d}+\frac {x}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sinh[c + d*x]^3)/(a + b*Cosh[c + d*x]),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

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

Rule 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 5562

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)])/(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 - b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 - b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 - b^2, 0]

Rule 5566

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_))/(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symb
ol] :> -Dist[a/b^2, Int[(e + f*x)^m*Sinh[c + d*x]^(n - 2), x], x] + (Dist[1/b, Int[(e + f*x)^m*Sinh[c + d*x]^(
n - 2)*Cosh[c + d*x], x], x] + Dist[(a^2 - b^2)/b^2, Int[((e + f*x)^m*Sinh[c + d*x]^(n - 2))/(a + b*Cosh[c + d
*x]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 1] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 2.96, size = 414, normalized size = 1.44 \[ \frac {4 \left (a^2-b^2\right ) \left (2 \text {Li}_2\left (\frac {b (\cosh (c+d x)+\sinh (c+d x))}{\sqrt {a^2-b^2}-a}\right )+2 \text {Li}_2\left (-\frac {b (\cosh (c+d x)+\sinh (c+d x))}{a+\sqrt {a^2-b^2}}\right )+2 (c+d x) \log \left (\frac {b (\sinh (c+d x)+\cosh (c+d x))}{a-\sqrt {a^2-b^2}}+1\right )+2 (c+d x) \log \left (\frac {b (\sinh (c+d x)+\cosh (c+d x))}{\sqrt {a^2-b^2}+a}+1\right )+\frac {4 a c \sqrt {-\left (a^2-b^2\right )^2} \tan ^{-1}\left (\frac {a+b \sinh (c+d x)+b \cosh (c+d x)}{\sqrt {b^2-a^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {4 a c \sqrt {-\left (a^2-b^2\right )^2} \tanh ^{-1}\left (\frac {a+b \sinh (c+d x)+b \cosh (c+d x)}{\sqrt {a^2-b^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-2 c \log (2 (\sinh (c+d x)+\cosh (c+d x)) (a+b \cosh (c+d x)))-(c+d x)^2+2 c (c+d x)\right )+8 a b \sinh (c+d x)-8 a b d x \cosh (c+d x)-b^2 \sinh (2 (c+d x))+2 b^2 d x \cosh (2 (c+d x))}{8 b^3 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sinh[c + d*x]^3)/(a + b*Cosh[c + d*x]),x]

[Out]

(-8*a*b*d*x*Cosh[c + d*x] + 2*b^2*d*x*Cosh[2*(c + d*x)] + 4*(a^2 - b^2)*(2*c*(c + d*x) - (c + d*x)^2 + (4*a*Sq
rt[-(a^2 - b^2)^2]*c*ArcTan[(a + b*Cosh[c + d*x] + b*Sinh[c + d*x])/Sqrt[-a^2 + b^2]])/(a^2 - b^2)^(3/2) + (4*
a*Sqrt[-(a^2 - b^2)^2]*c*ArcTanh[(a + b*Cosh[c + d*x] + b*Sinh[c + d*x])/Sqrt[a^2 - b^2]])/(-a^2 + b^2)^(3/2)
- 2*c*Log[2*(a + b*Cosh[c + d*x])*(Cosh[c + d*x] + Sinh[c + d*x])] + 2*(c + d*x)*Log[1 + (b*(Cosh[c + d*x] + S
inh[c + d*x]))/(a - Sqrt[a^2 - b^2])] + 2*(c + d*x)*Log[1 + (b*(Cosh[c + d*x] + Sinh[c + d*x]))/(a + Sqrt[a^2
- b^2])] + 2*PolyLog[2, (b*(Cosh[c + d*x] + Sinh[c + d*x]))/(-a + Sqrt[a^2 - b^2])] + 2*PolyLog[2, -((b*(Cosh[
c + d*x] + Sinh[c + d*x]))/(a + Sqrt[a^2 - b^2]))]) + 8*a*b*Sinh[c + d*x] - b^2*Sinh[2*(c + d*x)])/(8*b^3*d^2)

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fricas [B]  time = 0.53, size = 1196, normalized size = 4.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((2*b^2*d*x - b^2)*cosh(d*x + c)^4 + (2*b^2*d*x - b^2)*sinh(d*x + c)^4 + 2*b^2*d*x - 8*(a*b*d*x - a*b)*co
sh(d*x + c)^3 - 4*(2*a*b*d*x - 2*a*b - (2*b^2*d*x - b^2)*cosh(d*x + c))*sinh(d*x + c)^3 - 8*((a^2 - b^2)*d^2*x
^2 - 2*(a^2 - b^2)*c^2)*cosh(d*x + c)^2 - 2*(4*(a^2 - b^2)*d^2*x^2 - 8*(a^2 - b^2)*c^2 - 3*(2*b^2*d*x - b^2)*c
osh(d*x + c)^2 + 12*(a*b*d*x - a*b)*cosh(d*x + c))*sinh(d*x + c)^2 + b^2 - 8*(a*b*d*x + a*b)*cosh(d*x + c) + 1
6*((a^2 - b^2)*cosh(d*x + c)^2 + 2*(a^2 - b^2)*cosh(d*x + c)*sinh(d*x + c) + (a^2 - b^2)*sinh(d*x + c)^2)*dilo
g(-(a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1)
+ 16*((a^2 - b^2)*cosh(d*x + c)^2 + 2*(a^2 - b^2)*cosh(d*x + c)*sinh(d*x + c) + (a^2 - b^2)*sinh(d*x + c)^2)*d
ilog(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b +
1) - 16*((a^2 - b^2)*c*cosh(d*x + c)^2 + 2*(a^2 - b^2)*c*cosh(d*x + c)*sinh(d*x + c) + (a^2 - b^2)*c*sinh(d*x
+ c)^2)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) - 16*((a^2 - b^2)*c*cosh(
d*x + c)^2 + 2*(a^2 - b^2)*c*cosh(d*x + c)*sinh(d*x + c) + (a^2 - b^2)*c*sinh(d*x + c)^2)*log(2*b*cosh(d*x + c
) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) + 16*(((a^2 - b^2)*d*x + (a^2 - b^2)*c)*cosh(d*x + c)
^2 + 2*((a^2 - b^2)*d*x + (a^2 - b^2)*c)*cosh(d*x + c)*sinh(d*x + c) + ((a^2 - b^2)*d*x + (a^2 - b^2)*c)*sinh(
d*x + c)^2)*log((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2)
+ b)/b) + 16*(((a^2 - b^2)*d*x + (a^2 - b^2)*c)*cosh(d*x + c)^2 + 2*((a^2 - b^2)*d*x + (a^2 - b^2)*c)*cosh(d*
x + c)*sinh(d*x + c) + ((a^2 - b^2)*d*x + (a^2 - b^2)*c)*sinh(d*x + c)^2)*log((a*cosh(d*x + c) + a*sinh(d*x +
c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) - 4*(2*a*b*d*x - (2*b^2*d*x - b^2)*cosh
(d*x + c)^3 + 6*(a*b*d*x - a*b)*cosh(d*x + c)^2 + 2*a*b + 4*((a^2 - b^2)*d^2*x^2 - 2*(a^2 - b^2)*c^2)*cosh(d*x
+ c))*sinh(d*x + c))/(b^3*d^2*cosh(d*x + c)^2 + 2*b^3*d^2*cosh(d*x + c)*sinh(d*x + c) + b^3*d^2*sinh(d*x + c)
^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sinh(d*x + c)^3/(b*cosh(d*x + c) + a), x)

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maple [B]  time = 0.30, size = 860, normalized size = 2.99 \[ -\frac {\ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d b}-\frac {\ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} b}+\frac {c \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}+b \right )}{d^{2} b}-\frac {2 c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} b}+\frac {2 c x}{d b}-\frac {\ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d b}-\frac {\ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} b}-\frac {a^{2} c^{2}}{d^{2} b^{3}}-\frac {a \left (d x +1\right ) {\mathrm e}^{-d x -c}}{2 b^{2} d^{2}}-\frac {a \left (d x -1\right ) {\mathrm e}^{d x +c}}{2 b^{2} d^{2}}-\frac {x^{2} a^{2}}{2 b^{3}}+\frac {\left (2 d x +1\right ) {\mathrm e}^{-2 d x -2 c}}{16 b \,d^{2}}-\frac {\dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} b}-\frac {\dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} b}+\frac {a^{2} \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} b^{3}}+\frac {a^{2} \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} b^{3}}-\frac {2 a^{2} c x}{d \,b^{3}}+\frac {\ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) a^{2} x}{d \,b^{3}}+\frac {\ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) a^{2} c}{d^{2} b^{3}}+\frac {\ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) a^{2} x}{d \,b^{3}}+\frac {\ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) a^{2} c}{d^{2} b^{3}}+\frac {2 c \,a^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} b^{3}}-\frac {c \,a^{2} \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}+b \right )}{d^{2} b^{3}}+\frac {c^{2}}{d^{2} b}+\frac {\left (2 d x -1\right ) {\mathrm e}^{2 d x +2 c}}{16 b \,d^{2}}+\frac {x^{2}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(d*x+c)^3/(a+b*cosh(d*x+c)),x)

[Out]

-1/d/b*ln((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*x-1/d^2/b*ln((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/
(a+(a^2-b^2)^(1/2)))*c+1/d^2/b*c*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)+b)-2/d^2/b*c*ln(exp(d*x+c))+2/d/b*c*x-1/d/
b*ln((-b*exp(d*x+c)+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*x-1/d^2/b*ln((-b*exp(d*x+c)+(a^2-b^2)^(1/2)-a)/(-
a+(a^2-b^2)^(1/2)))*c-1/d^2/b^3*a^2*c^2-1/2*a*(d*x+1)/b^2/d^2*exp(-d*x-c)-1/2*a*(d*x-1)/b^2/d^2*exp(d*x+c)-1/2
*x^2/b^3*a^2+1/16*(2*d*x+1)/b/d^2*exp(-2*d*x-2*c)+1/d^2/b^3*a^2*dilog((-b*exp(d*x+c)+(a^2-b^2)^(1/2)-a)/(-a+(a
^2-b^2)^(1/2)))+1/d^2/b^3*a^2*dilog((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))-2/d/b^3*a^2*c*x+1/d/
b^3*ln((-b*exp(d*x+c)+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*a^2*x+1/d^2/b^3*ln((-b*exp(d*x+c)+(a^2-b^2)^(1/
2)-a)/(-a+(a^2-b^2)^(1/2)))*a^2*c+1/d/b^3*ln((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*a^2*x+1/d^2
/b^3*ln((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*a^2*c+2/d^2/b^3*c*a^2*ln(exp(d*x+c))-1/d^2/b^3*c
*a^2*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)+b)+1/d^2/b*c^2-1/d^2/b*dilog((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(a^2-
b^2)^(1/2)))-1/d^2/b*dilog((-b*exp(d*x+c)+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))+1/16*(2*d*x-1)/b/d^2*exp(2*
d*x+2*c)+1/2*x^2/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(8*(a^2*d^2*e^(2*c) - b^2*d^2*e^(2*c))*x^2 + (2*b^2*d*x*e^(4*c) - b^2*e^(4*c))*e^(2*d*x) - 8*(a*b*d*x*e^(
3*c) - a*b*e^(3*c))*e^(d*x) - 8*(a*b*d*x*e^c + a*b*e^c)*e^(-d*x) + (2*b^2*d*x + b^2)*e^(-2*d*x))*e^(-2*c)/(b^3
*d^2) - 1/8*integrate(16*((a^3*e^c - a*b^2*e^c)*x*e^(d*x) + (a^2*b - b^3)*x)/(b^4*e^(2*d*x + 2*c) + 2*a*b^3*e^
(d*x + c) + b^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sinh(c + d*x)^3)/(a + b*cosh(c + d*x)),x)

[Out]

int((x*sinh(c + d*x)^3)/(a + b*cosh(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(d*x+c)**3/(a+b*cosh(d*x+c)),x)

[Out]

Integral(x*sinh(c + d*x)**3/(a + b*cosh(c + d*x)), x)

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