3.364 \(\int \frac {(e+f x) \cosh (c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=278 \[ -\frac {a^2 (e+f x)^2}{2 b^3 f}+\frac {a^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^3 d^2}+\frac {a^2 f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^3 d^2}+\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^3 d}+\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^3 d}+\frac {a f \cosh (c+d x)}{b^2 d^2}-\frac {a (e+f x) \sinh (c+d x)}{b^2 d}-\frac {f \sinh (c+d x) \cosh (c+d x)}{4 b d^2}+\frac {(e+f x) \sinh ^2(c+d x)}{2 b d}+\frac {f x}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5579, 5446, 2635, 8, 3296, 2638, 5561, 2190, 2279, 2391} \[ \frac {a^2 f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^3 d^2}+\frac {a^2 f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b^3 d^2}+\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^3 d}+\frac {a^2 (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^3 d}-\frac {a^2 (e+f x)^2}{2 b^3 f}+\frac {a f \cosh (c+d x)}{b^2 d^2}-\frac {a (e+f x) \sinh (c+d x)}{b^2 d}-\frac {f \sinh (c+d x) \cosh (c+d x)}{4 b d^2}+\frac {(e+f x) \sinh ^2(c+d x)}{2 b d}+\frac {f x}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Cosh[c + d*x]*Sinh[c + d*x]^2)/(a + b*Sinh[c + d*x]),x]

[Out]

(f*x)/(4*b*d) - (a^2*(e + f*x)^2)/(2*b^3*f) + (a*f*Cosh[c + d*x])/(b^2*d^2) + (a^2*(e + f*x)*Log[1 + (b*E^(c +
d*x))/(a - Sqrt[a^2 + b^2])])/(b^3*d) + (a^2*(e + f*x)*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/(b^3*d
) + (a^2*f*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^3*d^2) + (a^2*f*PolyLog[2, -((b*E^(c + d*x
))/(a + Sqrt[a^2 + b^2]))])/(b^3*d^2) - (a*(e + f*x)*Sinh[c + d*x])/(b^2*d) - (f*Cosh[c + d*x]*Sinh[c + d*x])/
(4*b*d^2) + ((e + f*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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 5446

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

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), 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 5579

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

Rubi steps

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

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Mathematica [A]  time = 0.97, size = 423, normalized size = 1.52 \[ \frac {2 d e \left (\left (4 a^2+b^2\right ) \log (a+b \sinh (c+d x))-4 a b \sinh (c+d x)+2 b^2 \sinh ^2(c+d x)\right )+b^2 f \left (-2 \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+d x \left (-2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )-2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+d x\right )\right )+f \left (2 \left (4 a^2+b^2\right ) \left (\text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+\text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+(c+d x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+(c+d x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-c \log (a+b \sinh (c+d x))-\frac {1}{2} (c+d x)^2\right )-8 a b d x \sinh (c+d x)+8 a b \cosh (c+d x)-b^2 \sinh (2 (c+d x))+2 b^2 d x \cosh (2 (c+d x))\right )-2 b^2 d e \log (a+b \sinh (c+d x))}{8 b^3 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*Cosh[c + d*x]*Sinh[c + d*x]^2)/(a + b*Sinh[c + d*x]),x]

[Out]

(-2*b^2*d*e*Log[a + b*Sinh[c + d*x]] + b^2*f*(d*x*(d*x - 2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*
Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])]) - 2*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] - 2*Pol
yLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))]) + 2*d*e*((4*a^2 + b^2)*Log[a + b*Sinh[c + d*x]] - 4*a*b*Sin
h[c + d*x] + 2*b^2*Sinh[c + d*x]^2) + f*(8*a*b*Cosh[c + d*x] + 2*b^2*d*x*Cosh[2*(c + d*x)] + 2*(4*a^2 + b^2)*(
-1/2*(c + d*x)^2 + (c + d*x)*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] + (c + d*x)*Log[1 + (b*E^(c + d*x)
)/(a + Sqrt[a^2 + b^2])] - c*Log[a + b*Sinh[c + d*x]] + PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] + P
olyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))]) - 8*a*b*d*x*Sinh[c + d*x] - b^2*Sinh[2*(c + d*x)]))/(8*b^
3*d^2)

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fricas [B]  time = 0.50, size = 1248, normalized size = 4.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cosh(d*x+c)*sinh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(2*b^2*d*f*x + (2*b^2*d*f*x + 2*b^2*d*e - b^2*f)*cosh(d*x + c)^4 + (2*b^2*d*f*x + 2*b^2*d*e - b^2*f)*sinh
(d*x + c)^4 + 2*b^2*d*e - 8*(a*b*d*f*x + a*b*d*e - a*b*f)*cosh(d*x + c)^3 - 4*(2*a*b*d*f*x + 2*a*b*d*e - 2*a*b
*f - (2*b^2*d*f*x + 2*b^2*d*e - b^2*f)*cosh(d*x + c))*sinh(d*x + c)^3 + b^2*f - 8*(a^2*d^2*f*x^2 + 2*a^2*d^2*e
*x + 4*a^2*c*d*e - 2*a^2*c^2*f)*cosh(d*x + c)^2 - 2*(4*a^2*d^2*f*x^2 + 8*a^2*d^2*e*x + 16*a^2*c*d*e - 8*a^2*c^
2*f - 3*(2*b^2*d*f*x + 2*b^2*d*e - b^2*f)*cosh(d*x + c)^2 + 12*(a*b*d*f*x + a*b*d*e - a*b*f)*cosh(d*x + c))*si
nh(d*x + c)^2 + 8*(a*b*d*f*x + a*b*d*e + a*b*f)*cosh(d*x + c) + 16*(a^2*f*cosh(d*x + c)^2 + 2*a^2*f*cosh(d*x +
c)*sinh(d*x + c) + a^2*f*sinh(d*x + c)^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sin
h(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + 16*(a^2*f*cosh(d*x + c)^2 + 2*a^2*f*cosh(d*x + c)*sinh(d*x + c
) + a^2*f*sinh(d*x + c)^2)*dilog((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*d*e - a^2*c*f)*cosh(d*x + c)^2 + 2*(a^2*d*e - a^2*c*f)*cosh(d*x + c)*
sinh(d*x + c) + (a^2*d*e - a^2*c*f)*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*d*e - a^2*c*f)*cosh(d*x + c)^2 + 2*(a^2*d*e - a^2*c*f)*cosh(d*x + c)*sinh(d*x +
c) + (a^2*d*e - a^2*c*f)*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*d*f*x + a^2*c*f)*cosh(d*x + c)^2 + 2*(a^2*d*f*x + a^2*c*f)*cosh(d*x + c)*sinh(d*x + c) +
(a^2*d*f*x + a^2*c*f)*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*d*f*x + a^2*c*f)*cosh(d*x + c)^2 + 2*(a^2*d*f*x + a^2*c*f)*cos
h(d*x + c)*sinh(d*x + c) + (a^2*d*f*x + a^2*c*f)*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*f*x + 2*a*b*d*e + (2*b^2*d*f*x +
2*b^2*d*e - b^2*f)*cosh(d*x + c)^3 + 2*a*b*f - 6*(a*b*d*f*x + a*b*d*e - a*b*f)*cosh(d*x + c)^2 - 4*(a^2*d^2*f*
x^2 + 2*a^2*d^2*e*x + 4*a^2*c*d*e - 2*a^2*c^2*f)*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 \[ \int \frac {{\left (f x + e\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2}}{b \sinh \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cosh(d*x+c)*sinh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*cosh(d*x + c)*sinh(d*x + c)^2/(b*sinh(d*x + c) + a), x)

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maple [B]  time = 0.17, size = 565, normalized size = 2.03 \[ -\frac {a^{2} f \,x^{2}}{2 b^{3}}+\frac {a^{2} e x}{b^{3}}+\frac {\left (2 d f x +2 d e -f \right ) {\mathrm e}^{2 d x +2 c}}{16 d^{2} b}-\frac {a \left (d f x +d e -f \right ) {\mathrm e}^{d x +c}}{2 b^{2} d^{2}}+\frac {a \left (d f x +d e +f \right ) {\mathrm e}^{-d x -c}}{2 b^{2} d^{2}}+\frac {\left (2 d f x +2 d e +f \right ) {\mathrm e}^{-2 d x -2 c}}{16 d^{2} b}-\frac {a^{2} f c \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}{d^{2} b^{3}}+\frac {2 a^{2} f c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} b^{3}}+\frac {a^{2} e \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}{d \,b^{3}}-\frac {2 a^{2} e \ln \left ({\mathrm e}^{d x +c}\right )}{d \,b^{3}}+\frac {a^{2} f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \,b^{3}}+\frac {a^{2} f \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} b^{3}}+\frac {a^{2} f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \,b^{3}}+\frac {a^{2} f \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} b^{3}}+\frac {a^{2} f \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} b^{3}}+\frac {a^{2} f \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} b^{3}}-\frac {2 f \,a^{2} c x}{d \,b^{3}}-\frac {f \,a^{2} c^{2}}{d^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*cosh(d*x+c)*sinh(d*x+c)^2/(a+b*sinh(d*x+c)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cosh(d*x+c)*sinh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

1/8*e*(8*(d*x + c)*a^2/(b^3*d) - (4*a*e^(-d*x - c) - b)*e^(2*d*x + 2*c)/(b^2*d) + 8*a^2*log(-2*a*e^(-d*x - c)
+ b*e^(-2*d*x - 2*c) - b)/(b^3*d) + (4*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c))/(b^2*d)) + 1/16*f*((8*a^2*d^2*x^2*
e^(2*c) + (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) - 2*integrate(16*(a^3*x*e^(d*x + c
) - a^2*b*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 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\mathrm {sinh}\left (c+d\,x\right )}^2\,\left (e+f\,x\right )}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)*sinh(c + d*x)^2*(e + f*x))/(a + b*sinh(c + d*x)),x)

[Out]

int((cosh(c + d*x)*sinh(c + d*x)^2*(e + f*x))/(a + b*sinh(c + d*x)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cosh(d*x+c)*sinh(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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