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

Optimal. Leaf size=298 \[ \frac {f \left (a^2+b^2\right ) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^3 d^2}+\frac {f \left (a^2+b^2\right ) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^3 d^2}+\frac {\left (a^2+b^2\right ) (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^3 d}+\frac {\left (a^2+b^2\right ) (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^3 d}-\frac {\left (a^2+b^2\right ) (e+f x)^2}{2 b^3 f}+\frac {a f \cosh (c+d x)}{b^2 d^2}-\frac {a (e+f x) \sinh (c+d x)}{b^2 d}-\frac {f \sinh (c+d x) \cosh (c+d x)}{4 b d^2}+\frac {(e+f x) \sinh ^2(c+d x)}{2 b d}+\frac {f x}{4 b d} \]

[Out]

1/4*f*x/b/d-1/2*(a^2+b^2)*(f*x+e)^2/b^3/f+a*f*cosh(d*x+c)/b^2/d^2+(a^2+b^2)*(f*x+e)*ln(1+b*exp(d*x+c)/(a-(a^2+
b^2)^(1/2)))/b^3/d+(a^2+b^2)*(f*x+e)*ln(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/b^3/d+(a^2+b^2)*f*polylog(2,-b*exp
(d*x+c)/(a-(a^2+b^2)^(1/2)))/b^3/d^2+(a^2+b^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 = 0.36, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5565, 3296, 2638, 5446, 2635, 8, 5561, 2190, 2279, 2391} \[ \frac {f \left (a^2+b^2\right ) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^3 d^2}+\frac {f \left (a^2+b^2\right ) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b^3 d^2}+\frac {\left (a^2+b^2\right ) (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^3 d}+\frac {\left (a^2+b^2\right ) (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^3 d}-\frac {\left (a^2+b^2\right ) (e+f x)^2}{2 b^3 f}+\frac {a f \cosh (c+d x)}{b^2 d^2}-\frac {a (e+f x) \sinh (c+d x)}{b^2 d}-\frac {f \sinh (c+d x) \cosh (c+d x)}{4 b d^2}+\frac {(e+f x) \sinh ^2(c+d x)}{2 b d}+\frac {f x}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*x)/(4*b*d) - ((a^2 + b^2)*(e + f*x)^2)/(2*b^3*f) + (a*f*Cosh[c + d*x])/(b^2*d^2) + ((a^2 + b^2)*(e + f*x)*L
og[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(b^3*d) + ((a^2 + b^2)*(e + f*x)*Log[1 + (b*E^(c + d*x))/(a + S
qrt[a^2 + b^2])])/(b^3*d) + ((a^2 + b^2)*f*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^3*d^2) + (
(a^2 + b^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 5565

Int[(Cosh[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symb
ol] :> -Dist[a/b^2, Int[(e + f*x)^m*Cosh[c + d*x]^(n - 2), x], x] + (Dist[1/b, Int[(e + f*x)^m*Cosh[c + d*x]^(
n - 2)*Sinh[c + d*x], x], x] + Dist[(a^2 + b^2)/b^2, Int[((e + f*x)^m*Cosh[c + d*x]^(n - 2))/(a + b*Sinh[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 {(e+f x) \cosh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac {a \int (e+f x) \cosh (c+d x) \, dx}{b^2}+\frac {\int (e+f x) \cosh (c+d x) \sinh (c+d x) \, dx}{b}+\frac {\left (a^2+b^2\right ) \int \frac {(e+f x) \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{b^2}\\ &=-\frac {\left (a^2+b^2\right ) (e+f x)^2}{2 b^3 f}-\frac {a (e+f x) \sinh (c+d x)}{b^2 d}+\frac {(e+f x) \sinh ^2(c+d x)}{2 b d}+\frac {\left (a^2+b^2\right ) \int \frac {e^{c+d x} (e+f 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} (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 \sinh ^2(c+d x) \, dx}{2 b d}\\ &=-\frac {\left (a^2+b^2\right ) (e+f x)^2}{2 b^3 f}+\frac {a f \cosh (c+d x)}{b^2 d^2}+\frac {\left (a^2+b^2\right ) (e+f 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 ) (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 {f \int 1 \, dx}{4 b d}-\frac {\left (\left (a^2+b^2\right ) 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 (\left (a^2+b^2\right ) 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 {\left (a^2+b^2\right ) (e+f x)^2}{2 b^3 f}+\frac {a f \cosh (c+d x)}{b^2 d^2}+\frac {\left (a^2+b^2\right ) (e+f 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 ) (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 (\left (a^2+b^2\right ) 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 (\left (a^2+b^2\right ) 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 {\left (a^2+b^2\right ) (e+f x)^2}{2 b^3 f}+\frac {a f \cosh (c+d x)}{b^2 d^2}+\frac {\left (a^2+b^2\right ) (e+f 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 ) (e+f 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 ) f \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 ) 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 = 1.28, size = 251, normalized size = 0.84 \[ \frac {8 \left (a^2+b^2\right ) \left (f \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+f (c+d x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+d e \log (a+b \sinh (c+d x))-c f \log (a+b \sinh (c+d x))-\frac {1}{2} f (c+d x)^2\right )-8 a b d (e+f x) \sinh (c+d x)+8 a b f \cosh (c+d x)+2 b^2 d (e+f x) \cosh (2 (c+d x))-b^2 f \sinh (2 (c+d x))}{8 b^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*b*f*Cosh[c + d*x] + 2*b^2*d*(e + f*x)*Cosh[2*(c + d*x)] + 8*(a^2 + b^2)*(-1/2*(f*(c + d*x)^2) + f*(c + d*
x)*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] + f*(c + d*x)*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])]
+ d*e*Log[a + b*Sinh[c + d*x]] - c*f*Log[a + b*Sinh[c + d*x]] + f*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 +
b^2])] + f*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))]) - 8*a*b*d*(e + f*x)*Sinh[c + d*x] - b^2*f*Si
nh[2*(c + d*x)])/(8*b^3*d^2)

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fricas [B]  time = 0.50, size = 1416, normalized size = 4.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.21, size = 975, normalized size = 3.27 \[ -\frac {f \,x^{2}}{2 b}+\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 {2 f \,a^{2} c 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 ) 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 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 x}{b^{3}}+\frac {e x}{b}-\frac {a^{2} f \,x^{2}}{2 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 {\left (2 d f x +2 d e +f \right ) {\mathrm e}^{-2 d x -2 c}}{16 d^{2} b}-\frac {f \,c^{2}}{d^{2} b}+\frac {e \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}{d b}-\frac {2 e \ln \left ({\mathrm e}^{d x +c}\right )}{d b}+\frac {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}+\frac {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}+\frac {a \left (d f x +d e +f \right ) {\mathrm e}^{-d x -c}}{2 b^{2} d^{2}}-\frac {2 f c x}{d b}-\frac {f 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 f c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} b}+\frac {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}+\frac {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}+\frac {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}+\frac {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}-\frac {a \left (d f x +d e -f \right ) {\mathrm e}^{d x +c}}{2 b^{2} d^{2}}-\frac {f \,a^{2} c^{2}}{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 \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}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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