3.294 \(\int \frac {(e+f x)^3 \cosh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=527 \[ \frac {6 f^3 \sqrt {a^2+b^2} \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^4}-\frac {6 f^3 \sqrt {a^2+b^2} \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^4}-\frac {6 f^2 \sqrt {a^2+b^2} (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^3}+\frac {6 f^2 \sqrt {a^2+b^2} (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^3}+\frac {3 f \sqrt {a^2+b^2} (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {3 f \sqrt {a^2+b^2} (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^2}+\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^2 d}-\frac {a (e+f x)^4}{4 b^2 f}-\frac {6 f^3 \sinh (c+d x)}{b d^4}+\frac {6 f^2 (e+f x) \cosh (c+d x)}{b d^3}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}+\frac {(e+f x)^3 \cosh (c+d x)}{b d} \]

[Out]

-1/4*a*(f*x+e)^4/b^2/f+6*f^2*(f*x+e)*cosh(d*x+c)/b/d^3+(f*x+e)^3*cosh(d*x+c)/b/d-6*f^3*sinh(d*x+c)/b/d^4-3*f*(
f*x+e)^2*sinh(d*x+c)/b/d^2+(f*x+e)^3*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/b^2/d-(f*x+e)^3*ln
(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/b^2/d+3*f*(f*x+e)^2*polylog(2,-b*exp(d*x+c)/(a-(a^2+b^2)^
(1/2)))*(a^2+b^2)^(1/2)/b^2/d^2-3*f*(f*x+e)^2*polylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/b^2
/d^2-6*f^2*(f*x+e)*polylog(3,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/b^2/d^3+6*f^2*(f*x+e)*polylog(
3,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/b^2/d^3+6*f^3*polylog(4,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2))
)*(a^2+b^2)^(1/2)/b^2/d^4-6*f^3*polylog(4,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))*(a^2+b^2)^(1/2)/b^2/d^4

________________________________________________________________________________________

Rubi [A]  time = 0.91, antiderivative size = 527, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {5565, 32, 3296, 2637, 3322, 2264, 2190, 2531, 6609, 2282, 6589} \[ -\frac {6 f^2 \sqrt {a^2+b^2} (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^3}+\frac {6 f^2 \sqrt {a^2+b^2} (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b^2 d^3}+\frac {3 f \sqrt {a^2+b^2} (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {3 f \sqrt {a^2+b^2} (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b^2 d^2}+\frac {6 f^3 \sqrt {a^2+b^2} \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^4}-\frac {6 f^3 \sqrt {a^2+b^2} \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b^2 d^4}+\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b^2 d}-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cosh (c+d x)}{b d^3}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}-\frac {6 f^3 \sinh (c+d x)}{b d^4}+\frac {(e+f x)^3 \cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Cosh[c + d*x]^2)/(a + b*Sinh[c + d*x]),x]

[Out]

-(a*(e + f*x)^4)/(4*b^2*f) + (6*f^2*(e + f*x)*Cosh[c + d*x])/(b*d^3) + ((e + f*x)^3*Cosh[c + d*x])/(b*d) + (Sq
rt[a^2 + b^2]*(e + f*x)^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(b^2*d) - (Sqrt[a^2 + b^2]*(e + f*x)
^3*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/(b^2*d) + (3*Sqrt[a^2 + b^2]*f*(e + f*x)^2*PolyLog[2, -((b*
E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^2) - (3*Sqrt[a^2 + b^2]*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x)
)/(a + Sqrt[a^2 + b^2]))])/(b^2*d^2) - (6*Sqrt[a^2 + b^2]*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt
[a^2 + b^2]))])/(b^2*d^3) + (6*Sqrt[a^2 + b^2]*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]
))])/(b^2*d^3) + (6*Sqrt[a^2 + b^2]*f^3*PolyLog[4, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^4) - (6*S
qrt[a^2 + b^2]*f^3*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(b^2*d^4) - (6*f^3*Sinh[c + d*x])/(b*
d^4) - (3*f*(e + f*x)^2*Sinh[c + d*x])/(b*d^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 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]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \cosh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac {a \int (e+f x)^3 \, dx}{b^2}+\frac {\int (e+f x)^3 \sinh (c+d x) \, dx}{b}+\frac {\left (a^2+b^2\right ) \int \frac {(e+f x)^3}{a+b \sinh (c+d x)} \, dx}{b^2}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {(e+f x)^3 \cosh (c+d x)}{b d}+\frac {\left (2 \left (a^2+b^2\right )\right ) \int \frac {e^{c+d x} (e+f x)^3}{-b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{b^2}-\frac {(3 f) \int (e+f x)^2 \cosh (c+d x) \, dx}{b d}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {(e+f x)^3 \cosh (c+d x)}{b d}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}+\frac {\left (2 \sqrt {a^2+b^2}\right ) \int \frac {e^{c+d x} (e+f x)^3}{2 a-2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{b}-\frac {\left (2 \sqrt {a^2+b^2}\right ) \int \frac {e^{c+d x} (e+f x)^3}{2 a+2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{b}+\frac {\left (6 f^2\right ) \int (e+f x) \sinh (c+d x) \, dx}{b d^2}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cosh (c+d x)}{b d^3}+\frac {(e+f x)^3 \cosh (c+d x)}{b d}+\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}-\frac {\left (3 \sqrt {a^2+b^2} f\right ) \int (e+f x)^2 \log \left (1+\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{b^2 d}+\frac {\left (3 \sqrt {a^2+b^2} f\right ) \int (e+f x)^2 \log \left (1+\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{b^2 d}-\frac {\left (6 f^3\right ) \int \cosh (c+d x) \, dx}{b d^3}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cosh (c+d x)}{b d^3}+\frac {(e+f x)^3 \cosh (c+d x)}{b d}+\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d}+\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {6 f^3 \sinh (c+d x)}{b d^4}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}-\frac {\left (6 \sqrt {a^2+b^2} f^2\right ) \int (e+f x) \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{b^2 d^2}+\frac {\left (6 \sqrt {a^2+b^2} f^2\right ) \int (e+f x) \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{b^2 d^2}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cosh (c+d x)}{b d^3}+\frac {(e+f x)^3 \cosh (c+d x)}{b d}+\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d}+\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {6 \sqrt {a^2+b^2} f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^3}+\frac {6 \sqrt {a^2+b^2} f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^3}-\frac {6 f^3 \sinh (c+d x)}{b d^4}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}+\frac {\left (6 \sqrt {a^2+b^2} f^3\right ) \int \text {Li}_3\left (-\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{b^2 d^3}-\frac {\left (6 \sqrt {a^2+b^2} f^3\right ) \int \text {Li}_3\left (-\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{b^2 d^3}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cosh (c+d x)}{b d^3}+\frac {(e+f x)^3 \cosh (c+d x)}{b d}+\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d}+\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {6 \sqrt {a^2+b^2} f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^3}+\frac {6 \sqrt {a^2+b^2} f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^3}-\frac {6 f^3 \sinh (c+d x)}{b d^4}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}+\frac {\left (6 \sqrt {a^2+b^2} f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {b x}{-a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b^2 d^4}-\frac {\left (6 \sqrt {a^2+b^2} f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {b x}{a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b^2 d^4}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cosh (c+d x)}{b d^3}+\frac {(e+f x)^3 \cosh (c+d x)}{b d}+\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d}-\frac {\sqrt {a^2+b^2} (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d}+\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {3 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^2}-\frac {6 \sqrt {a^2+b^2} f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^3}+\frac {6 \sqrt {a^2+b^2} f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^3}+\frac {6 \sqrt {a^2+b^2} f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b^2 d^4}-\frac {6 \sqrt {a^2+b^2} f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b^2 d^4}-\frac {6 f^3 \sinh (c+d x)}{b d^4}-\frac {3 f (e+f x)^2 \sinh (c+d x)}{b d^2}\\ \end {align*}

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Mathematica [A]  time = 3.09, size = 933, normalized size = 1.77 \[ -\frac {a f^3 x^4 d^4+4 a e f^2 x^3 d^4+6 a e^2 f x^2 d^4+4 a e^3 x d^4+8 \sqrt {a^2+b^2} e^3 \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right ) d^3-4 b e^3 \cosh (c+d x) d^3-4 b f^3 x^3 \cosh (c+d x) d^3-12 b e f^2 x^2 \cosh (c+d x) d^3-12 b e^2 f x \cosh (c+d x) d^3-4 \sqrt {a^2+b^2} f^3 x^3 \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right ) d^3-12 \sqrt {a^2+b^2} e f^2 x^2 \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right ) d^3-12 \sqrt {a^2+b^2} e^2 f x \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right ) d^3+4 \sqrt {a^2+b^2} f^3 x^3 \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right ) d^3+12 \sqrt {a^2+b^2} e f^2 x^2 \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right ) d^3+12 \sqrt {a^2+b^2} e^2 f x \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right ) d^3-12 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right ) d^2+12 \sqrt {a^2+b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) d^2+12 b f^3 x^2 \sinh (c+d x) d^2+12 b e^2 f \sinh (c+d x) d^2+24 b e f^2 x \sinh (c+d x) d^2-24 b e f^2 \cosh (c+d x) d-24 b f^3 x \cosh (c+d x) d+24 \sqrt {a^2+b^2} e f^2 \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right ) d+24 \sqrt {a^2+b^2} f^3 x \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right ) d-24 \sqrt {a^2+b^2} e f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) d-24 \sqrt {a^2+b^2} f^3 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) d-24 \sqrt {a^2+b^2} f^3 \text {Li}_4\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+24 \sqrt {a^2+b^2} f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+24 b f^3 \sinh (c+d x)}{4 b^2 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Cosh[c + d*x]^2)/(a + b*Sinh[c + d*x]),x]

[Out]

-1/4*(4*a*d^4*e^3*x + 6*a*d^4*e^2*f*x^2 + 4*a*d^4*e*f^2*x^3 + a*d^4*f^3*x^4 + 8*Sqrt[a^2 + b^2]*d^3*e^3*ArcTan
h[(a + b*E^(c + d*x))/Sqrt[a^2 + b^2]] - 4*b*d^3*e^3*Cosh[c + d*x] - 24*b*d*e*f^2*Cosh[c + d*x] - 12*b*d^3*e^2
*f*x*Cosh[c + d*x] - 24*b*d*f^3*x*Cosh[c + d*x] - 12*b*d^3*e*f^2*x^2*Cosh[c + d*x] - 4*b*d^3*f^3*x^3*Cosh[c +
d*x] - 12*Sqrt[a^2 + b^2]*d^3*e^2*f*x*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 12*Sqrt[a^2 + b^2]*d^3*
e*f^2*x^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 4*Sqrt[a^2 + b^2]*d^3*f^3*x^3*Log[1 + (b*E^(c + d*x
))/(a - Sqrt[a^2 + b^2])] + 12*Sqrt[a^2 + b^2]*d^3*e^2*f*x*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] + 12
*Sqrt[a^2 + b^2]*d^3*e*f^2*x^2*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] + 4*Sqrt[a^2 + b^2]*d^3*f^3*x^3*
Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] - 12*Sqrt[a^2 + b^2]*d^2*f*(e + f*x)^2*PolyLog[2, (b*E^(c + d*x
))/(-a + Sqrt[a^2 + b^2])] + 12*Sqrt[a^2 + b^2]*d^2*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 +
b^2]))] + 24*Sqrt[a^2 + b^2]*d*e*f^2*PolyLog[3, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] + 24*Sqrt[a^2 + b^2]*
d*f^3*x*PolyLog[3, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] - 24*Sqrt[a^2 + b^2]*d*e*f^2*PolyLog[3, -((b*E^(c +
d*x))/(a + Sqrt[a^2 + b^2]))] - 24*Sqrt[a^2 + b^2]*d*f^3*x*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])
)] - 24*Sqrt[a^2 + b^2]*f^3*PolyLog[4, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] + 24*Sqrt[a^2 + b^2]*f^3*PolyLo
g[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))] + 12*b*d^2*e^2*f*Sinh[c + d*x] + 24*b*f^3*Sinh[c + d*x] + 24*b*
d^2*e*f^2*x*Sinh[c + d*x] + 12*b*d^2*f^3*x^2*Sinh[c + d*x])/(b^2*d^4)

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fricas [C]  time = 0.79, size = 2020, normalized size = 3.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(2*b*d^3*f^3*x^3 + 2*b*d^3*e^3 + 6*b*d^2*e^2*f + 12*b*d*e*f^2 + 12*b*f^3 + 6*(b*d^3*e*f^2 + b*d^2*f^3)*x^2
+ 2*(b*d^3*f^3*x^3 + b*d^3*e^3 - 3*b*d^2*e^2*f + 6*b*d*e*f^2 - 6*b*f^3 + 3*(b*d^3*e*f^2 - b*d^2*f^3)*x^2 + 3*
(b*d^3*e^2*f - 2*b*d^2*e*f^2 + 2*b*d*f^3)*x)*cosh(d*x + c)^2 + 2*(b*d^3*f^3*x^3 + b*d^3*e^3 - 3*b*d^2*e^2*f +
6*b*d*e*f^2 - 6*b*f^3 + 3*(b*d^3*e*f^2 - b*d^2*f^3)*x^2 + 3*(b*d^3*e^2*f - 2*b*d^2*e*f^2 + 2*b*d*f^3)*x)*sinh(
d*x + c)^2 + 12*((b*d^2*f^3*x^2 + 2*b*d^2*e*f^2*x + b*d^2*e^2*f)*cosh(d*x + c) + (b*d^2*f^3*x^2 + 2*b*d^2*e*f^
2*x + b*d^2*e^2*f)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^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) - 12*((b*d^2*f^3*x^2 + 2*b*d^2*e*f^2*x + b*d^2*e^2*
f)*cosh(d*x + c) + (b*d^2*f^3*x^2 + 2*b*d^2*e*f^2*x + b*d^2*e^2*f)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^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) - 4
*((b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*cosh(d*x + c) + (b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3
*b*c^2*d*e*f^2 - b*c^3*f^3)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^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) + 4*((b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*cosh(d*x + c)
+ (b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*log(2*b*co
sh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + 4*((b*d^3*f^3*x^3 + 3*b*d^3*e*f^2*x^2 + 3
*b*d^3*e^2*f*x + 3*b*c*d^2*e^2*f - 3*b*c^2*d*e*f^2 + b*c^3*f^3)*cosh(d*x + c) + (b*d^3*f^3*x^3 + 3*b*d^3*e*f^2
*x^2 + 3*b*d^3*e^2*f*x + 3*b*c*d^2*e^2*f - 3*b*c^2*d*e*f^2 + b*c^3*f^3)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*l
og(-(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
*((b*d^3*f^3*x^3 + 3*b*d^3*e*f^2*x^2 + 3*b*d^3*e^2*f*x + 3*b*c*d^2*e^2*f - 3*b*c^2*d*e*f^2 + b*c^3*f^3)*cosh(d
*x + c) + (b*d^3*f^3*x^3 + 3*b*d^3*e*f^2*x^2 + 3*b*d^3*e^2*f*x + 3*b*c*d^2*e^2*f - 3*b*c^2*d*e*f^2 + b*c^3*f^3
)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^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) + 24*(b*f^3*cosh(d*x + c) + b*f^3*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*po
lylog(4, (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) -
24*(b*f^3*cosh(d*x + c) + b*f^3*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*polylog(4, (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) - 24*((b*d*f^3*x + b*d*e*f^2)*cosh(d*x +
c) + (b*d*f^3*x + b*d*e*f^2)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*polylog(3, (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) + 24*((b*d*f^3*x + b*d*e*f^2)*cosh(d*x + c)
+ (b*d*f^3*x + b*d*e*f^2)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*polylog(3, (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) + 6*(b*d^3*e^2*f + 2*b*d^2*e*f^2 + 2*b*d*f^3)*x
- (a*d^4*f^3*x^4 + 4*a*d^4*e*f^2*x^3 + 6*a*d^4*e^2*f*x^2 + 4*a*d^4*e^3*x)*cosh(d*x + c) - (a*d^4*f^3*x^4 + 4*
a*d^4*e*f^2*x^3 + 6*a*d^4*e^2*f*x^2 + 4*a*d^4*e^3*x - 4*(b*d^3*f^3*x^3 + b*d^3*e^3 - 3*b*d^2*e^2*f + 6*b*d*e*f
^2 - 6*b*f^3 + 3*(b*d^3*e*f^2 - b*d^2*f^3)*x^2 + 3*(b*d^3*e^2*f - 2*b*d^2*e*f^2 + 2*b*d*f^3)*x)*cosh(d*x + c))
*sinh(d*x + c))/(b^2*d^4*cosh(d*x + c) + b^2*d^4*sinh(d*x + c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3} \cosh \left (d x + c\right )^{2}}{b \sinh \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*cosh(d*x + c)^2/(b*sinh(d*x + c) + a), x)

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maple [F]  time = 0.51, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \left (\cosh ^{2}\left (d x +c \right )\right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x)

[Out]

int((f*x+e)^3*cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, e^{3} {\left (\frac {2 \, {\left (d x + c\right )} a}{b^{2} d} - \frac {e^{\left (d x + c\right )}}{b d} - \frac {e^{\left (-d x - c\right )}}{b d} - \frac {2 \, \sqrt {a^{2} + b^{2}} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{b^{2} d}\right )} - \frac {{\left (a d^{4} f^{3} x^{4} e^{c} + 4 \, a d^{4} e f^{2} x^{3} e^{c} + 6 \, a d^{4} e^{2} f x^{2} e^{c} - 2 \, {\left (b d^{3} f^{3} x^{3} e^{\left (2 \, c\right )} + 3 \, {\left (d^{3} e f^{2} - d^{2} f^{3}\right )} b x^{2} e^{\left (2 \, c\right )} + 3 \, {\left (d^{3} e^{2} f - 2 \, d^{2} e f^{2} + 2 \, d f^{3}\right )} b x e^{\left (2 \, c\right )} - 3 \, {\left (d^{2} e^{2} f - 2 \, d e f^{2} + 2 \, f^{3}\right )} b e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} - 2 \, {\left (b d^{3} f^{3} x^{3} + 3 \, {\left (d^{3} e f^{2} + d^{2} f^{3}\right )} b x^{2} + 3 \, {\left (d^{3} e^{2} f + 2 \, d^{2} e f^{2} + 2 \, d f^{3}\right )} b x + 3 \, {\left (d^{2} e^{2} f + 2 \, d e f^{2} + 2 \, f^{3}\right )} b\right )} e^{\left (-d x\right )}\right )} e^{\left (-c\right )}}{4 \, b^{2} d^{4}} + \int \frac {2 \, {\left ({\left (a^{2} f^{3} e^{c} + b^{2} f^{3} e^{c}\right )} x^{3} + 3 \, {\left (a^{2} e f^{2} e^{c} + b^{2} e f^{2} e^{c}\right )} x^{2} + 3 \, {\left (a^{2} e^{2} f e^{c} + b^{2} e^{2} f e^{c}\right )} x\right )} e^{\left (d x\right )}}{b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b^{2} e^{\left (d x + c\right )} - b^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cosh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*e^3*(2*(d*x + c)*a/(b^2*d) - e^(d*x + c)/(b*d) - e^(-d*x - c)/(b*d) - 2*sqrt(a^2 + b^2)*log((b*e^(-d*x -
c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt(a^2 + b^2)))/(b^2*d)) - 1/4*(a*d^4*f^3*x^4*e^c + 4*a*d^4*
e*f^2*x^3*e^c + 6*a*d^4*e^2*f*x^2*e^c - 2*(b*d^3*f^3*x^3*e^(2*c) + 3*(d^3*e*f^2 - d^2*f^3)*b*x^2*e^(2*c) + 3*(
d^3*e^2*f - 2*d^2*e*f^2 + 2*d*f^3)*b*x*e^(2*c) - 3*(d^2*e^2*f - 2*d*e*f^2 + 2*f^3)*b*e^(2*c))*e^(d*x) - 2*(b*d
^3*f^3*x^3 + 3*(d^3*e*f^2 + d^2*f^3)*b*x^2 + 3*(d^3*e^2*f + 2*d^2*e*f^2 + 2*d*f^3)*b*x + 3*(d^2*e^2*f + 2*d*e*
f^2 + 2*f^3)*b)*e^(-d*x))*e^(-c)/(b^2*d^4) + integrate(2*((a^2*f^3*e^c + b^2*f^3*e^c)*x^3 + 3*(a^2*e*f^2*e^c +
b^2*e*f^2*e^c)*x^2 + 3*(a^2*e^2*f*e^c + b^2*e^2*f*e^c)*x)*e^(d*x)/(b^3*e^(2*d*x + 2*c) + 2*a*b^2*e^(d*x + c)
- b^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^2\,{\left (e+f\,x\right )}^3}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)^2*(e + f*x)^3)/(a + b*sinh(c + d*x)),x)

[Out]

int((cosh(c + d*x)^2*(e + f*x)^3)/(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)**3*cosh(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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