3.3.23 \(\int \frac {(e+f x)^3 \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [223]

Optimal. Leaf size=453 \[ \frac {(e+f x)^4}{4 b f}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}+\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d}-\frac {3 a f (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {3 a f (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {6 a f^2 (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {6 a f^2 (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {6 a f^3 \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^4}+\frac {6 a f^3 \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^4} \]

[Out]

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

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Rubi [A]
time = 0.58, antiderivative size = 453, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {5676, 32, 3403, 2296, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}+\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {a (e+f x)^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {a (e+f x)^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {(e+f x)^4}{4 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 3403

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 5676

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sinh[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[(e + f*x)^m*(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]

Rule 6724

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 6744

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1009\) vs. \(2(453)=906\).
time = 2.73, size = 1009, normalized size = 2.23 \begin {gather*} \frac {x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )}{4 b}+\frac {a \left (2 d^3 e^3 \sqrt {\left (a^2+b^2\right ) e^{2 c}} \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-3 \sqrt {a^2+b^2} d^3 e^2 e^c f x \log \left (1+\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-3 \sqrt {a^2+b^2} d^3 e e^c f^2 x^2 \log \left (1+\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-\sqrt {a^2+b^2} d^3 e^c f^3 x^3 \log \left (1+\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+3 \sqrt {a^2+b^2} d^3 e^2 e^c f x \log \left (1+\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+3 \sqrt {a^2+b^2} d^3 e e^c f^2 x^2 \log \left (1+\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+\sqrt {a^2+b^2} d^3 e^c f^3 x^3 \log \left (1+\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-3 \sqrt {a^2+b^2} d^2 e^c f (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+3 \sqrt {a^2+b^2} d^2 e^c f (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+6 \sqrt {a^2+b^2} d e e^c f^2 \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+6 \sqrt {a^2+b^2} d e^c f^3 x \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-6 \sqrt {a^2+b^2} d e e^c f^2 \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-6 \sqrt {a^2+b^2} d e^c f^3 x \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-6 \sqrt {a^2+b^2} e^c f^3 \text {PolyLog}\left (4,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+6 \sqrt {a^2+b^2} e^c f^3 \text {PolyLog}\left (4,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )\right )}{b \sqrt {a^2+b^2} d^4 \sqrt {\left (a^2+b^2\right ) e^{2 c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.53, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{3} \sinh \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1614 vs. \(2 (416) = 832\).
time = 0.39, size = 1614, normalized size = 3.56 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a^2 + b^2)*d^4*f^3*x^4 + 4*(a^2 + b^2)*d^4*f^2*x^3*cosh(1) + 6*(a^2 + b^2)*d^4*f*x^2*cosh(1)^2 + 4*(a^2
+ b^2)*d^4*x*cosh(1)^3 + 4*(a^2 + b^2)*d^4*x*sinh(1)^3 - 24*a*b*f^3*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*a*b*f^3*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) + 6*((a^2 + b^2)*d^4*f*x^2 + 2*(a^2 + b^2)*d^4*x*cosh(1))*sinh(1)^2 - 12*(a*b*d^2*f^3*x^2 + 2*a
*b*d^2*f^2*x*cosh(1) + a*b*d^2*f*cosh(1)^2 + a*b*d^2*f*sinh(1)^2 + 2*(a*b*d^2*f^2*x + a*b*d^2*f*cosh(1))*sinh(
1))*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*(a*b*d^2*f^3*x^2 + 2*a*b*d^2*f^2*x*cosh(1) + a*b*d^2*f*cosh(1)^2 + a*b*d^2*f
*sinh(1)^2 + 2*(a*b*d^2*f^2*x + a*b*d^2*f*cosh(1))*sinh(1))*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(d*x + c) + a*s
inh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - 4*(a*b*c^3*f^3 - 3*a*b*
c^2*d*f^2*cosh(1) + 3*a*b*c*d^2*f*cosh(1)^2 - a*b*d^3*cosh(1)^3 - a*b*d^3*sinh(1)^3 + 3*(a*b*c*d^2*f - a*b*d^3
*cosh(1))*sinh(1)^2 - 3*(a*b*c^2*d*f^2 - 2*a*b*c*d^2*f*cosh(1) + a*b*d^3*cosh(1)^2)*sinh(1))*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*(a*b*c^3*f^3 - 3*a*b*c^2
*d*f^2*cosh(1) + 3*a*b*c*d^2*f*cosh(1)^2 - a*b*d^3*cosh(1)^3 - a*b*d^3*sinh(1)^3 + 3*(a*b*c*d^2*f - a*b*d^3*co
sh(1))*sinh(1)^2 - 3*(a*b*c^2*d*f^2 - 2*a*b*c*d^2*f*cosh(1) + a*b*d^3*cosh(1)^2)*sinh(1))*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*(a*b*d^3*f^3*x^3 + a*b*c^3*
f^3 + 3*(a*b*d^3*f*x + a*b*c*d^2*f)*cosh(1)^2 + 3*(a*b*d^3*f*x + a*b*c*d^2*f)*sinh(1)^2 + 3*(a*b*d^3*f^2*x^2 -
 a*b*c^2*d*f^2)*cosh(1) + 3*(a*b*d^3*f^2*x^2 - a*b*c^2*d*f^2 + 2*(a*b*d^3*f*x + a*b*c*d^2*f)*cosh(1))*sinh(1))
*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) + 4*(a*b*d^3*f^3*x^3 + a*b*c^3*f^3 + 3*(a*b*d^3*f*x + a*b*c*d^2*f)*cosh(1)^2 + 3*(a*b*d^3
*f*x + a*b*c*d^2*f)*sinh(1)^2 + 3*(a*b*d^3*f^2*x^2 - a*b*c^2*d*f^2)*cosh(1) + 3*(a*b*d^3*f^2*x^2 - a*b*c^2*d*f
^2 + 2*(a*b*d^3*f*x + a*b*c*d^2*f)*cosh(1))*sinh(1))*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*(a*b*d*f^3*x + a*b*d*f^2*cosh(1)
 + a*b*d*f^2*sinh(1))*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*(a*b*d*f^3*x + a*b*d*f^2*cosh(1) + a*b*d*f^2*sinh(1))*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) + 4*((a^2 + b^2)*d^4*f^2*x^3 + 3*(a^2 + b^2)*d^4*f*x^2*cosh(1) + 3*(a^2 + b^2)*d^4*x*cosh(1)^2)
*sinh(1))/((a^2*b + b^3)*d^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e + f x\right )^{3} \sinh {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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