\(\int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx\) [222]

Optimal result

Integrand size = 22, antiderivative size = 327 \[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx=-\frac {x^4}{4 b}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^4}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx=-\frac {x^4}{4 b}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {6 \operatorname {PolyLog}\left (4,\frac {b e^{c+d x}}{-a+\sqrt {a^2-b^2}}\right )}{b d^4}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^4} \] Input:

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

Output:

-1/4*x^4/b + (x^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 - b^2])])/(b*d) + 
(x^3*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 - b^2])])/(b*d) + (3*x^2*PolyLo 
g[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 - b^2]))])/(b*d^2) + (3*x^2*PolyLog[2 
, -((b*E^(c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^2) - (6*x*PolyLog[3, -(( 
b*E^(c + d*x))/(a - Sqrt[a^2 - b^2]))])/(b*d^3) - (6*x*PolyLog[3, -((b*E^( 
c + d*x))/(a + Sqrt[a^2 - b^2]))])/(b*d^3) + (6*PolyLog[4, (b*E^(c + d*x)) 
/(-a + Sqrt[a^2 - b^2])])/(b*d^4) + (6*PolyLog[4, -((b*E^(c + d*x))/(a + S 
qrt[a^2 - b^2]))])/(b*d^4)
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6096, 2620, 3011, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6096

\(\displaystyle \int \frac {e^{c+d x} x^3}{a+b e^{c+d x}-\sqrt {a^2-b^2}}dx+\int \frac {e^{c+d x} x^3}{a+b e^{c+d x}+\sqrt {a^2-b^2}}dx-\frac {x^4}{4 b}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {3 \int x^2 \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2-b^2}}+1\right )dx}{b d}-\frac {3 \int x^2 \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2-b^2}}+1\right )dx}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {3 \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )dx}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}-\frac {3 \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )dx}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b}\)

\(\Big \downarrow \) 7163

\(\displaystyle -\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {\int \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {\int \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {\int e^{-c-d x} \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )de^{c+d x}}{d^2}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {\int e^{-c-d x} \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )de^{c+d x}}{d^2}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {\operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {\operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^4}{4 b}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2620

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] - Si 
mp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 6096

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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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 7163

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] - Simp[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]
 

Maple [F]

\[\int \frac {x^{3} \sinh \left (d x +c \right )}{a +b \cosh \left (d x +c \right )}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (299) = 598\).

Time = 0.11 (sec) , antiderivative size = 624, normalized size of antiderivative = 1.91 \[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(d^4*x^4 - 12*d^2*x^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*d^ 
2*x^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) + 4*c^3*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*c^3*log(2*b*c 
osh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) + 24*d 
*x*polylog(3, -(a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*s 
inh(d*x + c))*sqrt((a^2 - b^2)/b^2))/b) + 24*d*x*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*(d^3*x^3 + c^3)*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*(d^3* 
x^3 + c^3)*log((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*s 
inh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) - 24*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*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))/(b*d^4)
 

Sympy [F]

\[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx=\int \frac {x^{3} \sinh {\left (c + d x \right )}}{a + b \cosh {\left (c + d x \right )}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx=\int { \frac {x^{3} \sinh \left (d x + c\right )}{b \cosh \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx=\int { \frac {x^{3} \sinh \left (d x + c\right )}{b \cosh \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx=\int \frac {x^3\,\mathrm {sinh}\left (c+d\,x\right )}{a+b\,\mathrm {cosh}\left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx=e^{2 c} \left (\int \frac {e^{2 d x} x^{3}}{e^{2 d x +2 c} b +2 e^{d x +c} a +b}d x \right )-\left (\int \frac {x^{3}}{e^{2 d x +2 c} b +2 e^{d x +c} a +b}d x \right ) \] Input:

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

Output:

e**(2*c)*int((e**(2*d*x)*x**3)/(e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a + b) 
,x) - int(x**3/(e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a + b),x)