\(\int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx\) [322]

Optimal result

Integrand size = 26, antiderivative size = 234 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {2 f (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {2 f^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {2 f^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.86 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=-\frac {2 f \left (d \left (2 e \text {arctanh}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-f x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )+f x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )-f \operatorname {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )+f \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{b \sqrt {a^2+b^2} d^3}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))} \] Input:

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

Output:

(-2*f*(d*(2*e*ArcTanh[(a + b*E^(c + d*x))/Sqrt[a^2 + b^2]] - f*x*Log[1 + ( 
b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] + f*x*Log[1 + (b*E^(c + d*x))/(a + S 
qrt[a^2 + b^2])]) - 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]))]))/(b*Sqrt[a^2 + b^ 
2]*d^3) - (e + f*x)^2/(b*d*(a + b*Sinh[c + d*x]))
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {5987, 3042, 3803, 25, 2694, 27, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 5987

\(\displaystyle \frac {2 f \int \frac {e+f x}{a+b \sinh (c+d x)}dx}{b d}-\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {(e+f x)^2}{b d (a+b \sinh (c+d x))}+\frac {2 f \int \frac {e+f x}{a-i b \sin (i c+i d x)}dx}{b d}\)

\(\Big \downarrow \) 3803

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 2694

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]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3803

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])* 
(f_.)*(x_)]), x_Symbol] :> Simp[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 5987

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(490\) vs. \(2(214)=428\).

Time = 4.38 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.10

method	result	size
risch	\(-\frac {2 \left (x^{2} f^{2}+2 e f x +e^{2}\right ) {\mathrm e}^{d x +c}}{b d \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}-\frac {4 f e \,\operatorname {arctanh}\left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} b \sqrt {a^{2}+b^{2}}}+\frac {2 f^{2} \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d^{2} b \sqrt {a^{2}+b^{2}}}-\frac {2 f^{2} \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d^{2} b \sqrt {a^{2}+b^{2}}}+\frac {2 f^{2} \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{3} b \sqrt {a^{2}+b^{2}}}-\frac {2 f^{2} \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{3} b \sqrt {a^{2}+b^{2}}}+\frac {2 f^{2} \operatorname {dilog}\left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{3} b \sqrt {a^{2}+b^{2}}}-\frac {2 f^{2} \operatorname {dilog}\left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{3} b \sqrt {a^{2}+b^{2}}}+\frac {4 f^{2} c \,\operatorname {arctanh}\left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{3} b \sqrt {a^{2}+b^{2}}}\)	\(491\)

Input:

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

Output:

-2*(f^2*x^2+2*e*f*x+e^2)/b/d*exp(d*x+c)/(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b 
)-4/d^2/b*f*e/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^( 
1/2))+2/d^2/b*f^2/(a^2+b^2)^(1/2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a 
+(a^2+b^2)^(1/2)))*x-2/d^2/b*f^2/(a^2+b^2)^(1/2)*ln((b*exp(d*x+c)+(a^2+b^2 
)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*x+2/d^3/b*f^2/(a^2+b^2)^(1/2)*ln((-b*exp(d 
*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*c-2/d^3/b*f^2/(a^2+b^2)^(1/ 
2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*c+2/d^3/b*f^2/ 
(a^2+b^2)^(1/2)*dilog((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2 
)))-2/d^3/b*f^2/(a^2+b^2)^(1/2)*dilog((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+ 
(a^2+b^2)^(1/2)))+4/d^3/b*f^2*c/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c 
)+2*a)/(a^2+b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1378 vs. \(2 (212) = 424\).

Time = 0.12 (sec) , antiderivative size = 1378, normalized size of antiderivative = 5.89 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

2*((b^2*f^2*cosh(d*x + c)^2 + b^2*f^2*sinh(d*x + c)^2 + 2*a*b*f^2*cosh(d*x 
 + c) - b^2*f^2 + 2*(b^2*f^2*cosh(d*x + c) + a*b*f^2)*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) - (b^2*f^2*cosh(d* 
x + c)^2 + b^2*f^2*sinh(d*x + c)^2 + 2*a*b*f^2*cosh(d*x + c) - b^2*f^2 + 2 
*(b^2*f^2*cosh(d*x + c) + a*b*f^2)*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*di 
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 + 1) + (b^2*d*e*f - b^2*c*f^2 - (b^2*d*e*f 
 - b^2*c*f^2)*cosh(d*x + c)^2 - (b^2*d*e*f - b^2*c*f^2)*sinh(d*x + c)^2 - 
2*(a*b*d*e*f - a*b*c*f^2)*cosh(d*x + c) - 2*(a*b*d*e*f - a*b*c*f^2 + (b^2* 
d*e*f - b^2*c*f^2)*cosh(d*x + c))*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) 
- (b^2*d*e*f - b^2*c*f^2 - (b^2*d*e*f - b^2*c*f^2)*cosh(d*x + c)^2 - (b^2* 
d*e*f - b^2*c*f^2)*sinh(d*x + c)^2 - 2*(a*b*d*e*f - a*b*c*f^2)*cosh(d*x + 
c) - 2*(a*b*d*e*f - a*b*c*f^2 + (b^2*d*e*f - b^2*c*f^2)*cosh(d*x + c))*sin 
h(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) - (b^2*d*f^2*x + b^2*c*f^2 - (b^2*d*f 
^2*x + b^2*c*f^2)*cosh(d*x + c)^2 - (b^2*d*f^2*x + b^2*c*f^2)*sinh(d*x + c 
)^2 - 2*(a*b*d*f^2*x + a*b*c*f^2)*cosh(d*x + c) - 2*(a*b*d*f^2*x + a*b*c*f 
^2 + (b^2*d*f^2*x + b^2*c*f^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{{\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(e+f x)^2 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\text {too large to display} \] Input:

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

Output:

(4*e**(2*c + 2*d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a 
**2 + b**2))*a**3*b*f**2*i + 4*e**(2*c + 2*d*x)*sqrt(a**2 + b**2)*atan((e* 
*(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*a*b**3*d*e*f*i + 4*e**(2*c + 2*d* 
x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*a*b* 
*3*f**2*i + 8*e**(c + d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i) 
/sqrt(a**2 + b**2))*a**4*f**2*i + 8*e**(c + d*x)*sqrt(a**2 + b**2)*atan((e 
**(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*a**2*b**2*d*e*f*i + 8*e**(c + d* 
x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*a**2 
*b**2*f**2*i - 4*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a**2 
 + b**2))*a**3*b*f**2*i - 4*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i 
)/sqrt(a**2 + b**2))*a*b**3*d*e*f*i - 4*sqrt(a**2 + b**2)*atan((e**(c + d* 
x)*b*i + a*i)/sqrt(a**2 + b**2))*a*b**3*f**2*i - 8*e**(3*c + 2*d*x)*int((e 
**(d*x)*x)/(e**(4*c + 4*d*x)*b**2 + 4*e**(3*c + 3*d*x)*a*b + 4*e**(2*c + 2 
*d*x)*a**2 - 2*e**(2*c + 2*d*x)*b**2 - 4*e**(c + d*x)*a*b + b**2),x)*a**5* 
b**2*d**2*f**2 - 16*e**(3*c + 2*d*x)*int((e**(d*x)*x)/(e**(4*c + 4*d*x)*b* 
*2 + 4*e**(3*c + 3*d*x)*a*b + 4*e**(2*c + 2*d*x)*a**2 - 2*e**(2*c + 2*d*x) 
*b**2 - 4*e**(c + d*x)*a*b + b**2),x)*a**3*b**4*d**2*f**2 - 8*e**(3*c + 2* 
d*x)*int((e**(d*x)*x)/(e**(4*c + 4*d*x)*b**2 + 4*e**(3*c + 3*d*x)*a*b + 4* 
e**(2*c + 2*d*x)*a**2 - 2*e**(2*c + 2*d*x)*b**2 - 4*e**(c + d*x)*a*b + b** 
2),x)*a*b**6*d**2*f**2 + 2*e**(2*c + 2*d*x)*log(e**(2*c + 2*d*x)*b + 2*...