\(\int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx\) [19]

Optimal result

Integrand size = 31, antiderivative size = 261 \[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx=\frac {3 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{b (f+g)}-\frac {\sqrt {3} e^{\frac {5 (a-d)}{3}} \sqrt [3]{f-g} \arctan \left (\frac {1+\frac {2 e^{\frac {2}{3} (d+b x)} \sqrt [3]{f+g}}{\sqrt [3]{f-g}}}{\sqrt {3}}\right )}{b (f+g)^{4/3}}+\frac {e^{\frac {5 (a-d)}{3}} \sqrt [3]{f-g} \log \left (\sqrt [3]{f-g}-e^{\frac {2}{3} (d+b x)} \sqrt [3]{f+g}\right )}{b (f+g)^{4/3}}-\frac {e^{\frac {5 (a-d)}{3}} \sqrt [3]{f-g} \log \left ((f-g)^{2/3}+e^{\frac {2}{3} (d+b x)} \sqrt [3]{f-g} \sqrt [3]{f+g}+e^{\frac {4}{3} (d+b x)} (f+g)^{2/3}\right )}{2 b (f+g)^{4/3}} \] Output:

3*exp(5/3*a-d+2/3*b*x)/b/(f+g)-3^(1/2)*exp(5/3*a-5/3*d)*(f-g)^(1/3)*arctan 
(1/3*(1+2*exp(2/3*b*x+2/3*d)*(f+g)^(1/3)/(f-g)^(1/3))*3^(1/2))/b/(f+g)^(4/ 
3)+exp(5/3*a-5/3*d)*(f-g)^(1/3)*ln((f-g)^(1/3)-exp(2/3*b*x+2/3*d)*(f+g)^(1 
/3))/b/(f+g)^(4/3)-1/2*exp(5/3*a-5/3*d)*(f-g)^(1/3)*ln((f-g)^(2/3)+exp(2/3 
*b*x+2/3*d)*(f-g)^(1/3)*(f+g)^(1/3)+exp(4/3*b*x+4/3*d)*(f+g)^(2/3))/b/(f+g 
)^(4/3)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx=\frac {e^{\frac {5 (a-d)}{3}} \left (6 e^{\frac {2}{3} (d+b x)} \sqrt [3]{f+g}-2 \sqrt {3} \sqrt [3]{f-g} \arctan \left (\frac {1+\frac {2 e^{\frac {2}{3} (d+b x)} \sqrt [3]{f+g}}{\sqrt [3]{f-g}}}{\sqrt {3}}\right )+2 \sqrt [3]{f-g} \log \left (\sqrt [3]{f-g}-e^{\frac {2}{3} (d+b x)} \sqrt [3]{f+g}\right )-\sqrt [3]{f-g} \log \left ((f-g)^{2/3}+e^{\frac {2}{3} (d+b x)} \sqrt [3]{f-g} \sqrt [3]{f+g}+e^{\frac {4}{3} (d+b x)} (f+g)^{2/3}\right )\right )}{2 b (f+g)^{4/3}} \] Input:

Integrate[E^((5*(a + b*x))/3)/(g*Cosh[d + b*x] + f*Sinh[d + b*x]),x]
 

Output:

(E^((5*(a - d))/3)*(6*E^((2*(d + b*x))/3)*(f + g)^(1/3) - 2*Sqrt[3]*(f - g 
)^(1/3)*ArcTan[(1 + (2*E^((2*(d + b*x))/3)*(f + g)^(1/3))/(f - g)^(1/3))/S 
qrt[3]] + 2*(f - g)^(1/3)*Log[(f - g)^(1/3) - E^((2*(d + b*x))/3)*(f + g)^ 
(1/3)] - (f - g)^(1/3)*Log[(f - g)^(2/3) + E^((2*(d + b*x))/3)*(f - g)^(1/ 
3)*(f + g)^(1/3) + E^((4*(d + b*x))/3)*(f + g)^(2/3)]))/(2*b*(f + g)^(4/3) 
)
 

Rubi [A] (warning: unable to verify)

Time = 0.83 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.82, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {2720, 27, 807, 843, 750, 16, 1142, 27, 1082, 217, 1103}

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.

\(\displaystyle \int \frac {e^{\frac {5}{3} (a+b x)}}{f \sinh (b x+d)+g \cosh (b x+d)} \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {3 \int -\frac {2 e^{\frac {5 a}{3}+\frac {7 b x}{3}}}{f-g-e^{2 b x} (f+g)}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {6 e^{5 a/3} \int \frac {e^{\frac {7 b x}{3}}}{f-g-e^{2 b x} (f+g)}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {3 e^{5 a/3} \int \frac {e^{b x}}{f-g-e^{b x} (f+g)}de^{\frac {2 b x}{3}}}{b}\)

\(\Big \downarrow \) 843

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \int \frac {1}{f-g-e^{b x} (f+g)}de^{\frac {2 b x}{3}}}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

\(\Big \downarrow \) 750

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \left (\frac {\int \frac {1}{\sqrt [3]{f-g}-e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}de^{\frac {2 b x}{3}}}{3 (f-g)^{2/3}}+\frac {\int \frac {2 \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}}{3 (f-g)^{2/3}}\right )}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \left (\frac {\int \frac {2 \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}}{3 (f-g)^{2/3}}-\frac {\log \left (\sqrt [3]{f-g}-e^{\frac {2 b x}{3}} \sqrt [3]{f+g}\right )}{3 (f-g)^{2/3} \sqrt [3]{f+g}}\right )}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

\(\Big \downarrow \) 1142

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \left (\frac {\frac {3}{2} \sqrt [3]{f-g} \int \frac {1}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}+\frac {\int \frac {\sqrt [3]{f+g} \left (\sqrt [3]{f-g}+2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}\right )}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}}{2 \sqrt [3]{f+g}}}{3 (f-g)^{2/3}}-\frac {\log \left (\sqrt [3]{f-g}-e^{\frac {2 b x}{3}} \sqrt [3]{f+g}\right )}{3 (f-g)^{2/3} \sqrt [3]{f+g}}\right )}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \left (\frac {\frac {3}{2} \sqrt [3]{f-g} \int \frac {1}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}+\frac {1}{2} \int \frac {\sqrt [3]{f-g}+2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}}{3 (f-g)^{2/3}}-\frac {\log \left (\sqrt [3]{f-g}-e^{\frac {2 b x}{3}} \sqrt [3]{f+g}\right )}{3 (f-g)^{2/3} \sqrt [3]{f+g}}\right )}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{f-g}+2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}-\frac {3 \int \frac {1}{-\frac {2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{\sqrt [3]{f-g}}-4}d\left (\frac {2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{\sqrt [3]{f-g}}+1\right )}{\sqrt [3]{f+g}}}{3 (f-g)^{2/3}}-\frac {\log \left (\sqrt [3]{f-g}-e^{\frac {2 b x}{3}} \sqrt [3]{f+g}\right )}{3 (f-g)^{2/3} \sqrt [3]{f+g}}\right )}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{f-g}+2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{(f-g)^{2/3}+e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}}de^{\frac {2 b x}{3}}+\frac {\sqrt {3} \arctan \left (\frac {\frac {2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{\sqrt [3]{f-g}}+1}{\sqrt {3}}\right )}{\sqrt [3]{f+g}}}{3 (f-g)^{2/3}}-\frac {\log \left (\sqrt [3]{f-g}-e^{\frac {2 b x}{3}} \sqrt [3]{f+g}\right )}{3 (f-g)^{2/3} \sqrt [3]{f+g}}\right )}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {(f-g) \left (\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 e^{\frac {2 b x}{3}} \sqrt [3]{f+g}}{\sqrt [3]{f-g}}+1}{\sqrt {3}}\right )}{\sqrt [3]{f+g}}+\frac {\log \left (e^{\frac {2 b x}{3}} \sqrt [3]{f+g} \sqrt [3]{f-g}+e^{\frac {2 b x}{3}} (f+g)^{2/3}+(f-g)^{2/3}\right )}{2 \sqrt [3]{f+g}}}{3 (f-g)^{2/3}}-\frac {\log \left (\sqrt [3]{f-g}-e^{\frac {2 b x}{3}} \sqrt [3]{f+g}\right )}{3 (f-g)^{2/3} \sqrt [3]{f+g}}\right )}{f+g}-\frac {e^{\frac {2 b x}{3}}}{f+g}\right )}{b}\)

Input:

Int[E^((5*(a + b*x))/3)/(g*Cosh[d + b*x] + f*Sinh[d + b*x]),x]
 

Output:

(-3*E^((5*a)/3)*(-(E^((2*b*x)/3)/(f + g)) + ((f - g)*(-1/3*Log[(f - g)^(1/ 
3) - E^((2*b*x)/3)*(f + g)^(1/3)]/((f - g)^(2/3)*(f + g)^(1/3)) + ((Sqrt[3 
]*ArcTan[(1 + (2*E^((2*b*x)/3)*(f + g)^(1/3))/(f - g)^(1/3))/Sqrt[3]])/(f 
+ g)^(1/3) + Log[(f - g)^(2/3) + E^((2*b*x)/3)*(f - g)^(1/3)*(f + g)^(1/3) 
 + E^((2*b*x)/3)*(f + g)^(2/3)]/(2*(f + g)^(1/3)))/(3*(f - g)^(2/3))))/(f 
+ g)))/b
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*Rt[a, 3] - 
 Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; 
 FreeQ[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

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]]]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.44

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}}}{b \left (f +g \right )}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (b^{3} f^{4}+4 b^{3} f^{3} g +6 b^{3} f^{2} g^{2}+4 b^{3} f \,g^{3}+b^{3} g^{4}\right ) \textit {\_Z}^{3}-f +g \right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}+\left (-b f -b g \right ) \textit {\_R} \right )\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}\)	\(114\)

Input:

int(exp(5/3*b*x+5/3*a)/(g*cosh(b*x+d)+f*sinh(b*x+d)),x,method=_RETURNVERBO 
SE)
                                                                                    
                                                                                    
 

Output:

3*exp(5/3*a-d+2/3*b*x)/b/(f+g)+sum(_R*ln(exp(2/3*b*x+2/3*d)+(-b*f-b*g)*_R) 
,_R=RootOf((b^3*f^4+4*b^3*f^3*g+6*b^3*f^2*g^2+4*b^3*f*g^3+b^3*g^4)*_Z^3-f+ 
g))*exp(5/3*a-5/3*d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 563 vs. \(2 (202) = 404\).

Time = 0.11 (sec) , antiderivative size = 563, normalized size of antiderivative = 2.16 \[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx =\text {Too large to display} \] Input:

integrate(exp(5/3*b*x+5/3*a)/(g*cosh(b*x+d)+f*sinh(b*x+d)),x, algorithm="f 
ricas")
 

Output:

1/2*(6*cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) + 6*(cosh(-5/3*a + 5/3 
*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 - 6*cosh(1/3*b*x + 1/3 
*d)^2*sinh(-5/3*a + 5/3*d) + 2*(sqrt(3)*cosh(-5/3*a + 5/3*d) - sqrt(3)*sin 
h(-5/3*a + 5/3*d))*((f - g)/(f + g))^(1/3)*arctan(-1/3*(2*(sqrt(3)*(f + g) 
*cosh(1/3*b*x + 1/3*d)^2 + 2*sqrt(3)*(f + g)*cosh(1/3*b*x + 1/3*d)*sinh(1/ 
3*b*x + 1/3*d) + sqrt(3)*(f + g)*sinh(1/3*b*x + 1/3*d)^2)*((f - g)/(f + g) 
)^(2/3) + sqrt(3)*(f - g))/(f - g)) - ((f - g)/(f + g))^(1/3)*(cosh(-5/3*a 
 + 5/3*d) - sinh(-5/3*a + 5/3*d))*log(cosh(1/3*b*x + 1/3*d)^4 + 4*cosh(1/3 
*b*x + 1/3*d)^3*sinh(1/3*b*x + 1/3*d) + 6*cosh(1/3*b*x + 1/3*d)^2*sinh(1/3 
*b*x + 1/3*d)^2 + 4*cosh(1/3*b*x + 1/3*d)*sinh(1/3*b*x + 1/3*d)^3 + sinh(1 
/3*b*x + 1/3*d)^4 + (cosh(1/3*b*x + 1/3*d)^2 + 2*cosh(1/3*b*x + 1/3*d)*sin 
h(1/3*b*x + 1/3*d) + sinh(1/3*b*x + 1/3*d)^2)*((f - g)/(f + g))^(1/3) + (( 
f - g)/(f + g))^(2/3)) + 2*((f - g)/(f + g))^(1/3)*(cosh(-5/3*a + 5/3*d) - 
 sinh(-5/3*a + 5/3*d))*log(cosh(1/3*b*x + 1/3*d)^2 + 2*cosh(1/3*b*x + 1/3* 
d)*sinh(1/3*b*x + 1/3*d) + sinh(1/3*b*x + 1/3*d)^2 - ((f - g)/(f + g))^(1/ 
3)) + 12*(cosh(1/3*b*x + 1/3*d)*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3* 
d)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d))/(b*f + b*g)
 

Sympy [F]

\[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx=e^{\frac {5 a}{3}} \int \frac {e^{\frac {5 b x}{3}}}{f \sinh {\left (b x + d \right )} + g \cosh {\left (b x + d \right )}}\, dx \] Input:

integrate(exp(5/3*b*x+5/3*a)/(g*cosh(b*x+d)+f*sinh(b*x+d)),x)
 

Output:

exp(5*a/3)*Integral(exp(5*b*x/3)/(f*sinh(b*x + d) + g*cosh(b*x + d)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(exp(5/3*b*x+5/3*a)/(g*cosh(b*x+d)+f*sinh(b*x+d)),x, algorithm="m 
axima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(g-f>0)', see `assume?` for more 
details)Is
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx=\frac {{\left (\frac {2 \, {\left (f - g\right )} \left (\frac {f - g}{f e^{\left (2 \, d\right )} + g e^{\left (2 \, d\right )}}\right )^{\frac {1}{3}} \log \left ({\left | -\left (\frac {f - g}{f e^{\left (2 \, d\right )} + g e^{\left (2 \, d\right )}}\right )^{\frac {1}{3}} + e^{\left (\frac {2}{3} \, b x\right )} \right |}\right )}{f^{2} e^{\left (2 \, d\right )} - g^{2} e^{\left (2 \, d\right )}} - \frac {6 \, {\left (f^{3} e^{d} + f^{2} g e^{d} - f g^{2} e^{d} - g^{3} e^{d}\right )}^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (\frac {f - g}{f e^{\left (2 \, d\right )} + g e^{\left (2 \, d\right )}}\right )^{\frac {1}{3}} + 2 \, e^{\left (\frac {2}{3} \, b x\right )}\right )}}{3 \, \left (\frac {f - g}{f e^{\left (2 \, d\right )} + g e^{\left (2 \, d\right )}}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} f^{2} e^{\left (3 \, d\right )} + 2 \, \sqrt {3} f g e^{\left (3 \, d\right )} + \sqrt {3} g^{2} e^{\left (3 \, d\right )}} - \frac {{\left (f^{3} e^{d} + f^{2} g e^{d} - f g^{2} e^{d} - g^{3} e^{d}\right )}^{\frac {1}{3}} \log \left (\left (\frac {f - g}{f e^{\left (2 \, d\right )} + g e^{\left (2 \, d\right )}}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} \, b x\right )} + \left (\frac {f - g}{f e^{\left (2 \, d\right )} + g e^{\left (2 \, d\right )}}\right )^{\frac {2}{3}} + e^{\left (\frac {4}{3} \, b x\right )}\right )}{f^{2} e^{\left (3 \, d\right )} + 2 \, f g e^{\left (3 \, d\right )} + g^{2} e^{\left (3 \, d\right )}} + \frac {6 \, e^{\left (\frac {2}{3} \, b x\right )}}{f e^{\left (2 \, d\right )} + g e^{\left (2 \, d\right )}}\right )} e^{\left (\frac {5}{3} \, a + d\right )}}{2 \, b} \] Input:

integrate(exp(5/3*b*x+5/3*a)/(g*cosh(b*x+d)+f*sinh(b*x+d)),x, algorithm="g 
iac")
 

Output:

1/2*(2*(f - g)*((f - g)/(f*e^(2*d) + g*e^(2*d)))^(1/3)*log(abs(-((f - g)/( 
f*e^(2*d) + g*e^(2*d)))^(1/3) + e^(2/3*b*x)))/(f^2*e^(2*d) - g^2*e^(2*d)) 
- 6*(f^3*e^d + f^2*g*e^d - f*g^2*e^d - g^3*e^d)^(1/3)*arctan(1/3*sqrt(3)*( 
((f - g)/(f*e^(2*d) + g*e^(2*d)))^(1/3) + 2*e^(2/3*b*x))/((f - g)/(f*e^(2* 
d) + g*e^(2*d)))^(1/3))/(sqrt(3)*f^2*e^(3*d) + 2*sqrt(3)*f*g*e^(3*d) + sqr 
t(3)*g^2*e^(3*d)) - (f^3*e^d + f^2*g*e^d - f*g^2*e^d - g^3*e^d)^(1/3)*log( 
((f - g)/(f*e^(2*d) + g*e^(2*d)))^(1/3)*e^(2/3*b*x) + ((f - g)/(f*e^(2*d) 
+ g*e^(2*d)))^(2/3) + e^(4/3*b*x))/(f^2*e^(3*d) + 2*f*g*e^(3*d) + g^2*e^(3 
*d)) + 6*e^(2/3*b*x)/(f*e^(2*d) + g*e^(2*d)))*e^(5/3*a + d)/b
 

Mupad [B] (verification not implemented)

Time = 7.42 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx=\frac {3\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}}{b\,\left (f+g\right )}+\frac {\ln \left (\frac {2\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,{\left (f-g\right )}^{4/3}}{{\left (f+g\right )}^{7/3}}-\frac {2\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}\,\left (f-g\right )}{{\left (f+g\right )}^2}\right )\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,{\left (f-g\right )}^{1/3}}{b\,{\left (f+g\right )}^{4/3}}+\frac {{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-\frac {2\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}\,\left (f-g\right )}{{\left (f+g\right )}^2}+\frac {2\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (f-g\right )}^{4/3}}{{\left (f+g\right )}^{7/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (f-g\right )}^{1/3}}{b\,{\left (f+g\right )}^{4/3}}-\frac {{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-\frac {2\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}\,\left (f-g\right )}{{\left (f+g\right )}^2}-\frac {2\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (f-g\right )}^{4/3}}{{\left (f+g\right )}^{7/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (f-g\right )}^{1/3}}{b\,{\left (f+g\right )}^{4/3}} \] Input:

int(exp((5*a)/3 + (5*b*x)/3)/(g*cosh(d + b*x) + f*sinh(d + b*x)),x)
 

Output:

(3*exp((5*a)/3 - d + (2*b*x)/3))/(b*(f + g)) + (log((2*exp(5*a - 5*d)^(1/3 
)*(f - g)^(4/3))/(f + g)^(7/3) - (2*exp((5*a)/3 - d + (2*b*x)/3)*(f - g))/ 
(f + g)^2)*exp(5*a - 5*d)^(1/3)*(f - g)^(1/3))/(b*(f + g)^(4/3)) + (exp(5* 
a - 5*d)^(1/3)*log((2*exp(5*a - 5*d)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*(f - g)^ 
(4/3))/(f + g)^(7/3) - (2*exp((5*a)/3 - d + (2*b*x)/3)*(f - g))/(f + g)^2) 
*((3^(1/2)*1i)/2 - 1/2)*(f - g)^(1/3))/(b*(f + g)^(4/3)) - (exp(5*a - 5*d) 
^(1/3)*log(- (2*exp((5*a)/3 - d + (2*b*x)/3)*(f - g))/(f + g)^2 - (2*exp(5 
*a - 5*d)^(1/3)*((3^(1/2)*1i)/2 + 1/2)*(f - g)^(4/3))/(f + g)^(7/3))*((3^( 
1/2)*1i)/2 + 1/2)*(f - g)^(1/3))/(b*(f + g)^(4/3))
 

Reduce [F]

\[ \int \frac {e^{\frac {5}{3} (a+b x)}}{g \cosh (d+b x)+f \sinh (d+b x)} \, dx=\int \frac {e^{\frac {5 b x}{3}+\frac {5 a}{3}}}{\cosh \left (b x +d \right ) g +\sinh \left (b x +d \right ) f}d x \] Input:

int(exp(5/3*b*x+5/3*a)/(g*cosh(b*x+d)+f*sinh(b*x+d)),x)
 

Output:

int(e**((5*a + 5*b*x)/3)/(cosh(b*x + d)*g + sinh(b*x + d)*f),x)