\(\int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx\) [6]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F(-2)]
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 78 \[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx=-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}+\frac {\left (a^2+3 a b+3 b^2\right ) \cosh (x)}{b^3}-\frac {(a+3 b) \cosh ^3(x)}{3 b^2}+\frac {\cosh ^5(x)}{5 b} \] Output:

-(a+b)^3*arctan(b^(1/2)*cosh(x)/a^(1/2))/a^(1/2)/b^(7/2)+(a^2+3*a*b+3*b^2) 
*cosh(x)/b^3-1/3*(a+3*b)*cosh(x)^3/b^2+1/5*cosh(x)^5/b
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.90 \[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx=-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b}-i \sqrt {a+b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b}+i \sqrt {a+b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}+\frac {\left (8 a^2+22 a b+19 b^2\right ) \cosh (x)}{8 b^3}-\frac {(4 a+9 b) \cosh (3 x)}{48 b^2}+\frac {\cosh (5 x)}{80 b} \] Input:

Integrate[Sinh[x]^7/(a + b*Cosh[x]^2),x]
 

Output:

-(((a + b)^3*ArcTan[(Sqrt[b] - I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]])/(Sqrt[a] 
*b^(7/2))) - ((a + b)^3*ArcTan[(Sqrt[b] + I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a] 
])/(Sqrt[a]*b^(7/2)) + ((8*a^2 + 22*a*b + 19*b^2)*Cosh[x])/(8*b^3) - ((4*a 
 + 9*b)*Cosh[3*x])/(48*b^2) + Cosh[5*x]/(80*b)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 3669, 300, 2009}

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.

\(\displaystyle \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \cos \left (\frac {\pi }{2}+i x\right )^7}{a+b \sin \left (\frac {\pi }{2}+i x\right )^2}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\cos \left (i x+\frac {\pi }{2}\right )^7}{b \sin \left (i x+\frac {\pi }{2}\right )^2+a}dx\)

\(\Big \downarrow \) 3669

\(\displaystyle -\int \frac {\left (1-\cosh ^2(x)\right )^3}{b \cosh ^2(x)+a}d\cosh (x)\)

\(\Big \downarrow \) 300

\(\displaystyle -\int \left (-\frac {\cosh ^4(x)}{b}+\frac {(a+3 b) \cosh ^2(x)}{b^2}-\frac {a^2+3 b a+3 b^2}{b^3}+\frac {a^3+3 b a^2+3 b^2 a+b^3}{b^3 \left (b \cosh ^2(x)+a\right )}\right )d\cosh (x)\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (a^2+3 a b+3 b^2\right ) \cosh (x)}{b^3}-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}-\frac {(a+3 b) \cosh ^3(x)}{3 b^2}+\frac {\cosh ^5(x)}{5 b}\)

Input:

Int[Sinh[x]^7/(a + b*Cosh[x]^2),x]
 

Output:

-(((a + b)^3*ArcTan[(Sqrt[b]*Cosh[x])/Sqrt[a]])/(Sqrt[a]*b^(7/2))) + ((a^2 
 + 3*a*b + 3*b^2)*Cosh[x])/b^3 - ((a + 3*b)*Cosh[x]^3)/(3*b^2) + Cosh[x]^5 
/(5*b)
 

Defintions of rubi rules used

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 300
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3669
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   S 
ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] 
/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 
Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21

\[\frac {\frac {\cosh \left (x \right )^{5} b^{2}}{5}-\frac {a b \cosh \left (x \right )^{3}}{3}-b^{2} \cosh \left (x \right )^{3}+a^{2} \cosh \left (x \right )+3 a b \cosh \left (x \right )+3 b^{2} \cosh \left (x \right )}{b^{3}}+\frac {\left (-a^{3}-3 a^{2} b -3 b^{2} a -b^{3}\right ) \arctan \left (\frac {b \cosh \left (x \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\]

Input:

int(sinh(x)^7/(a+b*cosh(x)^2),x)
 

Output:

1/b^3*(1/5*cosh(x)^5*b^2-1/3*a*b*cosh(x)^3-b^2*cosh(x)^3+a^2*cosh(x)+3*a*b 
*cosh(x)+3*b^2*cosh(x))+(-a^3-3*a^2*b-3*a*b^2-b^3)/b^3/(a*b)^(1/2)*arctan( 
b*cosh(x)/(a*b)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1202 vs. \(2 (66) = 132\).

Time = 0.14 (sec) , antiderivative size = 2349, normalized size of antiderivative = 30.12 \[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(sinh(x)^7/(a+b*cosh(x)^2),x, algorithm="fricas")
 

Output:

[1/480*(3*a*b^3*cosh(x)^10 + 30*a*b^3*cosh(x)*sinh(x)^9 + 3*a*b^3*sinh(x)^ 
10 - 5*(4*a^2*b^2 + 9*a*b^3)*cosh(x)^8 + 5*(27*a*b^3*cosh(x)^2 - 4*a^2*b^2 
 - 9*a*b^3)*sinh(x)^8 + 40*(9*a*b^3*cosh(x)^3 - (4*a^2*b^2 + 9*a*b^3)*cosh 
(x))*sinh(x)^7 + 30*(8*a^3*b + 22*a^2*b^2 + 19*a*b^3)*cosh(x)^6 + 10*(63*a 
*b^3*cosh(x)^4 + 24*a^3*b + 66*a^2*b^2 + 57*a*b^3 - 14*(4*a^2*b^2 + 9*a*b^ 
3)*cosh(x)^2)*sinh(x)^6 + 4*(189*a*b^3*cosh(x)^5 - 70*(4*a^2*b^2 + 9*a*b^3 
)*cosh(x)^3 + 45*(8*a^3*b + 22*a^2*b^2 + 19*a*b^3)*cosh(x))*sinh(x)^5 + 30 
*(8*a^3*b + 22*a^2*b^2 + 19*a*b^3)*cosh(x)^4 + 10*(63*a*b^3*cosh(x)^6 - 35 
*(4*a^2*b^2 + 9*a*b^3)*cosh(x)^4 + 24*a^3*b + 66*a^2*b^2 + 57*a*b^3 + 45*( 
8*a^3*b + 22*a^2*b^2 + 19*a*b^3)*cosh(x)^2)*sinh(x)^4 + 3*a*b^3 + 40*(9*a* 
b^3*cosh(x)^7 - 7*(4*a^2*b^2 + 9*a*b^3)*cosh(x)^5 + 15*(8*a^3*b + 22*a^2*b 
^2 + 19*a*b^3)*cosh(x)^3 + 3*(8*a^3*b + 22*a^2*b^2 + 19*a*b^3)*cosh(x))*si 
nh(x)^3 - 5*(4*a^2*b^2 + 9*a*b^3)*cosh(x)^2 + 5*(27*a*b^3*cosh(x)^8 - 28*( 
4*a^2*b^2 + 9*a*b^3)*cosh(x)^6 + 90*(8*a^3*b + 22*a^2*b^2 + 19*a*b^3)*cosh 
(x)^4 - 4*a^2*b^2 - 9*a*b^3 + 36*(8*a^3*b + 22*a^2*b^2 + 19*a*b^3)*cosh(x) 
^2)*sinh(x)^2 - 240*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)^5 + 5*(a^3 + 
3*a^2*b + 3*a*b^2 + b^3)*cosh(x)^4*sinh(x) + 10*(a^3 + 3*a^2*b + 3*a*b^2 + 
 b^3)*cosh(x)^3*sinh(x)^2 + 10*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)^2*s 
inh(x)^3 + 5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(x)*sinh(x)^4 + (a^3 + 3* 
a^2*b + 3*a*b^2 + b^3)*sinh(x)^5)*sqrt(-a*b)*log((b*cosh(x)^4 + 4*b*cos...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx=\text {Timed out} \] Input:

integrate(sinh(x)**7/(a+b*cosh(x)**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx=\int { \frac {\sinh \left (x\right )^{7}}{b \cosh \left (x\right )^{2} + a} \,d x } \] Input:

integrate(sinh(x)^7/(a+b*cosh(x)^2),x, algorithm="maxima")
 

Output:

1/480*(3*b^2*e^(10*x) + 3*b^2 - 5*(4*a*b + 9*b^2)*e^(8*x) + 30*(8*a^2 + 22 
*a*b + 19*b^2)*e^(6*x) + 30*(8*a^2 + 22*a*b + 19*b^2)*e^(4*x) - 5*(4*a*b + 
 9*b^2)*e^(2*x))*e^(-5*x)/b^3 - 1/128*integrate(256*((a^3 + 3*a^2*b + 3*a* 
b^2 + b^3)*e^(3*x) - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^x)/(b^4*e^(4*x) + b 
^4 + 2*(2*a*b^3 + b^4)*e^(2*x)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sinh(x)^7/(a+b*cosh(x)^2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 2.60 (sec) , antiderivative size = 805, normalized size of antiderivative = 10.32 \[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx=\frac {{\mathrm {e}}^{-5\,x}}{160\,b}+\frac {{\mathrm {e}}^{5\,x}}{160\,b}+\frac {{\mathrm {e}}^{-x}\,\left (8\,a^2+22\,a\,b+19\,b^2\right )}{16\,b^3}-\frac {\left (2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,{\left (a+b\right )}^3\,\sqrt {a\,b^7}}{2\,a\,b^3\,\sqrt {{\left (a+b\right )}^6}}\right )-2\,\mathrm {atan}\left (\frac {2\,{\mathrm {e}}^{3\,x}\,\left (a^7\,\sqrt {a\,b^7}+b^7\,\sqrt {a\,b^7}+7\,a\,b^6\,\sqrt {a\,b^7}+7\,a^6\,b\,\sqrt {a\,b^7}+21\,a^2\,b^5\,\sqrt {a\,b^7}+35\,a^3\,b^4\,\sqrt {a\,b^7}+35\,a^4\,b^3\,\sqrt {a\,b^7}+21\,a^5\,b^2\,\sqrt {a\,b^7}\right )}{a\,b^3\,\sqrt {{\left (a+b\right )}^6}\,\left (4\,a^4+16\,a^3\,b+24\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}+\frac {a\,b^8\,{\mathrm {e}}^x\,\sqrt {a\,b^7}\,\left (\frac {4\,\left (2\,a\,b^7\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}+8\,a^2\,b^6\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}+12\,a^3\,b^5\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}+8\,a^4\,b^4\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}+2\,a^5\,b^3\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}\right )}{a^2\,b^{15}\,{\left (a+b\right )}^3}+\frac {2\,\left (a^7\,\sqrt {a\,b^7}+b^7\,\sqrt {a\,b^7}+7\,a\,b^6\,\sqrt {a\,b^7}+7\,a^6\,b\,\sqrt {a\,b^7}+21\,a^2\,b^5\,\sqrt {a\,b^7}+35\,a^3\,b^4\,\sqrt {a\,b^7}+35\,a^4\,b^3\,\sqrt {a\,b^7}+21\,a^5\,b^2\,\sqrt {a\,b^7}\right )}{a^2\,b^{11}\,\sqrt {a\,b^7}\,\sqrt {{\left (a+b\right )}^6}}\right )}{4\,a^4+16\,a^3\,b+24\,a^2\,b^2+16\,a\,b^3+4\,b^4}\right )\right )\,\sqrt {a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}}{2\,\sqrt {a\,b^7}}-\frac {{\mathrm {e}}^{-3\,x}\,\left (4\,a+9\,b\right )}{96\,b^2}-\frac {{\mathrm {e}}^{3\,x}\,\left (4\,a+9\,b\right )}{96\,b^2}+\frac {{\mathrm {e}}^x\,\left (8\,a^2+22\,a\,b+19\,b^2\right )}{16\,b^3} \] Input:

int(sinh(x)^7/(a + b*cosh(x)^2),x)
 

Output:

exp(-5*x)/(160*b) + exp(5*x)/(160*b) + (exp(-x)*(22*a*b + 8*a^2 + 19*b^2)) 
/(16*b^3) - ((2*atan((exp(x)*(a + b)^3*(a*b^7)^(1/2))/(2*a*b^3*((a + b)^6) 
^(1/2))) - 2*atan((2*exp(3*x)*(a^7*(a*b^7)^(1/2) + b^7*(a*b^7)^(1/2) + 7*a 
*b^6*(a*b^7)^(1/2) + 7*a^6*b*(a*b^7)^(1/2) + 21*a^2*b^5*(a*b^7)^(1/2) + 35 
*a^3*b^4*(a*b^7)^(1/2) + 35*a^4*b^3*(a*b^7)^(1/2) + 21*a^5*b^2*(a*b^7)^(1/ 
2)))/(a*b^3*((a + b)^6)^(1/2)*(16*a*b^3 + 16*a^3*b + 4*a^4 + 4*b^4 + 24*a^ 
2*b^2)) + (a*b^8*exp(x)*(a*b^7)^(1/2)*((4*(2*a*b^7*(6*a*b^5 + 6*a^5*b + a^ 
6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2)^(1/2) + 8*a^2*b^6*(6*a*b^5 
 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2)^(1/2) + 12* 
a^3*b^5*(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4* 
b^2)^(1/2) + 8*a^4*b^4*(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^ 
3*b^3 + 15*a^4*b^2)^(1/2) + 2*a^5*b^3*(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15* 
a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2)^(1/2)))/(a^2*b^15*(a + b)^3) + (2*(a^7* 
(a*b^7)^(1/2) + b^7*(a*b^7)^(1/2) + 7*a*b^6*(a*b^7)^(1/2) + 7*a^6*b*(a*b^7 
)^(1/2) + 21*a^2*b^5*(a*b^7)^(1/2) + 35*a^3*b^4*(a*b^7)^(1/2) + 35*a^4*b^3 
*(a*b^7)^(1/2) + 21*a^5*b^2*(a*b^7)^(1/2)))/(a^2*b^11*(a*b^7)^(1/2)*((a + 
b)^6)^(1/2))))/(16*a*b^3 + 16*a^3*b + 4*a^4 + 4*b^4 + 24*a^2*b^2)))*(6*a*b 
^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2)^(1/2))/(2 
*(a*b^7)^(1/2)) - (exp(-3*x)*(4*a + 9*b))/(96*b^2) - (exp(3*x)*(4*a + 9*b) 
)/(96*b^2) + (exp(x)*(22*a*b + 8*a^2 + 19*b^2))/(16*b^3)
 

Reduce [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 1512, normalized size of antiderivative = 19.38 \[ \int \frac {\sinh ^7(x)}{a+b \cosh ^2(x)} \, dx =\text {Too large to display} \] Input:

int(sinh(x)^7/(a+b*cosh(x)^2),x)
 

Output:

( - 480*e**(5*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 
2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*a* 
*3 - 1440*e**(5*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) 
+ 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))* 
a**2*b - 1440*e**(5*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + 
 b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b 
)))*a*b**2 - 480*e**(5*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt( 
a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a 
+ b)))*b**3 + 480*e**(5*x)*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*a 
tan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*a**4 + 1440* 
e**(5*x)*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt 
(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*a**3*b + 1440*e**(5*x)*sqrt(b) 
*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt( 
a)*sqrt(a + b) + 2*a + b)))*a**2*b**2 + 480*e**(5*x)*sqrt(b)*sqrt(2*sqrt(a 
)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
 + 2*a + b)))*a*b**3 - 240*e**(5*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqr 
t(a)*sqrt(a + b) - 2*a - b)*log( - sqrt(2*sqrt(a)*sqrt(a + b) - 2*a - b) + 
 e**x*sqrt(b))*a**3 - 720*e**(5*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt 
(a)*sqrt(a + b) - 2*a - b)*log( - sqrt(2*sqrt(a)*sqrt(a + b) - 2*a - b) + 
e**x*sqrt(b))*a**2*b - 720*e**(5*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*...