\(\int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx\) [127]

Optimal result

Integrand size = 17, antiderivative size = 207 \[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx=\frac {2}{15} (5 A b+3 a B) \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2 i \left (20 a A b+3 a^2 B-9 b^2 B\right ) E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {a+b \sinh (x)}}{15 b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i \left (a^2+b^2\right ) (5 A b+3 a B) \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{15 b \sqrt {a+b \sinh (x)}} \] Output:

2/15*(5*A*b+3*B*a)*cosh(x)*(a+b*sinh(x))^(1/2)+2/5*B*cosh(x)*(a+b*sinh(x)) 
^(3/2)+2/15*I*(20*A*a*b+3*B*a^2-9*B*b^2)*EllipticE(cos(1/4*Pi+1/2*I*x),2^( 
1/2)*(b/(I*a+b))^(1/2))*(a+b*sinh(x))^(1/2)/b/((a+b*sinh(x))/(a-I*b))^(1/2 
)+2/15*I*(a^2+b^2)*(5*A*b+3*B*a)*InverseJacobiAM(-1/4*Pi+1/2*I*x,2^(1/2)*( 
b/(I*a+b))^(1/2))*((a+b*sinh(x))/(a-I*b))^(1/2)/b/(a+b*sinh(x))^(1/2)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.95 \[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx=\frac {2 \left (\frac {i \left (b \left (15 a^2 A-5 A b^2-12 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),-\frac {2 i b}{a-i b}\right )+\left (20 a A b+3 a^2 B-9 b^2 B\right ) \left ((a-i b) E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )-a \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),-\frac {2 i b}{a-i b}\right )\right )\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{b}+\cosh (x) (a+b \sinh (x)) (5 A b+6 a B+3 b B \sinh (x))\right )}{15 \sqrt {a+b \sinh (x)}} \] Input:

Integrate[(a + b*Sinh[x])^(3/2)*(A + B*Sinh[x]),x]
 

Output:

(2*((I*(b*(15*a^2*A - 5*A*b^2 - 12*a*b*B)*EllipticF[(Pi - (2*I)*x)/4, ((-2 
*I)*b)/(a - I*b)] + (20*a*A*b + 3*a^2*B - 9*b^2*B)*((a - I*b)*EllipticE[(P 
i - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)] - a*EllipticF[(Pi - (2*I)*x)/4, ((-2 
*I)*b)/(a - I*b)]))*Sqrt[(a + b*Sinh[x])/(a - I*b)])/b + Cosh[x]*(a + b*Si 
nh[x])*(5*A*b + 6*a*B + 3*b*B*Sinh[x])))/(15*Sqrt[a + b*Sinh[x]])
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.882, Rules used = {3042, 3232, 27, 3042, 3232, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

\(\displaystyle \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a-i b \sin (i x))^{3/2} (A-i B \sin (i x))dx\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sinh (x)} (5 a A-3 b B+(5 A b+3 a B) \sinh (x))dx+\frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \sqrt {a+b \sinh (x)} (5 a A-3 b B+(5 A b+3 a B) \sinh (x))dx+\frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \int \sqrt {a-i b \sin (i x)} (5 a A-3 b B-i (5 A b+3 a B) \sin (i x))dx\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {15 A a^2-12 b B a-5 A b^2+\left (3 B a^2+20 A b a-9 b^2 B\right ) \sinh (x)}{2 \sqrt {a+b \sinh (x)}}dx+\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}\right )+\frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 A a^2-12 b B a-5 A b^2+\left (3 B a^2+20 A b a-9 b^2 B\right ) \sinh (x)}{\sqrt {a+b \sinh (x)}}dx+\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}\right )+\frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \int \frac {15 A a^2-12 b B a-5 A b^2-i \left (3 B a^2+20 A b a-9 b^2 B\right ) \sin (i x)}{\sqrt {a-i b \sin (i x)}}dx\right )\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (3 a^2 B+20 a A b-9 b^2 B\right ) \int \sqrt {a+b \sinh (x)}dx}{b}-\frac {\left (a^2+b^2\right ) (3 a B+5 A b) \int \frac {1}{\sqrt {a+b \sinh (x)}}dx}{b}\right )+\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}\right )+\frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {\left (3 a^2 B+20 a A b-9 b^2 B\right ) \int \sqrt {a-i b \sin (i x)}dx}{b}-\frac {\left (a^2+b^2\right ) (3 a B+5 A b) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {\left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}dx}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) (3 a B+5 A b) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {\left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}dx}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) (3 a B+5 A b) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) (3 a B+5 A b) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) (3 a B+5 A b) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}}dx}{b \sqrt {a+b \sinh (x)}}\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) (3 a B+5 A b) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}}dx}{b \sqrt {a+b \sinh (x)}}\right )\right )\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {2}{5} B \cosh (x) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) (3 a B+5 A b) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (3 a^2 B+20 a A b-9 b^2 B\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i \left (a^2+b^2\right ) (3 a B+5 A b) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right )}{b \sqrt {a+b \sinh (x)}}\right )\right )\)

Input:

Int[(a + b*Sinh[x])^(3/2)*(A + B*Sinh[x]),x]
 

Output:

(2*B*Cosh[x]*(a + b*Sinh[x])^(3/2))/5 + ((2*(5*A*b + 3*a*B)*Cosh[x]*Sqrt[a 
 + b*Sinh[x]])/3 + (((2*I)*(20*a*A*b + 3*a^2*B - 9*b^2*B)*EllipticE[Pi/4 - 
 (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/(b*Sqrt[(a + b*Sinh[x])/(a 
 - I*b)]) - ((2*I)*(a^2 + b^2)*(5*A*b + 3*a*B)*EllipticF[Pi/4 - (I/2)*x, ( 
2*b)/(I*a + b)]*Sqrt[(a + b*Sinh[x])/(a - I*b)])/(b*Sqrt[a + b*Sinh[x]]))/ 
3)/5
 

Defintions of rubi rules used

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 3042

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

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 3232

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[1/(m + 1)   Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* 
d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ 
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 
 0] && IntegerQ[2*m]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1032 vs. \(2 (182 ) = 364\).

Time = 2.50 (sec) , antiderivative size = 1033, normalized size of antiderivative = 4.99

method	result	size
default	\(\text {Expression too large to display}\)	\(1033\)
parts	\(\text {Expression too large to display}\)	\(1489\)

Input:

int((a+b*sinh(x))^(3/2)*(A+B*sinh(x)),x,method=_RETURNVERBOSE)
 

Output:

(cosh(x)^2*(a+b*sinh(x)))^(1/2)*(2*a^2*A*(1/b*a-I)*((-a-b*sinh(x))/(I*b-a) 
)^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*(b*(I+sinh(x))/(I*b-a))^(1/2)/(cosh( 
x)^2*(a+b*sinh(x)))^(1/2)*EllipticF(((-a-b*sinh(x))/(I*b-a))^(1/2),((a-I*b 
)/(I*b+a))^(1/2))+b*(A*b+2*B*a)*(2/3/b*(cosh(x)^2*(a+b*sinh(x)))^(1/2)-2/3 
*(1/b*a-I)*((-a-b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*(b 
*(I+sinh(x))/(I*b-a))^(1/2)/(cosh(x)^2*(a+b*sinh(x)))^(1/2)*EllipticF(((-a 
-b*sinh(x))/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))-4/3/b*a*(1/b*a-I)*((-a 
-b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*(b*(I+sinh(x))/(I 
*b-a))^(1/2)/(cosh(x)^2*(a+b*sinh(x)))^(1/2)*((-1/b*a-I)*EllipticE(((-a-b* 
sinh(x))/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+I*EllipticF(((-a-b*sinh(x 
))/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))))+2*a*(2*A*b+B*a)*(1/b*a-I)*((- 
a-b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*(b*(I+sinh(x))/( 
I*b-a))^(1/2)/(cosh(x)^2*(a+b*sinh(x)))^(1/2)*((-1/b*a-I)*EllipticE(((-a-b 
*sinh(x))/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+I*EllipticF(((-a-b*sinh( 
x))/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2)))+B*b^2*(2/5/b*sinh(x)*(cosh(x) 
^2*(a+b*sinh(x)))^(1/2)-8/15*a/b^2*(cosh(x)^2*(a+b*sinh(x)))^(1/2)-4/15/b* 
a*(1/b*a-I)*((-a-b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*( 
b*(I+sinh(x))/(I*b-a))^(1/2)/(cosh(x)^2*(a+b*sinh(x)))^(1/2)*EllipticF(((- 
a-b*sinh(x))/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+2*(-3/5+8/15*a^2/b^2) 
*(1/b*a-I)*((-a-b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 623 vs. \(2 (178) = 356\).

Time = 0.12 (sec) , antiderivative size = 623, normalized size of antiderivative = 3.01 \[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx =\text {Too large to display} \] Input:

integrate((a+b*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="fricas")
 

Output:

-1/90*(8*sqrt(1/2)*((6*B*a^3 - 5*A*a^2*b + 18*B*a*b^2 + 15*A*b^3)*cosh(x)^ 
2 + 2*(6*B*a^3 - 5*A*a^2*b + 18*B*a*b^2 + 15*A*b^3)*cosh(x)*sinh(x) + (6*B 
*a^3 - 5*A*a^2*b + 18*B*a*b^2 + 15*A*b^3)*sinh(x)^2)*sqrt(b)*weierstrassPI 
nverse(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, 1/3*(3*b*cosh 
(x) + 3*b*sinh(x) + 2*a)/b) + 24*sqrt(1/2)*((3*B*a^2*b + 20*A*a*b^2 - 9*B* 
b^3)*cosh(x)^2 + 2*(3*B*a^2*b + 20*A*a*b^2 - 9*B*b^3)*cosh(x)*sinh(x) + (3 
*B*a^2*b + 20*A*a*b^2 - 9*B*b^3)*sinh(x)^2)*sqrt(b)*weierstrassZeta(4/3*(4 
*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4 
*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, 1/3*(3*b*cosh(x) + 3*b*sin 
h(x) + 2*a)/b)) - 3*(3*B*b^3*cosh(x)^4 + 3*B*b^3*sinh(x)^4 - 3*B*b^3 + 2*( 
6*B*a*b^2 + 5*A*b^3)*cosh(x)^3 + 2*(6*B*b^3*cosh(x) + 6*B*a*b^2 + 5*A*b^3) 
*sinh(x)^3 - 4*(3*B*a^2*b + 20*A*a*b^2 - 9*B*b^3)*cosh(x)^2 + 2*(9*B*b^3*c 
osh(x)^2 - 6*B*a^2*b - 40*A*a*b^2 + 18*B*b^3 + 3*(6*B*a*b^2 + 5*A*b^3)*cos 
h(x))*sinh(x)^2 + 2*(6*B*a*b^2 + 5*A*b^3)*cosh(x) + 2*(6*B*b^3*cosh(x)^3 + 
 6*B*a*b^2 + 5*A*b^3 + 3*(6*B*a*b^2 + 5*A*b^3)*cosh(x)^2 - 4*(3*B*a^2*b + 
20*A*a*b^2 - 9*B*b^3)*cosh(x))*sinh(x))*sqrt(b*sinh(x) + a))/(b^2*cosh(x)^ 
2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2)
 

Sympy [F]

\[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx=\int \left (A + B \sinh {\left (x \right )}\right ) \left (a + b \sinh {\left (x \right )}\right )^{\frac {3}{2}}\, dx \] Input:

integrate((a+b*sinh(x))**(3/2)*(A+B*sinh(x)),x)
 

Output:

Integral((A + B*sinh(x))*(a + b*sinh(x))**(3/2), x)
 

Maxima [F]

\[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} {\left (b \sinh \left (x\right ) + a\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((a+b*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="maxima")
 

Output:

integrate((B*sinh(x) + A)*(b*sinh(x) + a)^(3/2), x)
 

Giac [F]

\[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} {\left (b \sinh \left (x\right ) + a\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((a+b*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="giac")
 

Output:

integrate((B*sinh(x) + A)*(b*sinh(x) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx=\int \left (A+B\,\mathrm {sinh}\left (x\right )\right )\,{\left (a+b\,\mathrm {sinh}\left (x\right )\right )}^{3/2} \,d x \] Input:

int((A + B*sinh(x))*(a + b*sinh(x))^(3/2),x)
 

Output:

int((A + B*sinh(x))*(a + b*sinh(x))^(3/2), x)
 

Reduce [F]

\[ \int (a+b \sinh (x))^{3/2} (A+B \sinh (x)) \, dx=\left (\int \sqrt {\sinh \left (x \right ) b +a}d x \right ) a^{2}+\left (\int \sqrt {\sinh \left (x \right ) b +a}\, \sinh \left (x \right )^{2}d x \right ) b^{2}+2 \left (\int \sqrt {\sinh \left (x \right ) b +a}\, \sinh \left (x \right )d x \right ) a b \] Input:

int((a+b*sinh(x))^(3/2)*(A+B*sinh(x)),x)
 

Output:

int(sqrt(sinh(x)*b + a),x)*a**2 + int(sqrt(sinh(x)*b + a)*sinh(x)**2,x)*b* 
*2 + 2*int(sqrt(sinh(x)*b + a)*sinh(x),x)*a*b