\(\int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx\) [126]

Optimal result

Integrand size = 17, antiderivative size = 259 \[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\frac {2}{105} \left (56 a A b+15 a^2 B-25 b^2 B\right ) \cosh (x) \sqrt {a+b \sinh (x)}+\frac {2}{35} (7 A b+5 a B) \cosh (x) (a+b \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {2 i \left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {a+b \sinh (x)}}{105 b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i \left (a^2+b^2\right ) \left (56 a A b+15 a^2 B-25 b^2 B\right ) \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{105 b \sqrt {a+b \sinh (x)}} \] Output:

2/105*(56*A*a*b+15*B*a^2-25*B*b^2)*cosh(x)*(a+b*sinh(x))^(1/2)+2/35*(7*A*b 
+5*B*a)*cosh(x)*(a+b*sinh(x))^(3/2)+2/7*B*cosh(x)*(a+b*sinh(x))^(5/2)+2/10 
5*I*(161*A*a^2*b-63*A*b^3+15*B*a^3-145*B*a*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)+2/105*I*(a^2+b^2)*(56*A*a*b+15*B*a^2-25*B*b^2)*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 = 0.59 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.93 \[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\frac {\frac {2 i \left (b \left (105 a^3 A-119 a A b^2-135 a^2 b B+25 b^3 B\right ) \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),-\frac {2 i b}{a-i b}\right )+\left (161 a^2 A b-63 A b^3+15 a^3 B-145 a b^2 B\right ) \left ((a-i b) E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )-a \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),-\frac {2 i b}{a-i b}\right )\right )\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{b}+\cosh (x) (a+b \sinh (x)) \left (154 a A b+90 a^2 B-65 b^2 B+15 b^2 B \cosh (2 x)+6 b (7 A b+15 a B) \sinh (x)\right )}{105 \sqrt {a+b \sinh (x)}} \] Input:

Integrate[(a + b*Sinh[x])^(5/2)*(A + B*Sinh[x]),x]
 

Output:

(((2*I)*(b*(105*a^3*A - 119*a*A*b^2 - 135*a^2*b*B + 25*b^3*B)*EllipticF[(P 
i - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)] + (161*a^2*A*b - 63*A*b^3 + 15*a^3*B 
 - 145*a*b^2*B)*((a - I*b)*EllipticE[(Pi - (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*Sin 
h[x])/(a - I*b)])/b + Cosh[x]*(a + b*Sinh[x])*(154*a*A*b + 90*a^2*B - 65*b 
^2*B + 15*b^2*B*Cosh[2*x] + 6*b*(7*A*b + 15*a*B)*Sinh[x]))/(105*Sqrt[a + b 
*Sinh[x]])
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.04, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.059, Rules used = {3042, 3232, 27, 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.

\(\displaystyle \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a-i b \sin (i x))^{5/2} (A-i B \sin (i x))dx\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {2}{7} \int \frac {1}{2} (a+b \sinh (x))^{3/2} (7 a A-5 b B+(7 A b+5 a B) \sinh (x))dx+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int (a+b \sinh (x))^{3/2} (7 a A-5 b B+(7 A b+5 a B) \sinh (x))dx+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \int (a-i b \sin (i x))^{3/2} (7 a A-5 b B-i (7 A b+5 a B) \sin (i x))dx\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sinh (x)} \left (35 A a^2-40 b B a-21 A b^2+\left (15 B a^2+56 A b a-25 b^2 B\right ) \sinh (x)\right )dx+\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {a+b \sinh (x)} \left (35 A a^2-40 b B a-21 A b^2+\left (15 B a^2+56 A b a-25 b^2 B\right ) \sinh (x)\right )dx+\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \int \sqrt {a-i b \sin (i x)} \left (35 A a^2-40 b B a-21 A b^2-i \left (15 B a^2+56 A b a-25 b^2 B\right ) \sin (i x)\right )dx\right )\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {105 A a^3-135 b B a^2-119 A b^2 a+25 b^3 B+\left (15 B a^3+161 A b a^2-145 b^2 B a-63 A b^3\right ) \sinh (x)}{2 \sqrt {a+b \sinh (x)}}dx+\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}\right )+\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {105 A a^3-135 b B a^2-119 A b^2 a+25 b^3 B+\left (15 B a^3+161 A b a^2-145 b^2 B a-63 A b^3\right ) \sinh (x)}{\sqrt {a+b \sinh (x)}}dx+\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}\right )+\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \int \frac {105 A a^3-135 b B a^2-119 A b^2 a+25 b^3 B-i \left (15 B a^3+161 A b a^2-145 b^2 B a-63 A b^3\right ) \sin (i x)}{\sqrt {a-i b \sin (i x)}}dx\right )\right )\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {\left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \int \sqrt {a+b \sinh (x)}dx}{b}-\frac {\left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \int \frac {1}{\sqrt {a+b \sinh (x)}}dx}{b}\right )+\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}\right )+\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {\left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \int \sqrt {a-i b \sin (i x)}dx}{b}-\frac {\left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\right )\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {\left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}dx}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {\left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}dx}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\right )\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{b}\right )\right )\right )\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}}dx}{b \sqrt {a+b \sinh (x)}}\right )\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}}dx}{b \sqrt {a+b \sinh (x)}}\right )\right )\right )\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {2}{7} B \cosh (x) (a+b \sinh (x))^{5/2}+\frac {1}{7} \left (\frac {2}{5} \cosh (x) (5 a B+7 A b) (a+b \sinh (x))^{3/2}+\frac {1}{5} \left (\frac {2}{3} \cosh (x) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {a+b \sinh (x)}+\frac {1}{3} \left (\frac {2 i \left (15 a^3 B+161 a^2 A b-145 a b^2 B-63 A b^3\right ) \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i \left (a^2+b^2\right ) \left (15 a^2 B+56 a A b-25 b^2 B\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right )}{b \sqrt {a+b \sinh (x)}}\right )\right )\right )\)

Input:

Int[(a + b*Sinh[x])^(5/2)*(A + B*Sinh[x]),x]
 

Output:

(2*B*Cosh[x]*(a + b*Sinh[x])^(5/2))/7 + ((2*(7*A*b + 5*a*B)*Cosh[x]*(a + b 
*Sinh[x])^(3/2))/5 + ((2*(56*a*A*b + 15*a^2*B - 25*b^2*B)*Cosh[x]*Sqrt[a + 
 b*Sinh[x]])/3 + (((2*I)*(161*a^2*A*b - 63*A*b^3 + 15*a^3*B - 145*a*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)*(56*a*A*b + 15*a^2*B - 25* 
b^2*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)/7
 

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. 1877 vs. \(2 (230 ) = 460\).

Time = 4.02 (sec) , antiderivative size = 1878, normalized size of antiderivative = 7.25

method	result	size
parts	\(\text {Expression too large to display}\)	\(1878\)
default	\(\text {Expression too large to display}\)	\(1893\)

Input:

int((a+b*sinh(x))^(5/2)*(A+B*sinh(x)),x,method=_RETURNVERBOSE)
 

Output:

2/15*A*(8*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)*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I 
*b-a)/(I*b+a))^(1/2))*a^3*b+8*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)*EllipticF((-(a+b*sinh(x)) 
/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a*b^3+15*(-(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)*Ellipt 
icF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^4+6*(-(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)*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2 
))*a^2*b^2-9*(-(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)*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-( 
I*b-a)/(I*b+a))^(1/2))*b^4-23*(-(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)*EllipticE((-(a+b*sinh(x))/( 
I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^4-14*(-(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)*EllipticE( 
(-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^2*b^2+9*(-(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)*EllipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2 
))*b^4+3*b^4*sinh(x)^4+14*a*b^3*sinh(x)^3+11*a^2*b^2*sinh(x)^2+3*b^4*sinh( 
x)^2+14*a*b^3*sinh(x)+11*a^2*b^2)/b/cosh(x)/(a+b*sinh(x))^(1/2)+2/21*B*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1121 vs. \(2 (226) = 452\).

Time = 0.13 (sec) , antiderivative size = 1121, normalized size of antiderivative = 4.33 \[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sinh(x))^(5/2)*(A+B*sinh(x)),x, algorithm="fricas")
 

Output:

-1/1260*(16*sqrt(1/2)*((30*B*a^4 + 7*A*a^3*b + 115*B*a^2*b^2 + 231*A*a*b^3 
 - 75*B*b^4)*cosh(x)^3 + 3*(30*B*a^4 + 7*A*a^3*b + 115*B*a^2*b^2 + 231*A*a 
*b^3 - 75*B*b^4)*cosh(x)^2*sinh(x) + 3*(30*B*a^4 + 7*A*a^3*b + 115*B*a^2*b 
^2 + 231*A*a*b^3 - 75*B*b^4)*cosh(x)*sinh(x)^2 + (30*B*a^4 + 7*A*a^3*b + 1 
15*B*a^2*b^2 + 231*A*a*b^3 - 75*B*b^4)*sinh(x)^3)*sqrt(b)*weierstrassPInve 
rse(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) + 48*sqrt(1/2)*((15*B*a^3*b + 161*A*a^2*b^2 - 145 
*B*a*b^3 - 63*A*b^4)*cosh(x)^3 + 3*(15*B*a^3*b + 161*A*a^2*b^2 - 145*B*a*b 
^3 - 63*A*b^4)*cosh(x)^2*sinh(x) + 3*(15*B*a^3*b + 161*A*a^2*b^2 - 145*B*a 
*b^3 - 63*A*b^4)*cosh(x)*sinh(x)^2 + (15*B*a^3*b + 161*A*a^2*b^2 - 145*B*a 
*b^3 - 63*A*b^4)*sinh(x)^3)*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*sinh(x) + 2*a)/b)) 
- 3*(15*B*b^4*cosh(x)^6 + 15*B*b^4*sinh(x)^6 + 6*(15*B*a*b^3 + 7*A*b^4)*co 
sh(x)^5 + 6*(15*B*b^4*cosh(x) + 15*B*a*b^3 + 7*A*b^4)*sinh(x)^5 + 15*B*b^4 
 + (180*B*a^2*b^2 + 308*A*a*b^3 - 115*B*b^4)*cosh(x)^4 + (225*B*b^4*cosh(x 
)^2 + 180*B*a^2*b^2 + 308*A*a*b^3 - 115*B*b^4 + 30*(15*B*a*b^3 + 7*A*b^4)* 
cosh(x))*sinh(x)^4 - 8*(15*B*a^3*b + 161*A*a^2*b^2 - 145*B*a*b^3 - 63*A*b^ 
4)*cosh(x)^3 + 4*(75*B*b^4*cosh(x)^3 - 30*B*a^3*b - 322*A*a^2*b^2 + 290*B* 
a*b^3 + 126*A*b^4 + 15*(15*B*a*b^3 + 7*A*b^4)*cosh(x)^2 + (180*B*a^2*b^...
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\text {Timed out} \] Input:

integrate((a+b*sinh(x))**(5/2)*(A+B*sinh(x)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} {\left (b \sinh \left (x\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+b*sinh(x))^(5/2)*(A+B*sinh(x)),x, algorithm="maxima")
 

Output:

integrate((B*sinh(x) + A)*(b*sinh(x) + a)^(5/2), x)
 

Giac [F]

\[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} {\left (b \sinh \left (x\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+b*sinh(x))^(5/2)*(A+B*sinh(x)),x, algorithm="giac")
 

Output:

integrate((B*sinh(x) + A)*(b*sinh(x) + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\int \left (A+B\,\mathrm {sinh}\left (x\right )\right )\,{\left (a+b\,\mathrm {sinh}\left (x\right )\right )}^{5/2} \,d x \] Input:

int((A + B*sinh(x))*(a + b*sinh(x))^(5/2),x)
 

Output:

int((A + B*sinh(x))*(a + b*sinh(x))^(5/2), x)
 

Reduce [F]

\[ \int (a+b \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\left (\int \sqrt {\sinh \left (x \right ) b +a}d x \right ) a^{3}+\left (\int \sqrt {\sinh \left (x \right ) b +a}\, \sinh \left (x \right )^{3}d x \right ) b^{3}+3 \left (\int \sqrt {\sinh \left (x \right ) b +a}\, \sinh \left (x \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sqrt {\sinh \left (x \right ) b +a}\, \sinh \left (x \right )d x \right ) a^{2} b \] Input:

int((a+b*sinh(x))^(5/2)*(A+B*sinh(x)),x)
 

Output:

int(sqrt(sinh(x)*b + a),x)*a**3 + int(sqrt(sinh(x)*b + a)*sinh(x)**3,x)*b* 
*3 + 3*int(sqrt(sinh(x)*b + a)*sinh(x)**2,x)*a*b**2 + 3*int(sqrt(sinh(x)*b 
 + a)*sinh(x),x)*a**2*b