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

Optimal result

Integrand size = 10, antiderivative size = 103 \[ \int \sinh ^{\frac {7}{2}}(a+b x) \, dx=-\frac {10 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{21 b \sqrt {\sinh (a+b x)}}-\frac {10 \cosh (a+b x) \sqrt {\sinh (a+b x)}}{21 b}+\frac {2 \cosh (a+b x) \sinh ^{\frac {5}{2}}(a+b x)}{7 b} \] Output:

-10/21*I*InverseJacobiAM(1/2*I*a-1/4*Pi+1/2*I*b*x,2^(1/2))*(I*sinh(b*x+a)) 
^(1/2)/b/sinh(b*x+a)^(1/2)-10/21*cosh(b*x+a)*sinh(b*x+a)^(1/2)/b+2/7*cosh( 
b*x+a)*sinh(b*x+a)^(5/2)/b
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.73 \[ \int \sinh ^{\frac {7}{2}}(a+b x) \, dx=\frac {40 i \operatorname {EllipticF}\left (\frac {1}{4} (-2 i a+\pi -2 i b x),2\right ) \sqrt {i \sinh (a+b x)}-26 \sinh (2 (a+b x))+3 \sinh (4 (a+b x))}{84 b \sqrt {\sinh (a+b x)}} \] Input:

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

Output:

((40*I)*EllipticF[((-2*I)*a + Pi - (2*I)*b*x)/4, 2]*Sqrt[I*Sinh[a + b*x]] 
- 26*Sinh[2*(a + b*x)] + 3*Sinh[4*(a + b*x)])/(84*b*Sqrt[Sinh[a + b*x]])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3115, 3042, 3115, 3042, 3121, 3042, 3120}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (-i \sin (i a+i b x))^{7/2}dx\)

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

\(\displaystyle \frac {2 \sinh ^{\frac {5}{2}}(a+b x) \cosh (a+b x)}{7 b}-\frac {5}{7} \left (\frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}-\frac {1}{3} \int \frac {1}{\sqrt {\sinh (a+b x)}}dx\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

\(\displaystyle \frac {2 \sinh ^{\frac {5}{2}}(a+b x) \cosh (a+b x)}{7 b}-\frac {5}{7} \left (\frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {i \sinh (a+b x)}}dx}{3 \sqrt {\sinh (a+b x)}}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 \sinh ^{\frac {5}{2}}(a+b x) \cosh (a+b x)}{7 b}-\frac {5}{7} \left (\frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}+\frac {2 i \sqrt {i \sinh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{3 b \sqrt {\sinh (a+b x)}}\right )\)

Input:

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

Output:

(-5*((((2*I)/3)*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]] 
)/(b*Sqrt[Sinh[a + b*x]]) + (2*Cosh[a + b*x]*Sqrt[Sinh[a + b*x]])/(3*b)))/ 
7 + (2*Cosh[a + b*x]*Sinh[a + b*x]^(5/2))/(7*b)
 

Defintions of rubi rules used

rule 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {\frac {5 i \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )}{21}+\frac {2 \cosh \left (b x +a \right )^{4} \sinh \left (b x +a \right )}{7}-\frac {16 \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{21}}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\)	\(116\)

Input:

int(sinh(b*x+a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

(5/21*I*(1-I*sinh(b*x+a))^(1/2)*2^(1/2)*(1+I*sinh(b*x+a))^(1/2)*(I*sinh(b* 
x+a))^(1/2)*EllipticF((1-I*sinh(b*x+a))^(1/2),1/2*2^(1/2))+2/7*cosh(b*x+a) 
^4*sinh(b*x+a)-16/21*cosh(b*x+a)^2*sinh(b*x+a))/cosh(b*x+a)/sinh(b*x+a)^(1 
/2)/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (77) = 154\).

Time = 0.09 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.17 \[ \int \sinh ^{\frac {7}{2}}(a+b x) \, dx=\frac {40 \, {\left (\sqrt {2} \cosh \left (b x + a\right )^{3} + 3 \, \sqrt {2} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sqrt {2} \sinh \left (b x + a\right )^{3}\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + {\left (3 \, \cosh \left (b x + a\right )^{6} + 18 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + 3 \, \sinh \left (b x + a\right )^{6} + {\left (45 \, \cosh \left (b x + a\right )^{2} - 23\right )} \sinh \left (b x + a\right )^{4} - 23 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (15 \, \cosh \left (b x + a\right )^{3} - 23 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + {\left (45 \, \cosh \left (b x + a\right )^{4} - 138 \, \cosh \left (b x + a\right )^{2} - 23\right )} \sinh \left (b x + a\right )^{2} - 23 \, \cosh \left (b x + a\right )^{2} + 2 \, {\left (9 \, \cosh \left (b x + a\right )^{5} - 46 \, \cosh \left (b x + a\right )^{3} - 23 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 3\right )} \sqrt {\sinh \left (b x + a\right )}}{84 \, {\left (b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b \sinh \left (b x + a\right )^{3}\right )}} \] Input:

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

Output:

1/84*(40*(sqrt(2)*cosh(b*x + a)^3 + 3*sqrt(2)*cosh(b*x + a)^2*sinh(b*x + a 
) + 3*sqrt(2)*cosh(b*x + a)*sinh(b*x + a)^2 + sqrt(2)*sinh(b*x + a)^3)*wei 
erstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a)) + (3*cosh(b*x + a)^6 
 + 18*cosh(b*x + a)*sinh(b*x + a)^5 + 3*sinh(b*x + a)^6 + (45*cosh(b*x + a 
)^2 - 23)*sinh(b*x + a)^4 - 23*cosh(b*x + a)^4 + 4*(15*cosh(b*x + a)^3 - 2 
3*cosh(b*x + a))*sinh(b*x + a)^3 + (45*cosh(b*x + a)^4 - 138*cosh(b*x + a) 
^2 - 23)*sinh(b*x + a)^2 - 23*cosh(b*x + a)^2 + 2*(9*cosh(b*x + a)^5 - 46* 
cosh(b*x + a)^3 - 23*cosh(b*x + a))*sinh(b*x + a) + 3)*sqrt(sinh(b*x + a)) 
)/(b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a)^2*sinh(b*x + a) + 3*b*cosh(b*x + 
a)*sinh(b*x + a)^2 + b*sinh(b*x + a)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sinh ^{\frac {7}{2}}(a+b x) \, dx=\int \sqrt {\sinh \left (b x +a \right )}\, \sinh \left (b x +a \right )^{3}d x \] Input:

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

Output:

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