\(\int (b \sinh (c+d x))^{7/2} \, dx\) [15]

Optimal result

Integrand size = 12, antiderivative size = 116 \[ \int (b \sinh (c+d x))^{7/2} \, dx=-\frac {10 i b^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{21 d \sqrt {b \sinh (c+d x)}}-\frac {10 b^3 \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{21 d}+\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{5/2}}{7 d} \] Output:

-10/21*I*b^4*InverseJacobiAM(1/2*I*c-1/4*Pi+1/2*I*d*x,2^(1/2))*(I*sinh(d*x 
+c))^(1/2)/d/(b*sinh(d*x+c))^(1/2)-10/21*b^3*cosh(d*x+c)*(b*sinh(d*x+c))^( 
1/2)/d+2/7*b*cosh(d*x+c)*(b*sinh(d*x+c))^(5/2)/d
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int (b \sinh (c+d x))^{7/2} \, dx=\frac {b^3 \left (-23 \cosh (c+d x)+3 \cosh (3 (c+d x))-\frac {20 \operatorname {EllipticF}\left (\frac {1}{4} (-2 i c+\pi -2 i d x),2\right )}{\sqrt {i \sinh (c+d x)}}\right ) \sqrt {b \sinh (c+d x)}}{42 d} \] Input:

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

Output:

(b^3*(-23*Cosh[c + d*x] + 3*Cosh[3*(c + d*x)] - (20*EllipticF[((-2*I)*c + 
Pi - (2*I)*d*x)/4, 2])/Sqrt[I*Sinh[c + d*x]])*Sqrt[b*Sinh[c + d*x]])/(42*d 
)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, 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.

\(\displaystyle \int (b \sinh (c+d x))^{7/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

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 = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {b^{4} \left (5 i \sqrt {-i \sinh \left (d x +c \right )+1}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-i \sinh \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \cosh \left (d x +c \right )^{4} \sinh \left (d x +c \right )-16 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{2}\right )}{21 \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\)	\(122\)

Input:

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

Output:

1/21*b^4*(5*I*(-I*sinh(d*x+c)+1)^(1/2)*2^(1/2)*(1+I*sinh(d*x+c))^(1/2)*(I* 
sinh(d*x+c))^(1/2)*EllipticF((-I*sinh(d*x+c)+1)^(1/2),1/2*2^(1/2))+6*cosh( 
d*x+c)^4*sinh(d*x+c)-16*sinh(d*x+c)*cosh(d*x+c)^2)/cosh(d*x+c)/(b*sinh(d*x 
+c))^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (90) = 180\).

Time = 0.12 (sec) , antiderivative size = 385, normalized size of antiderivative = 3.32 \[ \int (b \sinh (c+d x))^{7/2} \, dx=\frac {80 \, \sqrt {\frac {1}{2}} {\left (b^{3} \cosh \left (d x + c\right )^{3} + 3 \, b^{3} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + b^{3} \sinh \left (d x + c\right )^{3}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + {\left (3 \, b^{3} \cosh \left (d x + c\right )^{6} + 18 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 3 \, b^{3} \sinh \left (d x + c\right )^{6} - 23 \, b^{3} \cosh \left (d x + c\right )^{4} - 23 \, b^{3} \cosh \left (d x + c\right )^{2} + {\left (45 \, b^{3} \cosh \left (d x + c\right )^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (15 \, b^{3} \cosh \left (d x + c\right )^{3} - 23 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, b^{3} + {\left (45 \, b^{3} \cosh \left (d x + c\right )^{4} - 138 \, b^{3} \cosh \left (d x + c\right )^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (9 \, b^{3} \cosh \left (d x + c\right )^{5} - 46 \, b^{3} \cosh \left (d x + c\right )^{3} - 23 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}}{84 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + d \sinh \left (d x + c\right )^{3}\right )}} \] Input:

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

Output:

1/84*(80*sqrt(1/2)*(b^3*cosh(d*x + c)^3 + 3*b^3*cosh(d*x + c)^2*sinh(d*x + 
 c) + 3*b^3*cosh(d*x + c)*sinh(d*x + c)^2 + b^3*sinh(d*x + c)^3)*sqrt(b)*w 
eierstrassPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c)) + (3*b^3*cosh(d*x 
+ c)^6 + 18*b^3*cosh(d*x + c)*sinh(d*x + c)^5 + 3*b^3*sinh(d*x + c)^6 - 23 
*b^3*cosh(d*x + c)^4 - 23*b^3*cosh(d*x + c)^2 + (45*b^3*cosh(d*x + c)^2 - 
23*b^3)*sinh(d*x + c)^4 + 4*(15*b^3*cosh(d*x + c)^3 - 23*b^3*cosh(d*x + c) 
)*sinh(d*x + c)^3 + 3*b^3 + (45*b^3*cosh(d*x + c)^4 - 138*b^3*cosh(d*x + c 
)^2 - 23*b^3)*sinh(d*x + c)^2 + 2*(9*b^3*cosh(d*x + c)^5 - 46*b^3*cosh(d*x 
 + c)^3 - 23*b^3*cosh(d*x + c))*sinh(d*x + c))*sqrt(b*sinh(d*x + c)))/(d*c 
osh(d*x + c)^3 + 3*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*d*cosh(d*x + c)*sin 
h(d*x + c)^2 + d*sinh(d*x + c)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(b)*int(sqrt(sinh(c + d*x))*sinh(c + d*x)**3,x)*b**3