Integrand size = 12, antiderivative size = 116 \[ \int (b \sinh (c+d x))^{7/2} \, dx=-\frac {10 i b^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{21 d \sqrt {b \sinh (c+d x)}}-\frac {10 b^3 \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{21 d}+\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{5/2}}{7 d} \]
2/7*b*cosh(d*x+c)*(b*sinh(d*x+c))^(5/2)/d+10/21*I*b^4*(sin(1/2*I*c+1/4*Pi+ 1/2*I*d*x)^2)^(1/2)/sin(1/2*I*c+1/4*Pi+1/2*I*d*x)*EllipticF(cos(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
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int (b \sinh (c+d x))^{7/2} \, dx=\frac {b^3 \left (-23 \cosh (c+d x)+3 \cosh (3 (c+d x))-\frac {20 \operatorname {EllipticF}\left (\frac {1}{4} (-2 i c+\pi -2 i d x),2\right )}{\sqrt {i \sinh (c+d x)}}\right ) \sqrt {b \sinh (c+d x)}}{42 d} \]
(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 )
Time = 0.43 (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 |
\(\displaystyle \frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{5/2}}{7 d}-\frac {5}{7} b^2 \left (\frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}+\frac {2 i b^2 \sqrt {i \sinh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{3 d \sqrt {b \sinh (c+d x)}}\right )\) |
(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
3.1.15.3.1 Defintions of rubi rules used
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]
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]
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]
Time = 1.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {b^{4} \left (5 i \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+6 \cosh \left (d x +c \right )^{4} \sinh \left (d x +c \right )-16 \cosh \left (d x +c \right )^{2} \sinh \left (d x +c \right )\right )}{21 \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\) | \(122\) |
1/21*b^4*(5*I*(1-I*sinh(d*x+c))^(1/2)*2^(1/2)*(1+I*sinh(d*x+c))^(1/2)*(I*s inh(d*x+c))^(1/2)*EllipticF((1-I*sinh(d*x+c))^(1/2),1/2*2^(1/2))+6*cosh(d* x+c)^4*sinh(d*x+c)-16*cosh(d*x+c)^2*sinh(d*x+c))/cosh(d*x+c)/(b*sinh(d*x+c ))^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 394, normalized size of antiderivative = 3.40 \[ \int (b \sinh (c+d x))^{7/2} \, dx=\frac {40 \, {\left (\sqrt {2} b^{3} \cosh \left (d x + c\right )^{3} + 3 \, \sqrt {2} b^{3} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, \sqrt {2} b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + \sqrt {2} b^{3} \sinh \left (d x + c\right )^{3}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + {\left (3 \, b^{3} \cosh \left (d x + c\right )^{6} + 18 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 3 \, b^{3} \sinh \left (d x + c\right )^{6} - 23 \, b^{3} \cosh \left (d x + c\right )^{4} - 23 \, b^{3} \cosh \left (d x + c\right )^{2} + {\left (45 \, b^{3} \cosh \left (d x + c\right )^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (15 \, b^{3} \cosh \left (d x + c\right )^{3} - 23 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, b^{3} + {\left (45 \, b^{3} \cosh \left (d x + c\right )^{4} - 138 \, b^{3} \cosh \left (d x + c\right )^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (9 \, b^{3} \cosh \left (d x + c\right )^{5} - 46 \, b^{3} \cosh \left (d x + c\right )^{3} - 23 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}}{84 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + d \sinh \left (d x + c\right )^{3}\right )}} \]
1/84*(40*(sqrt(2)*b^3*cosh(d*x + c)^3 + 3*sqrt(2)*b^3*cosh(d*x + c)^2*sinh (d*x + c) + 3*sqrt(2)*b^3*cosh(d*x + c)*sinh(d*x + c)^2 + sqrt(2)*b^3*sinh (d*x + c)^3)*sqrt(b)*weierstrassPInverse(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*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + d*sinh(d*x + c)^3)
Timed out. \[ \int (b \sinh (c+d x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (b \sinh (c+d x))^{7/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
\[ \int (b \sinh (c+d x))^{7/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int (b \sinh (c+d x))^{7/2} \, dx=\int {\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{7/2} \,d x \]