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

Optimal result

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

[Out]

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)-1
0/21*b^3*cosh(d*x+c)*(b*sinh(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 2721, 2720} \[ \int (b \sinh (c+d x))^{7/2} \, dx=-\frac {10 i b^4 \sqrt {i \sinh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{21 d \sqrt {b \sinh (c+d x)}}-\frac {10 b^3 \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{21 d}+\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{5/2}}{7 d} \]

[In]

Int[(b*Sinh[c + d*x])^(7/2),x]

[Out]

(((-10*I)/21)*b^4*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2]*Sqrt[I*Sinh[c + d*x]])/(d*Sqrt[b*Sinh[c + d*x]]) - (10*
b^3*Cosh[c + d*x]*Sqrt[b*Sinh[c + d*x]])/(21*d) + (2*b*Cosh[c + d*x]*(b*Sinh[c + d*x])^(5/2))/(7*d)

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

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 2721

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{5/2}}{7 d}-\frac {1}{7} \left (5 b^2\right ) \int (b \sinh (c+d x))^{3/2} \, dx \\ & = -\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}+\frac {1}{21} \left (5 b^4\right ) \int \frac {1}{\sqrt {b \sinh (c+d x)}} \, dx \\ & = -\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}+\frac {\left (5 b^4 \sqrt {i \sinh (c+d x)}\right ) \int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx}{21 \sqrt {b \sinh (c+d x)}} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int (b \sinh (c+d x))^{7/2} \, dx=\frac {b^3 \left (-23 \cosh (c+d x)+3 \cosh (3 (c+d x))-\frac {20 \operatorname {EllipticF}\left (\frac {1}{4} (-2 i c+\pi -2 i d x),2\right )}{\sqrt {i \sinh (c+d x)}}\right ) \sqrt {b \sinh (c+d x)}}{42 d} \]

[In]

Integrate[(b*Sinh[c + d*x])^(7/2),x]

[Out]

(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)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05

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

[In]

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

[Out]

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*sinh(d*x+c))^(1/2)*EllipticF((1-I*sin
h(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

Fricas [C] (verification not implemented)

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 \[ \int (b \sinh (c+d x))^{7/2} \, dx=\frac {40 \, {\left (\sqrt {2} b^{3} \cosh \left (d x + c\right )^{3} + 3 \, \sqrt {2} b^{3} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, \sqrt {2} b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + \sqrt {2} b^{3} \sinh \left (d x + c\right )^{3}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + {\left (3 \, b^{3} \cosh \left (d x + c\right )^{6} + 18 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 3 \, b^{3} \sinh \left (d x + c\right )^{6} - 23 \, b^{3} \cosh \left (d x + c\right )^{4} - 23 \, b^{3} \cosh \left (d x + c\right )^{2} + {\left (45 \, b^{3} \cosh \left (d x + c\right )^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (15 \, b^{3} \cosh \left (d x + c\right )^{3} - 23 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, b^{3} + {\left (45 \, b^{3} \cosh \left (d x + c\right )^{4} - 138 \, b^{3} \cosh \left (d x + c\right )^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (9 \, b^{3} \cosh \left (d x + c\right )^{5} - 46 \, b^{3} \cosh \left (d x + c\right )^{3} - 23 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}}{84 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + d \sinh \left (d x + c\right )^{3}\right )}} \]

[In]

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

[Out]

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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(7/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(7/2), x)

Mupad [F(-1)]

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

[In]

int((b*sinh(c + d*x))^(7/2),x)

[Out]

int((b*sinh(c + d*x))^(7/2), x)