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

Optimal. Leaf size=116 \[ -\frac {10 i b^4 \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{21 d \sqrt {b \sinh (c+d x)}}-\frac {10 b^3 \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{21 d}+\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{5/2}}{7 d} \]

[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

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Rubi [A]  time = 0.06, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2635, 2642, 2641} \[ -\frac {10 b^3 \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{21 d}-\frac {10 i b^4 \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{21 d \sqrt {b \sinh (c+d x)}}+\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[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 2635

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] && Integer
Q[2*n]

Rule 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int (b \sinh (c+d x))^{7/2} \, dx &=\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 F\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\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*}

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Mathematica [A]  time = 0.28, size = 76, normalized size = 0.66 \[ \frac {b^3 \sqrt {b \sinh (c+d x)} \left (-23 \cosh (c+d x)+3 \cosh (3 (c+d x))-\frac {20 F\left (\left .\frac {1}{4} (-2 i c-2 i d x+\pi )\right |2\right )}{\sqrt {i \sinh (c+d x)}}\right )}{42 d} \]

Antiderivative was successfully verified.

[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)

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sinh \left (d x + c\right )} b^{3} \sinh \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sinh \left (d x + c\right )\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(7/2), x)

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maple [A]  time = 0.07, size = 122, normalized size = 1.05 \[ \frac {b^{4} \left (5 i \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+6 \sinh \left (d x +c \right ) \left (\cosh ^{4}\left (d x +c \right )\right )-16 \left (\cosh ^{2}\left (d x +c \right )\right ) \sinh \left (d x +c \right )\right )}{21 \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*sinh(d*x+c)*cosh(d*x+c)^4-16*cosh(d*x+c)^2*sinh(d*x+c))/cosh(d*x+c)/(b*sinh(d*x
+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sinh \left (d x + c\right )\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sinh(c + d*x))^(7/2),x)

[Out]

int((b*sinh(c + d*x))^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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