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

Optimal. Leaf size=103 \[ -\frac{10 i \sqrt{i \sinh (a+b x)} \text{EllipticF}\left (\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right ),2\right )}{21 b \sqrt{\sinh (a+b x)}}+\frac{2 \sinh ^{\frac{5}{2}}(a+b x) \cosh (a+b x)}{7 b}-\frac{10 \sqrt{\sinh (a+b x)} \cosh (a+b x)}{21 b} \]

[Out]

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

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Rubi [A]  time = 0.048612, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2635, 2642, 2641} \[ \frac{2 \sinh ^{\frac{5}{2}}(a+b x) \cosh (a+b x)}{7 b}-\frac{10 \sqrt{\sinh (a+b x)} \cosh (a+b x)}{21 b}-\frac{10 i \sqrt{i \sinh (a+b x)} F\left (\left .\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right )\right |2\right )}{21 b \sqrt{\sinh (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^(7/2),x]

[Out]

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

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 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]

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]

Rubi steps

\begin{align*} \int \sinh ^{\frac{7}{2}}(a+b x) \, dx &=\frac{2 \cosh (a+b x) \sinh ^{\frac{5}{2}}(a+b x)}{7 b}-\frac{5}{7} \int \sinh ^{\frac{3}{2}}(a+b x) \, dx\\ &=-\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}+\frac{5}{21} \int \frac{1}{\sqrt{\sinh (a+b x)}} \, dx\\ &=-\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}+\frac{\left (5 \sqrt{i \sinh (a+b x)}\right ) \int \frac{1}{\sqrt{i \sinh (a+b x)}} \, dx}{21 \sqrt{\sinh (a+b x)}}\\ &=-\frac{10 i F\left (\left .\frac{1}{2} \left (i a-\frac{\pi }{2}+i b x\right )\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}\\ \end{align*}

Mathematica [A]  time = 0.140299, size = 75, normalized size = 0.73 \[ \frac{40 i \sqrt{i \sinh (a+b x)} \text{EllipticF}\left (\frac{1}{4} (-2 i a-2 i b x+\pi ),2\right )-26 \sinh (2 (a+b x))+3 \sinh (4 (a+b x))}{84 b \sqrt{\sinh (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^(7/2),x]

[Out]

((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]])

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Maple [A]  time = 0.074, size = 116, normalized size = 1.1 \begin{align*}{\frac{1}{b\cosh \left ( bx+a \right ) } \left ({\frac{5\,i}{21}}\sqrt{1-i\sinh \left ( bx+a \right ) }\sqrt{2}\sqrt{1+i\sinh \left ( bx+a \right ) }\sqrt{i\sinh \left ( bx+a \right ) }{\it EllipticF} \left ( \sqrt{1-i\sinh \left ( bx+a \right ) },{\frac{\sqrt{2}}{2}} \right ) +{\frac{2\,\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}{7}}-{\frac{16\,\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{21}} \right ){\frac{1}{\sqrt{\sinh \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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*sinh(b*x+a)*cosh(b*x+a)^4-16/21*sinh(b*x+a)*cosh(b*x+a)^2)/cosh(b*x+a)/sinh(b*x+a)^
(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \left (b x + a\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sinh \left (b x + a\right )^{\frac{7}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(b*x + a)^(7/2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \left (b x + a\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^(7/2), x)