### 3.866 $$\int \frac{1}{(a+b \cosh (c+d x) \sinh (c+d x))^{5/2}} \, dx$$

Optimal. Leaf size=325 $-\frac{32 \sqrt{2} a b \cosh (2 c+2 d x)}{3 d \left (4 a^2+b^2\right )^2 \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{4 \sqrt{2} b \cosh (2 c+2 d x)}{3 d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^{3/2}}+\frac{4 i \sqrt{2} \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{3 d \left (4 a^2+b^2\right ) \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{32 i \sqrt{2} a \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{3 d \left (4 a^2+b^2\right )^2 \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}$

[Out]

(-4*Sqrt[2]*b*Cosh[2*c + 2*d*x])/(3*(4*a^2 + b^2)*d*(2*a + b*Sinh[2*c + 2*d*x])^(3/2)) - (32*Sqrt[2]*a*b*Cosh[
2*c + 2*d*x])/(3*(4*a^2 + b^2)^2*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]]) - (((32*I)/3)*Sqrt[2]*a*EllipticE[((2*I)*c
- Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*a + b)]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/((4*a^2 + b^2)^2*d*Sqrt[(2*a + b
*Sinh[2*c + 2*d*x])/(2*a - I*b)]) + (((4*I)/3)*Sqrt[2]*EllipticF[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*
a + b)]*Sqrt[(2*a + b*Sinh[2*c + 2*d*x])/(2*a - I*b)])/((4*a^2 + b^2)*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])

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Rubi [A]  time = 0.38078, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {2666, 2664, 2754, 2752, 2663, 2661, 2655, 2653} $-\frac{32 \sqrt{2} a b \cosh (2 c+2 d x)}{3 d \left (4 a^2+b^2\right )^2 \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{4 \sqrt{2} b \cosh (2 c+2 d x)}{3 d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^{3/2}}+\frac{4 i \sqrt{2} \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{3 d \left (4 a^2+b^2\right ) \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{32 i \sqrt{2} a \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{3 d \left (4 a^2+b^2\right )^2 \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-5/2),x]

[Out]

(-4*Sqrt[2]*b*Cosh[2*c + 2*d*x])/(3*(4*a^2 + b^2)*d*(2*a + b*Sinh[2*c + 2*d*x])^(3/2)) - (32*Sqrt[2]*a*b*Cosh[
2*c + 2*d*x])/(3*(4*a^2 + b^2)^2*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]]) - (((32*I)/3)*Sqrt[2]*a*EllipticE[((2*I)*c
- Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*a + b)]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/((4*a^2 + b^2)^2*d*Sqrt[(2*a + b
*Sinh[2*c + 2*d*x])/(2*a - I*b)]) + (((4*I)/3)*Sqrt[2]*EllipticF[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*
a + b)]*Sqrt[(2*a + b*Sinh[2*c + 2*d*x])/(2*a - I*b)])/((4*a^2 + b^2)*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+b \cosh (c+d x) \sinh (c+d x))^{5/2}} \, dx &=\int \frac{1}{\left (a+\frac{1}{2} b \sinh (2 c+2 d x)\right )^{5/2}} \, dx\\ &=-\frac{4 \sqrt{2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac{8 \int \frac{-\frac{3 a}{2}+\frac{1}{4} b \sinh (2 c+2 d x)}{\left (a+\frac{1}{2} b \sinh (2 c+2 d x)\right )^{3/2}} \, dx}{3 \left (4 a^2+b^2\right )}\\ &=-\frac{4 \sqrt{2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac{32 \sqrt{2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt{2 a+b \sinh (2 c+2 d x)}}+\frac{64 \int \frac{\frac{1}{16} \left (12 a^2-b^2\right )+\frac{1}{2} a b \sinh (2 c+2 d x)}{\sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}} \, dx}{3 \left (4 a^2+b^2\right )^2}\\ &=-\frac{4 \sqrt{2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac{32 \sqrt{2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt{2 a+b \sinh (2 c+2 d x)}}+\frac{(64 a) \int \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)} \, dx}{3 \left (4 a^2+b^2\right )^2}-\frac{4 \int \frac{1}{\sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}} \, dx}{3 \left (4 a^2+b^2\right )}\\ &=-\frac{4 \sqrt{2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac{32 \sqrt{2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt{2 a+b \sinh (2 c+2 d x)}}+\frac{\left (64 a \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}\right ) \int \sqrt{\frac{a}{a-\frac{i b}{2}}+\frac{b \sinh (2 c+2 d x)}{2 \left (a-\frac{i b}{2}\right )}} \, dx}{3 \left (4 a^2+b^2\right )^2 \sqrt{\frac{a+\frac{1}{2} b \sinh (2 c+2 d x)}{a-\frac{i b}{2}}}}-\frac{\left (4 \sqrt{\frac{a+\frac{1}{2} b \sinh (2 c+2 d x)}{a-\frac{i b}{2}}}\right ) \int \frac{1}{\sqrt{\frac{a}{a-\frac{i b}{2}}+\frac{b \sinh (2 c+2 d x)}{2 \left (a-\frac{i b}{2}\right )}}} \, dx}{3 \left (4 a^2+b^2\right ) \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}}\\ &=-\frac{4 \sqrt{2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac{32 \sqrt{2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{32 i \sqrt{2} a E\left (\frac{1}{2} \left (2 i c-\frac{\pi }{2}+2 i d x\right )|\frac{2 b}{2 i a+b}\right ) \sqrt{2 a+b \sinh (2 c+2 d x)}}{3 \left (4 a^2+b^2\right )^2 d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac{4 i \sqrt{2} F\left (\frac{1}{2} \left (2 i c-\frac{\pi }{2}+2 i d x\right )|\frac{2 b}{2 i a+b}\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}{3 \left (4 a^2+b^2\right ) d \sqrt{2 a+b \sinh (2 c+2 d x)}}\\ \end{align*}

Mathematica [A]  time = 1.59987, size = 237, normalized size = 0.73 $\frac{4 \sqrt{2} \left (-b \cosh (2 (c+d x)) \left (20 a^2+8 a b \sinh (2 (c+d x))+b^2\right )+(b-2 i a) (2 a-i b)^2 \left (\frac{2 a+b \sinh (2 (c+d x))}{2 a-i b}\right )^{3/2} F\left (\frac{1}{4} (-4 i c-4 i d x+\pi )|-\frac{2 i b}{2 a-i b}\right )+8 i a (2 a-i b)^2 \left (\frac{2 a+b \sinh (2 (c+d x))}{2 a-i b}\right )^{3/2} E\left (\frac{1}{4} (-4 i c-4 i d x+\pi )|-\frac{2 i b}{2 a-i b}\right )\right )}{3 d \left (4 a^2+b^2\right )^2 (2 a+b \sinh (2 (c+d x)))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-5/2),x]

[Out]

(4*Sqrt[2]*((8*I)*a*(2*a - I*b)^2*EllipticE[((-4*I)*c + Pi - (4*I)*d*x)/4, ((-2*I)*b)/(2*a - I*b)]*((2*a + b*S
inh[2*(c + d*x)])/(2*a - I*b))^(3/2) + (2*a - I*b)^2*((-2*I)*a + b)*EllipticF[((-4*I)*c + Pi - (4*I)*d*x)/4, (
(-2*I)*b)/(2*a - I*b)]*((2*a + b*Sinh[2*(c + d*x)])/(2*a - I*b))^(3/2) - b*Cosh[2*(c + d*x)]*(20*a^2 + b^2 + 8
*a*b*Sinh[2*(c + d*x)])))/(3*(4*a^2 + b^2)^2*d*(2*a + b*Sinh[2*(c + d*x)])^(3/2))

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Maple [A]  time = 0.98, size = 641, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^(5/2),x)

[Out]

4*((2*a+b*sinh(2*d*x+2*c))*cosh(2*d*x+2*c)^2)^(1/2)*(-2/3/b/(4*a^2+b^2)*((2*a+b*sinh(2*d*x+2*c))*cosh(2*d*x+2*
c)^2)^(1/2)/(sinh(2*d*x+2*c)+2*a/b)^2-16/3*b*cosh(2*d*x+2*c)^2/(4*a^2+b^2)^2*a/((2*a+b*sinh(2*d*x+2*c))*cosh(2
*d*x+2*c)^2)^(1/2)+2*(12*a^2-b^2)/(48*a^4+24*a^2*b^2+3*b^4)*(2*a/b-I)*((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/
2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((I+sinh(2*d*x+2*c))*b/(I*b-2*a))^(1/2)/((2*a+b*sinh(2*d*x+2*c))*c
osh(2*d*x+2*c)^2)^(1/2)*EllipticF(((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/2),((2*a-I*b)/(I*b+2*a))^(1/2))+16/3
*a*b/(4*a^2+b^2)^2*(2*a/b-I)*((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/
2)*((I+sinh(2*d*x+2*c))*b/(I*b-2*a))^(1/2)/((2*a+b*sinh(2*d*x+2*c))*cosh(2*d*x+2*c)^2)^(1/2)*((-2*a/b-I)*Ellip
ticE(((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/2),((2*a-I*b)/(I*b+2*a))^(1/2))+I*EllipticF(((-b*sinh(2*d*x+2*c)-
2*a)/(I*b-2*a))^(1/2),((2*a-I*b)/(I*b+2*a))^(1/2))))/cosh(2*d*x+2*c)/(4*a+2*b*sinh(2*d*x+2*c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cosh(d*x + c)*sinh(d*x + c) + a)^(-5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}}{b^{3} \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right )^{3} + 3 \, a b^{2} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 3 \, a^{2} b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cosh(d*x + c)*sinh(d*x + c) + a)/(b^3*cosh(d*x + c)^3*sinh(d*x + c)^3 + 3*a*b^2*cosh(d*x + c)^
2*sinh(d*x + c)^2 + 3*a^2*b*cosh(d*x + c)*sinh(d*x + c) + a^3), 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(1/(a+b*cosh(d*x+c)*sinh(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError