3.865 \(\int \frac{1}{(a+b \cosh (c+d x) \sinh (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac{2 \sqrt{2} b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{2 i \sqrt{2} \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{d \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}} \]

[Out]

(-2*Sqrt[2]*b*Cosh[2*c + 2*d*x])/((4*a^2 + b^2)*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]]) - ((2*I)*Sqrt[2]*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)*d*Sqrt[(2
*a + b*Sinh[2*c + 2*d*x])/(2*a - I*b)])

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Rubi [A]  time = 0.113346, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2666, 2664, 21, 2655, 2653} \[ -\frac{2 \sqrt{2} b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{2 i \sqrt{2} \sqrt{2 a+b \sinh (2 c+2 d x)} E\left (\frac{1}{2} \left (2 i c+2 i d x-\frac{\pi }{2}\right )|\frac{2 b}{2 i a+b}\right )}{d \left (4 a^2+b^2\right ) \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-3/2),x]

[Out]

(-2*Sqrt[2]*b*Cosh[2*c + 2*d*x])/((4*a^2 + b^2)*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]]) - ((2*I)*Sqrt[2]*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)*d*Sqrt[(2
*a + b*Sinh[2*c + 2*d*x])/(2*a - I*b)])

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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))^{3/2}} \, dx &=\int \frac{1}{\left (a+\frac{1}{2} b \sinh (2 c+2 d x)\right )^{3/2}} \, dx\\ &=-\frac{2 \sqrt{2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{8 \int \frac{-\frac{a}{2}-\frac{1}{4} b \sinh (2 c+2 d x)}{\sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)}} \, dx}{4 a^2+b^2}\\ &=-\frac{2 \sqrt{2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt{2 a+b \sinh (2 c+2 d x)}}+\frac{4 \int \sqrt{a+\frac{1}{2} b \sinh (2 c+2 d x)} \, dx}{4 a^2+b^2}\\ &=-\frac{2 \sqrt{2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt{2 a+b \sinh (2 c+2 d x)}}+\frac{\left (4 \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}{\left (4 a^2+b^2\right ) \sqrt{\frac{a+\frac{1}{2} b \sinh (2 c+2 d x)}{a-\frac{i b}{2}}}}\\ &=-\frac{2 \sqrt{2} b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d \sqrt{2 a+b \sinh (2 c+2 d x)}}-\frac{2 i \sqrt{2} 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)}}{\left (4 a^2+b^2\right ) d \sqrt{\frac{2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}\\ \end{align*}

Mathematica [A]  time = 0.473617, size = 119, normalized size = 0.75 \[ \frac{2 \left (-b \cosh (2 (c+d x))+(b+2 i a) \sqrt{\frac{2 a+b \sinh (2 (c+d x))}{2 a-i b}} E\left (\frac{1}{4} (-4 i c-4 i d x+\pi )|-\frac{2 i b}{2 a-i b}\right )\right )}{d \left (4 a^2+b^2\right ) \sqrt{a+\frac{1}{2} b \sinh (2 (c+d x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-3/2),x]

[Out]

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

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Maple [B]  time = 0.452, size = 630, normalized size = 4. \begin{align*} 4\,{\frac{1}{ \left ( 4\,{a}^{2}+{b}^{2} \right ) b\cosh \left ( 2\,dx+2\,c \right ) \sqrt{4\,a+2\,b\sinh \left ( 2\,dx+2\,c \right ) }d} \left ( 4\,\sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}}\sqrt{{\frac{ \left ( -\sinh \left ( 2\,dx+2\,c \right ) +i \right ) b}{ib+2\,a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( 2\,dx+2\,c \right ) \right ) b}{ib-2\,a}}}{\it EllipticF} \left ( \sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}},\sqrt{-{\frac{ib-2\,a}{ib+2\,a}}} \right ){a}^{2}+\sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}}\sqrt{{\frac{ \left ( -\sinh \left ( 2\,dx+2\,c \right ) +i \right ) b}{ib+2\,a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( 2\,dx+2\,c \right ) \right ) b}{ib-2\,a}}}{\it EllipticF} \left ( \sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}},\sqrt{-{\frac{ib-2\,a}{ib+2\,a}}} \right ){b}^{2}-4\,\sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}}\sqrt{{\frac{ \left ( -\sinh \left ( 2\,dx+2\,c \right ) +i \right ) b}{ib+2\,a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( 2\,dx+2\,c \right ) \right ) b}{ib-2\,a}}}{\it EllipticE} \left ( \sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}},\sqrt{-{\frac{ib-2\,a}{ib+2\,a}}} \right ){a}^{2}-\sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}}\sqrt{{\frac{ \left ( -\sinh \left ( 2\,dx+2\,c \right ) +i \right ) b}{ib+2\,a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( 2\,dx+2\,c \right ) \right ) b}{ib-2\,a}}}{\it EllipticE} \left ( \sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}},\sqrt{-{\frac{ib-2\,a}{ib+2\,a}}} \right ){b}^{2}-{b}^{2} \left ( \sinh \left ( 2\,dx+2\,c \right ) \right ) ^{2}-{b}^{2} \right ) } \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))^(3/2),x)

[Out]

4*(4*(-(2*a+b*sinh(2*d*x+2*c))/(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)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a^2+(-(2
*a+b*sinh(2*d*x+2*c))/(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)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*b^2-4*(-(2*a+b*si
nh(2*d*x+2*c))/(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)*EllipticE((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*a^2-(-(2*a+b*sinh(2*d*x+
2*c))/(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)*Ellipt
icE((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))*b^2-b^2*sinh(2*d*x+2*c)^2-b^2)/(4
*a^2+b^2)/b/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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*cosh(d*x + c)*sinh(d*x + c) + a)^(-3/2), 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^{2} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2}}, 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))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [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{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate((b*cosh(d*x + c)*sinh(d*x + c) + a)^(-3/2), x)