\(\int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx\) [594]

Optimal result

Integrand size = 13, antiderivative size = 112 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\frac {2 (b \cosh (x)+a \sinh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}+\frac {2 i E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}{\left (a^2-b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \]

[Out]

2*(b*cosh(x)+a*sinh(x))/(a^2-b^2)/(a*cosh(x)+b*sinh(x))^(1/2)+2*I*(cos(1/2*I*x-1/2*arctan(a,-I*b))^2)^(1/2)/co
s(1/2*I*x-1/2*arctan(a,-I*b))*EllipticE(sin(1/2*I*x-1/2*arctan(a,-I*b)),2^(1/2))*(a*cosh(x)+b*sinh(x))^(1/2)/(
a^2-b^2)/((a*cosh(x)+b*sinh(x))/(a^2-b^2)^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3155, 3157, 2719} \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\frac {2 (a \sinh (x)+b \cosh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}+\frac {2 i \sqrt {a \cosh (x)+b \sinh (x)} E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{\left (a^2-b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \]

[In]

Int[(a*Cosh[x] + b*Sinh[x])^(-3/2),x]

[Out]

(2*(b*Cosh[x] + a*Sinh[x]))/((a^2 - b^2)*Sqrt[a*Cosh[x] + b*Sinh[x]]) + ((2*I)*EllipticE[(I*x - ArcTan[a, (-I)
*b])/2, 2]*Sqrt[a*Cosh[x] + b*Sinh[x]])/((a^2 - b^2)*Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])

Rule 2719

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

Rule 3155

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

Rule 3157

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[c + d*x] +
b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2])^n, Int[Cos[c + d*x - ArcTan[a, b]]^n, x]
, x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {2 (b \cosh (x)+a \sinh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}+\frac {\int \sqrt {a \cosh (x)+b \sinh (x)} \, dx}{-a^2+b^2} \\ & = \frac {2 (b \cosh (x)+a \sinh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}+\frac {\sqrt {a \cosh (x)+b \sinh (x)} \int \sqrt {\cosh \left (x+i \tan ^{-1}(a,-i b)\right )} \, dx}{\left (-a^2+b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \\ & = \frac {2 (b \cosh (x)+a \sinh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}+\frac {2 i E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}{\left (a^2-b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\frac {b \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cosh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )\right ) \sinh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )-\sqrt {-\sinh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )} \left (2 a \sqrt {1-\frac {b^2}{a^2}} \cosh (x)-2 a \cosh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )+b \sinh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )\right )}{a b \sqrt {1-\frac {b^2}{a^2}} \sqrt {a \cosh (x)+b \sinh (x)} \sqrt {-\sinh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )}} \]

[In]

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

[Out]

(b*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cosh[x + ArcTanh[b/a]]^2]*Sinh[x + ArcTanh[b/a]] - Sqrt[-Sinh[x + Ar
cTanh[b/a]]^2]*(2*a*Sqrt[1 - b^2/a^2]*Cosh[x] - 2*a*Cosh[x + ArcTanh[b/a]] + b*Sinh[x + ArcTanh[b/a]]))/(a*b*S
qrt[1 - b^2/a^2]*Sqrt[a*Cosh[x] + b*Sinh[x]]*Sqrt[-Sinh[x + ArcTanh[b/a]]^2])

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.29

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\cosh \left (x \right )\right )}{\sqrt {a^{2}-b^{2}}\, \sqrt {-\sinh \left (x \right ) \sqrt {a^{2}-b^{2}}}}\)	\(33\)

[In]

int(1/(a*cosh(x)+b*sinh(x))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/(a^2-b^2)^(1/2)/(-sinh(x)*(a^2-b^2)^(1/2))^(1/2)*arctanh(cosh(x))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.98 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\frac {2 \, {\left ({\left (\sqrt {2} {\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \, \sqrt {2} {\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt {2} {\left (a + b\right )} \sinh \left (x\right )^{2} + \sqrt {2} {\left (a - b\right )}\right )} \sqrt {a + b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) + 2 \, {\left ({\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a + b\right )} \sinh \left (x\right )^{2}\right )} \sqrt {a \cosh \left (x\right ) + b \sinh \left (x\right )}\right )}}{a^{3} - a^{2} b - a b^{2} + b^{3} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \sinh \left (x\right )^{2}} \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^(3/2),x, algorithm="fricas")

[Out]

2*((sqrt(2)*(a + b)*cosh(x)^2 + 2*sqrt(2)*(a + b)*cosh(x)*sinh(x) + sqrt(2)*(a + b)*sinh(x)^2 + sqrt(2)*(a - b
))*sqrt(a + b)*weierstrassZeta(-4*(a - b)/(a + b), 0, weierstrassPInverse(-4*(a - b)/(a + b), 0, cosh(x) + sin
h(x))) + 2*((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2)*sqrt(a*cosh(x) + b*sinh(x)))/(a
^3 - a^2*b - a*b^2 + b^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2 + 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)*sinh(
x) + (a^3 + a^2*b - a*b^2 - b^3)*sinh(x)^2)

Sympy [F]

\[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\int \frac {1}{\left (a \cosh {\left (x \right )} + b \sinh {\left (x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))**(3/2),x)

[Out]

Integral((a*cosh(x) + b*sinh(x))**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\int \frac {1}{{\left (a\,\mathrm {cosh}\left (x\right )+b\,\mathrm {sinh}\left (x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(a*cosh(x) + b*sinh(x))^(3/2),x)

[Out]

int(1/(a*cosh(x) + b*sinh(x))^(3/2), x)