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

Optimal. Leaf size=77 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{2} \sqrt{a \cosh (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{\sinh (c+d x)}{2 d (a \cosh (c+d x)+a)^{3/2}} \]

[Out]

ArcTan[(Sqrt[a]*Sinh[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cosh[c + d*x]])]/(2*Sqrt[2]*a^(3/2)*d) + Sinh[c + d*x]/(2*d
*(a + a*Cosh[c + d*x])^(3/2))

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Rubi [A]  time = 0.0414038, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2650, 2649, 206} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{2} \sqrt{a \cosh (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{\sinh (c+d x)}{2 d (a \cosh (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cosh[c + d*x])^(-3/2),x]

[Out]

ArcTan[(Sqrt[a]*Sinh[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cosh[c + d*x]])]/(2*Sqrt[2]*a^(3/2)*d) + Sinh[c + d*x]/(2*d
*(a + a*Cosh[c + d*x])^(3/2))

Rule 2650

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

Rule 2649

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0813707, size = 63, normalized size = 0.82 \[ \frac{\cosh ^2\left (\frac{1}{2} (c+d x)\right ) \left (\tanh \left (\frac{1}{2} (c+d x)\right )+\cosh \left (\frac{1}{2} (c+d x)\right ) \tan ^{-1}\left (\sinh \left (\frac{1}{2} (c+d x)\right )\right )\right )}{d (a (\cosh (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[(c + d*x)/2]^2*(ArcTan[Sinh[(c + d*x)/2]]*Cosh[(c + d*x)/2] + Tanh[(c + d*x)/2]))/(d*(a*(1 + Cosh[c + d*
x]))^(3/2))

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Maple [B]  time = 0.045, size = 144, normalized size = 1.9 \begin{align*} -{\frac{\sqrt{2}}{4\,{a}^{2}d}\sqrt{ \left ( \sinh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a} \left ( \ln \left ( 2\,{\frac{\sqrt{ \left ( \sinh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}\sqrt{-a}-a}{\cosh \left ( 1/2\,dx+c/2 \right ) }} \right ) a \left ( \cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}-\sqrt{ \left ( \sinh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}\sqrt{-a} \right ) \left ( \cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-a}}} \left ( \sinh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{a \left ( \cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*cosh(d*x+c))^(3/2),x)

[Out]

-1/4*(sinh(1/2*d*x+1/2*c)^2*a)^(1/2)*(ln(2/cosh(1/2*d*x+1/2*c)*((sinh(1/2*d*x+1/2*c)^2*a)^(1/2)*(-a)^(1/2)-a))
*a*cosh(1/2*d*x+1/2*c)^2-(sinh(1/2*d*x+1/2*c)^2*a)^(1/2)*(-a)^(1/2))/a^2/cosh(1/2*d*x+1/2*c)/(-a)^(1/2)/sinh(1
/2*d*x+1/2*c)*2^(1/2)/(a*cosh(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [B]  time = 1.99841, size = 230, normalized size = 2.99 \begin{align*} \frac{1}{6} \, \sqrt{2}{\left (\frac{3 \, e^{\left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )} + 8 \, e^{\left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )} - 3 \, e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{{\left (a^{\frac{3}{2}} e^{\left (3 \, d x + 3 \, c\right )} + 3 \, a^{\frac{3}{2}} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a^{\frac{3}{2}} e^{\left (d x + c\right )} + a^{\frac{3}{2}}\right )} d} + \frac{3 \, \arctan \left (e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}\right )}{a^{\frac{3}{2}} d}\right )} - \frac{4 \, \sqrt{2} e^{\left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )}}{3 \,{\left (a^{\frac{3}{2}} d e^{\left (3 \, d x + 3 \, c\right )} + 3 \, a^{\frac{3}{2}} d e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a^{\frac{3}{2}} d e^{\left (d x + c\right )} + a^{\frac{3}{2}} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(2)*((3*e^(5/2*d*x + 5/2*c) + 8*e^(3/2*d*x + 3/2*c) - 3*e^(1/2*d*x + 1/2*c))/((a^(3/2)*e^(3*d*x + 3*c)
+ 3*a^(3/2)*e^(2*d*x + 2*c) + 3*a^(3/2)*e^(d*x + c) + a^(3/2))*d) + 3*arctan(e^(1/2*d*x + 1/2*c))/(a^(3/2)*d)
) - 4/3*sqrt(2)*e^(3/2*d*x + 3/2*c)/(a^(3/2)*d*e^(3*d*x + 3*c) + 3*a^(3/2)*d*e^(2*d*x + 2*c) + 3*a^(3/2)*d*e^(
d*x + c) + a^(3/2)*d)

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Fricas [B]  time = 1.90273, size = 632, normalized size = 8.21 \begin{align*} -\frac{\sqrt{2}{\left (\cosh \left (d x + c\right )^{2} + 2 \,{\left (\cosh \left (d x + c\right ) + 1\right )} \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) + 1\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{1}{2}} \sqrt{a} \sqrt{\frac{a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}}}{a}\right ) - 2 \, \sqrt{\frac{1}{2}}{\left (\cosh \left (d x + c\right )^{2} +{\left (2 \, \cosh \left (d x + c\right ) - 1\right )} \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - \cosh \left (d x + c\right )\right )} \sqrt{\frac{a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}}}{2 \,{\left (a^{2} d \cosh \left (d x + c\right )^{2} + a^{2} d \sinh \left (d x + c\right )^{2} + 2 \, a^{2} d \cosh \left (d x + c\right ) + a^{2} d + 2 \,{\left (a^{2} d \cosh \left (d x + c\right ) + a^{2} d\right )} \sinh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(2)*(cosh(d*x + c)^2 + 2*(cosh(d*x + c) + 1)*sinh(d*x + c) + sinh(d*x + c)^2 + 2*cosh(d*x + c) + 1)*
sqrt(a)*arctan(sqrt(2)*sqrt(1/2)*sqrt(a)*sqrt(a/(cosh(d*x + c) + sinh(d*x + c)))/a) - 2*sqrt(1/2)*(cosh(d*x +
c)^2 + (2*cosh(d*x + c) - 1)*sinh(d*x + c) + sinh(d*x + c)^2 - cosh(d*x + c))*sqrt(a/(cosh(d*x + c) + sinh(d*x
+ c))))/(a^2*d*cosh(d*x + c)^2 + a^2*d*sinh(d*x + c)^2 + 2*a^2*d*cosh(d*x + c) + a^2*d + 2*(a^2*d*cosh(d*x +
c) + a^2*d)*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \cosh{\left (c + d x \right )} + a\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cosh(d*x+c))**(3/2),x)

[Out]

Integral((a*cosh(c + d*x) + a)**(-3/2), x)

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Giac [A]  time = 1.25308, size = 99, normalized size = 1.29 \begin{align*} \frac{\frac{\sqrt{2} \arctan \left (e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}\right )}{\sqrt{a} d} + \frac{\sqrt{2}{\left (a^{\frac{3}{2}} e^{\left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )} - a^{\frac{3}{2}} e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}\right )}}{{\left (a e^{\left (d x + c\right )} + a\right )}^{2} d}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*arctan(e^(1/2*d*x + 1/2*c))/(sqrt(a)*d) + sqrt(2)*(a^(3/2)*e^(3/2*d*x + 3/2*c) - a^(3/2)*e^(1/2*d
*x + 1/2*c))/((a*e^(d*x + c) + a)^2*d))/a