3.47 \(\int \frac{1}{(a+a \cosh (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=107 \[ \frac{3 \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{2} \sqrt{a \cosh (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{3 \sinh (c+d x)}{16 a d (a \cosh (c+d x)+a)^{3/2}}+\frac{\sinh (c+d x)}{4 d (a \cosh (c+d x)+a)^{5/2}} \]

[Out]

(3*ArcTan[(Sqrt[a]*Sinh[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cosh[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Sinh[c + d*x]
/(4*d*(a + a*Cosh[c + d*x])^(5/2)) + (3*Sinh[c + d*x])/(16*a*d*(a + a*Cosh[c + d*x])^(3/2))

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Rubi [A]  time = 0.0624466, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2650, 2649, 206} \[ \frac{3 \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{2} \sqrt{a \cosh (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{3 \sinh (c+d x)}{16 a d (a \cosh (c+d x)+a)^{3/2}}+\frac{\sinh (c+d x)}{4 d (a \cosh (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cosh[c + d*x])^(-5/2),x]

[Out]

(3*ArcTan[(Sqrt[a]*Sinh[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cosh[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Sinh[c + d*x]
/(4*d*(a + a*Cosh[c + d*x])^(5/2)) + (3*Sinh[c + d*x])/(16*a*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))^{5/2}} \, dx &=\frac{\sinh (c+d x)}{4 d (a+a \cosh (c+d x))^{5/2}}+\frac{3 \int \frac{1}{(a+a \cosh (c+d x))^{3/2}} \, dx}{8 a}\\ &=\frac{\sinh (c+d x)}{4 d (a+a \cosh (c+d x))^{5/2}}+\frac{3 \sinh (c+d x)}{16 a d (a+a \cosh (c+d x))^{3/2}}+\frac{3 \int \frac{1}{\sqrt{a+a \cosh (c+d x)}} \, dx}{32 a^2}\\ &=\frac{\sinh (c+d x)}{4 d (a+a \cosh (c+d x))^{5/2}}+\frac{3 \sinh (c+d x)}{16 a d (a+a \cosh (c+d x))^{3/2}}+\frac{(3 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 )}{16 a^2 d}\\ &=\frac{3 \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{2} \sqrt{a+a \cosh (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{\sinh (c+d x)}{4 d (a+a \cosh (c+d x))^{5/2}}+\frac{3 \sinh (c+d x)}{16 a d (a+a \cosh (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.263268, size = 91, normalized size = 0.85 \[ \frac{\cosh ^5\left (\frac{1}{2} (c+d x)\right ) \left (32 \sinh ^5\left (\frac{1}{2} (c+d x)\right ) \text{csch}^4(c+d x)+3 \left (\tan ^{-1}\left (\sinh \left (\frac{1}{2} (c+d x)\right )\right )+\tanh \left (\frac{1}{2} (c+d x)\right ) \text{sech}\left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d (a (\cosh (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[(c + d*x)/2]^5*(32*Csch[c + d*x]^4*Sinh[(c + d*x)/2]^5 + 3*(ArcTan[Sinh[(c + d*x)/2]] + Sech[(c + d*x)/2
]*Tanh[(c + d*x)/2])))/(4*d*(a*(1 + Cosh[c + d*x]))^(5/2))

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Maple [B]  time = 0.046, size = 178, normalized size = 1.7 \begin{align*} -{\frac{\sqrt{2}}{32\,{a}^{3}d}\sqrt{ \left ( \sinh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a} \left ( 3\,\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 ( 1/2\,dx+c/2 \right ) \right ) ^{4}-3\,\sqrt{ \left ( \sinh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a} \left ( \cosh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\sqrt{-a}-2\,\sqrt{ \left ( \sinh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}\sqrt{-a} \right ) \left ( \cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\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))^(5/2),x)

[Out]

-1/32*(sinh(1/2*d*x+1/2*c)^2*a)^(1/2)*(3*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)^4-3*(sinh(1/2*d*x+1/2*c)^2*a)^(1/2)*cosh(1/2*d*x+1/2*c)^2*(-a)^(1/2)-2*(sinh(1/2*d*x
+1/2*c)^2*a)^(1/2)*(-a)^(1/2))/a^3/cosh(1/2*d*x+1/2*c)^3/(-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 = 1.89003, size = 338, normalized size = 3.16 \begin{align*} \frac{1}{80} \, \sqrt{2}{\left (\frac{15 \, e^{\left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )} + 70 \, e^{\left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )} + 128 \, e^{\left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )} - 70 \, e^{\left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )} - 15 \, e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{{\left (a^{\frac{5}{2}} e^{\left (5 \, d x + 5 \, c\right )} + 5 \, a^{\frac{5}{2}} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, a^{\frac{5}{2}} e^{\left (3 \, d x + 3 \, c\right )} + 10 \, a^{\frac{5}{2}} e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a^{\frac{5}{2}} e^{\left (d x + c\right )} + a^{\frac{5}{2}}\right )} d} + \frac{15 \, \arctan \left (e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}\right )}{a^{\frac{5}{2}} d}\right )} - \frac{8 \, \sqrt{2} e^{\left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )}}{5 \,{\left (a^{\frac{5}{2}} d e^{\left (5 \, d x + 5 \, c\right )} + 5 \, a^{\frac{5}{2}} d e^{\left (4 \, d x + 4 \, c\right )} + 10 \, a^{\frac{5}{2}} d e^{\left (3 \, d x + 3 \, c\right )} + 10 \, a^{\frac{5}{2}} d e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a^{\frac{5}{2}} d e^{\left (d x + c\right )} + a^{\frac{5}{2}} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*sqrt(2)*((15*e^(9/2*d*x + 9/2*c) + 70*e^(7/2*d*x + 7/2*c) + 128*e^(5/2*d*x + 5/2*c) - 70*e^(3/2*d*x + 3/2
*c) - 15*e^(1/2*d*x + 1/2*c))/((a^(5/2)*e^(5*d*x + 5*c) + 5*a^(5/2)*e^(4*d*x + 4*c) + 10*a^(5/2)*e^(3*d*x + 3*
c) + 10*a^(5/2)*e^(2*d*x + 2*c) + 5*a^(5/2)*e^(d*x + c) + a^(5/2))*d) + 15*arctan(e^(1/2*d*x + 1/2*c))/(a^(5/2
)*d)) - 8/5*sqrt(2)*e^(5/2*d*x + 5/2*c)/(a^(5/2)*d*e^(5*d*x + 5*c) + 5*a^(5/2)*d*e^(4*d*x + 4*c) + 10*a^(5/2)*
d*e^(3*d*x + 3*c) + 10*a^(5/2)*d*e^(2*d*x + 2*c) + 5*a^(5/2)*d*e^(d*x + c) + a^(5/2)*d)

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Fricas [B]  time = 1.96441, size = 1454, normalized size = 13.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*sqrt(2)*(cosh(d*x + c)^4 + 4*(cosh(d*x + c) + 1)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 4*cosh(d*x + c)^
3 + 6*(cosh(d*x + c)^2 + 2*cosh(d*x + c) + 1)*sinh(d*x + c)^2 + 6*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 + 3*cos
h(d*x + c)^2 + 3*cosh(d*x + c) + 1)*sinh(d*x + c) + 4*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)*(3*cosh(d*x + c)^4 + (12*cosh(d*x + c) + 11)*sinh
(d*x + c)^3 + 3*sinh(d*x + c)^4 + 11*cosh(d*x + c)^3 + (18*cosh(d*x + c)^2 + 33*cosh(d*x + c) - 11)*sinh(d*x +
c)^2 - 11*cosh(d*x + c)^2 + (12*cosh(d*x + c)^3 + 33*cosh(d*x + c)^2 - 22*cosh(d*x + c) - 3)*sinh(d*x + c) -
3*cosh(d*x + c))*sqrt(a/(cosh(d*x + c) + sinh(d*x + c))))/(a^3*d*cosh(d*x + c)^4 + a^3*d*sinh(d*x + c)^4 + 4*a
^3*d*cosh(d*x + c)^3 + 6*a^3*d*cosh(d*x + c)^2 + 4*a^3*d*cosh(d*x + c) + a^3*d + 4*(a^3*d*cosh(d*x + c) + a^3*
d)*sinh(d*x + c)^3 + 6*(a^3*d*cosh(d*x + c)^2 + 2*a^3*d*cosh(d*x + c) + a^3*d)*sinh(d*x + c)^2 + 4*(a^3*d*cosh
(d*x + c)^3 + 3*a^3*d*cosh(d*x + c)^2 + 3*a^3*d*cosh(d*x + c) + a^3*d)*sinh(d*x + c))

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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+a*cosh(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.37774, size = 138, normalized size = 1.29 \begin{align*} \frac{3 \, \sqrt{2} \arctan \left (e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}\right )}{16 \, a^{\frac{5}{2}} d} + \frac{\sqrt{2}{\left (3 \, a^{\frac{7}{2}} e^{\left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )} + 11 \, a^{\frac{7}{2}} e^{\left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )} - 11 \, a^{\frac{7}{2}} e^{\left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )} - 3 \, a^{\frac{7}{2}} e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}\right )}}{16 \,{\left (a e^{\left (d x + c\right )} + a\right )}^{4} a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*sqrt(2)*arctan(e^(1/2*d*x + 1/2*c))/(a^(5/2)*d) + 1/16*sqrt(2)*(3*a^(7/2)*e^(7/2*d*x + 7/2*c) + 11*a^(7/2
)*e^(5/2*d*x + 5/2*c) - 11*a^(7/2)*e^(3/2*d*x + 3/2*c) - 3*a^(7/2)*e^(1/2*d*x + 1/2*c))/((a*e^(d*x + c) + a)^4
*a^2*d)