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

Optimal. Leaf size=110 \[ -\frac{3 \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{2} \sqrt{a-a \cosh (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{3 \sinh (c+d x)}{16 a d (a-a \cosh (c+d x))^{3/2}}-\frac{\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}} \]

[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 = 0.062987, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2650, 2649, 206} \[ -\frac{3 \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{2} \sqrt{a-a \cosh (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{3 \sinh (c+d x)}{16 a d (a-a \cosh (c+d x))^{3/2}}-\frac{\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}} \]

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 = 0.193563, size = 115, normalized size = 1.05 \[ \frac{\sinh ^5\left (\frac{1}{2} (c+d x)\right ) \left (-\text{csch}^4\left (\frac{1}{4} (c+d x)\right )+6 \text{csch}^2\left (\frac{1}{4} (c+d x)\right )+\text{sech}^4\left (\frac{1}{4} (c+d x)\right )+6 \text{sech}^2\left (\frac{1}{4} (c+d x)\right )+24 \log \left (\tanh \left (\frac{1}{4} (c+d x)\right )\right )\right )}{32 a^2 d (\cosh (c+d x)-1)^2 \sqrt{a-a \cosh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*Csch[(c + d*x)/4]^2 - Csch[(c + d*x)/4]^4 + 24*Log[Tanh[(c + d*x)/4]] + 6*Sech[(c + d*x)/4]^2 + Sech[(c +
d*x)/4]^4)*Sinh[(c + d*x)/2]^5)/(32*a^2*d*(-1 + Cosh[c + d*x])^2*Sqrt[a - a*Cosh[c + d*x]])

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Maple [A]  time = 0.053, size = 137, normalized size = 1.3 \begin{align*} -{\frac{1}{32\,{a}^{2}d} \left ( -6\, \left ( \sinh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cosh \left ( 1/2\,dx+c/2 \right ) +4\,\cosh \left ( 1/2\,dx+c/2 \right ) + \left ( 3\,\ln \left ( 1+\cosh \left ( 1/2\,dx+c/2 \right ) \right ) -3\,\ln \left ( -1+\cosh \left ( 1/2\,dx+c/2 \right ) \right ) \right ) \left ( \sinh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4} \right ) \left ( 1+\cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1} \left ( -1+\cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1} \left ( \sinh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sinh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}}}} \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/a^2*(-6*sinh(1/2*d*x+1/2*c)^2*cosh(1/2*d*x+1/2*c)+4*cosh(1/2*d*x+1/2*c)+(3*ln(1+cosh(1/2*d*x+1/2*c))-3*l
n(-1+cosh(1/2*d*x+1/2*c)))*sinh(1/2*d*x+1/2*c)^4)/(1+cosh(1/2*d*x+1/2*c))/(-1+cosh(1/2*d*x+1/2*c))/sinh(1/2*d*
x+1/2*c)/(-2*sinh(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-a \cosh \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-a*cosh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.06401, size = 1605, normalized size = 14.59 \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]

-1/32*(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)*log(-(2*sqrt(2)*sqrt(1/2)*sq
rt(-a)*sqrt(-a/(cosh(d*x + c) + sinh(d*x + c)))*(cosh(d*x + c) + sinh(d*x + c)) + a*cosh(d*x + c) + a*sinh(d*x
+ c) + a)/(cosh(d*x + c) + sinh(d*x + c) - 1)) + 4*sqrt(1/2)*(3*cosh(d*x + c)^4 + (12*cosh(d*x + c) - 11)*sin
h(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*co
sh(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 = 1.37888, size = 227, normalized size = 2.06 \begin{align*} -\frac{3 \, \sqrt{2} \arctan \left (\frac{\sqrt{-a e^{\left (d x + c\right )}}}{\sqrt{a}}\right )}{16 \, a^{\frac{5}{2}} d \mathrm{sgn}\left (-e^{\left (d x + c\right )} + 1\right )} + \frac{\sqrt{2}{\left (3 \, \sqrt{-a e^{\left (d x + c\right )}} a^{3} e^{\left (3 \, d x + 3 \, c\right )} - 11 \, \sqrt{-a e^{\left (d x + c\right )}} a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 11 \, \sqrt{-a e^{\left (d x + c\right )}} a^{3} e^{\left (d x + c\right )} + 3 \, \sqrt{-a e^{\left (d x + c\right )}} a^{3}\right )}}{16 \,{\left (a e^{\left (d x + c\right )} - a\right )}^{4} a^{2} d \mathrm{sgn}\left (-e^{\left (d x + c\right )} + 1\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="giac")

[Out]

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