3.19 \(\int \frac {1}{(b \tanh (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}+\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {2}{3 b d (b \tanh (c+d x))^{3/2}} \]

[Out]

arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(5/2)/d+arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(5/2)/d-2/3/b/d/(b*ta
nh(d*x+c))^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3474, 3476, 329, 212, 206, 203} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}+\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {2}{3 b d (b \tanh (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tanh[c + d*x])^(-5/2),x]

[Out]

ArcTan[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]]/(b^(5/2)*d) + ArcTanh[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]]/(b^(5/2)*d) - 2/(3*
b*d*(b*Tanh[c + d*x])^(3/2))

Rule 203

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {1}{(b \tanh (c+d x))^{5/2}} \, dx &=-\frac {2}{3 b d (b \tanh (c+d x))^{3/2}}+\frac {\int \frac {1}{\sqrt {b \tanh (c+d x)}} \, dx}{b^2}\\ &=-\frac {2}{3 b d (b \tanh (c+d x))^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-b^2+x^2\right )} \, dx,x,b \tanh (c+d x)\right )}{b d}\\ &=-\frac {2}{3 b d (b \tanh (c+d x))^{3/2}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-b^2+x^4} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{b d}\\ &=-\frac {2}{3 b d (b \tanh (c+d x))^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{b^2 d}+\frac {\operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \tanh (c+d x)}\right )}{b^2 d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d}-\frac {2}{3 b d (b \tanh (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 38, normalized size = 0.48 \[ -\frac {2 \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\tanh ^2(c+d x)\right )}{3 b d (b \tanh (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tanh[c + d*x])^(-5/2),x]

[Out]

(-2*Hypergeometric2F1[-3/4, 1, 1/4, Tanh[c + d*x]^2])/(3*b*d*(b*Tanh[c + d*x])^(3/2))

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fricas [B]  time = 0.53, size = 1436, normalized size = 18.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(6*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x + c)^2 - 1)*sin
h(d*x + c)^2 - 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(-b)*arctan((cos
h(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c))/(
b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)) + 3*(cosh(d*x + c)^4 + 4*cosh(d*
x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 + 4*(
cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sin
h(d*x + c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(
cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x
+ c)) - 2*b)/(cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d
*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)) + 8*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x
+ c)^4 + 2*(3*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 + cosh(d*x + c))*s
inh(d*x + c) + 1)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(b^3*d*cosh(d*x + c)^4 + 4*b^3*d*cosh(d*x + c)*sinh(d*x
+ c)^3 + b^3*d*sinh(d*x + c)^4 - 2*b^3*d*cosh(d*x + c)^2 + b^3*d + 2*(3*b^3*d*cosh(d*x + c)^2 - b^3*d)*sinh(d
*x + c)^2 + 4*(b^3*d*cosh(d*x + c)^3 - b^3*d*cosh(d*x + c))*sinh(d*x + c)), -1/12*(6*(cosh(d*x + c)^4 + 4*cosh
(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 +
4*(cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(b)*arctan(sqrt(b)*sqrt(b*sinh(d*x + c)/cosh(d*x +
c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)) - 3*(cosh(d*x + c)^4 + 4*co
sh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2
+ 4*(cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(b)*log(2*b*cosh(d*x + c)^4 + 8*b*cosh(d*x + c)^3
*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*sinh(d*x + c)^3 + 2*b*sinh(d*x + c)^
4 + 2*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*cosh(d*x + c)^2 + 1)*sinh(d*x
+ c)^2 + cosh(d*x + c)^2 + 2*(2*cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c))*sqrt(b)*sqrt(b*sinh(d*x + c)/c
osh(d*x + c)) - b) + 8*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh(d*x +
c)^2 + 1)*sinh(d*x + c)^2 + 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(b*
sinh(d*x + c)/cosh(d*x + c)))/(b^3*d*cosh(d*x + c)^4 + 4*b^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*d*sinh(d*x
+ c)^4 - 2*b^3*d*cosh(d*x + c)^2 + b^3*d + 2*(3*b^3*d*cosh(d*x + c)^2 - b^3*d)*sinh(d*x + c)^2 + 4*(b^3*d*cosh
(d*x + c)^3 - b^3*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [B]  time = 0.62, size = 187, normalized size = 2.37 \[ -\frac {\frac {3 \, \pi - 3 \, \log \left ({\left | b \right |}\right ) - 8}{b^{\frac {5}{2}}} - \frac {12 \, \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} + \frac {6 \, \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right )}{b^{\frac {5}{2}}} - \frac {16 \, {\left (3 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{2} + b\right )}}{{\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} - \sqrt {b}\right )}^{3} b^{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((3*pi - 3*log(abs(b)) - 8)/b^(5/2) - 12*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))
/sqrt(b))/b^(5/2) + 6*log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sqrt(b*e^(4*d*x + 4*c) - b)))/b^(5/2) - 16*(3*(sqrt(b
)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^2 + b)/((sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b
) - sqrt(b))^3*b^2))/d

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maple [A]  time = 0.08, size = 64, normalized size = 0.81 \[ \frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}} d}+\frac {\arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}} d}-\frac {2}{3 b d \left (b \tanh \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tanh(d*x+c))^(5/2),x)

[Out]

arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(5/2)/d+arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(5/2)/d-2/3/b/d/(b*ta
nh(d*x+c))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tanh \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tanh(d*x + c))^(-5/2), x)

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mupad [B]  time = 1.36, size = 63, normalized size = 0.80 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{5/2}\,d}-\frac {2}{3\,b\,d\,{\left (b\,\mathrm {tanh}\left (c+d\,x\right )\right )}^{3/2}}+\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{5/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tanh(c + d*x))^(5/2),x)

[Out]

atan((b*tanh(c + d*x))^(1/2)/b^(1/2))/(b^(5/2)*d) - 2/(3*b*d*(b*tanh(c + d*x))^(3/2)) + atanh((b*tanh(c + d*x)
)^(1/2)/b^(1/2))/(b^(5/2)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tanh {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tanh(d*x+c))**(5/2),x)

[Out]

Integral((b*tanh(c + d*x))**(-5/2), x)

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