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[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))

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Rubi [A]  time = 0.0487133, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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]

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 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 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 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])

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*}

Mathematica [C]  time = 0.0768269, 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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Maple [A]  time = 0.02, size = 64, normalized size = 0.8 \begin{align*}{\frac{1}{d}\arctan \left ({\sqrt{b\tanh \left ( dx+c \right ) }{\frac{1}{\sqrt{b}}}} \right ){b}^{-{\frac{5}{2}}}}+{\frac{1}{d}{\it Artanh} \left ({\sqrt{b\tanh \left ( dx+c \right ) }{\frac{1}{\sqrt{b}}}} \right ){b}^{-{\frac{5}{2}}}}-{\frac{2}{3\,bd} \left ( b\tanh \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tanh \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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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Fricas [B]  time = 2.55463, size = 3934, normalized size = 49.8 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tanh{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError