3.1.14 \(\int (b \tanh (c+d x))^{5/2} \, dx\) [14]

Optimal. Leaf size=78 \[ -\frac {b^{5/2} \text {ArcTan}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 78, 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 = {3554, 3557, 335, 304, 209, 212} \begin {gather*} -\frac {b^{5/2} \text {ArcTan}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tanh[c + d*x])^(5/2),x]

[Out]

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

Rule 209

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

Rule 212

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

Rule 304

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

Rule 335

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 3554

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

Rule 3557

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

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Mathematica [A]
time = 0.15, size = 68, normalized size = 0.87 \begin {gather*} -\frac {(b \tanh (c+d x))^{5/2} \left (3 \text {ArcTan}\left (\sqrt {\tanh (c+d x)}\right )-3 \tanh ^{-1}\left (\sqrt {\tanh (c+d x)}\right )+2 \tanh ^{\frac {3}{2}}(c+d x)\right )}{3 d \tanh ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tanh[c + d*x])^(5/2),x]

[Out]

-1/3*((b*Tanh[c + d*x])^(5/2)*(3*ArcTan[Sqrt[Tanh[c + d*x]]] - 3*ArcTanh[Sqrt[Tanh[c + d*x]]] + 2*Tanh[c + d*x
]^(3/2)))/(d*Tanh[c + d*x]^(5/2))

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Maple [A]
time = 1.87, size = 60, normalized size = 0.77

method	result	size
derivativedivides	\(-\frac {2 b \left (\frac {\left (b \tanh \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {b^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}+\frac {b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}\right )}{d}\)	\(60\)
default	\(-\frac {2 b \left (\frac {\left (b \tanh \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {b^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}+\frac {b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}\right )}{d}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tanh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/d*b*(1/3*(b*tanh(d*x+c))^(3/2)-1/2*b^(3/2)*arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))+1/2*b^(3/2)*arctan((b*tan
h(d*x+c))^(1/2)/b^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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] Leaf count of result is larger than twice the leaf count of optimal. 463 vs. \(2 (62) = 124\).
time = 0.37, size = 980, normalized size = 12.56 \begin {gather*} \left [-\frac {6 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {-b} \arctan \left (\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b}\right ) - 3 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {-b} \log \left (-\frac {b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} - 2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1\right )} \sqrt {-b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}} - 2 \, b}{\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4}}\right ) + 8 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{12 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2} + d\right )}}, \frac {6 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b}\right ) + 3 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {b} \log \left (2 \, b \cosh \left (d x + c\right )^{4} + 8 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 12 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 8 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, b \sinh \left (d x + c\right )^{4} + 2 \, {\left (\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} + {\left (6 \, \cosh \left (d x + c\right )^{2} + 1\right )} \sinh \left (d x + c\right )^{2} + \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, \cosh \left (d x + c\right )^{3} + \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}} - b\right ) - 8 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{12 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2} + d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(6*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + b^2)*sqrt(-b)*arcta
n((cosh(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*(b^2*cosh(d*x + c)^2 +
2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + b^2)*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*
x + c)^3*sinh(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*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*
x + c) + b^2*sinh(d*x + c)^2 - b^2)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c
)*sinh(d*x + c) + d*sinh(d*x + c)^2 + d), 1/12*(6*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b
^2*sinh(d*x + c)^2 + b^2)*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*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x
+ c) + b^2*sinh(d*x + c)^2 + b^2)*sqrt(b)*log(2*b*cosh(d*x + c)^4 + 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*b*c
osh(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)/cosh(d*x + c)) - b) - 8*
(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 - b^2)*sqrt(b*sinh(d*x + c)/cos
h(d*x + c)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2 + d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tanh(c + d*x))**(5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (62) = 124\).
time = 0.51, size = 173, normalized size = 2.22 \begin {gather*} -\frac {6 \, b^{\frac {5}{2}} \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 ) + 3 \, b^{\frac {5}{2}} \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 ) + \frac {8 \, {\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^{3} + b^{4}\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}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.25, size = 62, normalized size = 0.79 \begin {gather*} \frac {b^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {2\,b\,{\left (b\,\mathrm {tanh}\left (c+d\,x\right )\right )}^{3/2}}{3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tanh(c + d*x))^(5/2),x)

[Out]

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

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