3.1.15 \(\int (b \tanh (c+d x))^{3/2} \, dx\) [15]

Optimal. Leaf size=75 \[ \frac {b^{3/2} \text {ArcTan}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 75, 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, 218, 212, 209} \begin {gather*} \frac {b^{3/2} \text {ArcTan}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tanh[c + d*x])^(3/2),x]

[Out]

(b^(3/2)*ArcTan[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]])/d + (b^(3/2)*ArcTanh[Sqrt[b*Tanh[c + d*x]]/Sqrt[b]])/d - (2*b*
Sqrt[b*Tanh[c + d*x]])/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 218

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

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Mathematica [A]
time = 0.06, size = 61, normalized size = 0.81 \begin {gather*} \frac {\left (\text {ArcTan}\left (\sqrt {\tanh (c+d x)}\right )+\tanh ^{-1}\left (\sqrt {\tanh (c+d x)}\right )-2 \sqrt {\tanh (c+d x)}\right ) (b \tanh (c+d x))^{3/2}}{d \tanh ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tanh[c + d*x])^(3/2),x]

[Out]

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

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Maple [A]
time = 1.74, size = 58, normalized size = 0.77

method	result	size
derivativedivides	\(-\frac {2 b \left (\sqrt {b \tanh \left (d x +c \right )}-\frac {\sqrt {b}\, \arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}-\frac {\sqrt {b}\, \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}\right )}{d}\)	\(58\)
default	\(-\frac {2 b \left (\sqrt {b \tanh \left (d x +c \right )}-\frac {\sqrt {b}\, \arctanh \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}-\frac {\sqrt {b}\, \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{2}\right )}{d}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tanh(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/d*b*((b*tanh(d*x+c))^(1/2)-1/2*b^(1/2)*arctanh((b*tanh(d*x+c))^(1/2)/b^(1/2))-1/2*b^(1/2)*arctan((b*tanh(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))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tanh(d*x + c))^(3/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (61) = 122\).
time = 0.51, size = 638, normalized size = 8.51 \begin {gather*} \left [-\frac {2 \, \sqrt {-b} 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 ) - \sqrt {-b} 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 \, b \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{4 \, d}, -\frac {2 \, b^{\frac {3}{2}} \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 ) - b^{\frac {3}{2}} \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 \, b \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}}}{4 \, d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tanh(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(2*sqrt(-b)*b*arctan((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)) -
sqrt(-b)*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
*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/d, -1/4*(2*b^(3/2)*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)) - b^(3/2)*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)/cosh(d*x + c)) - b) + 8*b*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/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 {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tanh(d*x+c))**(3/2),x)

[Out]

Integral((b*tanh(c + d*x))**(3/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (61) = 122\).
time = 0.45, size = 131, normalized size = 1.75 \begin {gather*} \frac {{\left (2 \, \sqrt {b} \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 ) - \sqrt {b} \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 \, b}{\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}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tanh(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.16, size = 61, normalized size = 0.81 \begin {gather*} \frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {2\,b\,\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{d}+\frac {b^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tanh(c + d*x))^(3/2),x)

[Out]

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

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