3.14 \(\int (b \tanh (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=78 \[ \frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b (b \tanh (c+d x))^{3/2}}{3 d} \]
[Out]
-((b^(5/2)*ArcTan[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)
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Rubi [A] time = 0.0487595, antiderivative size = 78, 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 =
{3473, 3476, 329, 298, 203, 206} \[ \frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \tanh (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b (b \tanh (c+d x))^{3/2}}{3 d} \]
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 3473
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 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 298
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 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])
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 \operatorname{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 ) \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \tanh (c+d x)}\right )}{d}-\frac{b^3 \operatorname{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*}
Mathematica [A] time = 0.205455, size = 68, normalized size = 0.87 \[ -\frac{(b \tanh (c+d x))^{5/2} \left (2 \tanh ^{\frac{3}{2}}(c+d x)-3 \tanh ^{-1}\left (\sqrt{\tanh (c+d x)}\right )+3 \tan ^{-1}\left (\sqrt{\tanh (c+d x)}\right )\right )}{3 d \tanh ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*Tanh[c + d*x])^(5/2),x]
[Out]
-((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)))/(3*d*Tanh[c + d*x]^(5/2))
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Maple [A] time = 0.018, size = 63, normalized size = 0.8 \begin{align*} -{\frac{1}{d}{b}^{{\frac{5}{2}}}\arctan \left ({\sqrt{b\tanh \left ( dx+c \right ) }{\frac{1}{\sqrt{b}}}} \right ) }+{\frac{1}{d}{b}^{{\frac{5}{2}}}{\it Artanh} \left ({\sqrt{b\tanh \left ( dx+c \right ) }{\frac{1}{\sqrt{b}}}} \right ) }-{\frac{2\,b}{3\,d} \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((b*tanh(d*x+c))^(5/2),x)
[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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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((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 = 2.55349, size = 2653, normalized size = 34.01 \begin{align*} \text{result too large to display} \end{align*}
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., size = 0, normalized size = 0. \begin{align*} \int \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((b*tanh(d*x+c))**(5/2),x)
[Out]
Integral((b*tanh(c + d*x))**(5/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((b*tanh(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((b*tanh(d*x + c))^(5/2), x)