3.143 \(\int \frac{\tanh ^3(c+d x)}{(a+b \text{sech}(c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=88 \[ \frac{2 \left (a^2-b^2\right )}{a b^2 d \sqrt{a+b \text{sech}(c+d x)}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{2 \sqrt{a+b \text{sech}(c+d x)}}{b^2 d} \]
[Out]
(2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (2*(a^2 - b^2))/(a*b^2*d*Sqrt[a + b*Sech[c + d*x]
]) + (2*Sqrt[a + b*Sech[c + d*x]])/(b^2*d)
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Rubi [A] time = 0.141404, antiderivative size = 88, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used =
{3885, 898, 1261, 206} \[ \frac{2 \left (a^2-b^2\right )}{a b^2 d \sqrt{a+b \text{sech}(c+d x)}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{2 \sqrt{a+b \text{sech}(c+d x)}}{b^2 d} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[c + d*x]^3/(a + b*Sech[c + d*x])^(3/2),x]
[Out]
(2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (2*(a^2 - b^2))/(a*b^2*d*Sqrt[a + b*Sech[c + d*x]
]) + (2*Sqrt[a + b*Sech[c + d*x]])/(b^2*d)
Rule 3885
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Rule 898
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]
Rule 1261
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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 \frac{\tanh ^3(c+d x)}{(a+b \text{sech}(c+d x))^{3/2}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{b^2-x^2}{x (a+x)^{3/2}} \, dx,x,b \text{sech}(c+d x)\right )}{b^2 d}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{-a^2+b^2+2 a x^2-x^4}{x^2 \left (-a+x^2\right )} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{b^2 d}\\ &=-\frac{2 \operatorname{Subst}\left (\int \left (-1+\frac{a^2-b^2}{a x^2}-\frac{b^2}{a \left (a-x^2\right )}\right ) \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{b^2 d}\\ &=\frac{2 \left (a^2-b^2\right )}{a b^2 d \sqrt{a+b \text{sech}(c+d x)}}+\frac{2 \sqrt{a+b \text{sech}(c+d x)}}{b^2 d}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{a d}\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{2 \left (a^2-b^2\right )}{a b^2 d \sqrt{a+b \text{sech}(c+d x)}}+\frac{2 \sqrt{a+b \text{sech}(c+d x)}}{b^2 d}\\ \end{align*}
Mathematica [A] time = 0.64969, size = 103, normalized size = 1.17 \[ \frac{2 \left (2 a^2+\frac{b^2 \sqrt{a \cosh (c+d x)+b} \tanh ^{-1}\left (\frac{\sqrt{a \cosh (c+d x)+b}}{\sqrt{a \cosh (c+d x)}}\right )}{\sqrt{a \cosh (c+d x)}}+a b \text{sech}(c+d x)-b^2\right )}{a b^2 d \sqrt{a+b \text{sech}(c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[c + d*x]^3/(a + b*Sech[c + d*x])^(3/2),x]
[Out]
(2*(2*a^2 - b^2 + (b^2*ArcTanh[Sqrt[b + a*Cosh[c + d*x]]/Sqrt[a*Cosh[c + d*x]]]*Sqrt[b + a*Cosh[c + d*x]])/Sqr
t[a*Cosh[c + d*x]] + a*b*Sech[c + d*x]))/(a*b^2*d*Sqrt[a + b*Sech[c + d*x]])
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Maple [F] time = 0.167, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tanh \left ( dx+c \right ) \right ) ^{3} \left ( a+b{\rm sech} \left (dx+c\right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x)
[Out]
int(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (d x + c\right )^{3}}{{\left (b \operatorname{sech}\left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate(tanh(d*x + c)^3/(b*sech(d*x + c) + a)^(3/2), x)
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Fricas [B] time = 11.7931, size = 2801, normalized size = 31.83 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
[1/2*((a*b^2*cosh(d*x + c)^2 + a*b^2*sinh(d*x + c)^2 + 2*b^3*cosh(d*x + c) + a*b^2 + 2*(a*b^2*cosh(d*x + c) +
b^3)*sinh(d*x + c))*sqrt(a)*log(-(2*a^2*cosh(d*x + c)^4 + 2*a^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^3 + 4*(2
*a^2*cosh(d*x + c) + a*b)*sinh(d*x + c)^3 + 4*a*b*cosh(d*x + c) + (4*a^2 + b^2)*cosh(d*x + c)^2 + (12*a^2*cosh
(d*x + c)^2 + 12*a*b*cosh(d*x + c) + 4*a^2 + b^2)*sinh(d*x + c)^2 + 2*a^2 + 2*(a*cosh(d*x + c)^4 + a*sinh(d*x
+ c)^4 + b*cosh(d*x + c)^3 + (4*a*cosh(d*x + c) + b)*sinh(d*x + c)^3 + 2*a*cosh(d*x + c)^2 + (6*a*cosh(d*x + c
)^2 + 3*b*cosh(d*x + c) + 2*a)*sinh(d*x + c)^2 + b*cosh(d*x + c) + (4*a*cosh(d*x + c)^3 + 3*b*cosh(d*x + c)^2
+ 4*a*cosh(d*x + c) + b)*sinh(d*x + c) + a)*sqrt(a)*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)) + 2*(4*a^2*cosh(
d*x + c)^3 + 6*a*b*cosh(d*x + c)^2 + 2*a*b + (4*a^2 + b^2)*cosh(d*x + c))*sinh(d*x + c))/(cosh(d*x + c)^2 + 2*
cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) + 4*(2*a^2*b*cosh(d*x + c) + 2*a^3 - a*b^2 + (2*a^3 - a*b^2)*c
osh(d*x + c)^2 + (2*a^3 - a*b^2)*sinh(d*x + c)^2 + 2*(a^2*b + (2*a^3 - a*b^2)*cosh(d*x + c))*sinh(d*x + c))*sq
rt((a*cosh(d*x + c) + b)/cosh(d*x + c)))/(a^3*b^2*d*cosh(d*x + c)^2 + a^3*b^2*d*sinh(d*x + c)^2 + 2*a^2*b^3*d*
cosh(d*x + c) + a^3*b^2*d + 2*(a^3*b^2*d*cosh(d*x + c) + a^2*b^3*d)*sinh(d*x + c)), -((a*b^2*cosh(d*x + c)^2 +
a*b^2*sinh(d*x + c)^2 + 2*b^3*cosh(d*x + c) + a*b^2 + 2*(a*b^2*cosh(d*x + c) + b^3)*sinh(d*x + c))*sqrt(-a)*a
rctan((a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + b*cosh(d*x + c) + (2*a*cosh(d*x + c) + b)*sinh(d*x + c) + a)*sq
rt(-a)*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c))/(a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x +
c) + a^2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c))) - 2*(2*a^2*b*cosh(d*x + c) + 2*a^3 - a*b^2 + (2*a^3 -
a*b^2)*cosh(d*x + c)^2 + (2*a^3 - a*b^2)*sinh(d*x + c)^2 + 2*(a^2*b + (2*a^3 - a*b^2)*cosh(d*x + c))*sinh(d*x
+ c))*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)))/(a^3*b^2*d*cosh(d*x + c)^2 + a^3*b^2*d*sinh(d*x + c)^2 + 2*a^
2*b^3*d*cosh(d*x + c) + a^3*b^2*d + 2*(a^3*b^2*d*cosh(d*x + c) + a^2*b^3*d)*sinh(d*x + c))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\left (c + d x \right )}}{\left (a + b \operatorname{sech}{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(d*x+c)**3/(a+b*sech(d*x+c))**(3/2),x)
[Out]
Integral(tanh(c + d*x)**3/(a + b*sech(c + d*x))**(3/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (d x + c\right )^{3}}{{\left (b \operatorname{sech}\left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate(tanh(d*x + c)^3/(b*sech(d*x + c) + a)^(3/2), x)