3.143 \(\int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=88 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{b^2 d} \]

[Out]

2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+2*(a^2-b^2)/a/b^2/d/(a+b*sech(d*x+c))^(1/2)+2*(a+b*sech(d
*x+c))^(1/2)/b^2/d

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Rubi [A]  time = 0.14, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.67, 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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fricas [B]  time = 1.24, size = 1107, normalized size = 12.58 \[ \text {result too large to display} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.56, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{3}\left (d x +c \right )}{\left (a +b \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tanh}\left (c+d\,x\right )}^3}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(c + d*x)^3/(a + b/cosh(c + d*x))^(3/2),x)

[Out]

int(tanh(c + d*x)^3/(a + b/cosh(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{3}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

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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