3.138 \(\int \frac{\tanh ^4(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx\)

Optimal. Leaf size=610 \[ \frac{2 \sqrt{a+b} \left (8 a^2-2 a b+9 b^2\right ) \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b^3 d}-\frac{4 \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{b d}+\frac{2 (a-b) \sqrt{a+b} \left (8 a^2+9 b^2\right ) \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^4 d}-\frac{8 a \tanh (c+d x) \sqrt{a+b \text{sech}(c+d x)}}{15 b^2 d}-\frac{4 (a-b) \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b^2 d}+\frac{2 \tanh (c+d x) \text{sech}(c+d x) \sqrt{a+b \text{sech}(c+d x)}}{5 b d}+\frac{2 \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d} \]

[Out]

(-4*(a - b)*Sqrt[a + b]*Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)
]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a
+ b]*(8*a^2 + 9*b^2)*Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S
qrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(15*b^4*d) - (4*Sqrt[a + b]*Cot
h[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x
]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(b*d) + (2*Sqrt[a + b]*(8*a^2 - 2*a*b + 9*b^2)*Coth[c +
d*x]*EllipticF[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(
a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(15*b^3*d) + (2*Sqrt[a + b]*Coth[c + d*x]*EllipticPi[(a + b)
/a, ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt
[-((b*(1 + Sech[c + d*x]))/(a - b))])/(a*d) - (8*a*Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x])/(15*b^2*d) + (2*Se
ch[c + d*x]*Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x])/(5*b*d)

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Rubi [A]  time = 0.790606, antiderivative size = 610, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3895, 3784, 3837, 3832, 4004, 3860, 4082, 4005} \[ \frac{2 \sqrt{a+b} \left (8 a^2-2 a b+9 b^2\right ) \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 (a-b) \sqrt{a+b} \left (8 a^2+9 b^2\right ) \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^4 d}-\frac{8 a \tanh (c+d x) \sqrt{a+b \text{sech}(c+d x)}}{15 b^2 d}-\frac{4 (a-b) \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b^2 d}+\frac{2 \tanh (c+d x) \text{sech}(c+d x) \sqrt{a+b \text{sech}(c+d x)}}{5 b d}-\frac{4 \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b d}+\frac{2 \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[c + d*x]^4/Sqrt[a + b*Sech[c + d*x]],x]

[Out]

(-4*(a - b)*Sqrt[a + b]*Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)
]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a
+ b]*(8*a^2 + 9*b^2)*Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S
qrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(15*b^4*d) - (4*Sqrt[a + b]*Cot
h[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x
]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(b*d) + (2*Sqrt[a + b]*(8*a^2 - 2*a*b + 9*b^2)*Coth[c +
d*x]*EllipticF[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(
a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(15*b^3*d) + (2*Sqrt[a + b]*Coth[c + d*x]*EllipticPi[(a + b)
/a, ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt
[-((b*(1 + Sech[c + d*x]))/(a - b))])/(a*d) - (8*a*Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x])/(15*b^2*d) + (2*Se
ch[c + d*x]*Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x])/(5*b*d)

Rule 3895

Int[cot[(c_.) + (d_.)*(x_)]^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandIntegrand
[(a + b*Csc[c + d*x])^n, (-1 + Csc[c + d*x]^2)^(m/2), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0]
&& IGtQ[m/2, 0] && IntegerQ[n - 1/2]

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3837

Int[csc[(e_.) + (f_.)*(x_)]^2/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> -Int[Csc[e + f*x]/Sqrt[
a + b*Csc[e + f*x]], x] + Int[(Csc[e + f*x]*(1 + Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x] /; FreeQ[{a, b, e
, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 3860

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*d^2
*Cos[e + f*x]*(d*Csc[e + f*x])^(n - 2)*Sqrt[a + b*Csc[e + f*x]])/(b*f*(2*n - 3)), x] + Dist[d^3/(b*(2*n - 3)),
Int[((d*Csc[e + f*x])^(n - 3)*Simp[2*a*(n - 3) + b*(2*n - 5)*Csc[e + f*x] - 2*a*(n - 2)*Csc[e + f*x]^2, x])/S
qrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n
]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rubi steps

\begin{align*} \int \frac{\tanh ^4(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx &=\int \left (\frac{1}{\sqrt{a+b \text{sech}(c+d x)}}-\frac{2 \text{sech}^2(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}}+\frac{\text{sech}^4(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}}\right ) \, dx\\ &=-\left (2 \int \frac{\text{sech}^2(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx\right )+\int \frac{1}{\sqrt{a+b \text{sech}(c+d x)}} \, dx+\int \frac{\text{sech}^4(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx\\ &=\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}+\frac{2 \text{sech}(c+d x) \sqrt{a+b \text{sech}(c+d x)} \tanh (c+d x)}{5 b d}+2 \int \frac{\text{sech}(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx-2 \int \frac{\text{sech}(c+d x) (1+\text{sech}(c+d x))}{\sqrt{a+b \text{sech}(c+d x)}} \, dx+\frac{\int \frac{\text{sech}(c+d x) \left (2 a+3 b \text{sech}(c+d x)-4 a \text{sech}^2(c+d x)\right )}{\sqrt{a+b \text{sech}(c+d x)}} \, dx}{5 b}\\ &=-\frac{4 (a-b) \sqrt{a+b} \coth (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{b^2 d}-\frac{4 \sqrt{a+b} \coth (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{b d}+\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{8 a \sqrt{a+b \text{sech}(c+d x)} \tanh (c+d x)}{15 b^2 d}+\frac{2 \text{sech}(c+d x) \sqrt{a+b \text{sech}(c+d x)} \tanh (c+d x)}{5 b d}+\frac{2 \int \frac{\text{sech}(c+d x) \left (a b+\frac{1}{2} \left (8 a^2+9 b^2\right ) \text{sech}(c+d x)\right )}{\sqrt{a+b \text{sech}(c+d x)}} \, dx}{15 b^2}\\ &=-\frac{4 (a-b) \sqrt{a+b} \coth (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{b^2 d}-\frac{4 \sqrt{a+b} \coth (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{b d}+\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{8 a \sqrt{a+b \text{sech}(c+d x)} \tanh (c+d x)}{15 b^2 d}+\frac{2 \text{sech}(c+d x) \sqrt{a+b \text{sech}(c+d x)} \tanh (c+d x)}{5 b d}+\frac{1}{15} \left (9+\frac{8 a^2}{b^2}\right ) \int \frac{\text{sech}(c+d x) (1+\text{sech}(c+d x))}{\sqrt{a+b \text{sech}(c+d x)}} \, dx-\frac{\left (8 a^2-2 a b+9 b^2\right ) \int \frac{\text{sech}(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx}{15 b^2}\\ &=-\frac{4 (a-b) \sqrt{a+b} \coth (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{b^2 d}+\frac{2 (a-b) \sqrt{a+b} \left (8 a^2+9 b^2\right ) \coth (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{15 b^4 d}-\frac{4 \sqrt{a+b} \coth (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{b d}+\frac{2 \sqrt{a+b} \left (8 a^2-2 a b+9 b^2\right ) \coth (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{15 b^3 d}+\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{8 a \sqrt{a+b \text{sech}(c+d x)} \tanh (c+d x)}{15 b^2 d}+\frac{2 \text{sech}(c+d x) \sqrt{a+b \text{sech}(c+d x)} \tanh (c+d x)}{5 b d}\\ \end{align*}

Mathematica [F]  time = 180.001, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[Tanh[c + d*x]^4/Sqrt[a + b*Sech[c + d*x]],x]

[Out]

$Aborted

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Maple [F]  time = 0.342, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tanh \left ( dx+c \right ) \right ) ^{4}{\frac{1}{\sqrt{a+b{\rm sech} \left (dx+c\right )}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x)

[Out]

int(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (d x + c\right )^{4}}{\sqrt{b \operatorname{sech}\left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(tanh(d*x + c)^4/sqrt(b*sech(d*x + c) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\tanh \left (d x + c\right )^{4}}{\sqrt{b \operatorname{sech}\left (d x + c\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(tanh(d*x + c)^4/sqrt(b*sech(d*x + c) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{4}{\left (c + d x \right )}}{\sqrt{a + b \operatorname{sech}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)**4/(a+b*sech(d*x+c))**(1/2),x)

[Out]

Integral(tanh(c + d*x)**4/sqrt(a + b*sech(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (d x + c\right )^{4}}{\sqrt{b \operatorname{sech}\left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(tanh(d*x + c)^4/sqrt(b*sech(d*x + c) + a), x)