3.149 \(\int \frac {1}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=347 \[ \frac {2 b^2 \tanh (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \sqrt {a+b} \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}-\frac {2 \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}} \]

[Out]

-2*coth(d*x+c)*EllipticE((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1
/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/a/d/(a+b)^(1/2)+2*coth(d*x+c)*EllipticF((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/
2),((a+b)/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/a/d/(a+b)^(1/2)+2*cot
h(d*x+c)*EllipticPi((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sech(d*
x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/a^2/d+2*b^2*tanh(d*x+c)/a/(a^2-b^2)/d/(a+b*sech(d*x+c))^(1
/2)

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Rubi [A]  time = 0.34, antiderivative size = 347, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3785, 4058, 3921, 3784, 3832, 4004} \[ \frac {2 b^2 \tanh (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \sqrt {a+b} \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}-\frac {2 \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x])^(-3/2),x]

[Out]

(-2*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))])/(a*Sqrt[a + b]*d) + (2*Coth[c + d*x]*EllipticF[A
rcSin[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*Sqrt[a + b]*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^2*d) + (2*b^2*Tanh[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sech[c + d*x]])

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 3785

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

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 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 4058

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b \text {sech}(c+d x))^{3/2}} \, dx &=\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}-\frac {2 \int \frac {\frac {1}{2} \left (-a^2+b^2\right )+\frac {1}{2} a b \text {sech}(c+d x)+\frac {1}{2} b^2 \text {sech}^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}-\frac {2 \int \frac {\frac {1}{2} \left (-a^2+b^2\right )+\left (\frac {a b}{2}-\frac {b^2}{2}\right ) \text {sech}(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a \left (a^2-b^2\right )}-\frac {b^2 \int \frac {\text {sech}(c+d x) (1+\text {sech}(c+d x))}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {2 \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}}}{a \sqrt {a+b} d}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}+\frac {\int \frac {1}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a}-\frac {b \int \frac {\text {sech}(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a (a+b)}\\ &=-\frac {2 \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}}}{a \sqrt {a+b} d}+\frac {2 \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}}}{a \sqrt {a+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^2 d}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 86.67, size = 0, normalized size = 0.00 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Sech[c + d*x])^(-3/2),x]

[Out]

Integrate[(a + b*Sech[c + d*x])^(-3/2), x]

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sech(d*x + c) + a)^(-3/2), x)

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sech(d*x+c))^(3/2),x)

[Out]

int(1/(a+b*sech(d*x+c))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sech(d*x + c) + a)^(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cosh(c + d*x))^(3/2),x)

[Out]

int(1/(a + b/cosh(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))**(3/2),x)

[Out]

Integral((a + b*sech(c + d*x))**(-3/2), x)

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