3.71 \(\int (d x)^m (a+b \text{sech}^{-1}(c x)) \, dx\)

Optimal. Leaf size=87 \[ \frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} (d x)^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right )}{d (m+1)^2}+\frac{(d x)^{m+1} \left (a+b \text{sech}^{-1}(c x)\right )}{d (m+1)} \]

[Out]

((d*x)^(1 + m)*(a + b*ArcSech[c*x]))/(d*(1 + m)) + (b*(d*x)^(1 + m)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Hyperge
ometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(d*(1 + m)^2)

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Rubi [A]  time = 0.0371767, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6283, 125, 364} \[ \frac{(d x)^{m+1} \left (a+b \text{sech}^{-1}(c x)\right )}{d (m+1)}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} (d x)^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right )}{d (m+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(a + b*ArcSech[c*x]),x]

[Out]

((d*x)^(1 + m)*(a + b*ArcSech[c*x]))/(d*(1 + m)) + (b*(d*x)^(1 + m)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Hyperge
ometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(d*(1 + m)^2)

Rule 6283

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSech[c*
x]))/(d*(m + 1)), x] + Dist[(b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)])/(m + 1), Int[(d*x)^m/(Sqrt[1 - c*x]*Sqrt[1 + c
*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rule 125

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[(a*c + b*d*x^2)
^m*(f*x)^p, x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 0] && GtQ[a, 0] && GtQ
[c, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (d x)^m \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=\frac{(d x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{d (1+m)}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{(d x)^m}{\sqrt{1-c x} \sqrt{1+c x}} \, dx}{1+m}\\ &=\frac{(d x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{d (1+m)}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{(d x)^m}{\sqrt{1-c^2 x^2}} \, dx}{1+m}\\ &=\frac{(d x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{d (1+m)}+\frac{b (d x)^{1+m} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};c^2 x^2\right )}{d (1+m)^2}\\ \end{align*}

Mathematica [A]  time = 0.147044, size = 97, normalized size = 1.11 \[ \frac{x (d x)^m \left ((m+1) (c x-1) \left (a+b \text{sech}^{-1}(c x)\right )-b \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right )\right )}{(m+1)^2 (c x-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*(a + b*ArcSech[c*x]),x]

[Out]

(x*(d*x)^m*((1 + m)*(-1 + c*x)*(a + b*ArcSech[c*x]) - b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*Hypergeome
tric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2]))/((1 + m)^2*(-1 + c*x))

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Maple [F]  time = 1.601, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a+b*arcsech(c*x)),x)

[Out]

int((d*x)^m*(a+b*arcsech(c*x)),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arcsech(c*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arcsech(c*x)),x, algorithm="fricas")

[Out]

integral((b*arcsech(c*x) + a)*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(a+b*asech(c*x)),x)

[Out]

Integral((d*x)**m*(a + b*asech(c*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arcsech(c*x)),x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)*(d*x)^m, x)