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

Optimal. Leaf size=87 \[ \frac {(d x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{d (m+1)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (d x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{d (m+1)^2} \]

[Out]

(d*x)^(1+m)*(a+b*arcsech(c*x))/d/(1+m)+b*(d*x)^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)*(1/(c*x+1
))^(1/2)*(c*x+1)^(1/2)/d/(1+m)^2

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Rubi [A]  time = 0.04, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.15, size = 97, normalized size = 1.11 \[ \frac {x (d x)^m \left ((m+1) (c x-1) \left (a+b \text {sech}^{-1}(c x)\right )-b \sqrt {\frac {1-c x}{c x+1}} \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )\right )}{(m+1)^2 (c x-1)} \]

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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fricas [F]  time = 1.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (d x\right )^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (d x\right )^{m}\,{d x} \]

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)

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maple [F]  time = 2.60, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{m} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (c^{2} d^{m} \int \frac {x^{2} x^{m}}{c^{2} {\left (m + 1\right )} x^{2} + {\left (c^{2} {\left (m + 1\right )} x^{2} - m - 1\right )} \sqrt {c x + 1} \sqrt {-c x + 1} - m - 1}\,{d x} + \frac {d^{m} x x^{m} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right ) - d^{m} x x^{m} \log \relax (x)}{m + 1} - \int \frac {{\left (c^{2} d^{m} {\left (m + 1\right )} x^{2} \log \relax (c) - d^{m} {\left (m + 1\right )} \log \relax (c) + d^{m}\right )} x^{m}}{c^{2} {\left (m + 1\right )} x^{2} - m - 1}\,{d x}\right )} b + \frac {\left (d x\right )^{m + 1} a}{d {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*d^m*integrate(x^2*x^m/(c^2*(m + 1)*x^2 + (c^2*(m + 1)*x^2 - m - 1)*sqrt(c*x + 1)*sqrt(-c*x + 1) - m - 1),
x) + (d^m*x*x^m*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1) - d^m*x*x^m*log(x))/(m + 1) - integrate((c^2*d^m*(m + 1
)*x^2*log(c) - d^m*(m + 1)*log(c) + d^m)*x^m/(c^2*(m + 1)*x^2 - m - 1), x))*b + (d*x)^(m + 1)*a/(d*(m + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a + b*acosh(1/(c*x))),x)

[Out]

int((d*x)^m*(a + b*acosh(1/(c*x))), x)

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

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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