3.59 $$\int \frac{e^{\text{csch}^{-1}(c x)} (d x)^m}{1+c^2 x^2} \, dx$$

Optimal. Leaf size=85 $\frac{(d x)^m \text{Hypergeometric2F1}\left (1,\frac{m}{2},\frac{m+2}{2},-c^2 x^2\right )}{c m}-\frac{d (d x)^{m-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-m}{2},\frac{3-m}{2},-\frac{1}{c^2 x^2}\right )}{c^2 (1-m)}$

[Out]

-((d*(d*x)^(-1 + m)*Hypergeometric2F1[1/2, (1 - m)/2, (3 - m)/2, -(1/(c^2*x^2))])/(c^2*(1 - m))) + ((d*x)^m*Hy
pergeometric2F1[1, m/2, (2 + m)/2, -(c^2*x^2)])/(c*m)

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Rubi [A]  time = 0.103734, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {6342, 339, 364} $\frac{(d x)^m \, _2F_1\left (1,\frac{m}{2};\frac{m+2}{2};-c^2 x^2\right )}{c m}-\frac{d (d x)^{m-1} \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};-\frac{1}{c^2 x^2}\right )}{c^2 (1-m)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(E^ArcCsch[c*x]*(d*x)^m)/(1 + c^2*x^2),x]

[Out]

-((d*(d*x)^(-1 + m)*Hypergeometric2F1[1/2, (1 - m)/2, (3 - m)/2, -(1/(c^2*x^2))])/(c^2*(1 - m))) + ((d*x)^m*Hy
pergeometric2F1[1, m/2, (2 + m)/2, -(c^2*x^2)])/(c*m)

Rule 6342

Int[(E^ArcCsch[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d^2/(a*c^2), Int[(d*x)
^(m - 2)/Sqrt[1 + 1/(c^2*x^2)], x], x] + Dist[d/c, Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d,
m}, x] && EqQ[b - a*c^2, 0]

Rule 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, Subst
[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]

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 \frac{e^{\text{csch}^{-1}(c x)} (d x)^m}{1+c^2 x^2} \, dx &=\frac{d \int \frac{(d x)^{-1+m}}{1+c^2 x^2} \, dx}{c}+\frac{d^2 \int \frac{(d x)^{-2+m}}{\sqrt{1+\frac{1}{c^2 x^2}}} \, dx}{c^2}\\ &=\frac{(d x)^m \, _2F_1\left (1,\frac{m}{2};\frac{2+m}{2};-c^2 x^2\right )}{c m}-\frac{\left (d \left (\frac{1}{x}\right )^{-1+m} (d x)^{-1+m}\right ) \operatorname{Subst}\left (\int \frac{x^{-m}}{\sqrt{1+\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{c^2}\\ &=-\frac{d (d x)^{-1+m} \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};-\frac{1}{c^2 x^2}\right )}{c^2 (1-m)}+\frac{(d x)^m \, _2F_1\left (1,\frac{m}{2};\frac{2+m}{2};-c^2 x^2\right )}{c m}\\ \end{align*}

Mathematica [A]  time = 0.243144, size = 88, normalized size = 1.04 $\frac{(d x)^m \left (\frac{x \sqrt{\frac{1}{c^2 x^2}+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2},\frac{m}{2}+1,-c^2 x^2\right )}{\sqrt{c^2 x^2+1}}+\frac{\text{Hypergeometric2F1}\left (1,\frac{m}{2},\frac{m}{2}+1,-c^2 x^2\right )}{c}\right )}{m}$

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcCsch[c*x]*(d*x)^m)/(1 + c^2*x^2),x]

[Out]

((d*x)^m*((Sqrt[1 + 1/(c^2*x^2)]*x*Hypergeometric2F1[1/2, m/2, 1 + m/2, -(c^2*x^2)])/Sqrt[1 + c^2*x^2] + Hyper
geometric2F1[1, m/2, 1 + m/2, -(c^2*x^2)]/c))/m

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Maple [F]  time = 0.382, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx \right ) ^{m}}{{c}^{2}{x}^{2}+1} \left ({\frac{1}{cx}}+\sqrt{1+{\frac{1}{{c}^{2}{x}^{2}}}} \right ) }\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(1+1/c^2/x^2)^(1/2))*(d*x)^m/(c^2*x^2+1),x)

[Out]

int((1/c/x+(1+1/c^2/x^2)^(1/2))*(d*x)^m/(c^2*x^2+1),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}{\left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} + \frac{1}{c x}\right )}}{c^{2} x^{2} + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*(d*x)^m/(c^2*x^2+1),x, algorithm="maxima")

[Out]

integrate((d*x)^m*(sqrt(1/(c^2*x^2) + 1) + 1/(c*x))/(c^2*x^2 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (d x\right )^{m} c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + \left (d x\right )^{m}}{c^{3} x^{3} + c x}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*(d*x)^m/(c^2*x^2+1),x, algorithm="fricas")

[Out]

integral(((d*x)^m*c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + (d*x)^m)/(c^3*x^3 + c*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (d x\right )^{m}}{c^{2} x^{3} + x}\, dx + \int \frac{c x \left (d x\right )^{m} \sqrt{1 + \frac{1}{c^{2} x^{2}}}}{c^{2} x^{3} + x}\, dx}{c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c**2/x**2)**(1/2))*(d*x)**m/(c**2*x**2+1),x)

[Out]

(Integral((d*x)**m/(c**2*x**3 + x), x) + Integral(c*x*(d*x)**m*sqrt(1 + 1/(c**2*x**2))/(c**2*x**3 + x), x))/c

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}{\left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} + \frac{1}{c x}\right )}}{c^{2} x^{2} + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*(d*x)^m/(c^2*x^2+1),x, algorithm="giac")

[Out]

integrate((d*x)^m*(sqrt(1/(c^2*x^2) + 1) + 1/(c*x))/(c^2*x^2 + 1), x)