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

Optimal. Leaf size=85 \[ \frac {(d x)^m \, _2F_1\left (1,\frac {m}{2};\frac {m+2}{2};-c^2 x^2\right )}{c m}-\frac {d (d x)^{m-1} \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};-\frac {1}{c^2 x^2}\right )}{c^2 (1-m)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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])

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]

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*}

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Mathematica [A]  time = 0.25, size = 88, normalized size = 1.04 \[ \frac {(d x)^m \left (\frac {x \sqrt {\frac {1}{c^2 x^2}+1} \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m}{2}+1;-c^2 x^2\right )}{\sqrt {c^2 x^2+1}}+\frac {\, _2F_1\left (1,\frac {m}{2};\frac {m}{2}+1;-c^2 x^2\right )}{c}\right )}{m} \]

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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fricas [F]  time = 2.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d x\right )^{m} c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + \left (d x\right )^{m}}{c^{3} x^{3} + c x}, x\right ) \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (\sqrt {\frac {1}{c^2\,x^2}+1}+\frac {1}{c\,x}\right )\,{\left (d\,x\right )}^m}{c^2\,x^2+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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