3.88 \(\int \frac {e^{\text {sech}^{-1}(c x)} (d x)^m}{1-c^2 x^2} \, dx\)

Optimal. Leaf size=89 \[ \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (d x)^m \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};c^2 x^2\right )}{c m}+\frac {(d x)^m \, _2F_1\left (1,\frac {m}{2};\frac {m+2}{2};c^2 x^2\right )}{c m} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6341, 6677, 125, 364} \[ \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (d x)^m \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};c^2 x^2\right )}{c m}+\frac {(d x)^m \, _2F_1\left (1,\frac {m}{2};\frac {m+2}{2};c^2 x^2\right )}{c m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6341

Int[(E^ArcSech[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/(a*c), Int[((d*x)^(m
- 1)*Sqrt[1/(1 + c*x)])/Sqrt[1 - c*x], 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 6677

Int[(u_)*((c_.)*((a_.) + (b_.)*(x_))^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c*(a + b*x)^n)^FracPart[p])/
(a + b*x)^(n*FracPart[p]), Int[u*(a + b*x)^(n*p), x], x] /; FreeQ[{a, b, c, n, p}, x] &&  !IntegerQ[p] &&  !Ma
tchQ[u, x^(n1_.)*(v_.) /; EqQ[n, n1 + 1]]

Rubi steps

\begin {align*} \int \frac {e^{\text {sech}^{-1}(c x)} (d x)^m}{1-c^2 x^2} \, dx &=\frac {d \int \frac {(d x)^{-1+m} \sqrt {\frac {1}{1+c x}}}{\sqrt {1-c x}} \, dx}{c}+\frac {d \int \frac {(d x)^{-1+m}}{1-c^2 x^2} \, dx}{c}\\ &=\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 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(d x)^{-1+m}}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{c}\\ &=\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 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(d x)^{-1+m}}{\sqrt {1-c^2 x^2}} \, dx}{c}\\ &=\frac {(d x)^m \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {2+m}{2};c^2 x^2\right )}{c 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 [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \int \frac {e^{\text {sech}^{-1}(c x)} (d x)^m}{1-c^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 1.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (d x\right )^{m} c x \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + \left (d x\right )^{m}}{c^{3} x^{3} - c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right ) \left (d x \right )^{m}}{-c^{2} x^{2}+1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(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 \[ -d^{m} \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{m}}{c^{3} x^{3} - c x}\,{d x} - d^{m} \int \frac {x^{m}}{2 \, {\left (c x + 1\right )}}\,{d x} - d^{m} \int \frac {x^{m}}{2 \, {\left (c x - 1\right )}}\,{d x} + \frac {d^{m} x^{m}}{c m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-int((((1/(c*x) - 1)^(1/2)*(1/(c*x) + 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 \[ - \frac {\int \frac {\left (d x\right )^{m}}{c^{2} x^{3} - x}\, dx + \int \frac {c x \left (d x\right )^{m} \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{3} - x}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)**(1/2)*(1+1/c/x)**(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*x))*sqrt(1 + 1/(c*x))/(c**2*x**
3 - x), x))/c

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