3.197 \(\int \frac {x^m \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx\)

Optimal. Leaf size=102 \[ \frac {x^{m+1} \sinh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-a^2 x^2\right )}{m^2+3 m+2} \]

[Out]

x^(1+m)*arcsinh(a*x)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],-a^2*x^2)/(1+m)-a*x^(2+m)*HypergeometricPFQ([1, 1+
1/2*m, 1+1/2*m],[3/2+1/2*m, 2+1/2*m],-a^2*x^2)/(m^2+3*m+2)

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Rubi [A]  time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {5762} \[ \frac {x^{m+1} \sinh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-a^2 x^2\right )}{m^2+3 m+2} \]

Antiderivative was successfully verified.

[In]

Int[(x^m*ArcSinh[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

(x^(1 + m)*ArcSinh[a*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)])/(1 + m) - (a*x^(2 + m)*Hyper
geometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(a^2*x^2)])/(2 + 3*m + m^2)

Rule 5762

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x
)^(m + 1)*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/(Sqrt[d]*f*(m + 1)),
x] - Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(c^2*x^2)])/(Sqrt
[d]*f^2*(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[d, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^m \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx &=\frac {x^{1+m} \sinh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};-a^2 x^2\right )}{1+m}-\frac {a x^{2+m} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-a^2 x^2\right )}{2+3 m+m^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 97, normalized size = 0.95 \[ \frac {x^{m+1} \left ((m+2) \sinh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-a^2 x^2\right )-a x \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-a^2 x^2\right )\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^m*ArcSinh[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

(x^(1 + m)*((2 + m)*ArcSinh[a*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)] - a*x*Hypergeometric
PFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(a^2*x^2)]))/((1 + m)*(2 + m))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m} \operatorname {arsinh}\left (a x\right )}{\sqrt {a^{2} x^{2} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m*arcsinh(a*x)/sqrt(a^2*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m*arcsinh(a*x)/sqrt(a^2*x^2 + 1), x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \arcsinh \left (a x \right )}{\sqrt {a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arcsinh(a*x)/(a^2*x^2+1)^(1/2),x)

[Out]

int(x^m*arcsinh(a*x)/(a^2*x^2+1)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \operatorname {arsinh}\left (a x\right )}{\sqrt {a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m*arcsinh(a*x)/sqrt(a^2*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*asinh(a*x))/(a^2*x^2 + 1)^(1/2),x)

[Out]

int((x^m*asinh(a*x))/(a^2*x^2 + 1)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*asinh(a*x)/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**m*asinh(a*x)/sqrt(a**2*x**2 + 1), x)

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