3.119 \(\int x^m \sinh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=137 \[ \frac {2 a^2 x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};-a^2 x^2\right )}{m^3+6 m^2+11 m+6}-\frac {2 a x^{m+2} \sinh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m^2+3 m+2}+\frac {x^{m+1} \sinh ^{-1}(a x)^2}{m+1} \]

[Out]

x^(1+m)*arcsinh(a*x)^2/(1+m)-2*a*x^(2+m)*arcsinh(a*x)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],-a^2*x^2)/(m^2+3*m+2)
+2*a^2*x^(3+m)*HypergeometricPFQ([1, 3/2+1/2*m, 3/2+1/2*m],[2+1/2*m, 5/2+1/2*m],-a^2*x^2)/(m^3+6*m^2+11*m+6)

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Rubi [A]  time = 0.10, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5661, 5762} \[ \frac {2 a^2 x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};-a^2 x^2\right )}{m^3+6 m^2+11 m+6}-\frac {2 a x^{m+2} \sinh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m^2+3 m+2}+\frac {x^{m+1} \sinh ^{-1}(a x)^2}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcSinh[a*x]^2,x]

[Out]

(x^(1 + m)*ArcSinh[a*x]^2)/(1 + m) - (2*a*x^(2 + m)*ArcSinh[a*x]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2,
-(a^2*x^2)])/(2 + 3*m + m^2) + (2*a^2*x^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m
/2}, -(a^2*x^2)])/(6 + 11*m + 6*m^2 + m^3)

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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 x^m \sinh ^{-1}(a x)^2 \, dx &=\frac {x^{1+m} \sinh ^{-1}(a x)^2}{1+m}-\frac {(2 a) \int \frac {x^{1+m} \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{1+m}\\ &=\frac {x^{1+m} \sinh ^{-1}(a x)^2}{1+m}-\frac {2 a x^{2+m} \sinh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-a^2 x^2\right )}{2+3 m+m^2}+\frac {2 a^2 x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};-a^2 x^2\right )}{6+11 m+6 m^2+m^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 123, normalized size = 0.90 \[ \frac {x^{m+1} \left (2 a^2 x^2 \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};-a^2 x^2\right )+(m+3) \sinh ^{-1}(a x) \left ((m+2) \sinh ^{-1}(a x)-2 a x \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )\right )\right )}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcSinh[a*x]^2,x]

[Out]

(x^(1 + m)*((3 + m)*ArcSinh[a*x]*((2 + m)*ArcSinh[a*x] - 2*a*x*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(
a^2*x^2)]) + 2*a^2*x^2*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, -(a^2*x^2)]))/((1 +
m)*(2 + m)*(3 + m))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \operatorname {arsinh}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m*arcsinh(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m*arcsinh(a*x)^2, x)

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maple [F]  time = 1.21, size = 0, normalized size = 0.00 \[ \int x^{m} \arcsinh \left (a x \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arcsinh(a*x)^2,x)

[Out]

int(x^m*arcsinh(a*x)^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*x^m*log(a*x + sqrt(a^2*x^2 + 1))^2/(m + 1) - integrate(2*(sqrt(a^2*x^2 + 1)*a^2*x^2*x^m + (a^3*x^3 + a*x)*x^
m)*log(a*x + sqrt(a^2*x^2 + 1))/(a^3*(m + 1)*x^3 + a*(m + 1)*x + (a^2*(m + 1)*x^2 + m + 1)*sqrt(a^2*x^2 + 1)),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*asinh(a*x)^2,x)

[Out]

int(x^m*asinh(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*asinh(a*x)**2,x)

[Out]

Integral(x**m*asinh(a*x)**2, x)

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