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

Optimal. Leaf size=137 \[ \frac{2 a^2 x^{m+3} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},-a^2 x^2\right )}{m^3+6 m^2+11 m+6}-\frac{2 a x^{m+2} \sinh ^{-1}(a x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )}{m^2+3 m+2}+\frac{x^{m+1} \sinh ^{-1}(a x)^2}{m+1} \]

[Out]

(x^(1 + m)*ArcSinh[a*x]^2)/(1 + m) - (2*a*x^(2 + m)*ArcSinh[a*x]*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)

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Rubi [A]  time = 0.0998867, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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*}

Mathematica [A]  time = 0.0382418, size = 123, normalized size = 0.9 \[ \frac{x^{m+1} \left (2 a^2 x^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},-a^2 x^2\right )+(m+3) \sinh ^{-1}(a x) \left ((m+2) \sinh ^{-1}(a x)-2 a x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )\right )\right )}{(m+1) (m+2) (m+3)} \]

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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Maple [F]  time = 0.55, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}\, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{m} \operatorname{arsinh}\left (a x\right )^{2}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \operatorname{asinh}^{2}{\left (a x \right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \operatorname{arsinh}\left (a x\right )^{2}\,{d x} \end{align*}

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)