3.194 \(\int \frac {x^m (a+b \sinh ^{-1}(c x))}{\sqrt {d+c^2 d x^2}} \, dx\)

Optimal. Leaf size=161 \[ \frac {\sqrt {c^2 x^2+1} x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{(m+1) \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {c^2 d x^2+d}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {5764, 5762} \[ \frac {\sqrt {c^2 x^2+1} x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{(m+1) \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/((
1 + m)*Sqrt[d + c^2*d*x^2]) - (b*c*x^(2 + m)*Sqrt[1 + c^2*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 +
m/2, 2 + m/2}, -(c^2*x^2)])/((2 + 3*m + m^2)*Sqrt[d + c^2*d*x^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]

Rule 5764

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 129, normalized size = 0.80 \[ \frac {\sqrt {c^2 x^2+1} x^{m+1} \left ((m+2) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-b c x \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )\right )}{(m+1) (m+2) \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Sqrt[1 + c^2*x^2]*((2 + m)*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*
x^2)] - b*c*x*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(c^2*x^2)]))/((1 + m)*(2 + m)*Sq
rt[d + c^2*d*x^2])

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{\sqrt {c^{2} d x^{2} + d}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsinh(c*x) + a)*x^m/sqrt(c^2*d*x^2 + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)*x^m/sqrt(c^2*d*x^2 + d), x)

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \left (a +b \arcsinh \left (c x \right )\right )}{\sqrt {c^{2} d \,x^{2}+d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)*x^m/sqrt(c^2*d*x^2 + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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