3.7.49 \(\int \frac {a+b \sinh ^{-1}(c x)}{(d+e x^2)^{3/2}} \, dx\) [649]

Optimal. Leaf size=70 \[ \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e}} \]

[Out]

-b*arctanh(e^(1/2)*(c^2*x^2+1)^(1/2)/c/(e*x^2+d)^(1/2))/d/e^(1/2)+x*(a+b*arcsinh(c*x))/d/(e*x^2+d)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {197, 5792, 12, 455, 65, 223, 212} \begin {gather*} \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2+1}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(d + e*x^2)^(3/2),x]

[Out]

(x*(a + b*ArcSinh[c*x]))/(d*Sqrt[d + e*x^2]) - (b*ArcTanh[(Sqrt[e]*Sqrt[1 + c^2*x^2])/(c*Sqrt[d + e*x^2])])/(d
*Sqrt[e])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 5792

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[e, c^2*d] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+e x^2\right )^{3/2}} \, dx &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-(b c) \int \frac {x}{d \sqrt {1+c^2 x^2} \sqrt {d+e x^2}} \, dx\\ &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {(b c) \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {d+e x^2}} \, dx}{d}\\ &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {(b c) \text {Subst}\left (\int \frac {1}{\sqrt {1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{2 d}\\ &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {d-\frac {e}{c^2}+\frac {e x^2}{c^2}}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{c d}\\ &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \text {Subst}\left (\int \frac {1}{1-\frac {e x^2}{c^2}} \, dx,x,\frac {\sqrt {1+c^2 x^2}}{\sqrt {d+e x^2}}\right )}{c d}\\ &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.08, size = 75, normalized size = 1.07 \begin {gather*} \frac {x \left (-b c x \sqrt {1+\frac {e x^2}{d}} F_1\left (1;\frac {1}{2},\frac {1}{2};2;-c^2 x^2,-\frac {e x^2}{d}\right )+2 \left (a+b \sinh ^{-1}(c x)\right )\right )}{2 d \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(d + e*x^2)^(3/2),x]

[Out]

(x*(-(b*c*x*Sqrt[1 + (e*x^2)/d]*AppellF1[1, 1/2, 1/2, 2, -(c^2*x^2), -((e*x^2)/d)]) + 2*(a + b*ArcSinh[c*x])))
/(2*d*Sqrt[d + e*x^2])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {a +b \arcsinh \left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/(e*x^2+d)^(3/2),x)

[Out]

int((a+b*arcsinh(c*x))/(e*x^2+d)^(3/2),x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c^2*d-%e>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (60) = 120\).
time = 0.36, size = 329, normalized size = 4.70 \begin {gather*} \frac {{\left (b x^{2} \cosh \left (1\right ) + b x^{2} \sinh \left (1\right ) + b d\right )} \sqrt {\cosh \left (1\right ) + \sinh \left (1\right )} \log \left (c^{4} d^{2} + {\left (8 \, c^{4} x^{4} + 8 \, c^{2} x^{2} + 1\right )} \cosh \left (1\right )^{2} + {\left (8 \, c^{4} x^{4} + 8 \, c^{2} x^{2} + 1\right )} \sinh \left (1\right )^{2} - 4 \, {\left (c^{3} d + {\left (2 \, c^{3} x^{2} + c\right )} \cosh \left (1\right ) + {\left (2 \, c^{3} x^{2} + c\right )} \sinh \left (1\right )\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} \sqrt {\cosh \left (1\right ) + \sinh \left (1\right )} + 2 \, {\left (4 \, c^{4} d x^{2} + 3 \, c^{2} d\right )} \cosh \left (1\right ) + 2 \, {\left (4 \, c^{4} d x^{2} + 3 \, c^{2} d + {\left (8 \, c^{4} x^{4} + 8 \, c^{2} x^{2} + 1\right )} \cosh \left (1\right )\right )} \sinh \left (1\right )\right ) + 4 \, {\left (b x \cosh \left (1\right ) + b x \sinh \left (1\right )\right )} \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 4 \, {\left (a x \cosh \left (1\right ) + a x \sinh \left (1\right )\right )} \sqrt {x^{2} \cosh \left (1\right ) + x^{2} \sinh \left (1\right ) + d}}{4 \, {\left (d x^{2} \cosh \left (1\right )^{2} + d x^{2} \sinh \left (1\right )^{2} + d^{2} \cosh \left (1\right ) + {\left (2 \, d x^{2} \cosh \left (1\right ) + d^{2}\right )} \sinh \left (1\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

1/4*((b*x^2*cosh(1) + b*x^2*sinh(1) + b*d)*sqrt(cosh(1) + sinh(1))*log(c^4*d^2 + (8*c^4*x^4 + 8*c^2*x^2 + 1)*c
osh(1)^2 + (8*c^4*x^4 + 8*c^2*x^2 + 1)*sinh(1)^2 - 4*(c^3*d + (2*c^3*x^2 + c)*cosh(1) + (2*c^3*x^2 + c)*sinh(1
))*sqrt(c^2*x^2 + 1)*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*sqrt(cosh(1) + sinh(1)) + 2*(4*c^4*d*x^2 + 3*c^2*d)*c
osh(1) + 2*(4*c^4*d*x^2 + 3*c^2*d + (8*c^4*x^4 + 8*c^2*x^2 + 1)*cosh(1))*sinh(1)) + 4*(b*x*cosh(1) + b*x*sinh(
1))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 4*(a*x*cosh(1) + a*x*sinh(1))*sqrt(x^2*
cosh(1) + x^2*sinh(1) + d))/(d*x^2*cosh(1)^2 + d*x^2*sinh(1)^2 + d^2*cosh(1) + (2*d*x^2*cosh(1) + d^2)*sinh(1)
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/(e*x**2+d)**(3/2),x)

[Out]

Integral((a + b*asinh(c*x))/(d + e*x**2)**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/(e*x^2 + d)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(d + e*x^2)^(3/2),x)

[Out]

int((a + b*asinh(c*x))/(d + e*x^2)^(3/2), x)

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