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

Optimal. Leaf size=120 \[ \frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{1}{9} b c d^2 \left (c^2 x^2+1\right )^{3/2}-\frac{5}{3} b c d^2 \sqrt{c^2 x^2+1}-b c d^2 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right ) \]

[Out]

(-5*b*c*d^2*Sqrt[1 + c^2*x^2])/3 - (b*c*d^2*(1 + c^2*x^2)^(3/2))/9 - (d^2*(a + b*ArcSinh[c*x]))/x + 2*c^2*d^2*
x*(a + b*ArcSinh[c*x]) + (c^4*d^2*x^3*(a + b*ArcSinh[c*x]))/3 - b*c*d^2*ArcTanh[Sqrt[1 + c^2*x^2]]

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Rubi [A]  time = 0.157219, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {270, 5730, 12, 1251, 897, 1153, 208} \[ \frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{1}{9} b c d^2 \left (c^2 x^2+1\right )^{3/2}-\frac{5}{3} b c d^2 \sqrt{c^2 x^2+1}-b c d^2 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*c*d^2*Sqrt[1 + c^2*x^2])/3 - (b*c*d^2*(1 + c^2*x^2)^(3/2))/9 - (d^2*(a + b*ArcSinh[c*x]))/x + 2*c^2*d^2*
x*(a + b*ArcSinh[c*x]) + (c^4*d^2*x^3*(a + b*ArcSinh[c*x]))/3 - b*c*d^2*ArcTanh[Sqrt[1 + c^2*x^2]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 5730

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(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, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 12

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

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d^2 \left (-3+6 c^2 x^2+c^4 x^4\right )}{3 x \sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{3} \left (b c d^2\right ) \int \frac{-3+6 c^2 x^2+c^4 x^4}{x \sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{6} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{-3+6 c^2 x+c^4 x^2}{x \sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{-8+4 x^2+x^4}{-\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{1+c^2 x^2}\right )}{3 c}\\ &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \left (5 c^2+c^2 x^2-\frac{3}{-\frac{1}{c^2}+\frac{x^2}{c^2}}\right ) \, dx,x,\sqrt{1+c^2 x^2}\right )}{3 c}\\ &=-\frac{5}{3} b c d^2 \sqrt{1+c^2 x^2}-\frac{1}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2}-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{1+c^2 x^2}\right )}{c}\\ &=-\frac{5}{3} b c d^2 \sqrt{1+c^2 x^2}-\frac{1}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2}-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+2 c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^4 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )-b c d^2 \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.123593, size = 124, normalized size = 1.03 \[ \frac{d^2 \left (3 a c^4 x^4+18 a c^2 x^2-9 a-b c^3 x^3 \sqrt{c^2 x^2+1}-16 b c x \sqrt{c^2 x^2+1}-9 b c x \log \left (\sqrt{c^2 x^2+1}+1\right )+3 b \left (c^4 x^4+6 c^2 x^2-3\right ) \sinh ^{-1}(c x)+9 b c x \log (x)\right )}{9 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-9*a + 18*a*c^2*x^2 + 3*a*c^4*x^4 - 16*b*c*x*Sqrt[1 + c^2*x^2] - b*c^3*x^3*Sqrt[1 + c^2*x^2] + 3*b*(-3 +
6*c^2*x^2 + c^4*x^4)*ArcSinh[c*x] + 9*b*c*x*Log[x] - 9*b*c*x*Log[1 + Sqrt[1 + c^2*x^2]]))/(9*x)

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Maple [A]  time = 0.009, size = 114, normalized size = 1. \begin{align*} c \left ({d}^{2}a \left ({\frac{{c}^{3}{x}^{3}}{3}}+2\,cx-{\frac{1}{cx}} \right ) +{d}^{2}b \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}}{3}}+2\,{\it Arcsinh} \left ( cx \right ) cx-{\frac{{\it Arcsinh} \left ( cx \right ) }{cx}}-{\frac{{c}^{2}{x}^{2}}{9}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{16}{9}\sqrt{{c}^{2}{x}^{2}+1}}-{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(d^2*a*(1/3*c^3*x^3+2*c*x-1/c/x)+d^2*b*(1/3*arcsinh(c*x)*c^3*x^3+2*arcsinh(c*x)*c*x-arcsinh(c*x)/c/x-1/9*c^2
*x^2*(c^2*x^2+1)^(1/2)-16/9*(c^2*x^2+1)^(1/2)-arctanh(1/(c^2*x^2+1)^(1/2))))

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Maxima [A]  time = 1.17115, size = 196, normalized size = 1.63 \begin{align*} \frac{1}{3} \, a c^{4} d^{2} x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{4} d^{2} + 2 \, a c^{2} d^{2} x + 2 \,{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} b c d^{2} -{\left (c \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arsinh}\left (c x\right )}{x}\right )} b d^{2} - \frac{a d^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*c^4*d^2*x^3 + 1/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*b*c^4*d
^2 + 2*a*c^2*d^2*x + 2*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*c*d^2 - (c*arcsinh(1/(sqrt(c^2)*abs(x))) + arc
sinh(c*x)/x)*b*d^2 - a*d^2/x

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Fricas [B]  time = 2.63919, size = 505, normalized size = 4.21 \begin{align*} \frac{3 \, a c^{4} d^{2} x^{4} + 18 \, a c^{2} d^{2} x^{2} - 9 \, b c d^{2} x \log \left (-c x + \sqrt{c^{2} x^{2} + 1} + 1\right ) + 9 \, b c d^{2} x \log \left (-c x + \sqrt{c^{2} x^{2} + 1} - 1\right ) - 3 \,{\left (b c^{4} + 6 \, b c^{2} - 3 \, b\right )} d^{2} x \log \left (-c x + \sqrt{c^{2} x^{2} + 1}\right ) - 9 \, a d^{2} + 3 \,{\left (b c^{4} d^{2} x^{4} + 6 \, b c^{2} d^{2} x^{2} -{\left (b c^{4} + 6 \, b c^{2} - 3 \, b\right )} d^{2} x - 3 \, b d^{2}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (b c^{3} d^{2} x^{3} + 16 \, b c d^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}{9 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*a*c^4*d^2*x^4 + 18*a*c^2*d^2*x^2 - 9*b*c*d^2*x*log(-c*x + sqrt(c^2*x^2 + 1) + 1) + 9*b*c*d^2*x*log(-c*x
+ sqrt(c^2*x^2 + 1) - 1) - 3*(b*c^4 + 6*b*c^2 - 3*b)*d^2*x*log(-c*x + sqrt(c^2*x^2 + 1)) - 9*a*d^2 + 3*(b*c^4
*d^2*x^4 + 6*b*c^2*d^2*x^2 - (b*c^4 + 6*b*c^2 - 3*b)*d^2*x - 3*b*d^2)*log(c*x + sqrt(c^2*x^2 + 1)) - (b*c^3*d^
2*x^3 + 16*b*c*d^2*x)*sqrt(c^2*x^2 + 1))/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int 2 a c^{2}\, dx + \int \frac{a}{x^{2}}\, dx + \int a c^{4} x^{2}\, dx + \int 2 b c^{2} \operatorname{asinh}{\left (c x \right )}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{x^{2}}\, dx + \int b c^{4} x^{2} \operatorname{asinh}{\left (c x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(2*a*c**2, x) + Integral(a/x**2, x) + Integral(a*c**4*x**2, x) + Integral(2*b*c**2*asinh(c*x), x
) + Integral(b*asinh(c*x)/x**2, x) + Integral(b*c**4*x**2*asinh(c*x), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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