3.173 \(\int \frac{a+b \sinh ^{-1}(c x)}{x^2 (d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=214 \[ -\frac{8 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 d x^2+d}}-\frac{4 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{d x \left (c^2 d x^2+d\right )^{3/2}}-\frac{b c \sqrt{c^2 d x^2+d}}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac{b c \log (x) \sqrt{c^2 d x^2+d}}{d^3 \sqrt{c^2 x^2+1}}+\frac{5 b c \sqrt{c^2 d x^2+d} \log \left (c^2 x^2+1\right )}{6 d^3 \sqrt{c^2 x^2+1}} \]

[Out]

-(b*c*Sqrt[d + c^2*d*x^2])/(6*d^3*(1 + c^2*x^2)^(3/2)) - (a + b*ArcSinh[c*x])/(d*x*(d + c^2*d*x^2)^(3/2)) - (4
*c^2*x*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2)) - (8*c^2*x*(a + b*ArcSinh[c*x]))/(3*d^2*Sqrt[d + c^2*
d*x^2]) + (b*c*Sqrt[d + c^2*d*x^2]*Log[x])/(d^3*Sqrt[1 + c^2*x^2]) + (5*b*c*Sqrt[d + c^2*d*x^2]*Log[1 + c^2*x^
2])/(6*d^3*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.213078, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {5747, 5690, 5687, 260, 261, 266, 44} \[ -\frac{8 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 d x^2+d}}-\frac{4 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{d x \left (c^2 d x^2+d\right )^{3/2}}-\frac{b c}{6 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{b c \sqrt{c^2 x^2+1} \log (x)}{d^2 \sqrt{c^2 d x^2+d}}+\frac{5 b c \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{6 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^2*(d + c^2*d*x^2)^(5/2)),x]

[Out]

-(b*c)/(6*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) - (a + b*ArcSinh[c*x])/(d*x*(d + c^2*d*x^2)^(3/2)) - (4*c
^2*x*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2)) - (8*c^2*x*(a + b*ArcSinh[c*x]))/(3*d^2*Sqrt[d + c^2*d*
x^2]) + (b*c*Sqrt[1 + c^2*x^2]*Log[x])/(d^2*Sqrt[d + c^2*d*x^2]) + (5*b*c*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/
(6*d^2*Sqrt[d + c^2*d*x^2])

Rule 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.310772, size = 227, normalized size = 1.06 \[ -\frac{\sqrt{c^2 d x^2+d} \left (16 a c^4 x^4 \sqrt{c^2 x^2+1}+24 a c^2 x^2 \sqrt{c^2 x^2+1}+6 a \sqrt{c^2 x^2+1}+b c^3 x^3-8 b c^5 x^5 \log \left (c^2 x^2+1\right )-16 b c^3 x^3 \log \left (c^2 x^2+1\right )+3 b c x \left (c^2 x^2+1\right )^2 \log \left (\frac{1}{c^2 x^2}+1\right )-8 b c x \log \left (c^2 x^2+1\right )+2 b \sqrt{c^2 x^2+1} \left (8 c^4 x^4+12 c^2 x^2+3\right ) \sinh ^{-1}(c x)+b c x\right )}{6 d^3 x \left (c^2 x^2+1\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^2*(d + c^2*d*x^2)^(5/2)),x]

[Out]

-(Sqrt[d + c^2*d*x^2]*(b*c*x + b*c^3*x^3 + 6*a*Sqrt[1 + c^2*x^2] + 24*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 16*a*c^4*x
^4*Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 + c^2*x^2]*(3 + 12*c^2*x^2 + 8*c^4*x^4)*ArcSinh[c*x] + 3*b*c*x*(1 + c^2*x^2)
^2*Log[1 + 1/(c^2*x^2)] - 8*b*c*x*Log[1 + c^2*x^2] - 16*b*c^3*x^3*Log[1 + c^2*x^2] - 8*b*c^5*x^5*Log[1 + c^2*x
^2]))/(6*d^3*x*(1 + c^2*x^2)^(5/2))

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Maple [B]  time = 0.151, size = 1257, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^2/(c^2*d*x^2+d)^(5/2),x)

[Out]

-a/d/x/(c^2*d*x^2+d)^(3/2)-4/3*a*c^2*x/d/(c^2*d*x^2+d)^(3/2)-8/3*a*c^2/d^2*x/(c^2*d*x^2+d)^(1/2)-16/3*b*(d*(c^
2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*arcsinh(c*x)*c-32/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2
*x^2+9)/d^3*x^9*c^10+32/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^7*(c^2*x^2+1)*c^8-
112/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^7*c^8+80/3*b*(d*(c^2*x^2+1))^(1/2)/(8*
c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^5*(c^2*x^2+1)*c^6-64/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26
*c^2*x^2+9)/d^3*x^5*arcsinh(c*x)*c^6+64/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^4*
arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^5-140/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^5*c
^6+20*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^3*(c^2*x^2+1)*c^4-56*b*(d*(c^2*x^2+1))
^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^3*arcsinh(c*x)*c^4+136/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+2
5*c^4*x^4+26*c^2*x^2+9)/d^3*x^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^3-24*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^
4*x^4+26*c^2*x^2+9)/d^3*x^3*c^4-4/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^2*c^3*(c
^2*x^2+1)^(1/2)+4*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x*(c^2*x^2+1)*c^2-44*b*(d*(c
^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x*arcsinh(c*x)*c^2+24*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*
x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c-4*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*
x^4+26*c^2*x^2+9)/d^3*x*c^2-3/2*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*c*(c^2*x^2+1)^
(1/2)-9*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3/x*arcsinh(c*x)+b*(d*(c^2*x^2+1))^(1/2)
/(c^2*x^2+1)^(1/2)/d^3*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)*c+5/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*ln(
1+(c*x+(c^2*x^2+1)^(1/2))^2)*c

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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((a+b*arcsinh(c*x))/x^2/(c^2*d*x^2+d)^(5/2),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 (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^2/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)/(c^6*d^3*x^8 + 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 + d^3*x^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x**2/(c**2*d*x**2+d)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^2/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^(5/2)*x^2), x)