3.164 \(\int \frac{a+b \sinh ^{-1}(c x)}{x^4 (d+c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=228 \[ \frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \sqrt{c^2 d x^2+d}}+\frac{4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d x \sqrt{c^2 d x^2+d}}-\frac{a+b \sinh ^{-1}(c x)}{3 d x^3 \sqrt{c^2 d x^2+d}}-\frac{b c \sqrt{c^2 d x^2+d}}{6 d^2 x^2 \sqrt{c^2 x^2+1}}-\frac{5 b c^3 \log (x) \sqrt{c^2 d x^2+d}}{3 d^2 \sqrt{c^2 x^2+1}}-\frac{b c^3 \sqrt{c^2 d x^2+d} \log \left (c^2 x^2+1\right )}{2 d^2 \sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.288848, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5747, 5687, 260, 266, 36, 29, 31, 44} \[ \frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \sqrt{c^2 d x^2+d}}+\frac{4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d x \sqrt{c^2 d x^2+d}}-\frac{a+b \sinh ^{-1}(c x)}{3 d x^3 \sqrt{c^2 d x^2+d}}-\frac{b c \sqrt{c^2 x^2+1}}{6 d x^2 \sqrt{c^2 d x^2+d}}-\frac{5 b c^3 \sqrt{c^2 x^2+1} \log (x)}{3 d \sqrt{c^2 d x^2+d}}-\frac{b c^3 \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 d \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^(3/2)),x]

[Out]

-(b*c*Sqrt[1 + c^2*x^2])/(6*d*x^2*Sqrt[d + c^2*d*x^2]) - (a + b*ArcSinh[c*x])/(3*d*x^3*Sqrt[d + c^2*d*x^2]) +
(4*c^2*(a + b*ArcSinh[c*x]))/(3*d*x*Sqrt[d + c^2*d*x^2]) + (8*c^4*x*(a + b*ArcSinh[c*x]))/(3*d*Sqrt[d + c^2*d*
x^2]) - (5*b*c^3*Sqrt[1 + c^2*x^2]*Log[x])/(3*d*Sqrt[d + c^2*d*x^2]) - (b*c^3*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^
2])/(2*d*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 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 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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

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^4 \left (d+c^2 d x^2\right )^{3/2}} \, dx &=-\frac{a+b \sinh ^{-1}(c x)}{3 d x^3 \sqrt{d+c^2 d x^2}}-\frac{1}{3} \left (4 c^2\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx+\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \int \frac{1}{x^3 \left (1+c^2 x^2\right )} \, dx}{3 d \sqrt{d+c^2 d x^2}}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{3 d x^3 \sqrt{d+c^2 d x^2}}+\frac{4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d x \sqrt{d+c^2 d x^2}}+\frac{1}{3} \left (8 c^4\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^{3/2}} \, dx+\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )}{6 d \sqrt{d+c^2 d x^2}}-\frac{\left (4 b c^3 \sqrt{1+c^2 x^2}\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx}{3 d \sqrt{d+c^2 d x^2}}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{3 d x^3 \sqrt{d+c^2 d x^2}}+\frac{4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d x \sqrt{d+c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \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^2}-\frac{c^2}{x}+\frac{c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d \sqrt{d+c^2 d x^2}}-\frac{\left (2 b c^3 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{3 d \sqrt{d+c^2 d x^2}}-\frac{\left (8 b c^5 \sqrt{1+c^2 x^2}\right ) \int \frac{x}{1+c^2 x^2} \, dx}{3 d \sqrt{d+c^2 d x^2}}\\ &=-\frac{b c \sqrt{1+c^2 x^2}}{6 d x^2 \sqrt{d+c^2 d x^2}}-\frac{a+b \sinh ^{-1}(c x)}{3 d x^3 \sqrt{d+c^2 d x^2}}+\frac{4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d x \sqrt{d+c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \sqrt{d+c^2 d x^2}}-\frac{b c^3 \sqrt{1+c^2 x^2} \log (x)}{3 d \sqrt{d+c^2 d x^2}}-\frac{7 b c^3 \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 d \sqrt{d+c^2 d x^2}}-\frac{\left (2 b c^3 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{3 d \sqrt{d+c^2 d x^2}}+\frac{\left (2 b c^5 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )}{3 d \sqrt{d+c^2 d x^2}}\\ &=-\frac{b c \sqrt{1+c^2 x^2}}{6 d x^2 \sqrt{d+c^2 d x^2}}-\frac{a+b \sinh ^{-1}(c x)}{3 d x^3 \sqrt{d+c^2 d x^2}}+\frac{4 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d x \sqrt{d+c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \sqrt{d+c^2 d x^2}}-\frac{5 b c^3 \sqrt{1+c^2 x^2} \log (x)}{3 d \sqrt{d+c^2 d x^2}}-\frac{b c^3 \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{2 d \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.299031, size = 216, normalized size = 0.95 \[ \frac{\sqrt{c^2 d x^2+d} \left (16 a c^4 x^4 \sqrt{c^2 x^2+1}+8 a c^2 x^2 \sqrt{c^2 x^2+1}-2 a \sqrt{c^2 x^2+1}-b c^3 x^3-8 b c^5 x^5 \log \left (c^2 x^2+1\right )+5 b c^3 x^3 \left (c^2 x^2+1\right ) \log \left (\frac{1}{c^2 x^2}+1\right )-8 b c^3 x^3 \log \left (c^2 x^2+1\right )+2 b \sqrt{c^2 x^2+1} \left (8 c^4 x^4+4 c^2 x^2-1\right ) \sinh ^{-1}(c x)-b c x\right )}{6 d^2 x^3 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.171, size = 965, normalized size = 4.2 \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^4/(c^2*d*x^2+d)^(3/2),x)

[Out]

-1/3*a/d/x^3/(c^2*d*x^2+d)^(1/2)+4/3*a*c^2/d/x/(c^2*d*x^2+d)^(1/2)+8/3*a*c^4/d*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^2*arcsinh(c*x)*c^3+32/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1
)/d^2*x^7*c^10-32/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x^5*(c^2*x^2+1)*c^8+16*b*(d*(c^2*x^2+1
))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x^5*c^8-16/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x^3*(c^2
*x^2+1)*c^6+64/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x^3*arcsinh(c*x)*c^6-64/3*b*(d*(c^2*x^2+1
))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^5+4*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x
^4+7*c^2*x^2-1)/d^2*x^3*c^6+4/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x*(c^2*x^2+1)*c^4+8*b*(d*(
c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x*arcsinh(c*x)*c^4+8/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*
x^2-1)/d^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^3-4/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*x*c^4-4/
3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2*c^3*(c^2*x^2+1)^(1/2)-4*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x
^4+7*c^2*x^2-1)/d^2/x*arcsinh(c*x)*c^2+1/6*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2/x^2*c*(c^2*x^2+
1)^(1/2)+1/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^4*x^4+7*c^2*x^2-1)/d^2/x^3*arcsinh(c*x)-b*(d*(c^2*x^2+1))^(1/2)/(c^2
*x^2+1)^(1/2)/d^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)*c^3-5/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*ln((c*
x+(c^2*x^2+1)^(1/2))^2-1)*c^3

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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^4/(c^2*d*x^2+d)^(3/2),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^{4} d^{2} x^{8} + 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^(3/2)*x^4), x)