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

Optimal. Leaf size=141 \[ \frac{2 c^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x}-\frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^3}-\frac{b c \sqrt{c^2 x^2+1}}{6 x^2 \sqrt{c^2 d x^2+d}}-\frac{2 b c^3 \sqrt{c^2 x^2+1} \log (x)}{3 \sqrt{c^2 d x^2+d}} \]

[Out]

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

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Rubi [A]  time = 0.18419, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5747, 5723, 29, 30} \[ \frac{2 c^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x}-\frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^3}-\frac{b c \sqrt{c^2 x^2+1}}{6 x^2 \sqrt{c^2 d x^2+d}}-\frac{2 b c^3 \sqrt{c^2 x^2+1} \log (x)}{3 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*Sqrt[1 + c^2*x^2])/(6*x^2*Sqrt[d + c^2*d*x^2]) - (Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3*d*x^3) +
(2*c^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3*d*x) - (2*b*c^3*Sqrt[1 + c^2*x^2]*Log[x])/(3*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 5723

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[(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*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 29

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.209244, size = 135, normalized size = 0.96 \[ \frac{2 a \left (2 c^4 x^4+c^2 x^2-1\right )+b c x \sqrt{c^2 x^2+1} \left (6 c^2 x^2-1\right )+2 b \left (2 c^4 x^4+c^2 x^2-1\right ) \sinh ^{-1}(c x)}{6 x^3 \sqrt{c^2 d x^2+d}}-\frac{2 b c^3 \log (x) \sqrt{d \left (c^2 x^2+1\right )}}{3 d \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.17, size = 791, normalized size = 5.6 \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)^(1/2),x)

[Out]

-1/3*a/d/x^3*(c^2*d*x^2+d)^(1/2)+2/3*a*c^2/d/x*(c^2*d*x^2+d)^(1/2)+4/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/
2)/d*arcsinh(c*x)*c^3+2/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*x^5*c^8-2/3*b*(d*(c^2*x^2+1))^(1/2
)/(3*c^4*x^4+2*c^2*x^2-1)/d*x^3*(c^2*x^2+1)*c^6+2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*x^3*arcsin
h(c*x)*c^6-2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*x^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^5+1/3*b*(d
*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*x^3*c^6+1/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*x*
(c^2*x^2+1)*c^4+1/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*x*arcsinh(c*x)*c^4+2/3*b*(d*(c^2*x^2+1))
^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^3-1/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c
^2*x^2-1)/d*x*c^4-1/2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*c^3*(c^2*x^2+1)^(1/2)-4/3*b*(d*(c^2*x^
2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d/x*arcsinh(c*x)*c^2+1/6*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d
/x^2*c*(c^2*x^2+1)^(1/2)+1/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d/x^3*arcsinh(c*x)-2/3*b*(d*(c^2*
x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.18487, size = 477, normalized size = 3.38 \begin{align*} \frac{2 \,{\left (2 \, b c^{4} x^{4} + b c^{2} x^{2} - b\right )} \sqrt{c^{2} d x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (b c^{5} x^{5} + b c^{3} x^{3}\right )} \sqrt{d} \log \left (\frac{c^{2} d x^{6} + c^{2} d x^{2} + d x^{4} - \sqrt{c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} + 1}{\left (x^{4} - 1\right )} \sqrt{d} + d}{c^{2} x^{4} + x^{2}}\right ) +{\left (4 \, a c^{4} x^{4} + 2 \, a c^{2} x^{2} +{\left (b c x^{3} - b c x\right )} \sqrt{c^{2} x^{2} + 1} - 2 \, a\right )} \sqrt{c^{2} d x^{2} + d}}{6 \,{\left (c^{2} d x^{5} + d x^{3}\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)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*(2*b*c^4*x^4 + b*c^2*x^2 - b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(b*c^5*x^5 + b*c^3*x
^3)*sqrt(d)*log((c^2*d*x^6 + c^2*d*x^2 + d*x^4 - sqrt(c^2*d*x^2 + d)*sqrt(c^2*x^2 + 1)*(x^4 - 1)*sqrt(d) + d)/
(c^2*x^4 + x^2)) + (4*a*c^4*x^4 + 2*a*c^2*x^2 + (b*c*x^3 - b*c*x)*sqrt(c^2*x^2 + 1) - 2*a)*sqrt(c^2*d*x^2 + d)
)/(c^2*d*x^5 + d*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asinh}{\left (c x \right )}}{x^{4} \sqrt{d \left (c^{2} x^{2} + 1\right )}}\, dx \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)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))/(x**4*sqrt(d*(c**2*x**2 + 1))), x)

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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}{\sqrt{c^{2} d x^{2} + d} 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)^(1/2),x, algorithm="giac")

[Out]

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