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

Optimal. Leaf size=147 \[ \frac{2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 d x^2+d}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}+\frac{b}{6 c d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 c d^2 \sqrt{c^2 d x^2+d}} \]

[Out]

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

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Rubi [A]  time = 0.0794904, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {5690, 5687, 260, 261} \[ \frac{2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 d x^2+d}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}+\frac{b}{6 c d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 c d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

Mathematica [A]  time = 0.193406, size = 143, normalized size = 0.97 \[ \frac{\sqrt{c^2 d x^2+d} \left (4 a c^3 x^3 \sqrt{c^2 x^2+1}+6 a c x \sqrt{c^2 x^2+1}+b c^2 x^2-2 b \left (c^2 x^2+1\right )^2 \log \left (c^2 x^2+1\right )+2 b c x \sqrt{c^2 x^2+1} \left (2 c^2 x^2+3\right ) \sinh ^{-1}(c x)+b\right )}{6 c d^3 \left (c^2 x^2+1\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.109, size = 1005, normalized size = 6.8 \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))/(c^2*d*x^2+d)^(5/2),x)

[Out]

1/3*a*x/d/(c^2*d*x^2+d)^(3/2)+2/3*a/d^2*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)/c/
d^3*arcsinh(c*x)-2/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*c^6/d^3*x^7+2/3*b*(d*(c^2*x^2
+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*c^4/d^3*(c^2*x^2+1)*x^5+2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10
*c^4*x^4+11*c^2*x^2+4)*c^4/d^3*arcsinh(c*x)*x^5-2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*
c^3/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^4-7/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*c^4
/d^3*x^5+5/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*c^2/d^3*(c^2*x^2+1)*x^3+17/3*b*(d*(c^
2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*c^2/d^3*arcsinh(c*x)*x^3-14/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c
^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*c/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^2-8/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^
6+10*c^4*x^4+11*c^2*x^2+4)*c^2/d^3*x^3+1/2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)*c/d^3*x
^2*(c^2*x^2+1)^(1/2)+b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)/d^3*(c^2*x^2+1)*x+4*b*(d*(c^2
*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)/d^3*arcsinh(c*x)*x-8/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+1
0*c^4*x^4+11*c^2*x^2+4)/c/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*
c^2*x^2+4)/d^3*x+2/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)/c/d^3*(c^2*x^2+1)^(1/2)-2/3*b
*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c/d^3*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)

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Maxima [A]  time = 1.28324, size = 170, normalized size = 1.16 \begin{align*} \frac{1}{6} \, b c{\left (\frac{1}{c^{4} d^{\frac{5}{2}} x^{2} + c^{2} d^{\frac{5}{2}}} - \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{\frac{5}{2}}}\right )} + \frac{1}{3} \, b{\left (\frac{2 \, x}{\sqrt{c^{2} d x^{2} + d} d^{2}} + \frac{x}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} d}\right )} \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \, a{\left (\frac{2 \, x}{\sqrt{c^{2} d x^{2} + d} d^{2}} + \frac{x}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(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)