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

Optimal. Leaf size=138 \[ -\frac {2 a b x \sqrt {c^2 x^2+1}}{c \sqrt {c^2 d x^2+d}}+\frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}+\frac {2 b^2 \left (c^2 x^2+1\right )}{c^2 \sqrt {c^2 d x^2+d}}-\frac {2 b^2 x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{c \sqrt {c^2 d x^2+d}} \]

[Out]

2*b^2*(c^2*x^2+1)/c^2/(c^2*d*x^2+d)^(1/2)-2*a*b*x*(c^2*x^2+1)^(1/2)/c/(c^2*d*x^2+d)^(1/2)-2*b^2*x*arcsinh(c*x)
*(c^2*x^2+1)^(1/2)/c/(c^2*d*x^2+d)^(1/2)+(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/c^2/d

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Rubi [A]  time = 0.12, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {5717, 5653, 261} \[ -\frac {2 a b x \sqrt {c^2 x^2+1}}{c \sqrt {c^2 d x^2+d}}+\frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}+\frac {2 b^2 \left (c^2 x^2+1\right )}{c^2 \sqrt {c^2 d x^2+d}}-\frac {2 b^2 x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{c \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5653

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

Rule 5717

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

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 127, normalized size = 0.92 \[ \frac {\sqrt {c^2 d x^2+d} \left (a^2 \sqrt {c^2 x^2+1}-2 b \sinh ^{-1}(c x) \left (b c x-a \sqrt {c^2 x^2+1}\right )-2 a b c x+2 b^2 \sqrt {c^2 x^2+1}+b^2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)^2\right )}{c^2 d \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 179, normalized size = 1.30 \[ \frac {{\left (b^{2} c^{2} x^{2} + b^{2}\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 2 \, {\left (a b c^{2} x^{2} - \sqrt {c^{2} x^{2} + 1} b^{2} c x + a b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left ({\left (a^{2} + 2 \, b^{2}\right )} c^{2} x^{2} - 2 \, \sqrt {c^{2} x^{2} + 1} a b c x + a^{2} + 2 \, b^{2}\right )} \sqrt {c^{2} d x^{2} + d}}{c^{4} d x^{2} + c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x}{\sqrt {c^{2} d x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.19, size = 296, normalized size = 2.14 \[ \frac {a^{2} \sqrt {c^{2} d \,x^{2}+d}}{c^{2} d}+b^{2} \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (\arcsinh \left (c x \right )^{2}-2 \arcsinh \left (c x \right )+2\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (\arcsinh \left (c x \right )^{2}+2 \arcsinh \left (c x \right )+2\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}\right )+2 a b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (-1+\arcsinh \left (c x \right )\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-c x \sqrt {c^{2} x^{2}+1}+1\right ) \left (1+\arcsinh \left (c x \right )\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x)

[Out]

a^2/c^2/d*(c^2*d*x^2+d)^(1/2)+b^2*(1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(arcsinh(c*x)^2
-2*arcsinh(c*x)+2)/c^2/d/(c^2*x^2+1)+1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(arcsinh(c*x)
^2+2*arcsinh(c*x)+2)/c^2/d/(c^2*x^2+1))+2*a*b*(1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(-1
+arcsinh(c*x))/c^2/d/(c^2*x^2+1)+1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(1+arcsinh(c*x))/
c^2/d/(c^2*x^2+1))

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maxima [A]  time = 0.51, size = 125, normalized size = 0.91 \[ -2 \, b^{2} {\left (\frac {x \operatorname {arsinh}\left (c x\right )}{c \sqrt {d}} - \frac {\sqrt {c^{2} x^{2} + 1}}{c^{2} \sqrt {d}}\right )} - \frac {2 \, a b x}{c \sqrt {d}} + \frac {\sqrt {c^{2} d x^{2} + d} b^{2} \operatorname {arsinh}\left (c x\right )^{2}}{c^{2} d} + \frac {2 \, \sqrt {c^{2} d x^{2} + d} a b \operatorname {arsinh}\left (c x\right )}{c^{2} d} + \frac {\sqrt {c^{2} d x^{2} + d} a^{2}}{c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

-2*b^2*(x*arcsinh(c*x)/(c*sqrt(d)) - sqrt(c^2*x^2 + 1)/(c^2*sqrt(d))) - 2*a*b*x/(c*sqrt(d)) + sqrt(c^2*d*x^2 +
 d)*b^2*arcsinh(c*x)^2/(c^2*d) + 2*sqrt(c^2*d*x^2 + d)*a*b*arcsinh(c*x)/(c^2*d) + sqrt(c^2*d*x^2 + d)*a^2/(c^2
*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{\sqrt {d\,c^2\,x^2+d}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2)^(1/2),x)

[Out]

int((x*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\sqrt {d \left (c^{2} x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asinh(c*x))**2/(c**2*d*x**2+d)**(1/2),x)

[Out]

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

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