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

Optimal. Leaf size=135 \[ -\frac {b d x \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac {3 b d x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}+\frac {d \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac {3 d \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^2}+\frac {1}{32} b^2 c^2 d x^4+\frac {5}{32} b^2 d x^2 \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5717, 5684, 5682, 5675, 30, 14} \[ -\frac {b d x \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac {3 b d x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}+\frac {d \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac {3 d \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^2}+\frac {1}{32} b^2 c^2 d x^4+\frac {5}{32} b^2 d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^2*d*x^2)/32 + (b^2*c^2*d*x^4)/32 - (3*b*d*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(16*c) - (b*d*x*(1 +
c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(8*c) - (3*d*(a + b*ArcSinh[c*x])^2)/(32*c^2) + (d*(1 + c^2*x^2)^2*(a + b
*ArcSinh[c*x])^2)/(4*c^2)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 5675

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

Rule 5682

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

Rule 5684

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*
(a + b*ArcSinh[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*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)^FracPart[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] && Gt
Q[n, 0] && GtQ[p, 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 x \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac {(b d) \int \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{2 c}\\ &=-\frac {b d x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac {d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{8} \left (b^2 d\right ) \int x \left (1+c^2 x^2\right ) \, dx-\frac {(3 b d) \int \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{8 c}\\ &=-\frac {3 b d x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}-\frac {b d x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac {d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{8} \left (b^2 d\right ) \int \left (x+c^2 x^3\right ) \, dx+\frac {1}{16} \left (3 b^2 d\right ) \int x \, dx-\frac {(3 b d) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{16 c}\\ &=\frac {5}{32} b^2 d x^2+\frac {1}{32} b^2 c^2 d x^4-\frac {3 b d x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}-\frac {b d x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac {3 d \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^2}+\frac {d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 155, normalized size = 1.15 \[ \frac {d \left (c x \left (8 a^2 c x \left (c^2 x^2+2\right )-2 a b \sqrt {c^2 x^2+1} \left (2 c^2 x^2+5\right )+b^2 c x \left (c^2 x^2+5\right )\right )+2 b \sinh ^{-1}(c x) \left (a \left (8 c^4 x^4+16 c^2 x^2+5\right )-b c x \sqrt {c^2 x^2+1} \left (2 c^2 x^2+5\right )\right )+b^2 \left (8 c^4 x^4+16 c^2 x^2+5\right ) \sinh ^{-1}(c x)^2\right )}{32 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(c*x*(8*a^2*c*x*(2 + c^2*x^2) + b^2*c*x*(5 + c^2*x^2) - 2*a*b*Sqrt[1 + c^2*x^2]*(5 + 2*c^2*x^2)) + 2*b*(-(b
*c*x*Sqrt[1 + c^2*x^2]*(5 + 2*c^2*x^2)) + a*(5 + 16*c^2*x^2 + 8*c^4*x^4))*ArcSinh[c*x] + b^2*(5 + 16*c^2*x^2 +
8*c^4*x^4)*ArcSinh[c*x]^2))/(32*c^2)

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fricas [A]  time = 0.75, size = 204, normalized size = 1.51 \[ \frac {{\left (8 \, a^{2} + b^{2}\right )} c^{4} d x^{4} + {\left (16 \, a^{2} + 5 \, b^{2}\right )} c^{2} d x^{2} + {\left (8 \, b^{2} c^{4} d x^{4} + 16 \, b^{2} c^{2} d x^{2} + 5 \, b^{2} d\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 2 \, {\left (8 \, a b c^{4} d x^{4} + 16 \, a b c^{2} d x^{2} + 5 \, a b d - {\left (2 \, b^{2} c^{3} d x^{3} + 5 \, b^{2} c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (2 \, a b c^{3} d x^{3} + 5 \, a b c d x\right )} \sqrt {c^{2} x^{2} + 1}}{32 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*((8*a^2 + b^2)*c^4*d*x^4 + (16*a^2 + 5*b^2)*c^2*d*x^2 + (8*b^2*c^4*d*x^4 + 16*b^2*c^2*d*x^2 + 5*b^2*d)*lo
g(c*x + sqrt(c^2*x^2 + 1))^2 + 2*(8*a*b*c^4*d*x^4 + 16*a*b*c^2*d*x^2 + 5*a*b*d - (2*b^2*c^3*d*x^3 + 5*b^2*c*d*
x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) - 2*(2*a*b*c^3*d*x^3 + 5*a*b*c*d*x)*sqrt(c^2*x^2 + 1))/c^2

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 191, normalized size = 1.41 \[ \frac {d \,a^{2} \left (\frac {1}{4} c^{4} x^{4}+\frac {1}{2} c^{2} x^{2}\right )+d \,b^{2} \left (\frac {\arcsinh \left (c x \right )^{2} \left (c^{2} x^{2}+1\right )^{2}}{4}-\frac {\arcsinh \left (c x \right ) c x \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}{8}-\frac {3 \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c x}{16}-\frac {3 \arcsinh \left (c x \right )^{2}}{32}+\frac {\left (c^{2} x^{2}+1\right )^{2}}{32}+\frac {3 c^{2} x^{2}}{32}+\frac {3}{32}\right )+2 d a b \left (\frac {\arcsinh \left (c x \right ) c^{4} x^{4}}{4}+\frac {\arcsinh \left (c x \right ) c^{2} x^{2}}{2}-\frac {c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {c^{2} x^{2}+1}}{32}+\frac {5 \arcsinh \left (c x \right )}{32}\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.37, size = 347, normalized size = 2.57 \[ \frac {1}{4} \, b^{2} c^{2} d x^{4} \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{4} \, a^{2} c^{2} d x^{4} + \frac {1}{2} \, b^{2} d x^{2} \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{16} \, {\left (8 \, x^{4} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac {3 \, \sqrt {c^{2} x^{2} + 1} x}{c^{4}} + \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5}}\right )} c\right )} a b c^{2} d + \frac {1}{32} \, {\left ({\left (\frac {x^{4}}{c^{2}} - \frac {3 \, x^{2}}{c^{4}} + \frac {3 \, \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{6}}\right )} c^{2} - 2 \, {\left (\frac {2 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac {3 \, \sqrt {c^{2} x^{2} + 1} x}{c^{4}} + \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5}}\right )} c \operatorname {arsinh}\left (c x\right )\right )} b^{2} c^{2} d + \frac {1}{2} \, a^{2} d x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{c^{3}}\right )}\right )} a b d + \frac {1}{4} \, {\left (c^{2} {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{4}}\right )} - 2 \, c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{c^{3}}\right )} \operatorname {arsinh}\left (c x\right )\right )} b^{2} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*c^2*d*x^4*arcsinh(c*x)^2 + 1/4*a^2*c^2*d*x^4 + 1/2*b^2*d*x^2*arcsinh(c*x)^2 + 1/16*(8*x^4*arcsinh(c*x)
- (2*sqrt(c^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c*x)/c^5)*c)*a*b*c^2*d + 1/32*((x^4/c^
2 - 3*x^2/c^4 + 3*log(c*x + sqrt(c^2*x^2 + 1))^2/c^6)*c^2 - 2*(2*sqrt(c^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 +
1)*x/c^4 + 3*arcsinh(c*x)/c^5)*c*arcsinh(c*x))*b^2*c^2*d + 1/2*a^2*d*x^2 + 1/2*(2*x^2*arcsinh(c*x) - c*(sqrt(c
^2*x^2 + 1)*x/c^2 - arcsinh(c*x)/c^3))*a*b*d + 1/4*(c^2*(x^2/c^2 - log(c*x + sqrt(c^2*x^2 + 1))^2/c^4) - 2*c*(
sqrt(c^2*x^2 + 1)*x/c^2 - arcsinh(c*x)/c^3)*arcsinh(c*x))*b^2*d

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

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),x)

[Out]

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

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sympy [A]  time = 3.33, size = 269, normalized size = 1.99 \[ \begin {cases} \frac {a^{2} c^{2} d x^{4}}{4} + \frac {a^{2} d x^{2}}{2} + \frac {a b c^{2} d x^{4} \operatorname {asinh}{\left (c x \right )}}{2} - \frac {a b c d x^{3} \sqrt {c^{2} x^{2} + 1}}{8} + a b d x^{2} \operatorname {asinh}{\left (c x \right )} - \frac {5 a b d x \sqrt {c^{2} x^{2} + 1}}{16 c} + \frac {5 a b d \operatorname {asinh}{\left (c x \right )}}{16 c^{2}} + \frac {b^{2} c^{2} d x^{4} \operatorname {asinh}^{2}{\left (c x \right )}}{4} + \frac {b^{2} c^{2} d x^{4}}{32} - \frac {b^{2} c d x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{8} + \frac {b^{2} d x^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{2} + \frac {5 b^{2} d x^{2}}{32} - \frac {5 b^{2} d x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{16 c} + \frac {5 b^{2} d \operatorname {asinh}^{2}{\left (c x \right )}}{32 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} d x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c**2*d*x**4/4 + a**2*d*x**2/2 + a*b*c**2*d*x**4*asinh(c*x)/2 - a*b*c*d*x**3*sqrt(c**2*x**2 + 1
)/8 + a*b*d*x**2*asinh(c*x) - 5*a*b*d*x*sqrt(c**2*x**2 + 1)/(16*c) + 5*a*b*d*asinh(c*x)/(16*c**2) + b**2*c**2*
d*x**4*asinh(c*x)**2/4 + b**2*c**2*d*x**4/32 - b**2*c*d*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)/8 + b**2*d*x**2*as
inh(c*x)**2/2 + 5*b**2*d*x**2/32 - 5*b**2*d*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(16*c) + 5*b**2*d*asinh(c*x)**2/(
32*c**2), Ne(c, 0)), (a**2*d*x**2/2, True))

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