3.7.15 \(\int (d+e x^2) (a+b \sinh ^{-1}(c x))^2 \, dx\) [615]

Optimal. Leaf size=153 \[ 2 b^2 d x-\frac {4 b^2 e x}{9 c^2}+\frac {2}{27} b^2 e x^3-\frac {2 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c}+\frac {4 b e \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3}-\frac {2 b e x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \]

[Out]

2*b^2*d*x-4/9*b^2*e*x/c^2+2/27*b^2*e*x^3+d*x*(a+b*arcsinh(c*x))^2+1/3*e*x^3*(a+b*arcsinh(c*x))^2-2*b*d*(a+b*ar
csinh(c*x))*(c^2*x^2+1)^(1/2)/c+4/9*b*e*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^3-2/9*b*e*x^2*(a+b*arcsinh(c*x)
)*(c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.20, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5793, 5772, 5798, 8, 5776, 5812, 30} \begin {gather*} -\frac {2 b d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c}-\frac {2 b e x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac {4 b e \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3}+d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {4 b^2 e x}{9 c^2}+2 b^2 d x+\frac {2}{27} b^2 e x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*b^2*d*x - (4*b^2*e*x)/(9*c^2) + (2*b^2*e*x^3)/27 - (2*b*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/c + (4*b*e
*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(9*c^3) - (2*b*e*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(9*c) +
d*x*(a + b*ArcSinh[c*x])^2 + (e*x^3*(a + b*ArcSinh[c*x])^2)/3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rule 5772

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 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5793

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcSinh[c*x])^n, (d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[e, c^2*d] && IntegerQ[p] &&
(p > 0 || IGtQ[n, 0])

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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]

Rule 5812

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

Rubi steps

\begin {align*} \int \left (d+e x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\int \left (d \left (a+b \sinh ^{-1}(c x)\right )^2+e x^2 \left (a+b \sinh ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx+e \int x^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )^2-(2 b c d) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx-\frac {1}{3} (2 b c e) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {2 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c}-\frac {2 b e x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )^2+\left (2 b^2 d\right ) \int 1 \, dx+\frac {1}{9} \left (2 b^2 e\right ) \int x^2 \, dx+\frac {(4 b e) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{9 c}\\ &=2 b^2 d x+\frac {2}{27} b^2 e x^3-\frac {2 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c}+\frac {4 b e \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3}-\frac {2 b e x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (4 b^2 e\right ) \int 1 \, dx}{9 c^2}\\ &=2 b^2 d x-\frac {4 b^2 e x}{9 c^2}+\frac {2}{27} b^2 e x^3-\frac {2 b d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c}+\frac {4 b e \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3}-\frac {2 b e x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 164, normalized size = 1.07 \begin {gather*} \frac {9 a^2 c^3 x \left (3 d+e x^2\right )-6 a b \sqrt {1+c^2 x^2} \left (-2 e+c^2 \left (9 d+e x^2\right )\right )+2 b^2 c x \left (-6 e+c^2 \left (27 d+e x^2\right )\right )-6 b \left (-3 a c^3 x \left (3 d+e x^2\right )+b \sqrt {1+c^2 x^2} \left (-2 e+c^2 \left (9 d+e x^2\right )\right )\right ) \sinh ^{-1}(c x)+9 b^2 c^3 x \left (3 d+e x^2\right ) \sinh ^{-1}(c x)^2}{27 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a^2*c^3*x*(3*d + e*x^2) - 6*a*b*Sqrt[1 + c^2*x^2]*(-2*e + c^2*(9*d + e*x^2)) + 2*b^2*c*x*(-6*e + c^2*(27*d
+ e*x^2)) - 6*b*(-3*a*c^3*x*(3*d + e*x^2) + b*Sqrt[1 + c^2*x^2]*(-2*e + c^2*(9*d + e*x^2)))*ArcSinh[c*x] + 9*b
^2*c^3*x*(3*d + e*x^2)*ArcSinh[c*x]^2)/(27*c^3)

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Maple [A]
time = 0.91, size = 206, normalized size = 1.35 \[\frac {\frac {a^{2} \left (\frac {1}{3} x^{3} c^{3} e +x \,c^{3} d \right )}{c^{2}}+\frac {b^{2} \left (\frac {\left (9 \arcsinh \left (c x \right )^{2} x^{2} c^{2} e +27 \arcsinh \left (c x \right )^{2} c^{2} d +2 c^{2} e \,x^{2}+54 c^{2} d -12 e \right ) x c}{27}-\frac {2 \arcsinh \left (c x \right ) \left (c^{2} e \,x^{2}+9 c^{2} d -2 e \right ) \sqrt {c^{2} x^{2}+1}}{9}\right )}{c^{2}}+\frac {2 a b \left (\frac {\arcsinh \left (c x \right ) x^{3} c^{3} e}{3}+\arcsinh \left (c x \right ) d \,c^{3} x -\frac {e \left (\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {c^{2} x^{2}+1}}{3}\right )}{3}-d \,c^{2} \sqrt {c^{2} x^{2}+1}\right )}{c^{2}}}{c}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a^2/c^2*(1/3*x^3*c^3*e+x*c^3*d)+b^2/c^2*(1/27*(9*arcsinh(c*x)^2*x^2*c^2*e+27*arcsinh(c*x)^2*c^2*d+2*c^2*e
*x^2+54*c^2*d-12*e)*x*c-2/9*arcsinh(c*x)*(c^2*e*x^2+9*c^2*d-2*e)*(c^2*x^2+1)^(1/2))+2*a*b/c^2*(1/3*arcsinh(c*x
)*x^3*c^3*e+arcsinh(c*x)*x*c^3*d-1/3*e*(1/3*c^2*x^2*(c^2*x^2+1)^(1/2)-2/3*(c^2*x^2+1)^(1/2))-d*c^2*(c^2*x^2+1)
^(1/2)))

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Maxima [A]
time = 0.27, size = 222, normalized size = 1.45 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} \operatorname {arsinh}\left (c x\right )^{2} e + b^{2} d x \operatorname {arsinh}\left (c x\right )^{2} + \frac {1}{3} \, a^{2} x^{3} e + 2 \, b^{2} d {\left (x - \frac {\sqrt {c^{2} x^{2} + 1} \operatorname {arsinh}\left (c x\right )}{c}\right )} + a^{2} d x + \frac {2}{9} \, {\left (3 \, x^{3} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} a b e - \frac {2}{27} \, {\left (3 \, c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {c^{2} x^{3} - 6 \, x}{c^{2}}\right )} b^{2} e + \frac {2 \, {\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} a b d}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*x^3*arcsinh(c*x)^2*e + b^2*d*x*arcsinh(c*x)^2 + 1/3*a^2*x^3*e + 2*b^2*d*(x - sqrt(c^2*x^2 + 1)*arcsinh
(c*x)/c) + a^2*d*x + 2/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*a*b*e
- 2/27*(3*c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4)*arcsinh(c*x) - (c^2*x^3 - 6*x)/c^2)*b^2*e +
2*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*a*b*d/c

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (142) = 284\).
time = 0.39, size = 304, normalized size = 1.99 \begin {gather*} \frac {27 \, {\left (a^{2} + 2 \, b^{2}\right )} c^{3} d x + 9 \, {\left (b^{2} c^{3} x^{3} \cosh \left (1\right ) + b^{2} c^{3} x^{3} \sinh \left (1\right ) + 3 \, b^{2} c^{3} d x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + {\left ({\left (9 \, a^{2} + 2 \, b^{2}\right )} c^{3} x^{3} - 12 \, b^{2} c x\right )} \cosh \left (1\right ) + 6 \, {\left (3 \, a b c^{3} x^{3} \cosh \left (1\right ) + 3 \, a b c^{3} x^{3} \sinh \left (1\right ) + 9 \, a b c^{3} d x - {\left (9 \, b^{2} c^{2} d + {\left (b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \cosh \left (1\right ) + {\left (b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \sinh \left (1\right )\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left ({\left (9 \, a^{2} + 2 \, b^{2}\right )} c^{3} x^{3} - 12 \, b^{2} c x\right )} \sinh \left (1\right ) - 6 \, {\left (9 \, a b c^{2} d + {\left (a b c^{2} x^{2} - 2 \, a b\right )} \cosh \left (1\right ) + {\left (a b c^{2} x^{2} - 2 \, a b\right )} \sinh \left (1\right )\right )} \sqrt {c^{2} x^{2} + 1}}{27 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(27*(a^2 + 2*b^2)*c^3*d*x + 9*(b^2*c^3*x^3*cosh(1) + b^2*c^3*x^3*sinh(1) + 3*b^2*c^3*d*x)*log(c*x + sqrt(
c^2*x^2 + 1))^2 + ((9*a^2 + 2*b^2)*c^3*x^3 - 12*b^2*c*x)*cosh(1) + 6*(3*a*b*c^3*x^3*cosh(1) + 3*a*b*c^3*x^3*si
nh(1) + 9*a*b*c^3*d*x - (9*b^2*c^2*d + (b^2*c^2*x^2 - 2*b^2)*cosh(1) + (b^2*c^2*x^2 - 2*b^2)*sinh(1))*sqrt(c^2
*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) + ((9*a^2 + 2*b^2)*c^3*x^3 - 12*b^2*c*x)*sinh(1) - 6*(9*a*b*c^2*d + (a
*b*c^2*x^2 - 2*a*b)*cosh(1) + (a*b*c^2*x^2 - 2*a*b)*sinh(1))*sqrt(c^2*x^2 + 1))/c^3

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Sympy [A]
time = 0.26, size = 279, normalized size = 1.82 \begin {gather*} \begin {cases} a^{2} d x + \frac {a^{2} e x^{3}}{3} + 2 a b d x \operatorname {asinh}{\left (c x \right )} + \frac {2 a b e x^{3} \operatorname {asinh}{\left (c x \right )}}{3} - \frac {2 a b d \sqrt {c^{2} x^{2} + 1}}{c} - \frac {2 a b e x^{2} \sqrt {c^{2} x^{2} + 1}}{9 c} + \frac {4 a b e \sqrt {c^{2} x^{2} + 1}}{9 c^{3}} + b^{2} d x \operatorname {asinh}^{2}{\left (c x \right )} + 2 b^{2} d x + \frac {b^{2} e x^{3} \operatorname {asinh}^{2}{\left (c x \right )}}{3} + \frac {2 b^{2} e x^{3}}{27} - \frac {2 b^{2} d \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{c} - \frac {2 b^{2} e x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{9 c} - \frac {4 b^{2} e x}{9 c^{2}} + \frac {4 b^{2} e \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{9 c^{3}} & \text {for}\: c \neq 0 \\a^{2} \left (d x + \frac {e x^{3}}{3}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*d*x + a**2*e*x**3/3 + 2*a*b*d*x*asinh(c*x) + 2*a*b*e*x**3*asinh(c*x)/3 - 2*a*b*d*sqrt(c**2*x**
2 + 1)/c - 2*a*b*e*x**2*sqrt(c**2*x**2 + 1)/(9*c) + 4*a*b*e*sqrt(c**2*x**2 + 1)/(9*c**3) + b**2*d*x*asinh(c*x)
**2 + 2*b**2*d*x + b**2*e*x**3*asinh(c*x)**2/3 + 2*b**2*e*x**3/27 - 2*b**2*d*sqrt(c**2*x**2 + 1)*asinh(c*x)/c
- 2*b**2*e*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(9*c) - 4*b**2*e*x/(9*c**2) + 4*b**2*e*sqrt(c**2*x**2 + 1)*asin
h(c*x)/(9*c**3), Ne(c, 0)), (a**2*(d*x + e*x**3/3), True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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