3.7.9 \(\int (d+e x^2) (a+b \sinh ^{-1}(c x)) \, dx\) [609]

Optimal. Leaf size=81 \[ -\frac {b \left (3 c^2 d-e\right ) \sqrt {1+c^2 x^2}}{3 c^3}-\frac {b e \left (1+c^2 x^2\right )^{3/2}}{9 c^3}+d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5792, 455, 45} \begin {gather*} d x \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {b \sqrt {c^2 x^2+1} \left (3 c^2 d-e\right )}{3 c^3}-\frac {b e \left (c^2 x^2+1\right )^{3/2}}{9 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 5792

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[e, c^2*d] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 71, normalized size = 0.88 \begin {gather*} \frac {1}{9} \left (3 a x \left (3 d+e x^2\right )-\frac {b \sqrt {1+c^2 x^2} \left (-2 e+c^2 \left (9 d+e x^2\right )\right )}{c^3}+3 b x \left (3 d+e x^2\right ) \sinh ^{-1}(c x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.60, size = 109, normalized size = 1.35

method	result	size
derivativedivides	\(\frac {\frac {a \left (d \,c^{3} x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+\frac {b \left (\arcsinh \left (c x \right ) d \,c^{3} x +\frac {\arcsinh \left (c x \right ) e \,c^{3} x^{3}}{3}-d \,c^{2} \sqrt {c^{2} x^{2}+1}-\frac {e \left (\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c}\)	\(109\)
default	\(\frac {\frac {a \left (d \,c^{3} x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+\frac {b \left (\arcsinh \left (c x \right ) d \,c^{3} x +\frac {\arcsinh \left (c x \right ) e \,c^{3} x^{3}}{3}-d \,c^{2} \sqrt {c^{2} x^{2}+1}-\frac {e \left (\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 93, normalized size = 1.15 \begin {gather*} \frac {1}{3} \, a x^{3} e + a d x + \frac {1}{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 )} b e + \frac {{\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 134, normalized size = 1.65 \begin {gather*} \frac {3 \, a c^{3} x^{3} \cosh \left (1\right ) + 3 \, a c^{3} x^{3} \sinh \left (1\right ) + 9 \, a c^{3} d x + 3 \, {\left (b c^{3} x^{3} \cosh \left (1\right ) + b c^{3} x^{3} \sinh \left (1\right ) + 3 \, b c^{3} d x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (9 \, b c^{2} d + {\left (b c^{2} x^{2} - 2 \, b\right )} \cosh \left (1\right ) + {\left (b c^{2} x^{2} - 2 \, b\right )} \sinh \left (1\right )\right )} \sqrt {c^{2} x^{2} + 1}}{9 \, 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)),x, algorithm="fricas")

[Out]

1/9*(3*a*c^3*x^3*cosh(1) + 3*a*c^3*x^3*sinh(1) + 9*a*c^3*d*x + 3*(b*c^3*x^3*cosh(1) + b*c^3*x^3*sinh(1) + 3*b*
c^3*d*x)*log(c*x + sqrt(c^2*x^2 + 1)) - (9*b*c^2*d + (b*c^2*x^2 - 2*b)*cosh(1) + (b*c^2*x^2 - 2*b)*sinh(1))*sq
rt(c^2*x^2 + 1))/c^3

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Sympy [A]
time = 0.15, size = 109, normalized size = 1.35 \begin {gather*} \begin {cases} a d x + \frac {a e x^{3}}{3} + b d x \operatorname {asinh}{\left (c x \right )} + \frac {b e x^{3} \operatorname {asinh}{\left (c x \right )}}{3} - \frac {b d \sqrt {c^{2} x^{2} + 1}}{c} - \frac {b e x^{2} \sqrt {c^{2} x^{2} + 1}}{9 c} + \frac {2 b e \sqrt {c^{2} x^{2} + 1}}{9 c^{3}} & \text {for}\: c \neq 0 \\a \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)),x)

[Out]

Piecewise((a*d*x + a*e*x**3/3 + b*d*x*asinh(c*x) + b*e*x**3*asinh(c*x)/3 - b*d*sqrt(c**2*x**2 + 1)/c - b*e*x**
2*sqrt(c**2*x**2 + 1)/(9*c) + 2*b*e*sqrt(c**2*x**2 + 1)/(9*c**3), Ne(c, 0)), (a*(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)),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 )\,\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))*(d + e*x^2),x)

[Out]

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

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