3.7.30 \(\int (d+e x^2) \sqrt {a+b \sinh ^{-1}(c x)} \, dx\) [630]

Optimal. Leaf size=322 \[ d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {\sqrt {b} d e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {\sqrt {b} e e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^3}+\frac {\sqrt {b} e e^{\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{48 c^3}-\frac {\sqrt {b} d e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {\sqrt {b} e e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^3}-\frac {\sqrt {b} e e^{-\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{48 c^3} \]

[Out]

1/144*e*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*b^(1/2)*3^(1/2)*Pi^(1/2)/c^3-1/144*e*erfi(3^(
1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*b^(1/2)*3^(1/2)*Pi^(1/2)/c^3/exp(3*a/b)+1/4*d*exp(a/b)*erf((a+b*arcsinh
(c*x))^(1/2)/b^(1/2))*b^(1/2)*Pi^(1/2)/c-1/16*e*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*b^(1/2)*Pi^(1/2
)/c^3-1/4*d*erfi((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*b^(1/2)*Pi^(1/2)/c/exp(a/b)+1/16*e*erfi((a+b*arcsinh(c*x))^
(1/2)/b^(1/2))*b^(1/2)*Pi^(1/2)/c^3/exp(a/b)+d*x*(a+b*arcsinh(c*x))^(1/2)+1/3*e*x^3*(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]
time = 0.61, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {5793, 5772, 5819, 3389, 2211, 2236, 2235, 5777, 3393} \begin {gather*} -\frac {\sqrt {\pi } \sqrt {b} e e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^3}+\frac {\sqrt {\frac {\pi }{3}} \sqrt {b} e e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{48 c^3}+\frac {\sqrt {\pi } \sqrt {b} e e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^3}-\frac {\sqrt {\frac {\pi }{3}} \sqrt {b} e e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{48 c^3}+\frac {\sqrt {\pi } \sqrt {b} d e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {\sqrt {\pi } \sqrt {b} d e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

d*x*Sqrt[a + b*ArcSinh[c*x]] + (e*x^3*Sqrt[a + b*ArcSinh[c*x]])/3 + (Sqrt[b]*d*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b
*ArcSinh[c*x]]/Sqrt[b]])/(4*c) - (Sqrt[b]*e*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(16*c^3) +
(Sqrt[b]*e*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(48*c^3) - (Sqrt[b]*d*Sqrt
[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(4*c*E^(a/b)) + (Sqrt[b]*e*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]
/Sqrt[b]])/(16*c^3*E^(a/b)) - (Sqrt[b]*e*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(48*c^3*
E^((3*a)/b))

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[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 5777

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

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 5819

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

Rubi steps

\begin {align*} \int \left (d+e x^2\right ) \sqrt {a+b \sinh ^{-1}(c x)} \, dx &=\int \left (d \sqrt {a+b \sinh ^{-1}(c x)}+e x^2 \sqrt {a+b \sinh ^{-1}(c x)}\right ) \, dx\\ &=d \int \sqrt {a+b \sinh ^{-1}(c x)} \, dx+e \int x^2 \sqrt {a+b \sinh ^{-1}(c x)} \, dx\\ &=d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {1}{2} (b c d) \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx-\frac {1}{6} (b c e) \int \frac {x^3}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx\\ &=d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {(b d) \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 c}-\frac {(b e) \text {Subst}\left (\int \frac {\sinh ^3(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{6 c^3}\\ &=d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {(b d) \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c}-\frac {(b d) \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c}-\frac {(i b e) \text {Subst}\left (\int \left (\frac {3 i \sinh (x)}{4 \sqrt {a+b x}}-\frac {i \sinh (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{6 c^3}\\ &=d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {d \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{2 c}-\frac {d \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{2 c}-\frac {(b e) \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{24 c^3}+\frac {(b e) \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^3}\\ &=d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {\sqrt {b} d e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {\sqrt {b} d e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {(b e) \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{48 c^3}-\frac {(b e) \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{48 c^3}-\frac {(b e) \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}+\frac {(b e) \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}\\ &=d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {\sqrt {b} d e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {\sqrt {b} d e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {e \text {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{24 c^3}-\frac {e \text {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{24 c^3}-\frac {e \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c^3}+\frac {e \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 c^3}\\ &=d x \sqrt {a+b \sinh ^{-1}(c x)}+\frac {1}{3} e x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {\sqrt {b} d e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {\sqrt {b} e e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^3}+\frac {\sqrt {b} e e^{\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{48 c^3}-\frac {\sqrt {b} d e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+\frac {\sqrt {b} e e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 c^3}-\frac {\sqrt {b} e e^{-\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{48 c^3}\\ \end {align*}

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Mathematica [A]
time = 1.97, size = 319, normalized size = 0.99 \begin {gather*} \frac {d e^{-\frac {a}{b}} \sqrt {a+b \sinh ^{-1}(c x)} \left (-\frac {e^{\frac {2 a}{b}} \Gamma \left (\frac {3}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )}{\sqrt {\frac {a}{b}+\sinh ^{-1}(c x)}}+\frac {\Gamma \left (\frac {3}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{\sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}}}\right )}{2 c}+\frac {e e^{-\frac {3 a}{b}} \sqrt {a+b \sinh ^{-1}(c x)} \left (9 e^{\frac {4 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {3}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )+\sqrt {3} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {3}{2},-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-9 e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {3}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )-\sqrt {3} e^{\frac {6 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {3}{2},\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{72 c^3 \sqrt {-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[a + b*ArcSinh[c*x]]*(-((E^((2*a)/b)*Gamma[3/2, a/b + ArcSinh[c*x]])/Sqrt[a/b + ArcSinh[c*x]]) + Gamma[
3/2, -((a + b*ArcSinh[c*x])/b)]/Sqrt[-((a + b*ArcSinh[c*x])/b)]))/(2*c*E^(a/b)) + (e*Sqrt[a + b*ArcSinh[c*x]]*
(9*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[3/2, a/b + ArcSinh[c*x]] + Sqrt[3]*Sqrt[a/b + ArcSinh[c*x
]]*Gamma[3/2, (-3*(a + b*ArcSinh[c*x]))/b] - 9*E^((2*a)/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[3/2, -((a + b*ArcSin
h[c*x])/b)] - Sqrt[3]*E^((6*a)/b)*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[3/2, (3*(a + b*ArcSinh[c*x]))/b]))/(72
*c^3*E^((3*a)/b)*Sqrt[-((a + b*ArcSinh[c*x])^2/b^2)])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (e \,x^{2}+d \right ) \sqrt {a +b \arcsinh \left (c x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \operatorname {asinh}{\left (c x \right )}} \left (d + e x^{2}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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