3.643 \(\int \frac {d+e x^2}{(a+b \sinh ^{-1}(c x))^{3/2}} \, dx\)

Optimal. Leaf size=349 \[ \frac {\sqrt {\pi } e e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}-\frac {\sqrt {3 \pi } e e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}-\frac {\sqrt {\pi } e e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}+\frac {\sqrt {3 \pi } e e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}-\frac {\sqrt {\pi } d e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c}+\frac {\sqrt {\pi } d e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c}-\frac {2 d \sqrt {c^2 x^2+1}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {2 e x^2 \sqrt {c^2 x^2+1}}{b c \sqrt {a+b \sinh ^{-1}(c x)}} \]

[Out]

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

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Rubi [A]  time = 0.69, antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5706, 5655, 5779, 3308, 2180, 2204, 2205, 5665} \[ \frac {\sqrt {\pi } e e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}-\frac {\sqrt {3 \pi } e e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}-\frac {\sqrt {\pi } e e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}+\frac {\sqrt {3 \pi } e e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 b^{3/2} c^3}-\frac {\sqrt {\pi } d e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c}+\frac {\sqrt {\pi } d e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c}-\frac {2 d \sqrt {c^2 x^2+1}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {2 e x^2 \sqrt {c^2 x^2+1}}{b c \sqrt {a+b \sinh ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(-2*d*Sqrt[1 + c^2*x^2])/(b*c*Sqrt[a + b*ArcSinh[c*x]]) - (2*e*x^2*Sqrt[1 + c^2*x^2])/(b*c*Sqrt[a + b*ArcSinh[
c*x]]) - (d*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(b^(3/2)*c) + (e*E^(a/b)*Sqrt[Pi]*Erf[Sqrt
[a + b*ArcSinh[c*x]]/Sqrt[b]])/(4*b^(3/2)*c^3) - (e*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x
]])/Sqrt[b]])/(4*b^(3/2)*c^3) + (d*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(b^(3/2)*c*E^(a/b)) - (e*S
qrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(4*b^(3/2)*c^3*E^(a/b)) + (e*Sqrt[3*Pi]*Erfi[(Sqrt[3]*Sqrt[a +
b*ArcSinh[c*x]])/Sqrt[b]])/(4*b^(3/2)*c^3*E^((3*a)/b))

Rule 2180

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] &&  !$UseGamma === True

Rule 2204

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 2205

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 3308

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 5655

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

Rule 5665

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

Rule 5706

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 5779

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

Rubi steps

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

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Mathematica [A]  time = 1.52, size = 303, normalized size = 0.87 \[ \frac {e^{-3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \left (\left (4 c^2 d-e\right ) e^{\frac {4 a}{b}+3 \sinh ^{-1}(c x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )+\left (4 c^2 d-e\right ) e^{\frac {2 a}{b}+3 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )+e^{\frac {3 a}{b}} \left (\sqrt {3} e e^{3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-\left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (4 c^2 d e^{2 \sinh ^{-1}(c x)}+e \left (e^{2 \sinh ^{-1}(c x)}-1\right )^2\right )\right )+\sqrt {3} e e^{3 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{4 b c^3 \sqrt {a+b \sinh ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x^2)/(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

((4*c^2*d - e)*E^((4*a)/b + 3*ArcSinh[c*x])*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, a/b + ArcSinh[c*x]] + Sqrt[3]*
e*E^(3*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-3*(a + b*ArcSinh[c*x]))/b] + (4*c^2*d - e)*E
^((2*a)/b + 3*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, -((a + b*ArcSinh[c*x])/b)] + E^((3*a)/b
)*(-((1 + E^(2*ArcSinh[c*x]))*(4*c^2*d*E^(2*ArcSinh[c*x]) + e*(-1 + E^(2*ArcSinh[c*x]))^2)) + Sqrt[3]*e*E^(3*(
a/b + ArcSinh[c*x]))*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (3*(a + b*ArcSinh[c*x]))/b]))/(4*b*c^3*E^(3*(a/b + Ar
cSinh[c*x]))*Sqrt[a + b*ArcSinh[c*x]])

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fricas [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((e*x^2+d)/(a+b*arcsinh(c*x))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{2} + d}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/(a+b*arcsinh(c*x))^(3/2),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)/(b*arcsinh(c*x) + a)^(3/2), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {e \,x^{2}+d}{\left (a +b \arcsinh \left (c x \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)/(a+b*arcsinh(c*x))^(3/2),x)

[Out]

int((e*x^2+d)/(a+b*arcsinh(c*x))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{2} + d}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/(a+b*arcsinh(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)/(b*arcsinh(c*x) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {e\,x^2+d}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)/(a + b*asinh(c*x))^(3/2),x)

[Out]

int((d + e*x^2)/(a + b*asinh(c*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d + e x^{2}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)/(a+b*asinh(c*x))**(3/2),x)

[Out]

Integral((d + e*x**2)/(a + b*asinh(c*x))**(3/2), x)

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