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

Optimal. Leaf size=217 \[ -\frac {\sqrt {\pi } e^3 e^{\frac {4 a}{b}} \text {erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{2}} e^3 e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\pi } e^3 e^{-\frac {4 a}{b}} \text {erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}-\frac {\sqrt {\frac {\pi }{2}} e^3 e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d} \]

[Out]

1/16*e^3*exp(2*a/b)*erf(2^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/d/b^(1/2)-1/16*e^3*erfi(2
^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/d/exp(2*a/b)/b^(1/2)-1/32*e^3*exp(4*a/b)*erf(2*(a+
b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d/b^(1/2)+1/32*e^3*erfi(2*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/
2)/d/exp(4*a/b)/b^(1/2)

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Rubi [A]  time = 0.46, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {5865, 12, 5669, 5448, 3308, 2180, 2204, 2205} \[ -\frac {\sqrt {\pi } e^3 e^{\frac {4 a}{b}} \text {Erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{2}} e^3 e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\pi } e^3 e^{-\frac {4 a}{b}} \text {Erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}-\frac {\sqrt {\frac {\pi }{2}} e^3 e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(32*Sqrt[b]*d) + (e^3*E^((2*a)/b)*Sq
rt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(8*Sqrt[b]*d) + (e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b
*ArcSinh[c + d*x]])/Sqrt[b]])/(32*Sqrt[b]*d*E^((4*a)/b)) - (e^3*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c
+ d*x]])/Sqrt[b]])/(8*Sqrt[b]*d*E^((2*a)/b))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5669

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

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

\begin {align*} \int \frac {(c e+d e x)^3}{\sqrt {a+b \sinh ^{-1}(c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^3 x^3}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh ^3(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \left (-\frac {\sinh (2 x)}{4 \sqrt {a+b x}}+\frac {\sinh (4 x)}{8 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}-\frac {e^3 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}\\ &=-\frac {e^3 \operatorname {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}+\frac {e^3 \operatorname {Subst}\left (\int \frac {e^{4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}+\frac {e^3 \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}-\frac {e^3 \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}\\ &=-\frac {e^3 \operatorname {Subst}\left (\int e^{\frac {4 a}{b}-\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b d}+\frac {e^3 \operatorname {Subst}\left (\int e^{-\frac {4 a}{b}+\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b d}+\frac {e^3 \operatorname {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{4 b d}-\frac {e^3 \operatorname {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{4 b d}\\ &=-\frac {e^3 e^{\frac {4 a}{b}} \sqrt {\pi } \text {erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {e^3 e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {e^3 e^{-\frac {4 a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}-\frac {e^3 e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 205, normalized size = 0.94 \[ \frac {e^3 e^{-\frac {4 a}{b}} \left (\sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-2 \sqrt {2} e^{\frac {2 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+e^{\frac {6 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (e^{\frac {2 a}{b}} \Gamma \left (\frac {1}{2},\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-2 \sqrt {2} \Gamma \left (\frac {1}{2},\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )\right )}{32 d \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e^3*(Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[1/2, (-4*(a + b*ArcSinh[c + d*x]))/b] - 2*Sqrt[2]*E^((2*a)/b)*
Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[1/2, (-2*(a + b*ArcSinh[c + d*x]))/b] + E^((6*a)/b)*Sqrt[a/b + ArcSi
nh[c + d*x]]*(-2*Sqrt[2]*Gamma[1/2, (2*(a + b*ArcSinh[c + d*x]))/b] + E^((2*a)/b)*Gamma[1/2, (4*(a + b*ArcSinh
[c + d*x]))/b])))/(32*d*E^((4*a)/b)*Sqrt[a + b*ArcSinh[c + d*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((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(1/2),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 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{\sqrt {b \operatorname {arsinh}\left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^3/(a + b*asinh(c + d*x))^(1/2),x)

[Out]

int((c*e + d*e*x)^3/(a + b*asinh(c + d*x))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3/(a+b*asinh(d*x+c))**(1/2),x)

[Out]

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

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