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

Optimal. Leaf size=150 \[ \frac {3 \sqrt {\pi } b^{3/2} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 d}+\frac {3 \sqrt {\pi } b^{3/2} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 d}-\frac {3 b \sqrt {(c+d x)^2+1} \sqrt {a+b \sinh ^{-1}(c+d x)}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{d} \]

[Out]

(d*x+c)*(a+b*arcsinh(d*x+c))^(3/2)/d+3/8*b^(3/2)*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d+3
/8*b^(3/2)*erfi((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d/exp(a/b)-3/2*b*(1+(d*x+c)^2)^(1/2)*(a+b*arcsinh
(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.25, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5863, 5653, 5717, 5657, 3307, 2180, 2205, 2204} \[ \frac {3 \sqrt {\pi } b^{3/2} e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 d}+\frac {3 \sqrt {\pi } b^{3/2} e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 d}-\frac {3 b \sqrt {(c+d x)^2+1} \sqrt {a+b \sinh ^{-1}(c+d x)}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^(3/2),x]

[Out]

(-3*b*Sqrt[1 + (c + d*x)^2]*Sqrt[a + b*ArcSinh[c + d*x]])/(2*d) + ((c + d*x)*(a + b*ArcSinh[c + d*x])^(3/2))/d
+ (3*b^(3/2)*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(8*d) + (3*b^(3/2)*Sqrt[Pi]*Erfi[Sqr
t[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(8*d*E^(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 3307

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

Rule 5653

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 5657

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

Rule 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 5863

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 272, normalized size = 1.81 \[ \frac {\sqrt {b} \left (\sqrt {\pi } (3 b-2 a) \left (\sinh \left (\frac {a}{b}\right )+\cosh \left (\frac {a}{b}\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )+\sqrt {\pi } (2 a+3 b) \left (\cosh \left (\frac {a}{b}\right )-\sinh \left (\frac {a}{b}\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )+4 \sqrt {b} \left (2 (c+d x) \sinh ^{-1}(c+d x)-3 \sqrt {(c+d x)^2+1}\right ) \sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 d}+\frac {a e^{-\frac {a}{b}} \sqrt {a+b \sinh ^{-1}(c+d x)} \left (\frac {\Gamma \left (\frac {3}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{\sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}}}-\frac {e^{\frac {2 a}{b}} \Gamma \left (\frac {3}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{\sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)}}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*Sqrt[a + b*ArcSinh[c + d*x]]*(-((E^((2*a)/b)*Gamma[3/2, a/b + ArcSinh[c + d*x]])/Sqrt[a/b + ArcSinh[c + d*x
]]) + Gamma[3/2, -((a + b*ArcSinh[c + d*x])/b)]/Sqrt[-((a + b*ArcSinh[c + d*x])/b)]))/(2*d*E^(a/b)) + (Sqrt[b]
*(4*Sqrt[b]*Sqrt[a + b*ArcSinh[c + d*x]]*(-3*Sqrt[1 + (c + d*x)^2] + 2*(c + d*x)*ArcSinh[c + d*x]) + (2*a + 3*
b)*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(Cosh[a/b] - Sinh[a/b]) + (-2*a + 3*b)*Sqrt[Pi]*Erf[Sqr
t[a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(Cosh[a/b] + Sinh[a/b])))/(8*d)

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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((a+b*arcsinh(d*x+c))^(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 \[ \int {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))^(3/2),x)

[Out]

int((a + b*asinh(c + d*x))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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