3.196 \(\int (a+b \sinh ^{-1}(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=179 \[ \frac {15 \sqrt {\pi } b^{5/2} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 d}-\frac {15 \sqrt {\pi } b^{5/2} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 d}+\frac {15 b^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{4 d}-\frac {5 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d} \]

[Out]

(d*x+c)*(a+b*arcsinh(d*x+c))^(5/2)/d+15/16*b^(5/2)*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d
-15/16*b^(5/2)*erfi((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d/exp(a/b)-5/2*b*(a+b*arcsinh(d*x+c))^(3/2)*(
1+(d*x+c)^2)^(1/2)/d+15/4*b^2*(d*x+c)*(a+b*arcsinh(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.40, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5863, 5653, 5717, 5779, 3308, 2180, 2204, 2205} \[ \frac {15 \sqrt {\pi } b^{5/2} e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 d}-\frac {15 \sqrt {\pi } b^{5/2} e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 d}+\frac {15 b^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{4 d}-\frac {5 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

(15*b^2*(c + d*x)*Sqrt[a + b*ArcSinh[c + d*x]])/(4*d) - (5*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3
/2))/(2*d) + ((c + d*x)*(a + b*ArcSinh[c + d*x])^(5/2))/d + (15*b^(5/2)*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSin
h[c + d*x]]/Sqrt[b]])/(16*d) - (15*b^(5/2)*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(16*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 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 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 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 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])

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 )^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {x \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=-\frac {5 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d}+\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int \sqrt {a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{4 d}\\ &=\frac {15 b^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{4 d}-\frac {5 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d}-\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{8 d}\\ &=\frac {15 b^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{4 d}-\frac {5 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d}-\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}\\ &=\frac {15 b^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{4 d}-\frac {5 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d}+\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}-\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}\\ &=\frac {15 b^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{4 d}-\frac {5 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d}+\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 d}-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 d}\\ &=\frac {15 b^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{4 d}-\frac {5 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{d}+\frac {15 b^{5/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 d}-\frac {15 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.50, size = 458, normalized size = 2.56 \[ \frac {\sqrt {b} \left (\sqrt {\pi } \left (4 a^2-12 a b+15 b^2\right ) \left (\sinh \left (\frac {a}{b}\right )+\cosh \left (\frac {a}{b}\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )+\sqrt {\pi } \left (4 a^2+12 a b+15 b^2\right ) \left (\sinh \left (\frac {a}{b}\right )-\cosh \left (\frac {a}{b}\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )+4 \sqrt {b} \sqrt {a+b \sinh ^{-1}(c+d x)} \left (2 \sqrt {(c+d x)^2+1} \left (a-5 b \sinh ^{-1}(c+d x)\right )+b (c+d x) \left (4 \sinh ^{-1}(c+d x)^2+15\right )\right )\right )+8 a^2 e^{-\frac {a}{b}} \sqrt {a+b \sinh ^{-1}(c+d x)} \left (\frac {\Gamma \left (\frac {3}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{\sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}}}-\frac {e^{\frac {2 a}{b}} \Gamma \left (\frac {3}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{\sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)}}\right )+4 a \sqrt {b} \left (\sqrt {\pi } (3 b-2 a) \left (\sinh \left (\frac {a}{b}\right )+\cosh \left (\frac {a}{b}\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )+\sqrt {\pi } (2 a+3 b) \left (\cosh \left (\frac {a}{b}\right )-\sinh \left (\frac {a}{b}\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )+4 \sqrt {b} \left (2 (c+d x) \sinh ^{-1}(c+d x)-3 \sqrt {(c+d x)^2+1}\right ) \sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{16 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

((8*a^2*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)]))/E^(a/b) + 4*a*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[S
qrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(Cosh[a/b] + Sinh[a/b])) + Sqrt[b]*(4*Sqrt[b]*Sqrt[a + b*ArcSinh[c + d*x]
]*(2*Sqrt[1 + (c + d*x)^2]*(a - 5*b*ArcSinh[c + d*x]) + b*(c + d*x)*(15 + 4*ArcSinh[c + d*x]^2)) + (4*a^2 + 12
*a*b + 15*b^2)*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(-Cosh[a/b] + Sinh[a/b]) + (4*a^2 - 12*a*b
+ 15*b^2)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(Cosh[a/b] + Sinh[a/b])))/(16*d)

________________________________________________________________________________________

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))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))^(5/2),x)

[Out]

int((a+b*arcsinh(d*x+c))^(5/2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))^(5/2),x)

[Out]

int((a + b*asinh(c + d*x))^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))**(5/2),x)

[Out]

Integral((a + b*asinh(c + d*x))**(5/2), x)

________________________________________________________________________________________