3.3.19 \(\int \frac {c e+d e x}{(a+b \sinh ^{-1}(c+d x))^{5/2}} \, dx\) [219]

Optimal. Leaf size=209 \[ -\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {8 e (c+d x)^2}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 e e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d} \]

[Out]

-2/3*e*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)/b^(5/2)/d+2/3*e*erfi(2^(1/2
)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(5/2)/d/exp(2*a/b)-2/3*e*(d*x+c)*(1+(d*x+c)^2)^(1/2)/
b/d/(a+b*arcsinh(d*x+c))^(3/2)-4/3*e/b^2/d/(a+b*arcsinh(d*x+c))^(1/2)-8/3*e*(d*x+c)^2/b^2/d/(a+b*arcsinh(d*x+c
))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.37, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {5859, 12, 5779, 5818, 5780, 5556, 3389, 2211, 2236, 2235, 5783} \begin {gather*} -\frac {2 \sqrt {2 \pi } e e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 \sqrt {2 \pi } e e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {8 e (c+d x)^2}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e \sqrt {(c+d x)^2+1} (c+d x)}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (4*e)/(3*b^2*d*Sqrt[a + b*ArcS
inh[c + d*x]]) - (8*e*(c + d*x)^2)/(3*b^2*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (2*e*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(S
qrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) + (2*e*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSin
h[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*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 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 5556

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 5779

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[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSinh[c*x])^(n +
1)/Sqrt[1 + c^2*x^2]), x], x] - Dist[m/(b*c*(n + 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] && LtQ[n, -2]

Rule 5780

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

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5818

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Dist[f*(m/
(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[(f*x)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x]
/; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -1]

Rule 5859

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

________________________________________________________________________________________

Mathematica [A]
time = 0.47, size = 227, normalized size = 1.09 \begin {gather*} \frac {e e^{-2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \left (-4 \sqrt {2} b e^{2 \sinh ^{-1}(c+d x)} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+e^{\frac {2 a}{b}} \left (-4 a+b-4 a e^{4 \sinh ^{-1}(c+d x)}-b e^{4 \sinh ^{-1}(c+d x)}-4 b \left (1+e^{4 \sinh ^{-1}(c+d x)}\right ) \sinh ^{-1}(c+d x)+4 \sqrt {2} e^{2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )\right )}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(-4*Sqrt[2]*b*E^(2*ArcSinh[c + d*x])*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-2*(a + b*ArcSinh[c
+ d*x]))/b] + E^((2*a)/b)*(-4*a + b - 4*a*E^(4*ArcSinh[c + d*x]) - b*E^(4*ArcSinh[c + d*x]) - 4*b*(1 + E^(4*Ar
cSinh[c + d*x]))*ArcSinh[c + d*x] + 4*Sqrt[2]*E^(2*(a/b + ArcSinh[c + d*x]))*Sqrt[a/b + ArcSinh[c + d*x]]*(a +
b*ArcSinh[c + d*x])*Gamma[1/2, (2*(a + b*ArcSinh[c + d*x]))/b])))/(6*b^2*d*E^(2*(a/b + ArcSinh[c + d*x]))*(a
+ b*ArcSinh[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {d e x +c e}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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((d*e*x+c*e)/(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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {c\,e+d\,e\,x}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________