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

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

[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))

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Rubi [A]  time = 0.54193, antiderivative size = 209, normalized size of antiderivative = 1., 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 = {5865, 12, 5667, 5774, 5669, 5448, 3308, 2180, 2204, 2205, 5675} \[ -\frac{2 \sqrt{2 \pi } e e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{2 \sqrt{2 \pi } e e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{8 e (c+d x)^2}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e \sqrt{(c+d x)^2+1} (c+d x)}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

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 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
]

Rule 12

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

Rule 5667

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 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && GtQ[d, 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 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 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 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 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rubi steps

\begin{align*} \int \frac{c e+d e x}{\left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e x}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{e \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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) \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 )}{3 b^3 d}+\frac{(8 e) \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 )}{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 = 0.673967, size = 227, normalized size = 1.09 \[ \frac{e e^{-2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )} \left (e^{\frac{2 a}{b}} \left (4 \sqrt{2} e^{2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Gamma}\left (\frac{1}{2},\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-4 a e^{4 \sinh ^{-1}(c+d x)}-4 a-b e^{4 \sinh ^{-1}(c+d x)}-4 b \left (e^{4 \sinh ^{-1}(c+d x)}+1\right ) \sinh ^{-1}(c+d x)+b\right )-4 \sqrt{2} b e^{2 \sinh ^{-1}(c+d x)} \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[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))

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{(dex+ce) \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d e x + c e}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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{align*}

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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d e x + c e}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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)