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

Optimal. Leaf size=148 \[ \frac{\sqrt{\frac{\pi }{2}} e e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}+\frac{\sqrt{\frac{\pi }{2}} e e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}-\frac{2 e \sqrt{(c+d x)^2+1} (c+d x)}{b d \sqrt{a+b \sinh ^{-1}(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.231014, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5865, 12, 5665, 3307, 2180, 2204, 2205} \[ \frac{\sqrt{\frac{\pi }{2}} e e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}+\frac{\sqrt{\frac{\pi }{2}} e e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{b^{3/2} d}-\frac{2 e \sqrt{(c+d x)^2+1} (c+d x)}{b d \sqrt{a+b \sinh ^{-1}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(b*d*Sqrt[a + b*ArcSinh[c + d*x]]) + (e*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqr
t[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(b^(3/2)*d) + (e*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c +
d*x]])/Sqrt[b]])/(b^(3/2)*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 5665

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.112226, size = 147, normalized size = 0.99 \[ \frac{e e^{-\frac{2 a}{b}} \left (\sqrt{2} \sqrt{-\frac{a+b \sinh ^{-1}(c+d x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-\sqrt{2} e^{\frac{4 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \text{Gamma}\left (\frac{1}{2},\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-2 e^{\frac{2 a}{b}} \sinh \left (2 \sinh ^{-1}(c+d x)\right )\right )}{2 b d \sqrt{a+b \sinh ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.105, size = 0, normalized size = 0. \begin{align*} \int{(dex+ce) \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{-{\frac{3}{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))^(3/2),x)

[Out]

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

[Out]

integrate((d*e*x + c*e)/(b*arcsinh(d*x + c) + a)^(3/2), 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))^(3/2),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 \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}} + b \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}} \operatorname{asinh}{\left (c + d x \right )}}\, dx + \int \frac{d x}{a \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}} + b \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}} \operatorname{asinh}{\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))**(3/2),x)

[Out]

e*(Integral(c/(a*sqrt(a + b*asinh(c + d*x)) + b*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)), x) + Integral(d*x/
(a*sqrt(a + b*asinh(c + d*x)) + b*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)), 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{3}{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))^(3/2),x, algorithm="giac")

[Out]

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