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

Optimal. Leaf size=254 \[ -\frac{3 \sqrt{\frac{\pi }{2}} d e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}+\frac{\sqrt{\frac{3 \pi }{2}} d e^{\frac{6 a}{b}} \text{Erf}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}-\frac{3 \sqrt{\frac{\pi }{2}} d e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}+\frac{\sqrt{\frac{3 \pi }{2}} d e^{-\frac{6 a}{b}} \text{Erfi}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}-\frac{2 d x^3 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

[Out]

(-2*d*x^3*(1 + c^2*x^2)^(3/2))/(b*c*Sqrt[a + b*ArcSinh[c*x]]) - (3*d*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[
a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^4) + (d*E^((6*a)/b)*Sqrt[(3*Pi)/2]*Erf[(Sqrt[6]*Sqrt[a + b*ArcSin
h[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^4) - (3*d*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^
(3/2)*c^4*E^((2*a)/b)) + (d*Sqrt[(3*Pi)/2]*Erfi[(Sqrt[6]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^4*E
^((6*a)/b))

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Rubi [A]  time = 1.30785, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 27, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {5777, 5779, 5448, 3307, 2180, 2204, 2205} \[ -\frac{3 \sqrt{\frac{\pi }{2}} d e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}+\frac{\sqrt{\frac{3 \pi }{2}} d e^{\frac{6 a}{b}} \text{Erf}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}-\frac{3 \sqrt{\frac{\pi }{2}} d e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}+\frac{\sqrt{\frac{3 \pi }{2}} d e^{-\frac{6 a}{b}} \text{Erfi}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}-\frac{2 d x^3 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*x^3*(1 + c^2*x^2)^(3/2))/(b*c*Sqrt[a + b*ArcSinh[c*x]]) - (3*d*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[
a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^4) + (d*E^((6*a)/b)*Sqrt[(3*Pi)/2]*Erf[(Sqrt[6]*Sqrt[a + b*ArcSin
h[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^4) - (3*d*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^
(3/2)*c^4*E^((2*a)/b)) + (d*Sqrt[(3*Pi)/2]*Erfi[(Sqrt[6]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^4*E
^((6*a)/b))

Rule 5777

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntP
art[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p -
1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x] - Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(
n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1), x],
x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

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 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 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{x^3 \left (d+c^2 d x^2\right )}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac{2 d x^3 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{(6 d) \int \frac{x^2 \sqrt{1+c^2 x^2}}{\sqrt{a+b \sinh ^{-1}(c x)}} \, dx}{b c}+\frac{(12 c d) \int \frac{x^4 \sqrt{1+c^2 x^2}}{\sqrt{a+b \sinh ^{-1}(c x)}} \, dx}{b}\\ &=-\frac{2 d x^3 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{(6 d) \operatorname{Subst}\left (\int \frac{\cosh ^2(x) \sinh ^2(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}+\frac{(12 d) \operatorname{Subst}\left (\int \frac{\cosh ^2(x) \sinh ^4(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{2 d x^3 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{(6 d) \operatorname{Subst}\left (\int \left (-\frac{1}{8 \sqrt{a+b x}}+\frac{\cosh (4 x)}{8 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}+\frac{(12 d) \operatorname{Subst}\left (\int \left (\frac{1}{16 \sqrt{a+b x}}-\frac{\cosh (2 x)}{32 \sqrt{a+b x}}-\frac{\cosh (4 x)}{16 \sqrt{a+b x}}+\frac{\cosh (6 x)}{32 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{2 d x^3 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}-\frac{(3 d) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^4}+\frac{(3 d) \operatorname{Subst}\left (\int \frac{\cosh (6 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^4}\\ &=-\frac{2 d x^3 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{(3 d) \operatorname{Subst}\left (\int \frac{e^{-6 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}-\frac{(3 d) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}-\frac{(3 d) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}+\frac{(3 d) \operatorname{Subst}\left (\int \frac{e^{6 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}\\ &=-\frac{2 d x^3 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{(3 d) \operatorname{Subst}\left (\int e^{\frac{6 a}{b}-\frac{6 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^4}-\frac{(3 d) \operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^4}-\frac{(3 d) \operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^4}+\frac{(3 d) \operatorname{Subst}\left (\int e^{-\frac{6 a}{b}+\frac{6 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^4}\\ &=-\frac{2 d x^3 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}-\frac{3 d e^{\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}+\frac{d e^{\frac{6 a}{b}} \sqrt{\frac{3 \pi }{2}} \text{erf}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}-\frac{3 d e^{-\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}+\frac{d e^{-\frac{6 a}{b}} \sqrt{\frac{3 \pi }{2}} \text{erfi}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^4}\\ \end{align*}

Mathematica [A]  time = 0.841677, size = 232, normalized size = 0.91 \[ \frac{d e^{-\frac{6 a}{b}} \left (\sqrt{6} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-3 \sqrt{2} e^{\frac{4 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+3 \sqrt{2} e^{\frac{8 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-\sqrt{6} e^{\frac{12 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-8 e^{\frac{6 a}{b}} \sinh ^3\left (2 \sinh ^{-1}(c x)\right )\right )}{32 b c^4 \sqrt{a+b \sinh ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d*(Sqrt[6]*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-6*(a + b*ArcSinh[c*x]))/b] - 3*Sqrt[2]*E^((4*a)/b)*Sq
rt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-2*(a + b*ArcSinh[c*x]))/b] + 3*Sqrt[2]*E^((8*a)/b)*Sqrt[a/b + ArcSi
nh[c*x]]*Gamma[1/2, (2*(a + b*ArcSinh[c*x]))/b] - Sqrt[6]*E^((12*a)/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (6*
(a + b*ArcSinh[c*x]))/b] - 8*E^((6*a)/b)*Sinh[2*ArcSinh[c*x]]^3))/(32*b*c^4*E^((6*a)/b)*Sqrt[a + b*ArcSinh[c*x
]])

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Maple [F]  time = 0.172, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ({c}^{2}d{x}^{2}+d \right ) \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )} x^{3}}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d \left (\int \frac{x^{3}}{a \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} + b \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} \operatorname{asinh}{\left (c x \right )}}\, dx + \int \frac{c^{2} x^{5}}{a \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} + b \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} \operatorname{asinh}{\left (c x \right )}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(Integral(x**3/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(c**2*x**5/(a*
sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )} x^{3}}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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