3.5.62 \(\int \frac {x^3 (d+c^2 d x^2)}{(a+b \sinh ^{-1}(c x))^{3/2}} \, dx\) [462]

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

[Out]

-3/32*d*exp(2*a/b)*erf(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/c^4-3/32*d*erfi(2^(1
/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/c^4/exp(2*a/b)+1/32*d*exp(6*a/b)*erf(6^(1/2)*(a
+b*arcsinh(c*x))^(1/2)/b^(1/2))*6^(1/2)*Pi^(1/2)/b^(3/2)/c^4+1/32*d*erfi(6^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1
/2))*6^(1/2)*Pi^(1/2)/b^(3/2)/c^4/exp(6*a/b)-2*d*x^3*(c^2*x^2+1)^(3/2)/b/c/(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]
time = 0.83, antiderivative size = 254, normalized size of antiderivative = 1.00, 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 = {5814, 5819, 5556, 3388, 2211, 2236, 2235} \begin {gather*} -\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)}} \end {gather*}

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 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 3388

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 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 5814

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/(b*c*(
n + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^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)/(b*f*(n + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^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[2*p, 0] && NeQ[m + 2*p + 1, 0] && IGtQ[m, -3]

Rule 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
0]

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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 1.41, size = 271, normalized size = 1.07 \begin {gather*} \frac {d e^{-\frac {6 a}{b}} \left (\sqrt {6} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \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}} \Gamma \left (\frac {1}{2},-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-e^{\frac {6 a}{b}} \left (64 c^3 x^3 \sqrt {1+c^2 x^2}-3 \sqrt {2} e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+\sqrt {6} e^{\frac {6 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+10 \sinh \left (2 \sinh ^{-1}(c x)\right )-8 \sinh \left (4 \sinh ^{-1}(c x)\right )+2 \sinh \left (6 \sinh ^{-1}(c x)\right )\right )\right )}{32 b c^4 \sqrt {a+b \sinh ^{-1}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[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] - E^((6*a)/b)*(64*c^3*x^3*Sqrt[1 + c^2*x
^2] - 3*Sqrt[2]*E^((2*a)/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (2*(a + b*ArcSinh[c*x]))/b] + Sqrt[6]*E^((6*a)
/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (6*(a + b*ArcSinh[c*x]))/b] + 10*Sinh[2*ArcSinh[c*x]] - 8*Sinh[4*ArcSi
nh[c*x]] + 2*Sinh[6*ArcSinh[c*x]])))/(32*b*c^4*E^((6*a)/b)*Sqrt[a + b*ArcSinh[c*x]])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (c^{2} d \,x^{2}+d \right )}{\left (a +b \arcsinh \left (c x \right )\right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(x^3*(c^2*d*x^2+d)/(a+b*arcsinh(c*x))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,\left (d\,c^2\,x^2+d\right )}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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