3.5.70 \(\int \frac {(d+c^2 d x^2)^2}{(a+b \sinh ^{-1}(c x))^{3/2}} \, dx\) [470]

Optimal. Leaf size=346 \[ -\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {5 d^2 e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c}-\frac {5 d^2 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}-\frac {d^2 e^{\frac {5 a}{b}} \sqrt {5 \pi } \text {Erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}+\frac {5 d^2 e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c}+\frac {5 d^2 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}+\frac {d^2 e^{-\frac {5 a}{b}} \sqrt {5 \pi } \text {Erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c} \]

[Out]

-5/8*d^2*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(3/2)/c+5/8*d^2*erfi((a+b*arcsinh(c*x))^(1/
2)/b^(1/2))*Pi^(1/2)/b^(3/2)/c/exp(a/b)-5/16*d^2*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1
/2)*Pi^(1/2)/b^(3/2)/c+5/16*d^2*erfi(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/b^(3/2)/c/exp(
3*a/b)-1/16*d^2*exp(5*a/b)*erf(5^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/b^(3/2)/c+1/16*d^2*e
rfi(5^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/b^(3/2)/c/exp(5*a/b)-2*d^2*(c^2*x^2+1)^(5/2)/b/
c/(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]
time = 0.45, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5790, 5819, 5556, 3389, 2211, 2236, 2235} \begin {gather*} -\frac {5 \sqrt {\pi } d^2 e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c}-\frac {5 \sqrt {3 \pi } d^2 e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}-\frac {\sqrt {5 \pi } d^2 e^{\frac {5 a}{b}} \text {Erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}+\frac {5 \sqrt {\pi } d^2 e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c}+\frac {5 \sqrt {3 \pi } d^2 e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}+\frac {\sqrt {5 \pi } d^2 e^{-\frac {5 a}{b}} \text {Erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}-\frac {2 d^2 \left (c^2 x^2+1\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*(1 + c^2*x^2)^(5/2))/(b*c*Sqrt[a + b*ArcSinh[c*x]]) - (5*d^2*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c
*x]]/Sqrt[b]])/(8*b^(3/2)*c) - (5*d^2*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/
(16*b^(3/2)*c) - (d^2*E^((5*a)/b)*Sqrt[5*Pi]*Erf[(Sqrt[5]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c) +
(5*d^2*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(8*b^(3/2)*c*E^(a/b)) + (5*d^2*Sqrt[3*Pi]*Erfi[(Sqrt[
3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c*E^((3*a)/b)) + (d^2*Sqrt[5*Pi]*Erfi[(Sqrt[5]*Sqrt[a + b*A
rcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c*E^((5*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 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 5790

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

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 {\left (d+c^2 d x^2\right )^2}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {\left (10 c d^2\right ) \int \frac {x \left (1+c^2 x^2\right )^{3/2}}{\sqrt {a+b \sinh ^{-1}(c x)}} \, dx}{b}\\ &=-\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {\left (10 d^2\right ) \text {Subst}\left (\int \frac {\cosh ^4(x) \sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}\\ &=-\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {\left (10 d^2\right ) \text {Subst}\left (\int \left (\frac {\sinh (x)}{8 \sqrt {a+b x}}+\frac {3 \sinh (3 x)}{16 \sqrt {a+b x}}+\frac {\sinh (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c}\\ &=-\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {\sinh (5 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c}+\frac {\left (15 d^2\right ) \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}\\ &=-\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {e^{5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}-\frac {\left (15 d^2\right ) \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}+\frac {\left (15 d^2\right ) \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}\\ &=-\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{\frac {5 a}{b}-\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{-\frac {5 a}{b}+\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c}-\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c}+\frac {\left (5 d^2\right ) \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c}-\frac {\left (15 d^2\right ) \text {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c}+\frac {\left (15 d^2\right ) \text {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c}\\ &=-\frac {2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {5 d^2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c}-\frac {5 d^2 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}-\frac {d^2 e^{\frac {5 a}{b}} \sqrt {5 \pi } \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}+\frac {5 d^2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c}+\frac {5 d^2 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}+\frac {d^2 e^{-\frac {5 a}{b}} \sqrt {5 \pi } \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c}\\ \end {align*}

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Mathematica [A]
time = 1.16, size = 440, normalized size = 1.27 \begin {gather*} \frac {d^2 e^{-5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \left (-e^{\frac {5 a}{b}}-5 e^{\frac {5 a}{b}+2 \sinh ^{-1}(c x)}-10 e^{\frac {5 a}{b}+4 \sinh ^{-1}(c x)}-10 e^{\frac {5 a}{b}+6 \sinh ^{-1}(c x)}-5 e^{\frac {5 a}{b}+8 \sinh ^{-1}(c x)}-e^{\frac {5 a}{b}+10 \sinh ^{-1}(c x)}+10 e^{\frac {6 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )+\sqrt {5} e^{5 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+5 \sqrt {3} e^{\frac {2 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+10 e^{\frac {4 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )+5 \sqrt {3} e^{\frac {8 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+\sqrt {5} e^{5 \left (\frac {2 a}{b}+\sinh ^{-1}(c x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{16 b c \sqrt {a+b \sinh ^{-1}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-E^((5*a)/b) - 5*E^((5*a)/b + 2*ArcSinh[c*x]) - 10*E^((5*a)/b + 4*ArcSinh[c*x]) - 10*E^((5*a)/b + 6*ArcS
inh[c*x]) - 5*E^((5*a)/b + 8*ArcSinh[c*x]) - E^((5*a)/b + 10*ArcSinh[c*x]) + 10*E^((6*a)/b + 5*ArcSinh[c*x])*S
qrt[a/b + ArcSinh[c*x]]*Gamma[1/2, a/b + ArcSinh[c*x]] + Sqrt[5]*E^(5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x]
)/b)]*Gamma[1/2, (-5*(a + b*ArcSinh[c*x]))/b] + 5*Sqrt[3]*E^((2*a)/b + 5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c
*x])/b)]*Gamma[1/2, (-3*(a + b*ArcSinh[c*x]))/b] + 10*E^((4*a)/b + 5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])
/b)]*Gamma[1/2, -((a + b*ArcSinh[c*x])/b)] + 5*Sqrt[3]*E^((8*a)/b + 5*ArcSinh[c*x])*Sqrt[a/b + ArcSinh[c*x]]*G
amma[1/2, (3*(a + b*ArcSinh[c*x]))/b] + Sqrt[5]*E^(5*((2*a)/b + ArcSinh[c*x]))*Sqrt[a/b + ArcSinh[c*x]]*Gamma[
1/2, (5*(a + b*ArcSinh[c*x]))/b]))/(16*b*c*E^(5*(a/b + ArcSinh[c*x]))*Sqrt[a + b*ArcSinh[c*x]])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((c^2*d*x^2 + d)^2/(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((c^2*d*x^2+d)^2/(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^{2} \left (\int \frac {2 c^{2} x^{2}}{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^{4} x^{4}}{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 {1}{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((c**2*d*x**2+d)**2/(a+b*asinh(c*x))**(3/2),x)

[Out]

d**2*(Integral(2*c**2*x**2/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(c**
4*x**4/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(1/(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 = 0.00, 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((c^2*d*x^2+d)^2/(a+b*arcsinh(c*x))^(3/2),x, algorithm="giac")

[Out]

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

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

[Out]

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

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