3.2.63 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^3 (d+c^2 d x^2)^{3/2}} \, dx\) [163]

Optimal. Leaf size=287 \[ -\frac {b c \sqrt {1+c^2 x^2}}{2 d x \sqrt {d+c^2 d x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}+\frac {b c^2 \sqrt {1+c^2 x^2} \text {ArcTan}(c x)}{d \sqrt {d+c^2 d x^2}}+\frac {3 c^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {d+c^2 d x^2}}+\frac {3 b c^2 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {3 b c^2 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {d+c^2 d x^2}} \]

[Out]

-3/2*c^2*(a+b*arcsinh(c*x))/d/(c^2*d*x^2+d)^(1/2)+1/2*(-a-b*arcsinh(c*x))/d/x^2/(c^2*d*x^2+d)^(1/2)-1/2*b*c*(c
^2*x^2+1)^(1/2)/d/x/(c^2*d*x^2+d)^(1/2)+b*c^2*arctan(c*x)*(c^2*x^2+1)^(1/2)/d/(c^2*d*x^2+d)^(1/2)+3*c^2*(a+b*a
rcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)/d/(c^2*d*x^2+d)^(1/2)+3/2*b*c^2*polylog(2,-c*x-(
c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)/d/(c^2*d*x^2+d)^(1/2)-3/2*b*c^2*polylog(2,c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+
1)^(1/2)/d/(c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5809, 5811, 5816, 4267, 2317, 2438, 209, 331} \begin {gather*} -\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {c^2 d x^2+d}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {c^2 d x^2+d}}+\frac {3 c^2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}+\frac {b c^2 \sqrt {c^2 x^2+1} \text {ArcTan}(c x)}{d \sqrt {c^2 d x^2+d}}+\frac {3 b c^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {c^2 d x^2+d}}-\frac {3 b c^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1}}{2 d x \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*c*Sqrt[1 + c^2*x^2])/(d*x*Sqrt[d + c^2*d*x^2]) - (3*c^2*(a + b*ArcSinh[c*x]))/(2*d*Sqrt[d + c^2*d*x^2]
) - (a + b*ArcSinh[c*x])/(2*d*x^2*Sqrt[d + c^2*d*x^2]) + (b*c^2*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(d*Sqrt[d + c^2
*d*x^2]) + (3*c^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/(d*Sqrt[d + c^2*d*x^2]) + (3
*b*c^2*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^ArcSinh[c*x]])/(2*d*Sqrt[d + c^2*d*x^2]) - (3*b*c^2*Sqrt[1 + c^2*x^2]*P
olyLog[2, E^ArcSinh[c*x]])/(2*d*Sqrt[d + c^2*d*x^2])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4267

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(Ar
cTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*
fz*x)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5809

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] + (-Dist[c^2*((m + 2*p + 3)/(f^2*
(m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && ILtQ[m, -1]

Rule 5811

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(-(f*x)^(m + 1))*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*f*(p + 1))), x] + (Dist[(m + 2*p + 3)/(2*d*(
p + 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*f*(p + 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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] ||
IntegerQ[p] || EqQ[n, 1])

Rule 5816

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

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^3 \left (d+c^2 d x^2\right )^{3/2}} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}-\frac {1}{2} \left (3 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^{3/2}} \, dx+\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 d \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 d x \sqrt {d+c^2 d x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}-\frac {\left (3 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {d+c^2 d x^2}} \, dx}{2 d}-\frac {\left (b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d \sqrt {d+c^2 d x^2}}+\frac {\left (3 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 d x \sqrt {d+c^2 d x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}+\frac {b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{d \sqrt {d+c^2 d x^2}}-\frac {\left (3 c^2 \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{2 d \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 d x \sqrt {d+c^2 d x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}+\frac {b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{d \sqrt {d+c^2 d x^2}}-\frac {\left (3 c^2 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 d x \sqrt {d+c^2 d x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}+\frac {b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{d \sqrt {d+c^2 d x^2}}+\frac {3 c^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {d+c^2 d x^2}}+\frac {\left (3 b c^2 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {\left (3 b c^2 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 d x \sqrt {d+c^2 d x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}+\frac {b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{d \sqrt {d+c^2 d x^2}}+\frac {3 c^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {d+c^2 d x^2}}+\frac {\left (3 b c^2 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {\left (3 b c^2 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 d x \sqrt {d+c^2 d x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \sqrt {d+c^2 d x^2}}+\frac {b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{d \sqrt {d+c^2 d x^2}}+\frac {3 c^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {d+c^2 d x^2}}+\frac {3 b c^2 \sqrt {1+c^2 x^2} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {d+c^2 d x^2}}-\frac {3 b c^2 \sqrt {1+c^2 x^2} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 d \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 5.66, size = 369, normalized size = 1.29 \begin {gather*} \frac {-\frac {4 a \left (1+3 c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{x^2+c^2 x^4}-12 a c^2 \sqrt {d} \log (x)+12 a c^2 \sqrt {d} \log \left (d+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+\frac {b c^2 d \left (-8 \sinh ^{-1}(c x)+16 \sqrt {1+c^2 x^2} \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-2 \sqrt {1+c^2 x^2} \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-12 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+12 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )-12 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+12 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-\sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )+2 \sqrt {1+c^2 x^2} \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{\sqrt {d+c^2 d x^2}}}{8 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a*(1 + 3*c^2*x^2)*Sqrt[d + c^2*d*x^2])/(x^2 + c^2*x^4) - 12*a*c^2*Sqrt[d]*Log[x] + 12*a*c^2*Sqrt[d]*Log[d
+ Sqrt[d]*Sqrt[d + c^2*d*x^2]] + (b*c^2*d*(-8*ArcSinh[c*x] + 16*Sqrt[1 + c^2*x^2]*ArcTan[Tanh[ArcSinh[c*x]/2]
] - 2*Sqrt[1 + c^2*x^2]*Coth[ArcSinh[c*x]/2] - Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 - 12*Sqrt
[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 12*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[
c*x])] - 12*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(-ArcSinh[c*x])] + 12*Sqrt[1 + c^2*x^2]*PolyLog[2, E^(-ArcSinh[c*x
])] - Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 2*Sqrt[1 + c^2*x^2]*Tanh[ArcSinh[c*x]/2]))/Sqrt[
d + c^2*d*x^2])/(8*d^2)

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Maple [A]
time = 3.23, size = 389, normalized size = 1.36

method	result	size
default	\(-\frac {a}{2 d \,x^{2} \sqrt {c^{2} d \,x^{2}+d}}-\frac {3 a \,c^{2}}{2 d \sqrt {c^{2} d \,x^{2}+d}}+\frac {3 a \,c^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c^{2} d \,x^{2}+d}}{x}\right )}{2 d^{\frac {3}{2}}}-\frac {3 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) c^{2}}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c}{2 d^{2} \sqrt {c^{2} x^{2}+1}\, x}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{2 d^{2} \left (c^{2} x^{2}+1\right ) x^{2}}+\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{\sqrt {c^{2} x^{2}+1}\, d^{2}}+\frac {3 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{2}}+\frac {3 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{2}}+\frac {3 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{2}}\)	\(389\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d/x^2/(c^2*d*x^2+d)^(1/2)-3/2*a*c^2/d/(c^2*d*x^2+d)^(1/2)+3/2*a*c^2/d^(3/2)*ln((2*d+2*d^(1/2)*(c^2*d*x^
2+d)^(1/2))/x)-3/2*b*(d*(c^2*x^2+1))^(1/2)/d^2/(c^2*x^2+1)*arcsinh(c*x)*c^2-1/2*b*(d*(c^2*x^2+1))^(1/2)/d^2/(c
^2*x^2+1)^(1/2)/x*c-1/2*b*(d*(c^2*x^2+1))^(1/2)/d^2/(c^2*x^2+1)/x^2*arcsinh(c*x)+2*b*(d*(c^2*x^2+1))^(1/2)/(c^
2*x^2+1)^(1/2)/d^2*arctan(c*x+(c^2*x^2+1)^(1/2))*c^2+3/2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*dilog(c
*x+(c^2*x^2+1)^(1/2))*c^2+3/2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*dilog(1+c*x+(c^2*x^2+1)^(1/2))*c^2
+3/2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))*c^2

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

[Out]

1/2*(3*c^2*arcsinh(1/(c*abs(x)))/d^(3/2) - 3*c^2/(sqrt(c^2*d*x^2 + d)*d) - 1/(sqrt(c^2*d*x^2 + d)*d*x^2))*a +
b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/((c^2*d*x^2 + d)^(3/2)*x^3), x)

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

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)/(c^4*d^2*x^7 + 2*c^2*d^2*x^5 + d^2*x^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{x^{3} \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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