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

Optimal. Leaf size=222 \[ -\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (c^2 x^2+1\right )^2}-\frac {15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac {7 b c}{8 d^3 \sqrt {c^2 x^2+1}}-\frac {b c}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {b c \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{d^3}+\frac {15 i b c \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac {15 i b c \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 d^3} \]

[Out]

-1/12*b*c/d^3/(c^2*x^2+1)^(3/2)+(-a-b*arcsinh(c*x))/d^3/x/(c^2*x^2+1)^2-5/4*c^2*x*(a+b*arcsinh(c*x))/d^3/(c^2*
x^2+1)^2-15/8*c^2*x*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)-15/4*c*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))
/d^3-b*c*arctanh((c^2*x^2+1)^(1/2))/d^3+15/8*I*b*c*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/d^3-15/8*I*b*c*polylo
g(2,I*(c*x+(c^2*x^2+1)^(1/2)))/d^3-7/8*b*c/d^3/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {5747, 5690, 5693, 4180, 2279, 2391, 261, 266, 51, 63, 208} \[ \frac {15 i b c \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac {15 i b c \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (c^2 x^2+1\right )^2}-\frac {15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac {7 b c}{8 d^3 \sqrt {c^2 x^2+1}}-\frac {b c}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {b c \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(12*d^3*(1 + c^2*x^2)^(3/2)) - (7*b*c)/(8*d^3*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])/(d^3*x*(1 + c^2
*x^2)^2) - (5*c^2*x*(a + b*ArcSinh[c*x]))/(4*d^3*(1 + c^2*x^2)^2) - (15*c^2*x*(a + b*ArcSinh[c*x]))/(8*d^3*(1
+ c^2*x^2)) - (15*c*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(4*d^3) - (b*c*ArcTanh[Sqrt[1 + c^2*x^2]])/d^
3 + (((15*I)/8)*b*c*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^3 - (((15*I)/8)*b*c*PolyLog[2, I*E^ArcSinh[c*x]])/d^3

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2279

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 2391

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 4180

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

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[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}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 5693

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

Rule 5747

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*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {1}{x \left (1+c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{2 d^3}+\frac {\left (5 b c^3\right ) \int \frac {x}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 d^3}-\frac {\left (15 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac {b c}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{2 d^3}+\frac {\left (15 b c^3\right ) \int \frac {x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{8 d^3}-\frac {\left (15 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{8 d^2}\\ &=-\frac {b c}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac {(15 c) \operatorname {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{2 d^3}\\ &=-\frac {b c}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {b \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{c d^3}+\frac {(15 i b c) \operatorname {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}-\frac {(15 i b c) \operatorname {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}\\ &=-\frac {b c}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {b c \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{d^3}+\frac {(15 i b c) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac {(15 i b c) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 d^3}\\ &=-\frac {b c}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac {5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac {15 c \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {b c \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{d^3}+\frac {15 i b c \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac {15 i b c \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}\\ \end {align*}

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Mathematica [C]  time = 1.05, size = 298, normalized size = 1.34 \[ -\frac {-\frac {15 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 x^3+x}-\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{x \left (c^2 x^2+1\right )^2}+\frac {45 \left (a+b \sinh ^{-1}(c x)\right )}{x}+45 a c \tan ^{-1}(c x)+\frac {2 b c \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};c^2 x^2+1\right )}{\left (c^2 x^2+1\right )^{3/2}}+\frac {15 b c \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};c^2 x^2+1\right )}{\sqrt {c^2 x^2+1}}-45 b \sqrt {-c^2} \text {Li}_2\left (\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+45 b \sqrt {-c^2} \text {Li}_2\left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+45 b c \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )+45 b \sqrt {-c^2} \sinh ^{-1}(c x) \log \left (\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}+1\right )-45 b \sqrt {-c^2} \sinh ^{-1}(c x) \log \left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )}{24 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/24*((45*(a + b*ArcSinh[c*x]))/x - (6*(a + b*ArcSinh[c*x]))/(x*(1 + c^2*x^2)^2) - (15*(a + b*ArcSinh[c*x]))/
(x + c^2*x^3) + 45*a*c*ArcTan[c*x] + 45*b*c*ArcTanh[Sqrt[1 + c^2*x^2]] + (2*b*c*Hypergeometric2F1[-3/2, 1, -1/
2, 1 + c^2*x^2])/(1 + c^2*x^2)^(3/2) + (15*b*c*Hypergeometric2F1[-1/2, 1, 1/2, 1 + c^2*x^2])/Sqrt[1 + c^2*x^2]
+ 45*b*Sqrt[-c^2]*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 45*b*Sqrt[-c^2]*ArcSinh[c*x]*Log[1 +
(Sqrt[-c^2]*E^ArcSinh[c*x])/c] - 45*b*Sqrt[-c^2]*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 45*b*Sqrt[-c^2]*P
olyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c])/d^3

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arsinh}\left (c x\right ) + a}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsinh(c*x) + a)/(c^6*d^3*x^8 + 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 + d^3*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 357, normalized size = 1.61 \[ -\frac {a}{d^{3} x}-\frac {7 a \,c^{4} x^{3}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}-\frac {9 a \,c^{2} x}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}-\frac {15 c a \arctan \left (c x \right )}{8 d^{3}}-\frac {b \arcsinh \left (c x \right )}{d^{3} x}-\frac {7 b \arcsinh \left (c x \right ) c^{4} x^{3}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}-\frac {9 b \arcsinh \left (c x \right ) c^{2} x}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}-\frac {15 c b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{8 d^{3}}-\frac {15 c b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}+\frac {15 c b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}+\frac {15 i c b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}-\frac {15 i c b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}-\frac {15 b \,c^{3} x^{2}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {47 b c}{24 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {b c}{d^{3} \sqrt {c^{2} x^{2}+1}}-\frac {c b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d^3/x-7/8*a/d^3*c^4*x^3/(c^2*x^2+1)^2-9/8*a/d^3*c^2*x/(c^2*x^2+1)^2-15/8*c*a/d^3*arctan(c*x)-b/d^3*arcsinh(
c*x)/x-7/8*b/d^3*arcsinh(c*x)*c^4*x^3/(c^2*x^2+1)^2-9/8*b/d^3*arcsinh(c*x)*c^2*x/(c^2*x^2+1)^2-15/8*c*b/d^3*ar
csinh(c*x)*arctan(c*x)-15/8*c*b/d^3*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+15/8*c*b/d^3*arctan(c*x)*l
n(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+15/8*I*c*b/d^3*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-15/8*I*c*b/d^3*dilog(
1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-15/8*b/d^3*c^3*x^2/(c^2*x^2+1)^(3/2)-47/24*b*c/d^3/(c^2*x^2+1)^(3/2)+b*c/d^3/
(c^2*x^2+1)^(1/2)-c*b/d^3*arctanh(1/(c^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{8} \, a {\left (\frac {15 \, c^{4} x^{4} + 25 \, c^{2} x^{2} + 8}{c^{4} d^{3} x^{5} + 2 \, c^{2} d^{3} x^{3} + d^{3} x} + \frac {15 \, c \arctan \left (c x\right )}{d^{3}}\right )} + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*((15*c^4*x^4 + 25*c^2*x^2 + 8)/(c^4*d^3*x^5 + 2*c^2*d^3*x^3 + d^3*x) + 15*c*arctan(c*x)/d^3) + b*integr
ate(log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^8 + 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 + d^3*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**6*x**8 + 3*c**4*x**6 + 3*c**2*x**4 + x**2), x) + Integral(b*asinh(c*x)/(c**6*x**8 + 3*c**4*x**
6 + 3*c**2*x**4 + x**2), x))/d**3

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