3.45 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^4 (d+c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=239 \[ \frac {5 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (c^2 x^2+1\right )}-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (c^2 x^2+1\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (c^2 x^2+1\right )}-\frac {5 i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac {5 i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}-\frac {b c}{6 d^2 x^2 \sqrt {c^2 x^2+1}}+\frac {b c^3}{3 d^2 \sqrt {c^2 x^2+1}}+\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{6 d^2} \]

[Out]

1/3*(-a-b*arcsinh(c*x))/d^2/x^3/(c^2*x^2+1)+5/3*c^2*(a+b*arcsinh(c*x))/d^2/x/(c^2*x^2+1)+5/2*c^4*x*(a+b*arcsin
h(c*x))/d^2/(c^2*x^2+1)+5*c^3*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))/d^2+13/6*b*c^3*arctanh((c^2*x^2
+1)^(1/2))/d^2-5/2*I*b*c^3*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/d^2+5/2*I*b*c^3*polylog(2,I*(c*x+(c^2*x^2+1)^
(1/2)))/d^2+1/3*b*c^3/d^2/(c^2*x^2+1)^(1/2)-1/6*b*c/d^2/x^2/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 264, normalized size of antiderivative = 1.10, number of steps used = 19, 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} \[ -\frac {5 i b c^3 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac {5 i b c^3 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (c^2 x^2+1\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (c^2 x^2+1\right )}-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (c^2 x^2+1\right )}+\frac {5 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac {5 b c^3}{6 d^2 \sqrt {c^2 x^2+1}}-\frac {b c \sqrt {c^2 x^2+1}}{2 d^2 x^2}+\frac {b c}{3 d^2 x^2 \sqrt {c^2 x^2+1}}+\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{6 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^2),x]

[Out]

(5*b*c^3)/(6*d^2*Sqrt[1 + c^2*x^2]) + (b*c)/(3*d^2*x^2*Sqrt[1 + c^2*x^2]) - (b*c*Sqrt[1 + c^2*x^2])/(2*d^2*x^2
) - (a + b*ArcSinh[c*x])/(3*d^2*x^3*(1 + c^2*x^2)) + (5*c^2*(a + b*ArcSinh[c*x]))/(3*d^2*x*(1 + c^2*x^2)) + (5
*c^4*x*(a + b*ArcSinh[c*x]))/(2*d^2*(1 + c^2*x^2)) + (5*c^3*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/d^2 +
(13*b*c^3*ArcTanh[Sqrt[1 + c^2*x^2]])/(6*d^2) - (((5*I)/2)*b*c^3*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^2 + (((5*
I)/2)*b*c^3*PolyLog[2, I*E^ArcSinh[c*x]])/d^2

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^4 \left (d+c^2 d x^2\right )^2} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}-\frac {1}{3} \left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )^2} \, dx+\frac {(b c) \int \frac {1}{x^3 \left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\left (5 c^4\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^2} \, dx+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}-\frac {\left (5 b c^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=\frac {b c}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (5 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}-\frac {\left (5 b c^5\right ) \int \frac {x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^2}+\frac {\left (5 c^4\right ) \int \frac {a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{2 d}\\ &=\frac {5 b c^3}{6 d^2 \sqrt {1+c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {b c \sqrt {1+c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (5 c^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2}-\frac {\left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{4 d^2}-\frac {\left (5 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{6 d^2}\\ &=\frac {5 b c^3}{6 d^2 \sqrt {1+c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {b c \sqrt {1+c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{2 d^2}-\frac {(5 b c) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{3 d^2}-\frac {\left (5 i b c^3\right ) \operatorname {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2}+\frac {\left (5 i b c^3\right ) \operatorname {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2}\\ &=\frac {5 b c^3}{6 d^2 \sqrt {1+c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {b c \sqrt {1+c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d^2}-\frac {\left (5 i b c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac {\left (5 i b c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d^2}\\ &=\frac {5 b c^3}{6 d^2 \sqrt {1+c^2 x^2}}+\frac {b c}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {b c \sqrt {1+c^2 x^2}}{2 d^2 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {13 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d^2}-\frac {5 i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}+\frac {5 i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 d^2}\\ \end {align*}

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Mathematica [C]  time = 0.58, size = 311, normalized size = 1.30 \[ \frac {5 a c^3 \tan ^{-1}(c x)+\frac {a}{c^2 x^5+x^3}+\frac {5 a c^2}{x}-\frac {5 a}{3 x^3}+5 b \left (-c^2\right )^{3/2} \text {Li}_2\left (\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-5 b \left (-c^2\right )^{3/2} \text {Li}_2\left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-\frac {5 b c \sqrt {c^2 x^2+1}}{6 x^2}+\frac {b \sinh ^{-1}(c x)}{c^2 x^5+x^3}+\frac {5 b c^2 \sinh ^{-1}(c x)}{x}-5 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}+1\right )+5 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )+\frac {b c^3 \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};c^2 x^2+1\right )}{\sqrt {c^2 x^2+1}}+\frac {35}{6} b c^3 \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )-\frac {5 b \sinh ^{-1}(c x)}{3 x^3}}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^2),x]

[Out]

((-5*a)/(3*x^3) + (5*a*c^2)/x - (5*b*c*Sqrt[1 + c^2*x^2])/(6*x^2) + a/(x^3 + c^2*x^5) - (5*b*ArcSinh[c*x])/(3*
x^3) + (5*b*c^2*ArcSinh[c*x])/x + (b*ArcSinh[c*x])/(x^3 + c^2*x^5) + 5*a*c^3*ArcTan[c*x] + (35*b*c^3*ArcTanh[S
qrt[1 + c^2*x^2]])/6 + (b*c^3*Hypergeometric2F1[-1/2, 2, 1/2, 1 + c^2*x^2])/Sqrt[1 + c^2*x^2] - 5*b*(-c^2)^(3/
2)*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 5*b*(-c^2)^(3/2)*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^A
rcSinh[c*x])/c] + 5*b*(-c^2)^(3/2)*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 5*b*(-c^2)^(3/2)*PolyLog[2, (Sq
rt[-c^2]*E^ArcSinh[c*x])/c])/(2*d^2)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arsinh}\left (c x\right ) + a}{c^{4} d^{2} x^{8} + 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{2} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^2,x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^2*x^4), x)

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maple [A]  time = 0.02, size = 332, normalized size = 1.39 \[ -\frac {a}{3 d^{2} x^{3}}+\frac {2 c^{2} a}{d^{2} x}+\frac {c^{4} a x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {5 c^{3} a \arctan \left (c x \right )}{2 d^{2}}-\frac {b \arcsinh \left (c x \right )}{3 d^{2} x^{3}}+\frac {2 c^{2} b \arcsinh \left (c x \right )}{d^{2} x}+\frac {c^{4} b \arcsinh \left (c x \right ) x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {5 c^{3} b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{2 d^{2}}+\frac {b \,c^{3}}{3 d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {13 c^{3} b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6 d^{2}}-\frac {b c}{6 d^{2} x^{2} \sqrt {c^{2} x^{2}+1}}+\frac {5 c^{3} b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {5 c^{3} b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {5 i c^{3} b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {5 i c^{3} b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^2,x)

[Out]

-1/3*a/d^2/x^3+2*c^2*a/d^2/x+1/2*c^4*a/d^2*x/(c^2*x^2+1)+5/2*c^3*a/d^2*arctan(c*x)-1/3*b/d^2*arcsinh(c*x)/x^3+
2*c^2*b/d^2*arcsinh(c*x)/x+1/2*c^4*b/d^2*arcsinh(c*x)*x/(c^2*x^2+1)+5/2*c^3*b/d^2*arcsinh(c*x)*arctan(c*x)+1/3
*b*c^3/d^2/(c^2*x^2+1)^(1/2)+13/6*c^3*b/d^2*arctanh(1/(c^2*x^2+1)^(1/2))-1/6*b*c/d^2/x^2/(c^2*x^2+1)^(1/2)+5/2
*c^3*b/d^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-5/2*c^3*b/d^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2
+1)^(1/2))-5/2*I*c^3*b/d^2*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+5/2*I*c^3*b/d^2*dilog(1-I*(1+I*c*x)/(c^2*x^2
+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, {\left (\frac {15 \, c^{3} \arctan \left (c x\right )}{d^{2}} + \frac {15 \, c^{4} x^{4} + 10 \, c^{2} x^{2} - 2}{c^{2} d^{2} x^{5} + d^{2} x^{3}}\right )} a + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{4} d^{2} x^{8} + 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

1/6*(15*c^3*arctan(c*x)/d^2 + (15*c^4*x^4 + 10*c^2*x^2 - 2)/(c^2*d^2*x^5 + d^2*x^3))*a + b*integrate(log(c*x +
sqrt(c^2*x^2 + 1))/(c^4*d^2*x^8 + 2*c^2*d^2*x^6 + d^2*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^4\,{\left (d\,c^2\,x^2+d\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^2),x)

[Out]

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x**4/(c**2*d*x**2+d)**2,x)

[Out]

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

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