3.1.54 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^4 (d+c^2 d x^2)^3} \, dx\) [54]

Optimal. Leaf size=295 \[ -\frac {b c^3}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c}{6 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {29 b c^3}{24 d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {19 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d^3}-\frac {35 i b c^3 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac {35 i b c^3 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 d^3} \]

[Out]

-1/12*b*c^3/d^3/(c^2*x^2+1)^(3/2)-1/6*b*c/d^3/x^2/(c^2*x^2+1)^(3/2)+1/3*(-a-b*arcsinh(c*x))/d^3/x^3/(c^2*x^2+1
)^2+7/3*c^2*(a+b*arcsinh(c*x))/d^3/x/(c^2*x^2+1)^2+35/12*c^4*x*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)^2+35/8*c^4*x
*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)+35/4*c^3*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))/d^3+19/6*b*c^3*a
rctanh((c^2*x^2+1)^(1/2))/d^3-35/8*I*b*c^3*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/d^3+35/8*I*b*c^3*polylog(2,I*
(c*x+(c^2*x^2+1)^(1/2)))/d^3+29/24*b*c^3/d^3/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 12, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5809, 5788, 5789, 4265, 2317, 2438, 267, 272, 53, 65, 214, 44} \begin {gather*} \frac {35 c^3 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (c^2 x^2+1\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (c^2 x^2+1\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (c^2 x^2+1\right )^2}-\frac {35 i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac {35 i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac {b c}{6 d^3 x^2 \left (c^2 x^2+1\right )^{3/2}}+\frac {29 b c^3}{24 d^3 \sqrt {c^2 x^2+1}}-\frac {b c^3}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac {19 b c^3 \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{6 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^3),x]

[Out]

-1/12*(b*c^3)/(d^3*(1 + c^2*x^2)^(3/2)) - (b*c)/(6*d^3*x^2*(1 + c^2*x^2)^(3/2)) + (29*b*c^3)/(24*d^3*Sqrt[1 +
c^2*x^2]) - (a + b*ArcSinh[c*x])/(3*d^3*x^3*(1 + c^2*x^2)^2) + (7*c^2*(a + b*ArcSinh[c*x]))/(3*d^3*x*(1 + c^2*
x^2)^2) + (35*c^4*x*(a + b*ArcSinh[c*x]))/(12*d^3*(1 + c^2*x^2)^2) + (35*c^4*x*(a + b*ArcSinh[c*x]))/(8*d^3*(1
+ c^2*x^2)) + (35*c^3*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(4*d^3) + (19*b*c^3*ArcTanh[Sqrt[1 + c^2*x
^2]])/(6*d^3) - (((35*I)/8)*b*c^3*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^3 + (((35*I)/8)*b*c^3*PolyLog[2, I*E^ArcS
inh[c*x]])/d^3

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 214

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

Rule 267

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 272

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 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 4265

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 5788

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/(2*(p + 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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &
& LtQ[p, -1] && NeQ[p, -3/2]

Rule 5789

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 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]

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^4 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}-\frac {1}{3} \left (7 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {1}{x^3 \left (1+c^2 x^2\right )^{5/2}} \, dx}{3 d^3}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {1}{3} \left (35 c^4\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^3} \, dx+\frac {(b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{6 d^3}-\frac {\left (7 b c^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )^{5/2}} \, dx}{3 d^3}\\ &=\frac {b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {(5 b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{18 d^3}-\frac {\left (7 b c^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{6 d^3}-\frac {\left (35 b c^5\right ) \int \frac {x}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{12 d^3}+\frac {\left (35 c^4\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^2} \, dx}{4 d}\\ &=\frac {7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {5 b c}{9 d^3 x^2 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac {(5 b c) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{6 d^3}-\frac {\left (7 b c^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^3}-\frac {\left (35 b c^5\right ) \int \frac {x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{8 d^3}+\frac {\left (35 c^4\right ) \int \frac {a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{8 d^2}\\ &=\frac {7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {49 b c^3}{24 d^3 \sqrt {1+c^2 x^2}}+\frac {5 b c}{9 d^3 x^2 \sqrt {1+c^2 x^2}}-\frac {5 b c \sqrt {1+c^2 x^2}}{6 d^3 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac {\left (35 c^3\right ) \text {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}-\frac {\left (5 b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{12 d^3}-\frac {\left (7 b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{6 d^3}\\ &=\frac {7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {49 b c^3}{24 d^3 \sqrt {1+c^2 x^2}}+\frac {5 b c}{9 d^3 x^2 \sqrt {1+c^2 x^2}}-\frac {5 b c \sqrt {1+c^2 x^2}}{6 d^3 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {(5 b c) \text {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{6 d^3}-\frac {(7 b c) \text {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^3}-\frac {\left (35 i b c^3\right ) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}+\frac {\left (35 i b c^3\right ) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}\\ &=\frac {7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {49 b c^3}{24 d^3 \sqrt {1+c^2 x^2}}+\frac {5 b c}{9 d^3 x^2 \sqrt {1+c^2 x^2}}-\frac {5 b c \sqrt {1+c^2 x^2}}{6 d^3 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {19 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d^3}-\frac {\left (35 i b c^3\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac {\left (35 i b c^3\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 d^3}\\ &=\frac {7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {49 b c^3}{24 d^3 \sqrt {1+c^2 x^2}}+\frac {5 b c}{9 d^3 x^2 \sqrt {1+c^2 x^2}}-\frac {5 b c \sqrt {1+c^2 x^2}}{6 d^3 x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {19 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d^3}-\frac {35 i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac {35 i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.64, size = 380, normalized size = 1.29 \begin {gather*} \frac {-\frac {70 a}{x^3}+\frac {210 a c^2}{x}+\frac {12 a}{x^3 \left (1+c^2 x^2\right )^2}-\frac {35 b c \sqrt {1+c^2 x^2}}{x^2}+\frac {42 a}{x^3+c^2 x^5}-\frac {70 b \sinh ^{-1}(c x)}{x^3}+\frac {210 b c^2 \sinh ^{-1}(c x)}{x}+\frac {12 b \sinh ^{-1}(c x)}{x^3 \left (1+c^2 x^2\right )^2}+\frac {42 b \sinh ^{-1}(c x)}{x^3+c^2 x^5}+210 a c^3 \text {ArcTan}(c x)+245 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )+\frac {4 b c^3 \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};1+c^2 x^2\right )}{\left (1+c^2 x^2\right )^{3/2}}+\frac {42 b c^3 \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};1+c^2 x^2\right )}{\sqrt {1+c^2 x^2}}-210 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+210 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+210 b \left (-c^2\right )^{3/2} \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-210 b \left (-c^2\right )^{3/2} \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{48 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^3),x]

[Out]

((-70*a)/x^3 + (210*a*c^2)/x + (12*a)/(x^3*(1 + c^2*x^2)^2) - (35*b*c*Sqrt[1 + c^2*x^2])/x^2 + (42*a)/(x^3 + c
^2*x^5) - (70*b*ArcSinh[c*x])/x^3 + (210*b*c^2*ArcSinh[c*x])/x + (12*b*ArcSinh[c*x])/(x^3*(1 + c^2*x^2)^2) + (
42*b*ArcSinh[c*x])/(x^3 + c^2*x^5) + 210*a*c^3*ArcTan[c*x] + 245*b*c^3*ArcTanh[Sqrt[1 + c^2*x^2]] + (4*b*c^3*H
ypergeometric2F1[-3/2, 2, -1/2, 1 + c^2*x^2])/(1 + c^2*x^2)^(3/2) + (42*b*c^3*Hypergeometric2F1[-1/2, 2, 1/2,
1 + c^2*x^2])/Sqrt[1 + c^2*x^2] - 210*b*(-c^2)^(3/2)*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 210
*b*(-c^2)^(3/2)*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 210*b*(-c^2)^(3/2)*PolyLog[2, (c*E^ArcSi
nh[c*x])/Sqrt[-c^2]] - 210*b*(-c^2)^(3/2)*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c])/(48*d^3)

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Maple [A]
time = 2.58, size = 406, normalized size = 1.38

method	result	size
derivativedivides	\(c^{3} \left (\frac {11 a \,c^{3} x^{3}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {13 a c x}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {35 a \arctan \left (c x \right )}{8 d^{3}}-\frac {a}{3 d^{3} c^{3} x^{3}}+\frac {3 a}{d^{3} c x}+\frac {11 b \arcsinh \left (c x \right ) c^{3} x^{3}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {13 b \arcsinh \left (c x \right ) c x}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {35 b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{8 d^{3}}-\frac {b \arcsinh \left (c x \right )}{3 d^{3} c^{3} x^{3}}+\frac {3 b \arcsinh \left (c x \right )}{d^{3} c x}+\frac {35 b \,c^{2} x^{2}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {103 b}{24 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {b}{6 d^{3} c^{2} x^{2} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {19 b}{6 d^{3} \sqrt {c^{2} x^{2}+1}}+\frac {19 b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6 d^{3}}+\frac {35 b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}-\frac {35 b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}-\frac {35 i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}+\frac {35 i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}\right )\)	\(406\)
default	\(c^{3} \left (\frac {11 a \,c^{3} x^{3}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {13 a c x}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {35 a \arctan \left (c x \right )}{8 d^{3}}-\frac {a}{3 d^{3} c^{3} x^{3}}+\frac {3 a}{d^{3} c x}+\frac {11 b \arcsinh \left (c x \right ) c^{3} x^{3}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {13 b \arcsinh \left (c x \right ) c x}{8 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {35 b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{8 d^{3}}-\frac {b \arcsinh \left (c x \right )}{3 d^{3} c^{3} x^{3}}+\frac {3 b \arcsinh \left (c x \right )}{d^{3} c x}+\frac {35 b \,c^{2} x^{2}}{8 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {103 b}{24 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {b}{6 d^{3} c^{2} x^{2} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {19 b}{6 d^{3} \sqrt {c^{2} x^{2}+1}}+\frac {19 b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6 d^{3}}+\frac {35 b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}-\frac {35 b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}-\frac {35 i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}+\frac {35 i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8 d^{3}}\right )\)	\(406\)

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)^3,x,method=_RETURNVERBOSE)

[Out]

c^3*(11/8*a/d^3/(c^2*x^2+1)^2*c^3*x^3+13/8*a/d^3/(c^2*x^2+1)^2*c*x+35/8*a/d^3*arctan(c*x)-1/3*a/d^3/c^3/x^3+3*
a/d^3/c/x+11/8*b/d^3*arcsinh(c*x)/(c^2*x^2+1)^2*c^3*x^3+13/8*b/d^3*arcsinh(c*x)/(c^2*x^2+1)^2*c*x+35/8*b/d^3*a
rcsinh(c*x)*arctan(c*x)-1/3*b/d^3*arcsinh(c*x)/c^3/x^3+3*b/d^3*arcsinh(c*x)/c/x+35/8*b/d^3*c^2*x^2/(c^2*x^2+1)
^(3/2)+103/24*b/d^3/(c^2*x^2+1)^(3/2)-1/6*b/d^3/c^2/x^2/(c^2*x^2+1)^(3/2)-19/6*b/d^3/(c^2*x^2+1)^(1/2)+19/6*b/
d^3*arctanh(1/(c^2*x^2+1)^(1/2))+35/8*b/d^3*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-35/8*b/d^3*arctan(
c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+35/8*I*b/d^3*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-35/8*I*b/d^3*dilo
g(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 \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^4/(c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

1/24*a*(105*c^3*arctan(c*x)/d^3 + (105*c^6*x^6 + 175*c^4*x^4 + 56*c^2*x^2 - 8)/(c^4*d^3*x^7 + 2*c^2*d^3*x^5 +
d^3*x^3)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3*x^4),
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^4/(c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*arcsinh(c*x) + a)/(c^6*d^3*x^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3*x^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a}{c^{6} x^{10} + 3 c^{4} x^{8} + 3 c^{2} x^{6} + x^{4}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{10} + 3 c^{4} x^{8} + 3 c^{2} x^{6} + x^{4}}\, dx}{d^{3}} \end {gather*}

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)**3,x)

[Out]

(Integral(a/(c**6*x**10 + 3*c**4*x**8 + 3*c**2*x**6 + x**4), x) + Integral(b*asinh(c*x)/(c**6*x**10 + 3*c**4*x
**8 + 3*c**2*x**6 + x**4), x))/d**3

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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^4/(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^4), 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^4\,{\left (d\,c^2\,x^2+d\right )}^3} \,d x \end {gather*}

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)^3),x)

[Out]

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^3), x)

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