\(\int \frac {a+b \text {arcsinh}(c x)}{x^4 (d+c^2 d x^2)} \, dx\) [36]

Optimal result

Integrand size = 24, antiderivative size = 156 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=-\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3}+\frac {c^2 (a+b \text {arcsinh}(c x))}{d x}+\frac {2 c^3 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{d}+\frac {7 b c^3 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right )}{6 d}-\frac {i b c^3 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{d}+\frac {i b c^3 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{d} \]

[Out]

1/3*(-a-b*arcsinh(c*x))/d/x^3+c^2*(a+b*arcsinh(c*x))/d/x+2*c^3*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2)
)/d+7/6*b*c^3*arctanh((c^2*x^2+1)^(1/2))/d-I*b*c^3*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/d+I*b*c^3*polylog(2,I
*(c*x+(c^2*x^2+1)^(1/2)))/d-1/6*b*c*(c^2*x^2+1)^(1/2)/d/x^2

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5809, 5789, 4265, 2317, 2438, 272, 65, 214, 44} \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=\frac {2 c^3 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{d}+\frac {c^2 (a+b \text {arcsinh}(c x))}{d x}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3}-\frac {i b c^3 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{d}+\frac {i b c^3 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{d}+\frac {7 b c^3 \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )}{6 d}-\frac {b c \sqrt {c^2 x^2+1}}{6 d x^2} \]

[In]

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

[Out]

-1/6*(b*c*Sqrt[1 + c^2*x^2])/(d*x^2) - (a + b*ArcSinh[c*x])/(3*d*x^3) + (c^2*(a + b*ArcSinh[c*x]))/(d*x) + (2*
c^3*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/d + (7*b*c^3*ArcTanh[Sqrt[1 + c^2*x^2]])/(6*d) - (I*b*c^3*Pol
yLog[2, (-I)*E^ArcSinh[c*x]])/d + (I*b*c^3*PolyLog[2, I*E^ArcSinh[c*x]])/d

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 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 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 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*} \text {integral}& = -\frac {a+b \text {arcsinh}(c x)}{3 d x^3}-c^2 \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )} \, dx+\frac {(b c) \int \frac {1}{x^3 \sqrt {1+c^2 x^2}} \, dx}{3 d} \\ & = -\frac {a+b \text {arcsinh}(c x)}{3 d x^3}+\frac {c^2 (a+b \text {arcsinh}(c x))}{d x}+c^4 \int \frac {a+b \text {arcsinh}(c x)}{d+c^2 d x^2} \, dx+\frac {(b c) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{6 d}-\frac {\left (b c^3\right ) \int \frac {1}{x \sqrt {1+c^2 x^2}} \, dx}{d} \\ & = -\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3}+\frac {c^2 (a+b \text {arcsinh}(c x))}{d x}+\frac {c^3 \text {Subst}(\int (a+b x) \text {sech}(x) \, dx,x,\text {arcsinh}(c x))}{d}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{12 d}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{2 d} \\ & = -\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3}+\frac {c^2 (a+b \text {arcsinh}(c x))}{d x}+\frac {2 c^3 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{d}-\frac {(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}-\frac {(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 )}{d}-\frac {\left (i b c^3\right ) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {arcsinh}(c x)\right )}{d}+\frac {\left (i b c^3\right ) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {arcsinh}(c x)\right )}{d} \\ & = -\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3}+\frac {c^2 (a+b \text {arcsinh}(c x))}{d x}+\frac {2 c^3 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{d}+\frac {7 b c^3 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right )}{6 d}-\frac {\left (i b c^3\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {arcsinh}(c x)}\right )}{d}+\frac {\left (i b c^3\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {arcsinh}(c x)}\right )}{d} \\ & = -\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3}+\frac {c^2 (a+b \text {arcsinh}(c x))}{d x}+\frac {2 c^3 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{d}+\frac {7 b c^3 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right )}{6 d}-\frac {i b c^3 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{d}+\frac {i b c^3 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.58 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=\frac {-2 a+6 a c^2 x^2-b c x \sqrt {1+c^2 x^2}-2 b \text {arcsinh}(c x)+6 b c^2 x^2 \text {arcsinh}(c x)+6 a c^3 x^3 \arctan (c x)+7 b c^3 x^3 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right )-6 b \left (-c^2\right )^{3/2} x^3 \text {arcsinh}(c x) \log \left (1+\frac {c e^{\text {arcsinh}(c x)}}{\sqrt {-c^2}}\right )+6 b \left (-c^2\right )^{3/2} x^3 \text {arcsinh}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\text {arcsinh}(c x)}}{c}\right )+6 b \left (-c^2\right )^{3/2} x^3 \operatorname {PolyLog}\left (2,\frac {c e^{\text {arcsinh}(c x)}}{\sqrt {-c^2}}\right )-6 b \left (-c^2\right )^{3/2} x^3 \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\text {arcsinh}(c x)}}{c}\right )}{6 d x^3} \]

[In]

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

[Out]

(-2*a + 6*a*c^2*x^2 - b*c*x*Sqrt[1 + c^2*x^2] - 2*b*ArcSinh[c*x] + 6*b*c^2*x^2*ArcSinh[c*x] + 6*a*c^3*x^3*ArcT
an[c*x] + 7*b*c^3*x^3*ArcTanh[Sqrt[1 + c^2*x^2]] - 6*b*(-c^2)^(3/2)*x^3*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x]
)/Sqrt[-c^2]] + 6*b*(-c^2)^(3/2)*x^3*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 6*b*(-c^2)^(3/2)*x^
3*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 6*b*(-c^2)^(3/2)*x^3*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c])/
(6*d*x^3)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(c^{3} \left (\frac {a \left (-\frac {1}{3 c^{3} x^{3}}+\frac {1}{c x}+\arctan \left (c x \right )\right )}{d}+\frac {b \left (-\frac {\operatorname {arcsinh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\operatorname {arcsinh}\left (c x \right )}{c x}+\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}+\frac {7 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}+\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d}\right )\)	\(215\)
default	\(c^{3} \left (\frac {a \left (-\frac {1}{3 c^{3} x^{3}}+\frac {1}{c x}+\arctan \left (c x \right )\right )}{d}+\frac {b \left (-\frac {\operatorname {arcsinh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\operatorname {arcsinh}\left (c x \right )}{c x}+\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}+\frac {7 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}+\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d}\right )\)	\(215\)
parts	\(\frac {a \left (c^{3} \arctan \left (c x \right )-\frac {1}{3 x^{3}}+\frac {c^{2}}{x}\right )}{d}+\frac {b \,c^{3} \left (-\frac {\operatorname {arcsinh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\operatorname {arcsinh}\left (c x \right )}{c x}+\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}+\frac {7 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}+\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d}\)	\(215\)

[In]

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

[Out]

c^3*(a/d*(-1/3/c^3/x^3+1/c/x+arctan(c*x))+b/d*(-1/3*arcsinh(c*x)/c^3/x^3+arcsinh(c*x)/c/x+arcsinh(c*x)*arctan(
c*x)-1/6/c^2/x^2*(c^2*x^2+1)^(1/2)+7/6*arctanh(1/(c^2*x^2+1)^(1/2))+arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(
1/2))-arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-I*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*dilog(1-I*(1+
I*c*x)/(c^2*x^2+1)^(1/2))))

Fricas [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )} x^{4}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=\frac {\int \frac {a}{c^{2} x^{6} + x^{4}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{6} + x^{4}}\, dx}{d} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )} x^{4}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )} x^{4}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^4\,\left (d\,c^2\,x^2+d\right )} \,d x \]

[In]

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

[Out]

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