3.1.26 \(\int \frac {(d+c^2 d x^2)^3 (a+b \sinh ^{-1}(c x))}{x^3} \, dx\) [26]

Optimal. Leaf size=249 \[ -\frac {3}{32} b c^3 d^3 x \sqrt {1+c^2 x^2}+\frac {7}{16} b c^3 d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}-\frac {3}{32} b c^2 d^3 \sinh ^{-1}(c x)+\frac {3}{2} c^2 d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )-\frac {3}{2} b c^2 d^3 \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right ) \]

[Out]

7/16*b*c^3*d^3*x*(c^2*x^2+1)^(3/2)-1/2*b*c*d^3*(c^2*x^2+1)^(5/2)/x-3/32*b*c^2*d^3*arcsinh(c*x)+3/2*c^2*d^3*(c^
2*x^2+1)*(a+b*arcsinh(c*x))+3/4*c^2*d^3*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))-1/2*d^3*(c^2*x^2+1)^3*(a+b*arcsinh(c*
x))/x^2+3/2*c^2*d^3*(a+b*arcsinh(c*x))^2/b+3*c^2*d^3*(a+b*arcsinh(c*x))*ln(1-1/(c*x+(c^2*x^2+1)^(1/2))^2)-3/2*
b*c^2*d^3*polylog(2,1/(c*x+(c^2*x^2+1)^(1/2))^2)-3/32*b*c^3*d^3*x*(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5802, 283, 201, 221, 5801, 5775, 3797, 2221, 2317, 2438} \begin {gather*} -\frac {d^3 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {3}{4} c^2 d^3 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{2} c^2 d^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+3 c^2 d^3 \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {3}{2} b c^2 d^3 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c x)}\right )-\frac {b c d^3 \left (c^2 x^2+1\right )^{5/2}}{2 x}-\frac {3}{32} b c^2 d^3 \sinh ^{-1}(c x)+\frac {7}{16} b c^3 d^3 x \left (c^2 x^2+1\right )^{3/2}-\frac {3}{32} b c^3 d^3 x \sqrt {c^2 x^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*c^3*d^3*x*Sqrt[1 + c^2*x^2])/32 + (7*b*c^3*d^3*x*(1 + c^2*x^2)^(3/2))/16 - (b*c*d^3*(1 + c^2*x^2)^(5/2))
/(2*x) - (3*b*c^2*d^3*ArcSinh[c*x])/32 + (3*c^2*d^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/2 + (3*c^2*d^3*(1 + c^
2*x^2)^2*(a + b*ArcSinh[c*x]))/4 - (d^3*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*x]))/(2*x^2) + (3*c^2*d^3*(a + b*ArcS
inh[c*x])^2)/(2*b) + 3*c^2*d^3*(a + b*ArcSinh[c*x])*Log[1 - E^(-2*ArcSinh[c*x])] - (3*b*c^2*d^3*PolyLog[2, E^(
-2*ArcSinh[c*x])])/2

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 221

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

Rule 283

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

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5801

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

Rule 5802

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

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{x^3} \, dx &=-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\left (3 c^2 d\right ) \int \frac {\left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx+\frac {1}{2} \left (b c d^3\right ) \int \frac {\left (1+c^2 x^2\right )^{5/2}}{x^2} \, dx\\ &=-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\left (3 c^2 d^2\right ) \int \frac {\left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac {1}{4} \left (3 b c^3 d^3\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx+\frac {1}{2} \left (5 b c^3 d^3\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx\\ &=\frac {7}{16} b c^3 d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}+\frac {3}{2} c^2 d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\left (3 c^2 d^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x} \, dx-\frac {1}{16} \left (9 b c^3 d^3\right ) \int \sqrt {1+c^2 x^2} \, dx-\frac {1}{2} \left (3 b c^3 d^3\right ) \int \sqrt {1+c^2 x^2} \, dx+\frac {1}{8} \left (15 b c^3 d^3\right ) \int \sqrt {1+c^2 x^2} \, dx\\ &=-\frac {3}{32} b c^3 d^3 x \sqrt {1+c^2 x^2}+\frac {7}{16} b c^3 d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}+\frac {3}{2} c^2 d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\left (3 c^2 d^3\right ) \text {Subst}\left (\int (a+b x) \coth (x) \, dx,x,\sinh ^{-1}(c x)\right )-\frac {1}{32} \left (9 b c^3 d^3\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx-\frac {1}{4} \left (3 b c^3 d^3\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx+\frac {1}{16} \left (15 b c^3 d^3\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {3}{32} b c^3 d^3 x \sqrt {1+c^2 x^2}+\frac {7}{16} b c^3 d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}-\frac {3}{32} b c^2 d^3 \sinh ^{-1}(c x)+\frac {3}{2} c^2 d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}-\left (6 c^2 d^3\right ) \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {3}{32} b c^3 d^3 x \sqrt {1+c^2 x^2}+\frac {7}{16} b c^3 d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}-\frac {3}{32} b c^2 d^3 \sinh ^{-1}(c x)+\frac {3}{2} c^2 d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\left (3 b c^2 d^3\right ) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {3}{32} b c^3 d^3 x \sqrt {1+c^2 x^2}+\frac {7}{16} b c^3 d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}-\frac {3}{32} b c^2 d^3 \sinh ^{-1}(c x)+\frac {3}{2} c^2 d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\frac {1}{2} \left (3 b c^2 d^3\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )\\ &=-\frac {3}{32} b c^3 d^3 x \sqrt {1+c^2 x^2}+\frac {7}{16} b c^3 d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {b c d^3 \left (1+c^2 x^2\right )^{5/2}}{2 x}-\frac {3}{32} b c^2 d^3 \sinh ^{-1}(c x)+\frac {3}{2} c^2 d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {3}{4} c^2 d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+3 c^2 d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+\frac {3}{2} b c^2 d^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 210, normalized size = 0.84 \begin {gather*} \frac {d^3 \left (-16 a+48 a c^4 x^4+8 a c^6 x^6-16 b c x \sqrt {1+c^2 x^2}-21 b c^3 x^3 \sqrt {1+c^2 x^2}-2 b c^5 x^5 \sqrt {1+c^2 x^2}+48 b c^2 x^2 \sinh ^{-1}(c x)^2+21 b c^2 x^2 \tanh ^{-1}\left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )+8 b \sinh ^{-1}(c x) \left (-2+6 c^4 x^4+c^6 x^6+12 c^2 x^2 \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )\right )+96 a c^2 x^2 \log (x)-48 b c^2 x^2 \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )\right )}{32 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(-16*a + 48*a*c^4*x^4 + 8*a*c^6*x^6 - 16*b*c*x*Sqrt[1 + c^2*x^2] - 21*b*c^3*x^3*Sqrt[1 + c^2*x^2] - 2*b*c
^5*x^5*Sqrt[1 + c^2*x^2] + 48*b*c^2*x^2*ArcSinh[c*x]^2 + 21*b*c^2*x^2*ArcTanh[(c*x)/Sqrt[1 + c^2*x^2]] + 8*b*A
rcSinh[c*x]*(-2 + 6*c^4*x^4 + c^6*x^6 + 12*c^2*x^2*Log[1 - E^(-2*ArcSinh[c*x])]) + 96*a*c^2*x^2*Log[x] - 48*b*
c^2*x^2*PolyLog[2, E^(-2*ArcSinh[c*x])]))/(32*x^2)

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Maple [A]
time = 6.02, size = 299, normalized size = 1.20

method	result	size
derivativedivides	\(c^{2} \left (\frac {a \,d^{3} c^{4} x^{4}}{4}+\frac {3 a \,d^{3} c^{2} x^{2}}{2}-\frac {a \,d^{3}}{2 c^{2} x^{2}}+3 a \,d^{3} \ln \left (c x \right )+3 d^{3} b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+3 d^{3} b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+\frac {d^{3} b}{2}+3 d^{3} b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )+3 d^{3} b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {3 d^{3} b \arcsinh \left (c x \right )^{2}}{2}+\frac {21 b \,d^{3} \arcsinh \left (c x \right )}{32}+\frac {d^{3} b \arcsinh \left (c x \right ) c^{4} x^{4}}{4}+\frac {3 d^{3} b \arcsinh \left (c x \right ) c^{2} x^{2}}{2}-\frac {d^{3} b \arcsinh \left (c x \right )}{2 c^{2} x^{2}}-\frac {d^{3} b \sqrt {c^{2} x^{2}+1}}{2 c x}-\frac {d^{3} b \,c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{16}-\frac {21 b c \,d^{3} x \sqrt {c^{2} x^{2}+1}}{32}\right )\)	\(299\)
default	\(c^{2} \left (\frac {a \,d^{3} c^{4} x^{4}}{4}+\frac {3 a \,d^{3} c^{2} x^{2}}{2}-\frac {a \,d^{3}}{2 c^{2} x^{2}}+3 a \,d^{3} \ln \left (c x \right )+3 d^{3} b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+3 d^{3} b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+\frac {d^{3} b}{2}+3 d^{3} b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )+3 d^{3} b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {3 d^{3} b \arcsinh \left (c x \right )^{2}}{2}+\frac {21 b \,d^{3} \arcsinh \left (c x \right )}{32}+\frac {d^{3} b \arcsinh \left (c x \right ) c^{4} x^{4}}{4}+\frac {3 d^{3} b \arcsinh \left (c x \right ) c^{2} x^{2}}{2}-\frac {d^{3} b \arcsinh \left (c x \right )}{2 c^{2} x^{2}}-\frac {d^{3} b \sqrt {c^{2} x^{2}+1}}{2 c x}-\frac {d^{3} b \,c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{16}-\frac {21 b c \,d^{3} x \sqrt {c^{2} x^{2}+1}}{32}\right )\)	\(299\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*(1/4*a*d^3*c^4*x^4+3/2*a*d^3*c^2*x^2-1/2*a*d^3/c^2/x^2+3*a*d^3*ln(c*x)+3*d^3*b*arcsinh(c*x)*ln(1-c*x-(c^2*
x^2+1)^(1/2))+3*d^3*b*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))+1/2*d^3*b+3*d^3*b*polylog(2,c*x+(c^2*x^2+1)^(1/
2))+3*d^3*b*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-3/2*d^3*b*arcsinh(c*x)^2+21/32*b*d^3*arcsinh(c*x)+1/4*d^3*b*arcs
inh(c*x)*c^4*x^4+3/2*d^3*b*arcsinh(c*x)*c^2*x^2-1/2*d^3*b*arcsinh(c*x)/c^2/x^2-1/2*d^3*b/c/x*(c^2*x^2+1)^(1/2)
-1/16*d^3*b*c^3*x^3*(c^2*x^2+1)^(1/2)-21/32*b*c*d^3*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((c^2*d*x^2+d)^3*(a+b*arcsinh(c*x))/x^3,x, algorithm="maxima")

[Out]

1/4*a*c^6*d^3*x^4 + 3/2*a*c^4*d^3*x^2 + 3*a*c^2*d^3*log(x) - 1/2*b*d^3*(sqrt(c^2*x^2 + 1)*c/x + arcsinh(c*x)/x
^2) - 1/2*a*d^3/x^2 + integrate(b*c^6*d^3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) + 3*b*c^4*d^3*x*log(c*x + sqrt(c^2*
x^2 + 1)) + 3*b*c^2*d^3*log(c*x + sqrt(c^2*x^2 + 1))/x, 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((c^2*d*x^2+d)^3*(a+b*arcsinh(c*x))/x^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*(Integral(a/x**3, x) + Integral(3*a*c**2/x, x) + Integral(3*a*c**4*x, x) + Integral(a*c**6*x**3, x) + Int
egral(b*asinh(c*x)/x**3, x) + Integral(3*b*c**2*asinh(c*x)/x, x) + Integral(3*b*c**4*x*asinh(c*x), x) + Integr
al(b*c**6*x**3*asinh(c*x), x))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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