3.1.24 \(\int \frac {(d+c^2 d x^2)^3 (a+b \sinh ^{-1}(c x))}{x} \, dx\) [24]

Optimal. Leaf size=221 \[ -\frac {19}{48} b c d^3 x \sqrt {1+c^2 x^2}-\frac {7}{72} b c d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}-\frac {19}{48} b d^3 \sinh ^{-1}(c x)+\frac {1}{2} d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )-\frac {1}{2} b d^3 \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right ) \]

[Out]

-7/72*b*c*d^3*x*(c^2*x^2+1)^(3/2)-1/36*b*c*d^3*x*(c^2*x^2+1)^(5/2)-19/48*b*d^3*arcsinh(c*x)+1/2*d^3*(c^2*x^2+1
)*(a+b*arcsinh(c*x))+1/4*d^3*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))+1/6*d^3*(c^2*x^2+1)^3*(a+b*arcsinh(c*x))+1/2*d^3
*(a+b*arcsinh(c*x))^2/b+d^3*(a+b*arcsinh(c*x))*ln(1-1/(c*x+(c^2*x^2+1)^(1/2))^2)-1/2*b*d^3*polylog(2,1/(c*x+(c
^2*x^2+1)^(1/2))^2)-19/48*b*c*d^3*x*(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5801, 5775, 3797, 2221, 2317, 2438, 201, 221} \begin {gather*} \frac {1}{6} d^3 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{2} d^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d^3 \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{36} b c d^3 x \left (c^2 x^2+1\right )^{5/2}-\frac {7}{72} b c d^3 x \left (c^2 x^2+1\right )^{3/2}-\frac {19}{48} b c d^3 x \sqrt {c^2 x^2+1}-\frac {1}{2} b d^3 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c x)}\right )-\frac {19}{48} b d^3 \sinh ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-19*b*c*d^3*x*Sqrt[1 + c^2*x^2])/48 - (7*b*c*d^3*x*(1 + c^2*x^2)^(3/2))/72 - (b*c*d^3*x*(1 + c^2*x^2)^(5/2))/
36 - (19*b*d^3*ArcSinh[c*x])/48 + (d^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/2 + (d^3*(1 + c^2*x^2)^2*(a + b*Arc
Sinh[c*x]))/4 + (d^3*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*x]))/6 + (d^3*(a + b*ArcSinh[c*x])^2)/(2*b) + d^3*(a + b
*ArcSinh[c*x])*Log[1 - E^(-2*ArcSinh[c*x])] - (b*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 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]

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx &=\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )+d \int \frac {\left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac {1}{6} \left (b c d^3\right ) \int \left (1+c^2 x^2\right )^{5/2} \, dx\\ &=-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )+d^2 \int \frac {\left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac {1}{36} \left (5 b c d^3\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx-\frac {1}{4} \left (b c d^3\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx\\ &=-\frac {7}{72} b c d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}+\frac {1}{2} d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )+d^3 \int \frac {a+b \sinh ^{-1}(c x)}{x} \, dx-\frac {1}{48} \left (5 b c d^3\right ) \int \sqrt {1+c^2 x^2} \, dx-\frac {1}{16} \left (3 b c d^3\right ) \int \sqrt {1+c^2 x^2} \, dx-\frac {1}{2} \left (b c d^3\right ) \int \sqrt {1+c^2 x^2} \, dx\\ &=-\frac {19}{48} b c d^3 x \sqrt {1+c^2 x^2}-\frac {7}{72} b c d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}+\frac {1}{2} d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )+d^3 \text {Subst}\left (\int (a+b x) \coth (x) \, dx,x,\sinh ^{-1}(c x)\right )-\frac {1}{96} \left (5 b c d^3\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx-\frac {1}{32} \left (3 b c d^3\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx-\frac {1}{4} \left (b c d^3\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {19}{48} b c d^3 x \sqrt {1+c^2 x^2}-\frac {7}{72} b c d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}-\frac {19}{48} b d^3 \sinh ^{-1}(c x)+\frac {1}{2} d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}-\left (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 {19}{48} b c d^3 x \sqrt {1+c^2 x^2}-\frac {7}{72} b c d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}-\frac {19}{48} b d^3 \sinh ^{-1}(c x)+\frac {1}{2} d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\left (b d^3\right ) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {19}{48} b c d^3 x \sqrt {1+c^2 x^2}-\frac {7}{72} b c d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}-\frac {19}{48} b d^3 \sinh ^{-1}(c x)+\frac {1}{2} d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\frac {1}{2} \left (b d^3\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )\\ &=-\frac {19}{48} b c d^3 x \sqrt {1+c^2 x^2}-\frac {7}{72} b c d^3 x \left (1+c^2 x^2\right )^{3/2}-\frac {1}{36} b c d^3 x \left (1+c^2 x^2\right )^{5/2}-\frac {19}{48} b d^3 \sinh ^{-1}(c x)+\frac {1}{2} d^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{4} d^3 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d^3 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+\frac {1}{2} b d^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 189, normalized size = 0.86 \begin {gather*} \frac {1}{144} d^3 \left (216 a c^2 x^2+108 a c^4 x^4+24 a c^6 x^6-75 b c x \sqrt {1+c^2 x^2}-22 b c^3 x^3 \sqrt {1+c^2 x^2}-4 b c^5 x^5 \sqrt {1+c^2 x^2}-72 b \sinh ^{-1}(c x)^2+144 a \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+3 \sinh ^{-1}(c x) \left (-48 a+b \left (25+72 c^2 x^2+36 c^4 x^4+8 c^6 x^6\right )+48 b \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )\right )+72 b \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(216*a*c^2*x^2 + 108*a*c^4*x^4 + 24*a*c^6*x^6 - 75*b*c*x*Sqrt[1 + c^2*x^2] - 22*b*c^3*x^3*Sqrt[1 + c^2*x^
2] - 4*b*c^5*x^5*Sqrt[1 + c^2*x^2] - 72*b*ArcSinh[c*x]^2 + 144*a*Log[1 - E^(2*ArcSinh[c*x])] + 3*ArcSinh[c*x]*
(-48*a + b*(25 + 72*c^2*x^2 + 36*c^4*x^4 + 8*c^6*x^6) + 48*b*Log[1 - E^(2*ArcSinh[c*x])]) + 72*b*PolyLog[2, E^
(2*ArcSinh[c*x])]))/144

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Maple [A]
time = 3.60, size = 284, normalized size = 1.29

method	result	size
derivativedivides	\(\frac {a \,d^{3} c^{6} x^{6}}{6}+\frac {3 a \,d^{3} c^{4} x^{4}}{4}+\frac {3 a \,d^{3} c^{2} x^{2}}{2}+a \,d^{3} \ln \left (c x \right )+d^{3} b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+d^{3} b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )+\frac {25 b \,d^{3} \arcsinh \left (c x \right )}{48}-\frac {d^{3} b \arcsinh \left (c x \right )^{2}}{2}+\frac {3 d^{3} b \arcsinh \left (c x \right ) c^{4} x^{4}}{4}+\frac {3 d^{3} b \arcsinh \left (c x \right ) c^{2} x^{2}}{2}+\frac {d^{3} b \arcsinh \left (c x \right ) c^{6} x^{6}}{6}+d^{3} b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+d^{3} b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {d^{3} b \,c^{5} x^{5} \sqrt {c^{2} x^{2}+1}}{36}-\frac {11 d^{3} b \,c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{72}-\frac {25 b c \,d^{3} x \sqrt {c^{2} x^{2}+1}}{48}\)	\(284\)
default	\(\frac {a \,d^{3} c^{6} x^{6}}{6}+\frac {3 a \,d^{3} c^{4} x^{4}}{4}+\frac {3 a \,d^{3} c^{2} x^{2}}{2}+a \,d^{3} \ln \left (c x \right )+d^{3} b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+d^{3} b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )+\frac {25 b \,d^{3} \arcsinh \left (c x \right )}{48}-\frac {d^{3} b \arcsinh \left (c x \right )^{2}}{2}+\frac {3 d^{3} b \arcsinh \left (c x \right ) c^{4} x^{4}}{4}+\frac {3 d^{3} b \arcsinh \left (c x \right ) c^{2} x^{2}}{2}+\frac {d^{3} b \arcsinh \left (c x \right ) c^{6} x^{6}}{6}+d^{3} b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+d^{3} b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {d^{3} b \,c^{5} x^{5} \sqrt {c^{2} x^{2}+1}}{36}-\frac {11 d^{3} b \,c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{72}-\frac {25 b c \,d^{3} x \sqrt {c^{2} x^{2}+1}}{48}\)	\(284\)

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,x,method=_RETURNVERBOSE)

[Out]

1/6*a*d^3*c^6*x^6+3/4*a*d^3*c^4*x^4+3/2*a*d^3*c^2*x^2+a*d^3*ln(c*x)+d^3*b*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+d^
3*b*polylog(2,c*x+(c^2*x^2+1)^(1/2))+25/48*b*d^3*arcsinh(c*x)-1/2*d^3*b*arcsinh(c*x)^2+3/4*d^3*b*arcsinh(c*x)*
c^4*x^4+3/2*d^3*b*arcsinh(c*x)*c^2*x^2+1/6*d^3*b*arcsinh(c*x)*c^6*x^6+d^3*b*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^
(1/2))+d^3*b*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))-1/36*d^3*b*c^5*x^5*(c^2*x^2+1)^(1/2)-11/72*d^3*b*c^3*x^3
*(c^2*x^2+1)^(1/2)-25/48*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,x, algorithm="maxima")

[Out]

1/6*a*c^6*d^3*x^6 + 3/4*a*c^4*d^3*x^4 + 3/2*a*c^2*d^3*x^2 + a*d^3*log(x) + integrate(b*c^6*d^3*x^5*log(c*x + s
qrt(c^2*x^2 + 1)) + 3*b*c^4*d^3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) + 3*b*c^2*d^3*x*log(c*x + sqrt(c^2*x^2 + 1))
+ b*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,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, 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}\, dx + \int 3 a c^{2} x\, dx + \int 3 a c^{4} x^{3}\, dx + \int a c^{6} x^{5}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x}\, dx + \int 3 b c^{2} x \operatorname {asinh}{\left (c x \right )}\, dx + \int 3 b c^{4} x^{3} \operatorname {asinh}{\left (c x \right )}\, dx + \int b c^{6} x^{5} \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,x)

[Out]

d**3*(Integral(a/x, x) + Integral(3*a*c**2*x, x) + Integral(3*a*c**4*x**3, x) + Integral(a*c**6*x**5, x) + Int
egral(b*asinh(c*x)/x, x) + Integral(3*b*c**2*x*asinh(c*x), x) + Integral(3*b*c**4*x**3*asinh(c*x), x) + Integr
al(b*c**6*x**5*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,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} \,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,x)

[Out]

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

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