3.1.17 \(\int \frac {(d+c^2 d x^2)^2 (a+b \sinh ^{-1}(c x))}{x^3} \, dx\) [17]

Optimal. Leaf size=187 \[ \frac {1}{4} b c^3 d^2 x \sqrt {1+c^2 x^2}-\frac {b c d^2 \left (1+c^2 x^2\right )^{3/2}}{2 x}+\frac {1}{4} b c^2 d^2 \sinh ^{-1}(c x)+c^2 d^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{b}+2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )-b c^2 d^2 \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 12, 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*} c^2 d^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{b}+2 c^2 d^2 \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-b c^2 d^2 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c x)}\right )-\frac {b c d^2 \left (c^2 x^2+1\right )^{3/2}}{2 x}+\frac {1}{4} b c^2 d^2 \sinh ^{-1}(c x)+\frac {1}{4} b c^3 d^2 x \sqrt {c^2 x^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c^3*d^2*x*Sqrt[1 + c^2*x^2])/4 - (b*c*d^2*(1 + c^2*x^2)^(3/2))/(2*x) + (b*c^2*d^2*ArcSinh[c*x])/4 + c^2*d^2
*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]) - (d^2*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(2*x^2) + (c^2*d^2*(a + b*Arc
Sinh[c*x])^2)/b + 2*c^2*d^2*(a + b*ArcSinh[c*x])*Log[1 - E^(-2*ArcSinh[c*x])] - b*c^2*d^2*PolyLog[2, E^(-2*Arc
Sinh[c*x])]

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

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Mathematica [A]
time = 0.13, size = 170, normalized size = 0.91 \begin {gather*} \frac {d^2 \left (-2 a+2 a c^4 x^4-2 b c x \sqrt {1+c^2 x^2}-b c^3 x^3 \sqrt {1+c^2 x^2}+4 b c^2 x^2 \sinh ^{-1}(c x)^2+b c^2 x^2 \tanh ^{-1}\left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )+2 b \sinh ^{-1}(c x) \left (-1+c^4 x^4+4 c^2 x^2 \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )\right )+8 a c^2 x^2 \log (x)-4 b c^2 x^2 \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-2*a + 2*a*c^4*x^4 - 2*b*c*x*Sqrt[1 + c^2*x^2] - b*c^3*x^3*Sqrt[1 + c^2*x^2] + 4*b*c^2*x^2*ArcSinh[c*x]^
2 + b*c^2*x^2*ArcTanh[(c*x)/Sqrt[1 + c^2*x^2]] + 2*b*ArcSinh[c*x]*(-1 + c^4*x^4 + 4*c^2*x^2*Log[1 - E^(-2*ArcS
inh[c*x])]) + 8*a*c^2*x^2*Log[x] - 4*b*c^2*x^2*PolyLog[2, E^(-2*ArcSinh[c*x])]))/(4*x^2)

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Maple [A]
time = 5.10, size = 248, normalized size = 1.33

method	result	size
derivativedivides	\(c^{2} \left (\frac {a \,d^{2} c^{2} x^{2}}{2}-\frac {a \,d^{2}}{2 c^{2} x^{2}}+2 a \,d^{2} \ln \left (c x \right )-d^{2} b \arcsinh \left (c x \right )^{2}+\frac {d^{2} b \arcsinh \left (c x \right ) c^{2} x^{2}}{2}-\frac {b c \,d^{2} x \sqrt {c^{2} x^{2}+1}}{4}+\frac {b \,d^{2} \arcsinh \left (c x \right )}{4}+\frac {d^{2} b}{2}-\frac {d^{2} b \sqrt {c^{2} x^{2}+1}}{2 c x}-\frac {d^{2} b \arcsinh \left (c x \right )}{2 c^{2} x^{2}}+2 d^{2} b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+2 d^{2} b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+2 d^{2} b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+2 d^{2} b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\right )\)	\(248\)
default	\(c^{2} \left (\frac {a \,d^{2} c^{2} x^{2}}{2}-\frac {a \,d^{2}}{2 c^{2} x^{2}}+2 a \,d^{2} \ln \left (c x \right )-d^{2} b \arcsinh \left (c x \right )^{2}+\frac {d^{2} b \arcsinh \left (c x \right ) c^{2} x^{2}}{2}-\frac {b c \,d^{2} x \sqrt {c^{2} x^{2}+1}}{4}+\frac {b \,d^{2} \arcsinh \left (c x \right )}{4}+\frac {d^{2} b}{2}-\frac {d^{2} b \sqrt {c^{2} x^{2}+1}}{2 c x}-\frac {d^{2} b \arcsinh \left (c x \right )}{2 c^{2} x^{2}}+2 d^{2} b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+2 d^{2} b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+2 d^{2} b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+2 d^{2} b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\right )\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

[Out]

d**2*(Integral(a/x**3, x) + Integral(2*a*c**2/x, x) + Integral(a*c**4*x, x) + Integral(b*asinh(c*x)/x**3, x) +
Integral(2*b*c**2*asinh(c*x)/x, x) + Integral(b*c**4*x*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)^2*(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 = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^2}{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)^2)/x^3,x)

[Out]

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

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