3.1.37 \(\int \frac {x^4 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^2} \, dx\) [37]

Optimal. Leaf size=171 \[ \frac {b}{2 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b \sqrt {1+c^2 x^2}}{c^5 d^2}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {3 \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}+\frac {3 i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2}-\frac {3 i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2} \]

[Out]

3/2*x*(a+b*arcsinh(c*x))/c^4/d^2-1/2*x^3*(a+b*arcsinh(c*x))/c^2/d^2/(c^2*x^2+1)-3*(a+b*arcsinh(c*x))*arctan(c*
x+(c^2*x^2+1)^(1/2))/c^5/d^2+3/2*I*b*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d^2-3/2*I*b*polylog(2,I*(c*x+(c
^2*x^2+1)^(1/2)))/c^5/d^2+1/2*b/c^5/d^2/(c^2*x^2+1)^(1/2)-b*(c^2*x^2+1)^(1/2)/c^5/d^2

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Rubi [A]
time = 0.18, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5810, 5812, 5789, 4265, 2317, 2438, 267, 272, 45} \begin {gather*} -\frac {3 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {3 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2}-\frac {3 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2}-\frac {b \sqrt {c^2 x^2+1}}{c^5 d^2}+\frac {b}{2 c^5 d^2 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

b/(2*c^5*d^2*Sqrt[1 + c^2*x^2]) - (b*Sqrt[1 + c^2*x^2])/(c^5*d^2) + (3*x*(a + b*ArcSinh[c*x]))/(2*c^4*d^2) - (
x^3*(a + b*ArcSinh[c*x]))/(2*c^2*d^2*(1 + c^2*x^2)) - (3*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(c^5*d^2
) + (((3*I)/2)*b*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^5*d^2) - (((3*I)/2)*b*PolyLog[2, I*E^ArcSinh[c*x]])/(c^5*
d^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 5810

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

Rule 5812

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Dist[f^2*((m - 1)/(c^2*
(m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*p + 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] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0
]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^2} \, dx &=-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {b \int \frac {x^3}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 c d^2}+\frac {3 \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx}{2 c^2 d}\\ &=\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {(3 b) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{2 c^3 d^2}+\frac {b \text {Subst}\left (\int \frac {x}{\left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{4 c d^2}-\frac {3 \int \frac {a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{2 c^4 d}\\ &=-\frac {3 b \sqrt {1+c^2 x^2}}{2 c^5 d^2}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {3 \text {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^5 d^2}+\frac {b \text {Subst}\left (\int \left (-\frac {1}{c^2 \left (1+c^2 x\right )^{3/2}}+\frac {1}{c^2 \sqrt {1+c^2 x}}\right ) \, dx,x,x^2\right )}{4 c d^2}\\ &=\frac {b}{2 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b \sqrt {1+c^2 x^2}}{c^5 d^2}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}+\frac {(3 i b) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^5 d^2}-\frac {(3 i b) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^5 d^2}\\ &=\frac {b}{2 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b \sqrt {1+c^2 x^2}}{c^5 d^2}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}+\frac {(3 i b) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2}-\frac {(3 i b) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2}\\ &=\frac {b}{2 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b \sqrt {1+c^2 x^2}}{c^5 d^2}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}+\frac {3 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2}-\frac {3 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 268, normalized size = 1.57 \begin {gather*} \frac {3 a c x+2 a c^3 x^3-b \sqrt {1+c^2 x^2}-2 b c^2 x^2 \sqrt {1+c^2 x^2}+3 b c x \sinh ^{-1}(c x)+2 b c^3 x^3 \sinh ^{-1}(c x)-3 a \text {ArcTan}(c x)-3 a c^2 x^2 \text {ArcTan}(c x)-3 i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-3 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+3 i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+3 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+3 i b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )-3 i b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c^5 d^2 \left (1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*c*x + 2*a*c^3*x^3 - b*Sqrt[1 + c^2*x^2] - 2*b*c^2*x^2*Sqrt[1 + c^2*x^2] + 3*b*c*x*ArcSinh[c*x] + 2*b*c^3*
x^3*ArcSinh[c*x] - 3*a*ArcTan[c*x] - 3*a*c^2*x^2*ArcTan[c*x] - (3*I)*b*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]]
- (3*I)*b*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] + (3*I)*b*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] + (3
*I)*b*c^2*x^2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] + (3*I)*b*(1 + c^2*x^2)*PolyLog[2, (-I)*E^ArcSinh[c*x]] -
(3*I)*b*(1 + c^2*x^2)*PolyLog[2, I*E^ArcSinh[c*x]])/(2*c^5*d^2*(1 + c^2*x^2))

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Maple [A]
time = 2.52, size = 260, normalized size = 1.52

method	result	size
derivativedivides	\(\frac {\frac {a c x}{d^{2}}+\frac {a c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {3 a \arctan \left (c x \right )}{2 d^{2}}+\frac {b \arcsinh \left (c x \right ) c x}{d^{2}}+\frac {b \arcsinh \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {3 b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{2 d^{2}}-\frac {b \,c^{2} x^{2}}{d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {b}{2 d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {3 b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {3 b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {3 i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {3 i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}}{c^{5}}\)	\(260\)
default	\(\frac {\frac {a c x}{d^{2}}+\frac {a c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {3 a \arctan \left (c x \right )}{2 d^{2}}+\frac {b \arcsinh \left (c x \right ) c x}{d^{2}}+\frac {b \arcsinh \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {3 b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{2 d^{2}}-\frac {b \,c^{2} x^{2}}{d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {b}{2 d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {3 b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {3 b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {3 i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {3 i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}}{c^{5}}\)	\(260\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^5*(a/d^2*c*x+1/2*a/d^2*c*x/(c^2*x^2+1)-3/2*a/d^2*arctan(c*x)+b/d^2*arcsinh(c*x)*c*x+1/2*b/d^2*arcsinh(c*x)
*c*x/(c^2*x^2+1)-3/2*b/d^2*arcsinh(c*x)*arctan(c*x)-b/d^2*c^2*x^2/(c^2*x^2+1)^(1/2)-1/2*b/d^2/(c^2*x^2+1)^(1/2
)-3/2*b/d^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*b/d^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)
^(1/2))+3/2*I*b/d^2*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*I*b/d^2*dilog(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(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

1/2*a*(x/(c^6*d^2*x^2 + c^4*d^2) + 2*x/(c^4*d^2) - 3*arctan(c*x)/(c^5*d^2)) + b*integrate(x^4*log(c*x + sqrt(c
^2*x^2 + 1))/(c^4*d^2*x^4 + 2*c^2*d^2*x^2 + d^2), 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(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x^{4}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x**4/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b*x**4*asinh(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1),
x))/d**2

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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(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,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 {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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