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

Optimal. Leaf size=156 \[ \frac {2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}-\frac {i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {b \left (c^2 x^2+1\right )^{3/2}}{9 c^5 d}+\frac {4 b \sqrt {c^2 x^2+1}}{3 c^5 d} \]

[Out]

-1/9*b*(c^2*x^2+1)^(3/2)/c^5/d-x*(a+b*arcsinh(c*x))/c^4/d+1/3*x^3*(a+b*arcsinh(c*x))/c^2/d+2*(a+b*arcsinh(c*x)
)*arctan(c*x+(c^2*x^2+1)^(1/2))/c^5/d-I*b*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d+I*b*polylog(2,I*(c*x+(c^
2*x^2+1)^(1/2)))/c^5/d+4/3*b*(c^2*x^2+1)^(1/2)/c^5/d

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Rubi [A]  time = 0.24, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5767, 5693, 4180, 2279, 2391, 261, 266, 43} \[ -\frac {i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac {2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}-\frac {b \left (c^2 x^2+1\right )^{3/2}}{9 c^5 d}+\frac {4 b \sqrt {c^2 x^2+1}}{3 c^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*Sqrt[1 + c^2*x^2])/(3*c^5*d) - (b*(1 + c^2*x^2)^(3/2))/(9*c^5*d) - (x*(a + b*ArcSinh[c*x]))/(c^4*d) + (x^
3*(a + b*ArcSinh[c*x]))/(3*c^2*d) + (2*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(c^5*d) - (I*b*PolyLog[2,
(-I)*E^ArcSinh[c*x]])/(c^5*d) + (I*b*PolyLog[2, I*E^ArcSinh[c*x]])/(c^5*d)

Rule 43

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 261

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 266

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 2279

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 2391

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 4180

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 5693

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 5767

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*d^IntPart[p]*(d
+ e*x^2)^FracPart[p])/(c*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[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] && GtQ[m
, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 170, normalized size = 1.09 \[ \frac {3 a c^3 x^3-9 a c x+9 a \tan ^{-1}(c x)+3 b c^3 x^3 \sinh ^{-1}(c x)-b c^2 x^2 \sqrt {c^2 x^2+1}+11 b \sqrt {c^2 x^2+1}-9 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )+9 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )-9 b c x \sinh ^{-1}(c x)+9 i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-9 i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{9 c^5 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-9*a*c*x + 3*a*c^3*x^3 + 11*b*Sqrt[1 + c^2*x^2] - b*c^2*x^2*Sqrt[1 + c^2*x^2] - 9*b*c*x*ArcSinh[c*x] + 3*b*c^
3*x^3*ArcSinh[c*x] + 9*a*ArcTan[c*x] + (9*I)*b*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - (9*I)*b*ArcSinh[c*x]*L
og[1 + I*E^ArcSinh[c*x]] - (9*I)*b*PolyLog[2, (-I)*E^ArcSinh[c*x]] + (9*I)*b*PolyLog[2, I*E^ArcSinh[c*x]])/(9*
c^5*d)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{4} \operatorname {arsinh}\left (c x\right ) + a x^{4}}{c^{2} d x^{2} + d}, x\right ) \]

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),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.14, size = 266, normalized size = 1.71 \[ \frac {a \,x^{3}}{3 c^{2} d}-\frac {a x}{c^{4} d}+\frac {a \arctan \left (c x \right )}{c^{5} d}+\frac {b \arcsinh \left (c x \right ) x^{3}}{3 c^{2} d}-\frac {b \arcsinh \left (c x \right ) x}{c^{4} d}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{c^{5} d}-\frac {b \,x^{2} \sqrt {c^{2} x^{2}+1}}{9 c^{3} d}+\frac {11 b \sqrt {c^{2} x^{2}+1}}{9 c^{5} d}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{c^{5} d}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{c^{5} d}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{c^{5} d}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{c^{5} d} \]

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),x)

[Out]

1/3/c^2*a/d*x^3-1/c^4*a/d*x+1/c^5*a/d*arctan(c*x)+1/3/c^2*b/d*arcsinh(c*x)*x^3-1/c^4*b/d*arcsinh(c*x)*x+1/c^5*
b/d*arcsinh(c*x)*arctan(c*x)-1/9/c^3*b/d*x^2*(c^2*x^2+1)^(1/2)+11/9*b*(c^2*x^2+1)^(1/2)/c^5/d+1/c^5*b/d*arctan
(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/c^5*b/d*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-I/c^5*b/d*
dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I/c^5*b/d*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 \[ \frac {1}{3} \, a {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4} d} + \frac {3 \, \arctan \left (c x\right )}{c^{5} d}\right )} + b \int \frac {x^{4} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{2} d x^{2} + d}\,{d x} \]

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),x, algorithm="maxima")

[Out]

1/3*a*((c^2*x^3 - 3*x)/(c^4*d) + 3*arctan(c*x)/(c^5*d)) + b*integrate(x^4*log(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*
x^2 + d), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{d\,c^2\,x^2+d} \,d x \]

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),x)

[Out]

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

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

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),x)

[Out]

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

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