3.302 \(\int \frac {x^3 (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=383 \[ -\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt {c^2 d x^2+d}}+\frac {2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {4 b \sqrt {c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {4 a b x \sqrt {c^2 x^2+1}}{c^3 d \sqrt {c^2 d x^2+d}}+\frac {2 b x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d \sqrt {c^2 d x^2+d}}+\frac {2 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {2 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^4 d \sqrt {c^2 d x^2+d}}+\frac {2 b^2 \left (c^2 x^2+1\right )}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {4 b^2 x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{c^3 d \sqrt {c^2 d x^2+d}} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {5751, 5717, 5653, 261, 5767, 5693, 4180, 2279, 2391} \[ \frac {2 i b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {2 i b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c^4 d \sqrt {c^2 d x^2+d}}+\frac {2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {4 a b x \sqrt {c^2 x^2+1}}{c^3 d \sqrt {c^2 d x^2+d}}+\frac {2 b x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d \sqrt {c^2 d x^2+d}}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt {c^2 d x^2+d}}-\frac {4 b \sqrt {c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d \sqrt {c^2 d x^2+d}}+\frac {2 b^2 \left (c^2 x^2+1\right )}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {4 b^2 x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{c^3 d \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 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 5717

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

Rule 5751

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*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[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] && Gt
Q[m, 1]

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

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Mathematica [A]  time = 0.47, size = 318, normalized size = 0.83 \[ \frac {a^2 c^2 x^2+2 a^2-2 a b c x \sqrt {c^2 x^2+1}+2 a b c^2 x^2 \sinh ^{-1}(c x)-4 a b \sqrt {c^2 x^2+1} \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+4 a b \sinh ^{-1}(c x)+2 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )+2 b^2 c^2 x^2+b^2 c^2 x^2 \sinh ^{-1}(c x)^2-2 b^2 c x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+2 i b^2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-2 i b^2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+2 b^2 \sinh ^{-1}(c x)^2+2 b^2}{c^4 d \sqrt {c^2 d x^2+d}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{3} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{3} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{3}\right )} \sqrt {c^{2} d x^{2} + d}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^3*arcsinh(c*x)^2 + 2*a*b*x^3*arcsinh(c*x) + a^2*x^3)*sqrt(c^2*d*x^2 + d)/(c^4*d^2*x^4 + 2*c^2*
d^2*x^2 + d^2), 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^3*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(3/2),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 = 0.40, size = 703, normalized size = 1.84 \[ \frac {a^{2} x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2 a^{2}}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}+\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2} x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {2 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x}{c^{3} d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {2 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {2 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2}}{c^{4} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {2 b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}}{c^{4} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {2 i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1-i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}+\frac {2 i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}-\frac {2 i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1-i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}+\frac {2 i a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}+\frac {2 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {2 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{c^{3} d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {4 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{c^{4} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {2 i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1+i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}}-\frac {2 i a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -2 \, a b c {\left (\frac {x}{c^{4} d^{\frac {3}{2}}} + \frac {\arctan \left (c x\right )}{c^{5} d^{\frac {3}{2}}}\right )} + 2 \, a b {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} \operatorname {arsinh}\left (c x\right ) + a^{2} {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} + b^{2} {\left (\frac {{\left (c^{2} x^{2} + 2\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{\sqrt {c^{2} x^{2} + 1} c^{4} d^{\frac {3}{2}}} - \int \frac {2 \, {\left (c^{4} x^{4} + 3 \, c^{2} x^{2} + {\left (c^{3} x^{3} + 2 \, c x\right )} \sqrt {c^{2} x^{2} + 1} + 2\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (c^{5} d^{\frac {3}{2}} x^{2} + c^{3} d^{\frac {3}{2}}\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (c^{6} d^{\frac {3}{2}} x^{3} + c^{4} d^{\frac {3}{2}} x\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*asinh(c*x))**2/(d*(c**2*x**2 + 1))**(3/2), x)

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