3.4.14 \(\int \frac {x (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^{5/2}} \, dx\) [314]

Optimal. Leaf size=270 \[ \frac {b^2}{3 c^2 d^2 \sqrt {d+c^2 d x^2}}+\frac {b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt {d+c^2 d x^2}}-\frac {i b^2 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt {d+c^2 d x^2}}+\frac {i b^2 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt {d+c^2 d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5798, 5788, 5789, 4265, 2317, 2438, 267} \begin {gather*} \frac {2 b \sqrt {c^2 x^2+1} \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d^2 \sqrt {c^2 d x^2+d}}+\frac {b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt {c^2 d x^2+d}}+\frac {i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c^2 d^2 \sqrt {c^2 d x^2+d}}+\frac {b^2}{3 c^2 d^2 \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(5/2),x]

[Out]

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

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 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 5788

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(
p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*(p + 1))), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a
+ b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2
)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &
& LtQ[p, -1] && NeQ[p, -3/2]

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 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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]

Rubi steps

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

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Mathematica [A]
time = 0.64, size = 254, normalized size = 0.94 \begin {gather*} \frac {-a^2+a b \left (-2 \sinh ^{-1}(c x)+\sqrt {1+c^2 x^2} \left (c x+2 \left (1+c^2 x^2\right ) \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )\right )+b^2 \left (1+c^2 x^2+c x \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)-\sinh ^{-1}(c x)^2-i \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+i \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-i \left (1+c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+i \left (1+c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(5/2),x]

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (261 ) = 522\).
time = 1.44, size = 591, normalized size = 2.19

method	result	size
default	\(-\frac {a^{2}}{3 c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x}{3 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c}+\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, x^{2}}{3 d^{3} \left (c^{2} x^{2}+1\right )^{2}}-\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2}}{3 d^{3} \left (c^{2} x^{2}+1\right )^{2} c^{2}}+\frac {b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}}{3 d^{3} \left (c^{2} x^{2}+1\right )^{2} c^{2}}-\frac {i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{3 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}+\frac {i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1-i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{3 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}-\frac {i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1+i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{3 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}+\frac {i b^{2} \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1-i \left (c x +\sqrt {c^{2} x^{2}+1}\right )\right )}{3 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}+\frac {a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{3 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c}-\frac {2 a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{3 d^{3} \left (c^{2} x^{2}+1\right )^{2} c^{2}}+\frac {i a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{3 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}-\frac {i a b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{3 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}\)	\(591\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*a^2/c^2/d/(c^2*d*x^2+d)^(3/2)+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c*arcsinh(c*x)*x+1/3*b^
2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2*x^2-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^2*arcsinh(c*x)
^2+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^2-1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d
^3*arcsinh(c*x)*ln(1+I*(c*x+(c^2*x^2+1)^(1/2)))+1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*arcs
inh(c*x)*ln(1-I*(c*x+(c^2*x^2+1)^(1/2)))-1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*dilog(1+I*(
c*x+(c^2*x^2+1)^(1/2)))+1/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*dilog(1-I*(c*x+(c^2*x^2+1)^(
1/2)))+1/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c*x-2/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2
/c^2*arcsinh(c*x)+1/3*I*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*ln(c*x+(c^2*x^2+1)^(1/2)+I)-1/3*I*
a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b^2*x*arcsinh(c*x)^2 + 2*a*b*x*arcsinh(c*x) + a^2*x)/(c^6*d^3*x^6 + 3*c^4*d^3*x^
4 + 3*c^2*d^3*x^2 + d^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asinh(c*x))**2/(c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x*(a + b*asinh(c*x))**2/(d*(c**2*x**2 + 1))**(5/2), x)

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Giac [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*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2*x/(c^2*d*x^2 + d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2)^(5/2),x)

[Out]

int((x*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2)^(5/2), x)

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