3.1.75 \(\int \frac {(\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x))}{x} \, dx\) [75]

Optimal. Leaf size=179 \[ -\frac {23}{15} b c \pi ^{5/2} x-\frac {11}{45} b c^3 \pi ^{5/2} x^3-\frac {1}{25} b c^5 \pi ^{5/2} x^5+\pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-2 \pi ^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-b \pi ^{5/2} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+b \pi ^{5/2} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \]

[Out]

-23/15*b*c*Pi^(5/2)*x-11/45*b*c^3*Pi^(5/2)*x^3-1/25*b*c^5*Pi^(5/2)*x^5+1/3*Pi*(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsi
nh(c*x))+1/5*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))-2*Pi^(5/2)*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1
/2))-b*Pi^(5/2)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+b*Pi^(5/2)*polylog(2,c*x+(c^2*x^2+1)^(1/2))+Pi^2*(a+b*arcsin
h(c*x))*(Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]
time = 0.27, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5808, 5806, 5816, 4267, 2317, 2438, 8, 200} \begin {gather*} \frac {1}{5} \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} \pi \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\pi ^2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-2 \pi ^{5/2} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{25} \pi ^{5/2} b c^5 x^5-\frac {11}{45} \pi ^{5/2} b c^3 x^3-\pi ^{5/2} b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+\pi ^{5/2} b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )-\frac {23}{15} \pi ^{5/2} b c x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-23*b*c*Pi^(5/2)*x)/15 - (11*b*c^3*Pi^(5/2)*x^3)/45 - (b*c^5*Pi^(5/2)*x^5)/25 + Pi^2*Sqrt[Pi + c^2*Pi*x^2]*(a
+ b*ArcSinh[c*x]) + (Pi*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/3 + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*Arc
Sinh[c*x]))/5 - 2*Pi^(5/2)*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]] - b*Pi^(5/2)*PolyLog[2, -E^ArcSinh[c*x
]] + b*Pi^(5/2)*PolyLog[2, E^ArcSinh[c*x]]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 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 4267

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(Ar
cTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*
fz*x)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5806

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

Rule 5808

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((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])^n/(f*(m + 2*p + 1))), x] + (Dist[2*d*(p/(m + 2*p + 1)), Int
[(f*x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]

Rule 5816

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(1/c^(m
+ 1))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx &=\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\pi \int \frac {\left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac {\left (b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right )^2 \, dx}{5 \sqrt {1+c^2 x^2}}\\ &=\frac {1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\pi ^2 \int \frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac {\left (b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx}{5 \sqrt {1+c^2 x^2}}-\frac {\left (b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right ) \, dx}{3 \sqrt {1+c^2 x^2}}\\ &=-\frac {8 b c \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}{15 \sqrt {1+c^2 x^2}}-\frac {11 b c^3 \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2}}{45 \sqrt {1+c^2 x^2}}-\frac {b c^5 \pi ^2 x^5 \sqrt {\pi +c^2 \pi x^2}}{25 \sqrt {1+c^2 x^2}}+\pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (\pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{\sqrt {1+c^2 x^2}}-\frac {\left (b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int 1 \, dx}{\sqrt {1+c^2 x^2}}\\ &=-\frac {23 b c \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}{15 \sqrt {1+c^2 x^2}}-\frac {11 b c^3 \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2}}{45 \sqrt {1+c^2 x^2}}-\frac {b c^5 \pi ^2 x^5 \sqrt {\pi +c^2 \pi x^2}}{25 \sqrt {1+c^2 x^2}}+\pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (\pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}\\ &=-\frac {23 b c \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}{15 \sqrt {1+c^2 x^2}}-\frac {11 b c^3 \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2}}{45 \sqrt {1+c^2 x^2}}-\frac {b c^5 \pi ^2 x^5 \sqrt {\pi +c^2 \pi x^2}}{25 \sqrt {1+c^2 x^2}}+\pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2 \pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (b \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (b \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}\\ &=-\frac {23 b c \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}{15 \sqrt {1+c^2 x^2}}-\frac {11 b c^3 \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2}}{45 \sqrt {1+c^2 x^2}}-\frac {b c^5 \pi ^2 x^5 \sqrt {\pi +c^2 \pi x^2}}{25 \sqrt {1+c^2 x^2}}+\pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2 \pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (b \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (b \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}\\ &=-\frac {23 b c \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}{15 \sqrt {1+c^2 x^2}}-\frac {11 b c^3 \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2}}{45 \sqrt {1+c^2 x^2}}-\frac {b c^5 \pi ^2 x^5 \sqrt {\pi +c^2 \pi x^2}}{25 \sqrt {1+c^2 x^2}}+\pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2 \pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {b \pi ^2 \sqrt {\pi +c^2 \pi x^2} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {b \pi ^2 \sqrt {\pi +c^2 \pi x^2} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 257, normalized size = 1.44 \begin {gather*} \frac {1}{225} \pi ^{5/2} \left (-345 b c x-55 b c^3 x^3-9 b c^5 x^5+345 a \sqrt {1+c^2 x^2}+165 a c^2 x^2 \sqrt {1+c^2 x^2}+45 a c^4 x^4 \sqrt {1+c^2 x^2}+345 b \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+165 b c^2 x^2 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+45 b c^4 x^4 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+225 b \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-225 b \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )+225 a \log (x)-225 a \log \left (\pi \left (1+\sqrt {1+c^2 x^2}\right )\right )+225 b \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-225 b \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(5/2)*(-345*b*c*x - 55*b*c^3*x^3 - 9*b*c^5*x^5 + 345*a*Sqrt[1 + c^2*x^2] + 165*a*c^2*x^2*Sqrt[1 + c^2*x^2]
+ 45*a*c^4*x^4*Sqrt[1 + c^2*x^2] + 345*b*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 165*b*c^2*x^2*Sqrt[1 + c^2*x^2]*Arc
Sinh[c*x] + 45*b*c^4*x^4*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 225*b*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - 225*
b*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 225*a*Log[x] - 225*a*Log[Pi*(1 + Sqrt[1 + c^2*x^2])] + 225*b*PolyL
og[2, -E^(-ArcSinh[c*x])] - 225*b*PolyLog[2, E^(-ArcSinh[c*x])]))/225

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Maple [A]
time = 5.79, size = 284, normalized size = 1.59

method	result	size
default	\(a \left (\frac {\left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{5}+\pi \left (\frac {\left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{3}+\pi \left (\sqrt {\pi \,c^{2} x^{2}+\pi }-\sqrt {\pi }\, \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )\right )\right )\right )-\frac {b \,c^{5} \pi ^{\frac {5}{2}} x^{5}}{25}-\frac {11 b \,c^{3} \pi ^{\frac {5}{2}} x^{3}}{45}+\frac {23 b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{15}-b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+\frac {11 b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{2} c^{2}}{15}-\frac {23 b c \,\pi ^{\frac {5}{2}} x}{15}+\frac {b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{4} c^{4}}{5}+b \,\pi ^{\frac {5}{2}} \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )-b \,\pi ^{\frac {5}{2}} \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(1/5*(Pi*c^2*x^2+Pi)^(5/2)+Pi*(1/3*(Pi*c^2*x^2+Pi)^(3/2)+Pi*((Pi*c^2*x^2+Pi)^(1/2)-Pi^(1/2)*arctanh(Pi^(1/2)
/(Pi*c^2*x^2+Pi)^(1/2)))))-1/25*b*c^5*Pi^(5/2)*x^5-11/45*b*c^3*Pi^(5/2)*x^3+23/15*b*Pi^(5/2)*arcsinh(c*x)*(c^2
*x^2+1)^(1/2)-b*Pi^(5/2)*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))+b*Pi^(5/2)*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)
^(1/2))+11/15*b*Pi^(5/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^2*c^2-23/15*b*c*Pi^(5/2)*x+1/5*b*Pi^(5/2)*arcsinh(c*
x)*(c^2*x^2+1)^(1/2)*x^4*c^4+b*Pi^(5/2)*polylog(2,c*x+(c^2*x^2+1)^(1/2))-b*Pi^(5/2)*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((pi*c^2*x^2+pi)^(5/2)*(a+b*arcsinh(c*x))/x,x, algorithm="maxima")

[Out]

-1/15*(15*pi^(5/2)*arcsinh(1/(c*abs(x))) - 15*pi^2*sqrt(pi + pi*c^2*x^2) - 5*pi*(pi + pi*c^2*x^2)^(3/2) - 3*(p
i + pi*c^2*x^2)^(5/2))*a + b*integrate((pi + pi*c^2*x^2)^(5/2)*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((pi*c^2*x^2+pi)^(5/2)*(a+b*arcsinh(c*x))/x,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \pi ^{\frac {5}{2}} \left (\int \frac {a \sqrt {c^{2} x^{2} + 1}}{x}\, dx + \int 2 a c^{2} x \sqrt {c^{2} x^{2} + 1}\, dx + \int a c^{4} x^{3} \sqrt {c^{2} x^{2} + 1}\, dx + \int \frac {b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{x}\, dx + \int 2 b c^{2} x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}\, dx + \int b c^{4} x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

pi**(5/2)*(Integral(a*sqrt(c**2*x**2 + 1)/x, x) + Integral(2*a*c**2*x*sqrt(c**2*x**2 + 1), x) + Integral(a*c**
4*x**3*sqrt(c**2*x**2 + 1), x) + Integral(b*sqrt(c**2*x**2 + 1)*asinh(c*x)/x, x) + Integral(2*b*c**2*x*sqrt(c*
*2*x**2 + 1)*asinh(c*x), x) + Integral(b*c**4*x**3*sqrt(c**2*x**2 + 1)*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((pi*c^2*x^2+pi)^(5/2)*(a+b*arcsinh(c*x))/x,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 (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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