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

Optimal. Leaf size=262 \[ \frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {c^2 d x^2+d}}-\frac {2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt {c^2 d x^2+d}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {b \sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {c^2 d x^2+d}}+\frac {b \sqrt {c^2 x^2+1} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {c^2 d x^2+d}}-\frac {b c x}{6 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {7 b \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{6 d^2 \sqrt {c^2 d x^2+d}} \]

[Out]

1/3*(a+b*arcsinh(c*x))/d/(c^2*d*x^2+d)^(3/2)+(a+b*arcsinh(c*x))/d^2/(c^2*d*x^2+d)^(1/2)-1/6*b*c*x/d^2/(c^2*x^2
+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-7/6*b*arctan(c*x)*(c^2*x^2+1)^(1/2)/d^2/(c^2*d*x^2+d)^(1/2)-2*(a+b*arcsinh(c*x))
*arctanh(c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)/d^2/(c^2*d*x^2+d)^(1/2)-b*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*
(c^2*x^2+1)^(1/2)/d^2/(c^2*d*x^2+d)^(1/2)+b*polylog(2,c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)/d^2/(c^2*d*x^2+
d)^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5755, 5764, 5760, 4182, 2279, 2391, 203, 199} \[ -\frac {b \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {c^2 d x^2+d}}+\frac {b \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {c^2 d x^2+d}}+\frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {c^2 d x^2+d}}-\frac {2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt {c^2 d x^2+d}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {b c x}{6 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {7 b \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{6 d^2 \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*x)/(6*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) + (a + b*ArcSinh[c*x])/(3*d*(d + c^2*d*x^2)^(3/2)) + (a
+ b*ArcSinh[c*x])/(d^2*Sqrt[d + c^2*d*x^2]) - (7*b*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(6*d^2*Sqrt[d + c^2*d*x^2])
- (2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/(d^2*Sqrt[d + c^2*d*x^2]) - (b*Sqrt[1 +
c^2*x^2]*PolyLog[2, -E^ArcSinh[c*x]])/(d^2*Sqrt[d + c^2*d*x^2]) + (b*Sqrt[1 + c^2*x^2]*PolyLog[2, E^ArcSinh[c*
x]])/(d^2*Sqrt[d + c^2*d*x^2])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 4182

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 5755

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 + 1)*(a + b*ArcSinh[c*x])^n)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(2*f*(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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rule 5760

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

Rule 5764

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

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^{5/2}} \, dx &=\frac {a+b \sinh ^{-1}(c x)}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^{3/2}} \, dx}{d}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {d+c^2 d x^2}}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {d+c^2 d x^2}} \, dx}{d^2}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {d+c^2 d x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {1+c^2 x^2} \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {d+c^2 d x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {1+c^2 x^2} \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {d+c^2 d x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {d+c^2 d x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{d^2 \sqrt {d+c^2 d x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}-\frac {b \sqrt {1+c^2 x^2} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}+\frac {b \sqrt {1+c^2 x^2} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.20, size = 247, normalized size = 0.94 \[ \frac {\frac {2 a \left (3 c^2 x^2+4\right ) \sqrt {c^2 d x^2+d}}{\left (c^2 x^2+1\right )^2}-6 a \sqrt {d} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+d\right )+6 a \sqrt {d} \log (x)+\frac {b d^2 \left (c^2 x^2+1\right )^{3/2} \left (-\frac {c x}{c^2 x^2+1}+\frac {6 \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+\frac {2 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}+6 \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-6 \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+6 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-6 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )-14 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{\left (c^2 d x^2+d\right )^{3/2}}}{6 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*a*(4 + 3*c^2*x^2)*Sqrt[d + c^2*d*x^2])/(1 + c^2*x^2)^2 + 6*a*Sqrt[d]*Log[x] - 6*a*Sqrt[d]*Log[d + Sqrt[d]*
Sqrt[d + c^2*d*x^2]] + (b*d^2*(1 + c^2*x^2)^(3/2)*(-((c*x)/(1 + c^2*x^2)) + (2*ArcSinh[c*x])/(1 + c^2*x^2)^(3/
2) + (6*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - 14*ArcTan[Tanh[ArcSinh[c*x]/2]] + 6*ArcSinh[c*x]*Log[1 - E^(-ArcSinh
[c*x])] - 6*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 6*PolyLog[2, -E^(-ArcSinh[c*x])] - 6*PolyLog[2, E^(-ArcS
inh[c*x])]))/(d + c^2*d*x^2)^(3/2))/(6*d^3)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} d x^{2} + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{c^{6} d^{3} x^{7} + 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} + d^{3} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)/(c^6*d^3*x^7 + 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 + d^3*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 364, normalized size = 1.39 \[ \frac {a}{3 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {a}{d^{2} \sqrt {c^{2} d \,x^{2}+d}}-\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c^{2} d \,x^{2}+d}}{x}\right )}{d^{\frac {5}{2}}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{2} c^{2}}{d^{3} \left (c^{2} x^{2}+1\right )^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c x}{6 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {4 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{3 d^{3} \left (c^{2} x^{2}+1\right )^{2}}-\frac {7 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{3 \sqrt {c^{2} x^{2}+1}\, d^{3}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{\sqrt {c^{2} x^{2}+1}\, d^{3}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\sqrt {c^{2} x^{2}+1}\, d^{3}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\sqrt {c^{2} x^{2}+1}\, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a/d/(c^2*d*x^2+d)^(3/2)+a/d^2/(c^2*d*x^2+d)^(1/2)-a/d^(5/2)*ln((2*d+2*d^(1/2)*(c^2*d*x^2+d)^(1/2))/x)+b*(d
*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2*arcsinh(c*x)*x^2*c^2-1/6*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)*c
*x+4/3*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2*arcsinh(c*x)-7/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^
3*arctan(c*x+(c^2*x^2+1)^(1/2))-b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*dilog(c*x+(c^2*x^2+1)^(1/2))-b*(
d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*dilog(1+c*x+(c^2*x^2+1)^(1/2))-b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^
(1/2)/d^3*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(3*arcsinh(1/(c*abs(x)))/d^(5/2) - 3/(sqrt(c^2*d*x^2 + d)*d^2) - 1/((c^2*d*x^2 + d)^(3/2)*d)) + b*integ
rate(log(c*x + sqrt(c^2*x^2 + 1))/((c^2*d*x^2 + d)^(5/2)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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