\(\int \frac {a+b \text {arcsinh}(c x)}{x^2 (d+c^2 d x^2)^{3/2}} \, dx\) [162]

Optimal result

Integrand size = 26, antiderivative size = 143 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {a+b \text {arcsinh}(c x)}{d x \sqrt {d+c^2 d x^2}}-\frac {2 c^2 x (a+b \text {arcsinh}(c x))}{d \sqrt {d+c^2 d x^2}}+\frac {b c \sqrt {d+c^2 d x^2} \log (x)}{d^2 \sqrt {1+c^2 x^2}}+\frac {b c \sqrt {d+c^2 d x^2} \log \left (1+c^2 x^2\right )}{2 d^2 \sqrt {1+c^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {277, 197, 5804, 12, 457, 78} \[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {2 c^2 x (a+b \text {arcsinh}(c x))}{d \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{d x \sqrt {c^2 d x^2+d}}+\frac {b c \log (x) \sqrt {c^2 d x^2+d}}{d^2 \sqrt {c^2 x^2+1}}+\frac {b c \sqrt {c^2 d x^2+d} \log \left (c^2 x^2+1\right )}{2 d^2 \sqrt {c^2 x^2+1}} \]

[In]

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

[Out]

-((a + b*ArcSinh[c*x])/(d*x*Sqrt[d + c^2*d*x^2])) - (2*c^2*x*(a + b*ArcSinh[c*x]))/(d*Sqrt[d + c^2*d*x^2]) + (
b*c*Sqrt[d + c^2*d*x^2]*Log[x])/(d^2*Sqrt[1 + c^2*x^2]) + (b*c*Sqrt[d + c^2*d*x^2]*Log[1 + c^2*x^2])/(2*d^2*Sq
rt[1 + c^2*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5804

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p -
1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \text {arcsinh}(c x)}{d x \sqrt {d+c^2 d x^2}}-\frac {2 c^2 x (a+b \text {arcsinh}(c x))}{d \sqrt {d+c^2 d x^2}}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \int \frac {-1-2 c^2 x^2}{d^2 x \left (1+c^2 x^2\right )} \, dx}{\sqrt {1+c^2 x^2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{d x \sqrt {d+c^2 d x^2}}-\frac {2 c^2 x (a+b \text {arcsinh}(c x))}{d \sqrt {d+c^2 d x^2}}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \int \frac {-1-2 c^2 x^2}{x \left (1+c^2 x^2\right )} \, dx}{d^2 \sqrt {1+c^2 x^2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{d x \sqrt {d+c^2 d x^2}}-\frac {2 c^2 x (a+b \text {arcsinh}(c x))}{d \sqrt {d+c^2 d x^2}}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \frac {-1-2 c^2 x}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d^2 \sqrt {1+c^2 x^2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{d x \sqrt {d+c^2 d x^2}}-\frac {2 c^2 x (a+b \text {arcsinh}(c x))}{d \sqrt {d+c^2 d x^2}}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \text {Subst}\left (\int \left (-\frac {1}{x}-\frac {c^2}{1+c^2 x}\right ) \, dx,x,x^2\right )}{2 d^2 \sqrt {1+c^2 x^2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{d x \sqrt {d+c^2 d x^2}}-\frac {2 c^2 x (a+b \text {arcsinh}(c x))}{d \sqrt {d+c^2 d x^2}}+\frac {b c \sqrt {d+c^2 d x^2} \log (x)}{d^2 \sqrt {1+c^2 x^2}}+\frac {b c \sqrt {d+c^2 d x^2} \log \left (1+c^2 x^2\right )}{2 d^2 \sqrt {1+c^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {\sqrt {d+c^2 d x^2} \left (2 a \sqrt {1+c^2 x^2}+4 a c^2 x^2 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (1+2 c^2 x^2\right ) \text {arcsinh}(c x)+b c x \left (1+c^2 x^2\right ) \log \left (1+\frac {1}{c^2 x^2}\right )-2 b c x \log \left (1+c^2 x^2\right )-2 b c^3 x^3 \log \left (1+c^2 x^2\right )\right )}{2 d^2 x \left (1+c^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

-1/2*(Sqrt[d + c^2*d*x^2]*(2*a*Sqrt[1 + c^2*x^2] + 4*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 + c^2*x^2]*(1 +
2*c^2*x^2)*ArcSinh[c*x] + b*c*x*(1 + c^2*x^2)*Log[1 + 1/(c^2*x^2)] - 2*b*c*x*Log[1 + c^2*x^2] - 2*b*c^3*x^3*Lo
g[1 + c^2*x^2]))/(d^2*x*(1 + c^2*x^2)^(3/2))

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.67

method	result	size
default	\(a \left (-\frac {1}{d x \sqrt {c^{2} d \,x^{2}+d}}-\frac {2 c^{2} x}{d \sqrt {c^{2} d \,x^{2}+d}}\right )-\frac {b \left (2 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{4} c^{4}-2 \sqrt {c^{2} x^{2}+1}\, \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{3} c^{3}+2 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{2} c^{2}-\sqrt {c^{2} x^{2}+1}\, \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x c +\operatorname {arcsinh}\left (c x \right )\right ) \left (2 c^{2} x^{2}+1+2 c x \sqrt {c^{2} x^{2}+1}\right ) \sqrt {d \left (c^{2} x^{2}+1\right )}}{x \,d^{2} \left (c^{2} x^{2}+1\right )}\)	\(239\)
parts	\(a \left (-\frac {1}{d x \sqrt {c^{2} d \,x^{2}+d}}-\frac {2 c^{2} x}{d \sqrt {c^{2} d \,x^{2}+d}}\right )-\frac {b \left (2 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{4} c^{4}-2 \sqrt {c^{2} x^{2}+1}\, \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{3} c^{3}+2 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{2} c^{2}-\sqrt {c^{2} x^{2}+1}\, \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x c +\operatorname {arcsinh}\left (c x \right )\right ) \left (2 c^{2} x^{2}+1+2 c x \sqrt {c^{2} x^{2}+1}\right ) \sqrt {d \left (c^{2} x^{2}+1\right )}}{x \,d^{2} \left (c^{2} x^{2}+1\right )}\)	\(239\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{x^{2} \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {1}{2} \, b c {\left (\frac {\log \left (c^{2} x^{2} + 1\right )}{d^{\frac {3}{2}}} + \frac {2 \, \log \left (x\right )}{d^{\frac {3}{2}}}\right )} - {\left (\frac {2 \, c^{2} x}{\sqrt {c^{2} d x^{2} + d} d} + \frac {1}{\sqrt {c^{2} d x^{2} + d} d x}\right )} b \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, c^{2} x}{\sqrt {c^{2} d x^{2} + d} d} + \frac {1}{\sqrt {c^{2} d x^{2} + d} d x}\right )} a \]

[In]

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

[Out]

1/2*b*c*(log(c^2*x^2 + 1)/d^(3/2) + 2*log(x)/d^(3/2)) - (2*c^2*x/(sqrt(c^2*d*x^2 + d)*d) + 1/(sqrt(c^2*d*x^2 +
d)*d*x))*b*arcsinh(c*x) - (2*c^2*x/(sqrt(c^2*d*x^2 + d)*d) + 1/(sqrt(c^2*d*x^2 + d)*d*x))*a

Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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