\(\int \frac {a+b \text {arcsinh}(c x)}{(d+c^2 d x^2)^{5/2}} \, dx\) [171]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {b}{6 c d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {x (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 c d^2 \sqrt {d+c^2 d x^2}} \]

[Out]

1/3*x*(a+b*arcsinh(c*x))/d/(c^2*d*x^2+d)^(3/2)+2/3*x*(a+b*arcsinh(c*x))/d^2/(c^2*d*x^2+d)^(1/2)+1/6*b/c/d^2/(c
^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-1/3*b*ln(c^2*x^2+1)*(c^2*x^2+1)^(1/2)/c/d^2/(c^2*d*x^2+d)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5788, 5787, 266, 267} \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {b}{6 c d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 c d^2 \sqrt {c^2 d x^2+d}} \]

[In]

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

[Out]

b/(6*c*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) + (x*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2)) + (2*
x*(a + b*ArcSinh[c*x]))/(3*d^2*Sqrt[d + c^2*d*x^2]) - (b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(3*c*d^2*Sqrt[d +
c^2*d*x^2])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 5787

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

Rubi steps \begin{align*} \text {integral}& = \frac {x (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{3 d}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {x}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}} \\ & = \frac {b}{6 c d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {x (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (2 b c \sqrt {1+c^2 x^2}\right ) \int \frac {x}{1+c^2 x^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}} \\ & = \frac {b}{6 c d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {x (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 c d^2 \sqrt {d+c^2 d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.97 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (b+b c^2 x^2+6 a c x \sqrt {1+c^2 x^2}+4 a c^3 x^3 \sqrt {1+c^2 x^2}+2 b c x \sqrt {1+c^2 x^2} \left (3+2 c^2 x^2\right ) \text {arcsinh}(c x)-2 b \left (1+c^2 x^2\right )^2 \log \left (1+c^2 x^2\right )\right )}{6 c d^3 \left (1+c^2 x^2\right )^{5/2}} \]

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(b + b*c^2*x^2 + 6*a*c*x*Sqrt[1 + c^2*x^2] + 4*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 2*b*c*x*Sqrt
[1 + c^2*x^2]*(3 + 2*c^2*x^2)*ArcSinh[c*x] - 2*b*(1 + c^2*x^2)^2*Log[1 + c^2*x^2]))/(6*c*d^3*(1 + c^2*x^2)^(5/
2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(424\) vs. \(2(127)=254\).

Time = 0.28 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.89

method	result	size
default	\(a \left (\frac {x}{3 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {c^{2} d \,x^{2}+d}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+3 c x +2 \sqrt {c^{2} x^{2}+1}\right ) \left (-8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{6} c^{6}+8 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{5} c^{5}-24 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}+20 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{3} c^{3}+2 c^{4} x^{4}-2 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+6 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}-24 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}+12 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x c +4 c^{2} x^{2}-3 c x \sqrt {c^{2} x^{2}+1}+8 \,\operatorname {arcsinh}\left (c x \right )-8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )+2\right )}{6 \left (3 c^{6} x^{6}+10 c^{4} x^{4}+11 c^{2} x^{2}+4\right ) c \,d^{3}}\)	\(425\)
parts	\(a \left (\frac {x}{3 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {c^{2} d \,x^{2}+d}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+3 c x +2 \sqrt {c^{2} x^{2}+1}\right ) \left (-8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{6} c^{6}+8 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{5} c^{5}-24 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}+20 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{3} c^{3}+2 c^{4} x^{4}-2 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+6 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}-24 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}+12 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x c +4 c^{2} x^{2}-3 c x \sqrt {c^{2} x^{2}+1}+8 \,\operatorname {arcsinh}\left (c x \right )-8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )+2\right )}{6 \left (3 c^{6} x^{6}+10 c^{4} x^{4}+11 c^{2} x^{2}+4\right ) c \,d^{3}}\)	\(425\)

[In]

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

[Out]

a*(1/3/d*x/(c^2*d*x^2+d)^(3/2)+2/3/d^2*x/(c^2*d*x^2+d)^(1/2))+1/6*b*(d*(c^2*x^2+1))^(1/2)*(2*c^3*x^3+2*c^2*x^2
*(c^2*x^2+1)^(1/2)+3*c*x+2*(c^2*x^2+1)^(1/2))*(-8*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)*x^6*c^6+8*(c^2*x^2+1)^(1/2)*
ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)*x^5*c^5-24*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)*x^4*c^4+20*(c^2*x^2+1)^(1/2)*ln(1+(
c*x+(c^2*x^2+1)^(1/2))^2)*x^3*c^3+2*c^4*x^4-2*c^3*x^3*(c^2*x^2+1)^(1/2)+6*arcsinh(c*x)*c^2*x^2-24*ln(1+(c*x+(c
^2*x^2+1)^(1/2))^2)*x^2*c^2+12*(c^2*x^2+1)^(1/2)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)*x*c+4*c^2*x^2-3*c*x*(c^2*x^2+
1)^(1/2)+8*arcsinh(c*x)-8*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+2)/(3*c^6*x^6+10*c^4*x^4+11*c^2*x^2+4)/c/d^3

Fricas [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {1}{6} \, b c {\left (\frac {1}{c^{4} d^{\frac {5}{2}} x^{2} + c^{2} d^{\frac {5}{2}}} - \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {2 \, x}{\sqrt {c^{2} d x^{2} + d} d^{2}} + \frac {x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, x}{\sqrt {c^{2} d x^{2} + d} d^{2}} + \frac {x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d}\right )} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+b*arcsinh(c*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)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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