\(\int \frac {x (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [170]

Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {b x}{6 c d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {a+b \text {arcsinh}(c x)}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {b \sqrt {1+c^2 x^2} \arctan (c x)}{6 c^2 d^2 \sqrt {d+c^2 d x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5798, 205, 209} \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {a+b \text {arcsinh}(c x)}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {b \sqrt {c^2 x^2+1} \arctan (c x)}{6 c^2 d^2 \sqrt {c^2 d x^2+d}}+\frac {b x}{6 c d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}} \]

[In]

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

[Out]

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

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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*} \text {integral}& = -\frac {a+b \text {arcsinh}(c x)}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d+c^2 d x^2}} \\ & = \frac {b x}{6 c d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {a+b \text {arcsinh}(c x)}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{6 c d^2 \sqrt {d+c^2 d x^2}} \\ & = \frac {b x}{6 c d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {a+b \text {arcsinh}(c x)}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {b \sqrt {1+c^2 x^2} \arctan (c x)}{6 c^2 d^2 \sqrt {d+c^2 d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90 \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (b c x+b c^3 x^3-2 a \sqrt {1+c^2 x^2}-2 b \sqrt {1+c^2 x^2} \text {arcsinh}(c x)+b \left (1+c^2 x^2\right )^2 \arctan (c x)\right )}{6 c^2 d^3 \left (1+c^2 x^2\right )^{5/2}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.74

method	result	size
default	\(-\frac {a}{3 c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{2}}+\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}-\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}\)	\(198\)
parts	\(-\frac {a}{3 c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{2}}+\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}-\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{2} d^{3}}\)	\(198\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.46 \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {{\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 4 \, \sqrt {c^{2} d x^{2} + d} b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, \sqrt {c^{2} d x^{2} + d} {\left (\sqrt {c^{2} x^{2} + 1} b c x - 2 \, a\right )}}{12 \, {\left (c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

-1/12*((b*c^4*x^4 + 2*b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(c^2*d*x^2 + d)*sqrt(c^2*x^2 + 1)*c*sqrt(d)*x/(c^4*d
*x^4 - d)) + 4*sqrt(c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 + 1)) - 2*sqrt(c^2*d*x^2 + d)*(sqrt(c^2*x^2 + 1)*b
*c*x - 2*a))/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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