\(\int \frac {x}{(a+b x^2)^2 \sqrt {c+d x^2}} \, dx\) [761]

Optimal result

Integrand size = 22, antiderivative size = 87 \[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {c+d x^2}}{2 (b c-a d) \left (a+b x^2\right )}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 \sqrt {b} (b c-a d)^{3/2}} \]

[Out]

1/2*d*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(3/2)/b^(1/2)-1/2*(d*x^2+c)^(1/2)/(-a*d+b*c
)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {455, 44, 65, 214} \[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 \sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x^2}}{2 \left (a+b x^2\right ) (b c-a d)} \]

[In]

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

[Out]

-1/2*Sqrt[c + d*x^2]/((b*c - a*d)*(a + b*x^2)) + (d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*Sqr
t[b]*(b*c - a*d)^(3/2))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {c+d x^2}}{2 (b c-a d) \left (a+b x^2\right )}-\frac {d \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 (b c-a d)} \\ & = -\frac {\sqrt {c+d x^2}}{2 (b c-a d) \left (a+b x^2\right )}-\frac {\text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 (b c-a d)} \\ & = -\frac {\sqrt {c+d x^2}}{2 (b c-a d) \left (a+b x^2\right )}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 \sqrt {b} (b c-a d)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {1}{2} \left (-\frac {\sqrt {c+d x^2}}{(b c-a d) \left (a+b x^2\right )}+\frac {d \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} (-b c+a d)^{3/2}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(\frac {d \left (b \,x^{2}+a \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {d \,x^{2}+c}\, \sqrt {\left (a d -b c \right ) b}}{2 \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right ) \left (b \,x^{2}+a \right )}\)	\(90\)
default	\(-\frac {\sqrt {-a b}\, \left (\frac {b \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{\left (a d -b c \right ) \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {d \sqrt {-a b}\, \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 a \,b^{2}}+\frac {\sqrt {-a b}\, \left (\frac {b \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{\left (a d -b c \right ) \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {d \sqrt {-a b}\, \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 a \,b^{2}}\)	\(524\)

[In]

int(x/(b*x^2+a)^2/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (71) = 142\).

Time = 0.27 (sec) , antiderivative size = 404, normalized size of antiderivative = 4.64 \[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\left [-\frac {{\left (b d x^{2} + a d\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2}\right )}}, \frac {{\left (b d x^{2} + a d\right )} \sqrt {-b^{2} c + a b d} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2}\right )}}\right ] \]

[In]

integrate(x/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*((b*d*x^2 + a*d)*sqrt(b^2*c - a*b*d)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d -
3*a*b*d^2)*x^2 - 4*(b*d*x^2 + 2*b*c - a*d)*sqrt(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2))
+ 4*(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2 + (b^4*c^2 - 2*a*b^3*c*d + a^2*b^2
*d^2)*x^2), 1/4*((b*d*x^2 + a*d)*sqrt(-b^2*c + a*b*d)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(-b^2*c + a*b*d)
*sqrt(d*x^2 + c)/(b^2*c^2 - a*b*c*d + (b^2*c*d - a*b*d^2)*x^2)) - 2*(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(a*b^3*c^
2 - 2*a^2*b^2*c*d + a^3*b*d^2 + (b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^2)]

Sympy [F]

\[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\int \frac {x}{\left (a + b x^{2}\right )^{2} \sqrt {c + d x^{2}}}\, dx \]

[In]

integrate(x/(b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

Integral(x/((a + b*x**2)**2*sqrt(c + d*x**2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=-\frac {d \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} {\left (b c - a d\right )}} - \frac {\sqrt {d x^{2} + c} d}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} {\left (b c - a d\right )}} \]

[In]

integrate(x/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/2*d*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*(b*c - a*d)) - 1/2*sqrt(d*x^2 + c)
*d/(((d*x^2 + c)*b - b*c + a*d)*(b*c - a*d))

Mupad [B] (verification not implemented)

Time = 5.52 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int \frac {x}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx=\frac {d\,\sqrt {d\,x^2+c}}{2\,\left (a\,d-b\,c\right )\,\left (b\,\left (d\,x^2+c\right )+a\,d-b\,c\right )}+\frac {d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )}{2\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{3/2}} \]

[In]

int(x/((a + b*x^2)^2*(c + d*x^2)^(1/2)),x)

[Out]

(d*(c + d*x^2)^(1/2))/(2*(a*d - b*c)*(b*(c + d*x^2) + a*d - b*c)) + (d*atan((b^(1/2)*(c + d*x^2)^(1/2))/(a*d -
b*c)^(1/2)))/(2*b^(1/2)*(a*d - b*c)^(3/2))