\(\int \frac {x \sqrt {c+d x^2}}{(a+b x^2)^2} \, dx\) [734]

Optimal result

Integrand size = 22, antiderivative size = 80 \[ \int \frac {x \sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=-\frac {\sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}-\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{3/2} \sqrt {b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 80, 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, 43, 65, 214} \[ \int \frac {x \sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=-\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2}}{2 b \left (a+b x^2\right )} \]

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[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 {\sqrt {c+d x}}{(a+b x)^2} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {c+d x^2}}{2 b \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} \\ & = -\frac {\sqrt {c+d x^2}}{2 b \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} \\ & = -\frac {\sqrt {c+d x^2}}{2 b \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 b^{3/2} \sqrt {b c-a d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {x \sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=-\frac {\sqrt {c+d x^2}}{2 b \left (a+b x^2\right )}+\frac {d \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{2 b^{3/2} \sqrt {-b c+a d}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.99 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {d \,x^{2}+c}}{b \,x^{2}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{2 b}\)	\(65\)
default	\(\text {Expression too large to display}\)	\(1320\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (64) = 128\).

Time = 0.27 (sec) , antiderivative size = 356, normalized size of antiderivative = 4.45 \[ \int \frac {x \sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=\left [\frac {{\left (b d x^{2} + a d\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{3} c - a^{2} b^{2} d + {\left (b^{4} c - a b^{3} d\right )} x^{2}\right )}}, -\frac {{\left (b d x^{2} + a d\right )} \sqrt {-b^{2} c + a b d} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{3} c - a^{2} b^{2} d + {\left (b^{4} c - a b^{3} d\right )} x^{2}\right )}}\right ] \]

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(x*(d*x^2+c)^(1/2)/(b*x^2+a)^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 = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99 \[ \int \frac {x \sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=\frac {d \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b} - \frac {\sqrt {d x^{2} + c} d}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {x \sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=\frac {d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )}{2\,b^{3/2}\,\sqrt {a\,d-b\,c}}-\frac {d\,\sqrt {d\,x^2+c}}{2\,\left (d\,b^2\,x^2+a\,d\,b\right )} \]

[In]

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

[Out]

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