\(\int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx\) [679]

Optimal result

Integrand size = 22, antiderivative size = 65 \[ \int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2}}{b}-\frac {\sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, 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, 52, 65, 214} \[ \int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2}}{b}-\frac {\sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{3/2}} \]

[In]

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

[Out]

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

Rule 52

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2}}{b}-\frac {\sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.96 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(\frac {\sqrt {d \,x^{2}+c}-\frac {\left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{b}\)	\(61\)
risch	\(\frac {\sqrt {d \,x^{2}+c}}{b}-\frac {\left (a d -b c \right ) \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}\right )}{b}\)	\(327\)
default	\(\frac {\sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}+\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {\frac {d \sqrt {-a b}}{b}+d \left (x -\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}\right )}{b}+\frac {\left (a d -b c \right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}}{2 b}+\frac {\sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}-\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {-\frac {d \sqrt {-a b}}{b}+d \left (x +\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}\right )}{b}+\frac {\left (a d -b c \right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}}{2 b}\)	\(647\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.92 \[ \int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx=\left [\frac {\sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, \sqrt {d x^{2} + c}}{4 \, b}, -\frac {\sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, \sqrt {d x^{2} + c}}{2 \, b}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.35 \[ \int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx=\begin {cases} \frac {2 \left (\frac {d \sqrt {c + d x^{2}}}{2 b} - \frac {d \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 b^{2} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\begin {cases} \frac {x^{2}}{2 a} & \text {for}\: b = 0 \\\frac {\log {\left (2 a + 2 b x^{2} \right )}}{2 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(d*sqrt(c + d*x**2)/(2*b) - d*(a*d - b*c)*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*b**2*sqrt
((a*d - b*c)/b)))/d, Ne(d, 0)), (sqrt(c)*Piecewise((x**2/(2*a), Eq(b, 0)), (log(2*a + 2*b*x**2)/(2*b), True)),
True))

Maxima [F(-2)]

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

[In]

integrate(x*(d*x^2+c)^(1/2)/(b*x^2+a),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 = 64, normalized size of antiderivative = 0.98 \[ \int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx=\frac {{\left (b c - a d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b} + \frac {\sqrt {d x^{2} + c}}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int \frac {x \sqrt {c+d x^2}}{a+b x^2} \, dx=\frac {\sqrt {d\,x^2+c}}{b}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )\,\sqrt {a\,d-b\,c}}{b^{3/2}} \]

[In]

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

[Out]

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