\(\int \frac {\sqrt {c+d x^2}}{x (a+b x^2)} \, dx\) [681]

Optimal result

Integrand size = 24, antiderivative size = 80 \[ \int \frac {\sqrt {c+d x^2}}{x \left (a+b x^2\right )} \, dx=-\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {\sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {457, 85, 65, 214} \[ \int \frac {\sqrt {c+d x^2}}{x \left (a+b x^2\right )} \, dx=\frac {\sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}}-\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a} \]

[In]

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

[Out]

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

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 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[(b*e - a*f)/(b*c
- a*d), Int[(e + f*x)^(p - 1)/(a + b*x), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[(e + f*x)^(p - 1)/(c + d*x
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]

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 457

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {c+d x^2}}{x \left (a+b x^2\right )} \, dx=\frac {\frac {\sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {b}}-\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {\frac {\left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )}{a}\)	\(70\)
default	\(\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{a}-\frac {\sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}+\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {\frac {d \sqrt {-a b}}{b}+d \left (x -\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}\right )}{b}+\frac {\left (a d -b c \right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}}{2 a}-\frac {\sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}-\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {-\frac {d \sqrt {-a b}}{b}+d \left (x +\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}\right )}{b}+\frac {\left (a d -b c \right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}}{2 a}\)	\(689\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 578, normalized size of antiderivative = 7.22 \[ \int \frac {\sqrt {c+d x^2}}{x \left (a+b x^2\right )} \, dx=\left [\frac {\sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right )}{4 \, a}, \frac {4 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, a}, \frac {\sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right )}{2 \, a}, \frac {\sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right )}{2 \, a}\right ] \]

[In]

integrate((d*x^2+c)^(1/2)/x/(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)) + 2*sqrt(c)
*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2))/a, 1/4*(4*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + sq
rt((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)))/a, 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)) + sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2))/a, 1/2*(sqrt(-(b*c - a*d)/b)*arcta
n(-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(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)))/a]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (66) = 132\).

Time = 3.65 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.89 \[ \int \frac {\sqrt {c+d x^2}}{x \left (a+b x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {c d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{2 a \sqrt {- c}} + \frac {d \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 a b \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\\sqrt {c} \left (- \frac {b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{2}}{a} & \text {for}\: b = 0 \\- \frac {\log {\left (a - 2 b \left (\frac {a}{2 b} + x^{2}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{a} - \frac {b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{2}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + 2 b \left (\frac {a}{2 b} + x^{2}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{a}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {c+d x^2}}{x \left (a+b x^2\right )} \, dx=-\frac {{\left (b c - a d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a} + \frac {c \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {c+d x^2}}{x \left (a+b x^2\right )} \, dx=\frac {\mathrm {atanh}\left (\frac {2\,a\,b^2\,c\,d^3\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{2\,a\,b^3\,c^2\,d^3-2\,a^2\,b^2\,c\,d^4}\right )\,\sqrt {b^2\,c-a\,b\,d}}{a\,b}-\frac {\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{a} \]

[In]

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

[Out]

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