\(\int \frac {x}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\) [971]

Optimal result

Integrand size = 24, antiderivative size = 45 \[ \int \frac {x}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {455, 65, 223, 212} \[ \int \frac {x}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

[In]

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

[Out]

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

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(88\) vs. \(2(33)=66\).

Time = 3.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.98

method	result	size
default	\(\frac {\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{2 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}\)	\(89\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{2 \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \sqrt {b d}}\)	\(89\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (33) = 66\).

Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 4.31 \[ \int \frac {x}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\left [\frac {\sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right )}{4 \, b d}, -\frac {\sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right )}{2 \, b d}\right ] \]

[In]

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

[Out]

[1/4*sqrt(b*d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d*x^2 +
b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d))/(b*d), -1/2*sqrt(-b*d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d
)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2))/(b*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(x/(b*x^2+a)^(1/2)/(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 = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {x}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {b \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}{\sqrt {-b\,d}\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}\right )}{\sqrt {-b\,d}} \]

[In]

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

[Out]

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