\(\int \frac {x}{\sqrt {a-b x^2} \sqrt {c+d x^2}} \, dx\) [990]

Optimal result

Integrand size = 25, antiderivative size = 47 \[ \int \frac {x}{\sqrt {a-b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {d} \sqrt {a-b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

[Out]

-arctan(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 = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {455, 65, 223, 209} \[ \int \frac {x}{\sqrt {a-b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {d} \sqrt {a-b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

[In]

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

[Out]

-(ArcTan[(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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 {\tan ^{-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 = 0.35 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {x}{\sqrt {a-b x^2} \sqrt {c+d x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a-b x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(91\) vs. \(2(35)=70\).

Time = 3.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.96

method	result	size
default	\(-\frac {\arctan \left (\frac {\sqrt {b d}\, \left (-2 b d \,x^{2}+a d -b c \right )}{2 b d \sqrt {\left (-b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}\right ) \sqrt {-b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{2 \sqrt {b d}\, \sqrt {\left (-b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}\)	\(92\)
elliptic	\(\frac {\sqrt {\left (-b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \arctan \left (\frac {\sqrt {b d}\, \left (x^{2}-\frac {a d -b c}{2 b d}\right )}{\sqrt {-b d \,x^{4}+\left (a d -b c \right ) x^{2}+a c}}\right )}{2 \sqrt {-b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \sqrt {b d}}\)	\(97\)

[In]

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

[Out]

-1/2*arctan(1/2*(b*d)^(1/2)*(-2*b*d*x^2+a*d-b*c)/b/d/((-b*x^2+a)*(d*x^2+c))^(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. 92 vs. \(2 (35) = 70\).

Time = 0.30 (sec) , antiderivative size = 201, normalized size of antiderivative = 4.28 \[ \int \frac {x}{\sqrt {a-b x^2} \sqrt {c+d x^2}} \, dx=\left [-\frac {\sqrt {-b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d - a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b d x^{2} + b c - a d\right )} \sqrt {-b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}\right )}{4 \, b d}, -\frac {\sqrt {b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c - a d\right )} \sqrt {-b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}}{2 \, {\left (b^{2} d^{2} x^{4} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right )}{2 \, b d}\right ] \]

[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 = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int \frac {x}{\sqrt {a-b x^2} \sqrt {c+d x^2}} \, dx=\frac {b \log \left ({\left | -\sqrt {-b x^{2} + a} \sqrt {-b d} + \sqrt {b^{2} c + {\left (b x^{2} - a\right )} b d + a b d} \right |}\right )}{\sqrt {-b d} {\left | b \right |}} \]

[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 = 5.68 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02 \[ \int \frac {x}{\sqrt {a-b x^2} \sqrt {c+d x^2}} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {a-b\,x^2}-\sqrt {a}\right )}{\sqrt {b\,d}\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}\right )}{\sqrt {b\,d}} \]

[In]

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

[Out]

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