\(\int \frac {x}{(a+b x^2)^{3/2} \sqrt {c+d x^2}} \, dx\) [982]

Optimal result

Integrand size = 24, antiderivative size = 34 \[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {c+d x^2}}{(b c-a d) \sqrt {a+b x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {455, 37} \[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {c+d x^2}}{\sqrt {a+b x^2} (b c-a d)} \]

[In]

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

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {c+d x^2}}{(b c-a d) \sqrt {a+b x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {\sqrt {d \,x^{2}+c}}{\sqrt {b \,x^{2}+a}\, \left (a d -b c \right )}\)	\(30\)
default	\(\frac {\sqrt {d \,x^{2}+c}}{\sqrt {b \,x^{2}+a}\, \left (a d -b c \right )}\)	\(30\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {d \,x^{2}+c}}{\sqrt {b \,x^{2}+a}\, \left (a d -b c \right ) \sqrt {b d \,x^{4}+a d \,x^{2}+c b \,x^{2}+a c}}\)	\(71\)

[In]

int(x/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}} \]

[In]

integrate(x/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=\int \frac {x}{\left (a + b x^{2}\right )^{\frac {3}{2}} \sqrt {c + d x^{2}}}\, dx \]

[In]

integrate(x/(b*x**2+a)**(3/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x/((a + b*x**2)**(3/2)*sqrt(c + d*x**2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x/(b*x^2+a)^(3/2)/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (30) = 60\).

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.06 \[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=-\frac {2 \, \sqrt {b d} b}{{\left (b^{2} c - a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )} {\left | b \right |}} \]

[In]

integrate(x/(b*x^2+a)^(3/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32 \[ \int \frac {x}{\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \, dx=\frac {d\,x^2+c}{\left (a\,d\,\sqrt {d\,x^2+c}-b\,c\,\sqrt {d\,x^2+c}\right )\,\sqrt {b\,x^2+a}} \]

[In]

int(x/((a + b*x^2)^(3/2)*(c + d*x^2)^(1/2)),x)

[Out]

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