\(\int \frac {x^2}{(a+b x^2) (c+d x^2)^{3/2}} \, dx\) [715]

Optimal result

Integrand size = 24, antiderivative size = 74 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx=\frac {x}{(b c-a d) \sqrt {c+d x^2}}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, 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 = {482, 12, 385, 211} \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx=\frac {x}{\sqrt {c+d x^2} (b c-a d)}-\frac {\sqrt {a} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{3/2}} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 385

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

Rule 482

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(n*(b*c - a*d)*(p + 1))), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(74)=148\).

Time = 1.38 (sec) , antiderivative size = 319, normalized size of antiderivative = 4.31 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {(b c-a d) x}{\sqrt {c+d x^2}}+\frac {\left (-b c+a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{\sqrt {a} d}+\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \left (b c-a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{\sqrt {a} d}}{(b c-a d)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18

method	result	size
pseudoelliptic	\(\frac {a \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right ) \sqrt {d \,x^{2}+c}-x \sqrt {\left (a d -b c \right ) a}}{\left (a d -b c \right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (a d -b c \right ) a}}\)	\(87\)
default	\(\frac {x}{b c \sqrt {d \,x^{2}+c}}-\frac {a \left (-\frac {b}{\left (a d -b c \right ) \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 {2 d \sqrt {-a b}\, \left (2 d \left (x -\frac {\sqrt {-a b}}{b}\right )+\frac {2 d \sqrt {-a b}}{b}\right )}{\left (a d -b c \right ) \left (-\frac {4 d \left (a d -b c \right )}{b}+\frac {4 d^{2} a}{b}\right ) \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 {b \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 )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{2 \sqrt {-a b}\, b}+\frac {a \left (-\frac {b}{\left (a d -b c \right ) \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 {2 d \sqrt {-a b}\, \left (2 d \left (x +\frac {\sqrt {-a b}}{b}\right )-\frac {2 d \sqrt {-a b}}{b}\right )}{\left (a d -b c \right ) \left (-\frac {4 d \left (a d -b c \right )}{b}+\frac {4 d^{2} a}{b}\right ) \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 {b \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 )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{2 \sqrt {-a b}\, b}\)	\(758\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 334, normalized size of antiderivative = 4.51 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (d x^{2} + c\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, \sqrt {d x^{2} + c} x}{4 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}, \frac {{\left (d x^{2} + c\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) + 2 \, \sqrt {d x^{2} + c} x}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ] \]

[In]

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

[Out]

[-1/4*((d*x^2 + c)*sqrt(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 -
4*a^2*c*d)*x^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)*sqrt(d*x^2 + c)*sqrt(-a/(b*
c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*sqrt(d*x^2 + c)*x)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2), 1/2*((d*
x^2 + c)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^
3 + a*c*x)) + 2*sqrt(d*x^2 + c)*x)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.39 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx=-\frac {a \sqrt {d} \arctan \left (-\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{\sqrt {a b c d - a^{2} d^{2}} {\left (b c - a d\right )}} + \frac {x}{\sqrt {d x^{2} + c} {\left (b c - a d\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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