\(\int \frac {x}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\) [725]

Optimal result

Integrand size = 22, antiderivative size = 98 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \]

[Out]

1/3/(-a*d+b*c)/(d*x^2+c)^(3/2)-b^(3/2)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(5/2)+b/(-
a*d+b*c)^2/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {455, 53, 65, 214} \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}}+\frac {b}{\sqrt {c+d x^2} (b c-a d)^2}+\frac {1}{3 \left (c+d x^2\right )^{3/2} (b c-a d)} \]

[In]

Int[x/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

1/(3*(b*c - a*d)*(c + d*x^2)^(3/2)) + b/((b*c - a*d)^2*Sqrt[c + d*x^2]) - (b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d
*x^2])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

Rule 53

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] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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) (c+d x)^{5/2}} \, dx,x,x^2\right ) \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b \text {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{2 (b c-a d)} \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}+\frac {b^2 \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 (b c-a d)^2} \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}+\frac {b^2 \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d (b c-a d)^2} \\ & = \frac {1}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {b}{(b c-a d)^2 \sqrt {c+d x^2}}-\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {4 b c-a d+3 b d x^2}{3 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}} \]

[In]

Integrate[x/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

(4*b*c - a*d + 3*b*d*x^2)/(3*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + (b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt
[-(b*c) + a*d]])/(-(b*c) + a*d)^(5/2)

Maple [A] (verified)

Time = 2.96 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(-\frac {-3 b^{2} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \left (d \,x^{2}+c \right )^{\frac {3}{2}}+\sqrt {\left (a d -b c \right ) b}\, \left (\left (-3 d \,x^{2}-4 c \right ) b +a d \right )}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right )^{2}}\)	\(102\)
default	\(\text {Expression too large to display}\)	\(1390\)

[In]

int(x/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (82) = 164\).

Time = 0.33 (sec) , antiderivative size = 511, normalized size of antiderivative = 5.21 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b d^{2} x^{4} + 2 \, b c d x^{2} + b c^{2}\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (3 \, b d x^{2} + 4 \, b c - a d\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (b d^{2} x^{4} + 2 \, b c d x^{2} + b c^{2}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) + 2 \, {\left (3 \, b d x^{2} + 4 \, b c - a d\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}\right ] \]

[In]

integrate(x/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b*d^2*x^4 + 2*b*c*d*x^2 + b*c^2)*sqrt(b/(b*c - a*d))*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*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2
+ c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(3*b*d*x^2 + 4*b*c - a*d)*sqrt(d*x^2 + c))/(b^2*c^
4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2
*c*d^3)*x^2), 1/6*(3*(b*d^2*x^4 + 2*b*c*d*x^2 + b*c^2)*sqrt(-b/(b*c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)
*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) + 2*(3*b*d*x^2 + 4*b*c - a*d)*sqrt(d*x^2 + c))/(b^2*c^4
- 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*
c*d^3)*x^2)]

Sympy [A] (verification not implemented)

Time = 7.41 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.42 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b d}{2 \sqrt {c + d x^{2}} \left (a d - b c\right )^{2}} + \frac {b d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )^{2}} - \frac {d}{6 \left (c + d x^{2}\right )^{\frac {3}{2}} \left (a d - b c\right )}\right )}{d} & \text {for}\: d \neq 0 \\\begin {cases} \frac {x^{2}}{2 a c^{\frac {5}{2}}} & \text {for}\: b = 0 \\\tilde {\infty } x^{2} & \text {for}\: c^{\frac {5}{2}} = 0 \\\frac {\log {\left (2 a c^{\frac {5}{2}} + 2 b c^{\frac {5}{2}} x^{2} \right )}}{2 b c^{\frac {5}{2}}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]

[In]

integrate(x/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Piecewise((2*(b*d/(2*sqrt(c + d*x**2)*(a*d - b*c)**2) + b*d*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*sqrt
((a*d - b*c)/b)*(a*d - b*c)**2) - d/(6*(c + d*x**2)**(3/2)*(a*d - b*c)))/d, Ne(d, 0)), (Piecewise((x**2/(2*a*c
**(5/2)), Eq(b, 0)), (zoo*x**2, Eq(c**(5/2), 0)), (log(2*a*c**(5/2) + 2*b*c**(5/2)*x**2)/(2*b*c**(5/2)), True)
), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x/(b*x^2+a)/(d*x^2+c)^(5/2),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 = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.20 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {3 \, {\left (d x^{2} + c\right )} b + b c - a d}{3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \]

[In]

integrate(x/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.61 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05 \[ \int \frac {x}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )}{{\left (a\,d-b\,c\right )}^{5/2}}-\frac {\frac {1}{3\,\left (a\,d-b\,c\right )}-\frac {b\,\left (d\,x^2+c\right )}{{\left (a\,d-b\,c\right )}^2}}{{\left (d\,x^2+c\right )}^{3/2}} \]

[In]

int(x/((a + b*x^2)*(c + d*x^2)^(5/2)),x)

[Out]

(b^(3/2)*atan((b^(1/2)*(c + d*x^2)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(a*d - b*c)^(5/2)))/(a*d - b*c)^(5/2
) - (1/(3*(a*d - b*c)) - (b*(c + d*x^2))/(a*d - b*c)^2)/(c + d*x^2)^(3/2)