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

Optimal. Leaf size=113 \[ -\frac {3 d}{2 \sqrt {c+d x^2} (b c-a d)^2}-\frac {1}{2 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}+\frac {3 \sqrt {b} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 (b c-a d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {444, 51, 63, 208} \[ -\frac {3 d}{2 \sqrt {c+d x^2} (b c-a d)^2}-\frac {1}{2 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}+\frac {3 \sqrt {b} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 (b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

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 63

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 208

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

Rule 444

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*} \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {(3 d) \operatorname {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 (b c-a d)}\\ &=-\frac {3 d}{2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {(3 b d) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 (b c-a d)^2}\\ &=-\frac {3 d}{2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 (b c-a d)^2}\\ &=-\frac {3 d}{2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {3 \sqrt {b} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.46 \[ -\frac {d \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {b \left (d x^2+c\right )}{a d-b c}\right )}{\sqrt {c+d x^2} (a d-b c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Hypergeometric2F1[-1/2, 2, 1/2, -((b*(c + d*x^2))/(-(b*c) + a*d))])/((-(b*c) + a*d)^2*Sqrt[c + d*x^2]))

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fricas [B]  time = 0.62, size = 537, normalized size = 4.75 \[ \left [\frac {3 \, {\left (b d^{2} x^{4} + a c d + {\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (3 \, b d x^{2} + b c + 2 \, a d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}, -\frac {3 \, {\left (b d^{2} x^{4} + a c d + {\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) + 2 \, {\left (3 \, b d x^{2} + b c + 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.39, size = 153, normalized size = 1.35 \[ -\frac {3 \, b d \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} - \frac {3 \, {\left (d x^{2} + c\right )} b d - 2 \, b c d + 2 \, a d^{2}}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left ({\left (d x^{2} + c\right )}^{\frac {3}{2}} b - \sqrt {d x^{2} + c} b c + \sqrt {d x^{2} + c} a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 989, normalized size = 8.75 \[ \frac {3 d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{4 \left (a d -b c \right )^{2} \sqrt {-\frac {a d -b c}{b}}}+\frac {3 d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{4 \left (a d -b c \right )^{2} \sqrt {-\frac {a d -b c}{b}}}+\frac {3 \sqrt {-a b}\, d^{2} x}{4 \left (a d -b c \right )^{2} \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, b c}-\frac {3 \sqrt {-a b}\, d^{2} x}{4 \left (a d -b c \right )^{2} \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, b c}-\frac {3 d}{4 \left (a d -b c \right )^{2} \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}-\frac {3 d}{4 \left (a d -b c \right )^{2} \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}-\frac {\sqrt {-a b}\, d x}{2 \left (a d -b c \right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a b c}+\frac {\sqrt {-a b}\, d x}{2 \left (a d -b c \right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a b c}-\frac {\sqrt {-a b}}{4 \left (a d -b c \right ) \left (x -\frac {\sqrt {-a b}}{b}\right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a b}+\frac {\sqrt {-a b}}{4 \left (a d -b c \right ) \left (x +\frac {\sqrt {-a b}}{b}\right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a)^2/(d*x^2+c)^(3/2),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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.99, size = 130, normalized size = 1.15 \[ -\frac {\frac {d}{a\,d-b\,c}+\frac {3\,b\,d\,\left (d\,x^2+c\right )}{2\,{\left (a\,d-b\,c\right )}^2}}{b\,{\left (d\,x^2+c\right )}^{3/2}+\sqrt {d\,x^2+c}\,\left (a\,d-b\,c\right )}-\frac {3\,\sqrt {b}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )}{2\,{\left (a\,d-b\,c\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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