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

Optimal. Leaf size=72 \[ \frac {1}{\sqrt {c+d x^2} (b c-a d)}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {444, 51, 63, 208} \begin {gather*} \frac {1}{\sqrt {c+d x^2} (b c-a d)}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.02, size = 50, normalized size = 0.69 \begin {gather*} -\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^2+c\right )}{b c-a d}\right )}{\sqrt {c+d x^2} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 81, normalized size = 1.12 \begin {gather*} \frac {1}{\sqrt {c+d x^2} (b c-a d)}+\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{(a d-b c)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 323, normalized size = 4.49 \begin {gather*} \left [-\frac {{\left (d x^{2} + c\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 \, \sqrt {d x^{2} + c}}{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 {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 \, \sqrt {d x^{2} + c}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((d*x^2 + c)*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*sqrt(d*x^2 + c))/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2), 1/2*((d*x^2 + c)*s
qrt(-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*sqrt(d*x^2 + c))/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)]

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giac [A]  time = 0.37, size = 71, normalized size = 0.99 \begin {gather*} \frac {b \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} {\left (b c - a d\right )}} + \frac {1}{\sqrt {d x^{2} + c} {\left (b c - a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 618, normalized size = 8.58 \begin {gather*} \frac {\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 )}{2 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}+\frac {\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 )}{2 \left (a d -b c \right ) \sqrt {-\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}}\, 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}}\, b c}-\frac {1}{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}}}-\frac {1}{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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a)/(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 = 0.74, size = 61, normalized size = 0.85 \begin {gather*} -\frac {1}{\sqrt {d\,x^2+c}\,\left (a\,d-b\,c\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}}{\sqrt {a\,d-b\,c}}\right )}{{\left (a\,d-b\,c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 19.96, size = 61, normalized size = 0.85 \begin {gather*} - \frac {1}{\sqrt {c + d x^{2}} \left (a d - b c\right )} - \frac {\operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{\sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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