3.7.93 \(\int \frac {1}{x^2 (a+b x^2) \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac {b \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2}}{a c x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {480, 12, 377, 205} \begin {gather*} -\frac {b \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2}}{a c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 205

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

Rule 377

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 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 3.94, size = 177, normalized size = 2.39 \begin {gather*} -\frac {\left (\frac {d x^2}{c}+1\right ) \left (\frac {4 x^2 \left (c+d x^2\right ) (b c-a d) \, _2F_1\left (2,2;\frac {5}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )}{3 c^2 \left (a+b x^2\right )}+\frac {\left (c+2 d x^2\right ) \sin ^{-1}\left (\sqrt {\frac {x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{c \sqrt {\frac {a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}}}\right )}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((1 + (d*x^2)/c)*(((c + 2*d*x^2)*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c*Sqrt[(a*(b*c - a*d)*x^2
*(c + d*x^2))/(c^2*(a + b*x^2)^2)]) + (4*(b*c - a*d)*x^2*(c + d*x^2)*Hypergeometric2F1[2, 2, 5/2, ((b*c - a*d)
*x^2)/(c*(a + b*x^2))])/(3*c^2*(a + b*x^2))))/(x*(a + b*x^2)*Sqrt[c + d*x^2]))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.21, size = 137, normalized size = 1.85 \begin {gather*} -\frac {b \sqrt {b c-a d} \tan ^{-1}\left (\frac {b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}-\frac {b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{a^{3/2} (a d-b c)}-\frac {\sqrt {c+d x^2}}{a c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.31, size = 324, normalized size = 4.38 \begin {gather*} \left [-\frac {\sqrt {-a b c + a^{2} d} b c x \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 c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x}, -\frac {\sqrt {a b c - a^{2} d} b c x \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c}}{2 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.32, size = 111, normalized size = 1.50 \begin {gather*} d^{\frac {3}{2}} {\left (\frac {b \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}} a d} + \frac {2}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )} a d}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.02, size = 334, normalized size = 4.51 \begin {gather*} \frac {b \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 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, a}-\frac {b \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 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, a}-\frac {\sqrt {d \,x^{2}+c}}{a c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,\left (b\,x^2+a\right )\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (a + b x^{2}\right ) \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*(a + b*x**2)*sqrt(c + d*x**2)), x)

________________________________________________________________________________________