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

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {469, 583, 12, 377, 205} \[ -\frac {(3 b c-2 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{5/2} \sqrt {b c-a d}}-\frac {3 \sqrt {c+d x^2}}{2 a^2 x}+\frac {\sqrt {c+d x^2}}{2 a x \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[c + d*x^2])/(2*a^2*x) + Sqrt[c + d*x^2]/(2*a*x*(a + b*x^2)) - ((3*b*c - 2*a*d)*ArcTan[(Sqrt[b*c - a*d
]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(5/2)*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 469

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)/(a*e*n*(p + 1)), x] + Dist[1/(a*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)
^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m + n*(p + 1) + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{
a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c
, d, e, m, n, p, q, x]

Rule 583

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

Rubi steps

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

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Mathematica [A]  time = 5.07, size = 101, normalized size = 0.89 \[ \frac {(2 a d-3 b c) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{5/2} \sqrt {b c-a d}}+\left (-\frac {b x}{2 a^2 \left (a+b x^2\right )}-\frac {1}{a^2 x}\right ) \sqrt {c+d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.25, size = 458, normalized size = 4.05 \[ \left [\frac {{\left ({\left (3 \, b^{2} c - 2 \, a b d\right )} x^{3} + {\left (3 \, a b c - 2 \, a^{2} d\right )} x\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (2 \, a^{2} b c - 2 \, a^{3} d + 3 \, {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left ({\left (a^{3} b^{2} c - a^{4} b d\right )} x^{3} + {\left (a^{4} b c - a^{5} d\right )} x\right )}}, -\frac {{\left ({\left (3 \, b^{2} c - 2 \, a b d\right )} x^{3} + {\left (3 \, a b c - 2 \, a^{2} d\right )} x\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (2 \, a^{2} b c - 2 \, a^{3} d + 3 \, {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left ({\left (a^{3} b^{2} c - a^{4} b d\right )} x^{3} + {\left (a^{4} b c - a^{5} d\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(((3*b^2*c - 2*a*b*d)*x^3 + (3*a*b*c - 2*a^2*d)*x)*sqrt(-a*b*c + a^2*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*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*(2*a^2*b*c - 2*a^3*d + 3*(a*b^2*c - a^2*b*d)*x^2)*sqrt(d*x^2 + c)
)/((a^3*b^2*c - a^4*b*d)*x^3 + (a^4*b*c - a^5*d)*x), -1/4*(((3*b^2*c - 2*a*b*d)*x^3 + (3*a*b*c - 2*a^2*d)*x)*s
qrt(a*b*c - a^2*d)*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*(2*a^2*b*c - 2*a^3*d + 3*(a*b^2*c - a^2*b*d)*x^2)*sqrt(d*x^2 + c))/((a^3*
b^2*c - a^4*b*d)*x^3 + (a^4*b*c - a^5*d)*x)]

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giac [B]  time = 4.07, size = 329, normalized size = 2.91 \[ \frac {{\left (3 \, b c \sqrt {d} - 2 \, a d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} a^{2}} + \frac {3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} \sqrt {d} + 10 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c d^{\frac {3}{2}} + 3 \, b c^{3} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c d - b c^{3}\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*b*c*sqrt(d) - 2*a*d^(3/2))*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^2) + (3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c*sqrt(d) - 2*(sqrt(d)*x - sqrt
(d*x^2 + c))^4*a*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^2*sqrt(d) + 10*(sqrt(d)*x - sqrt(d*x^2 + c))^
2*a*c*d^(3/2) + 3*b*c^3*sqrt(d))/(((sqrt(d)*x - sqrt(d*x^2 + c))^6*b - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c +
4*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*d + 3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^2 - 4*(sqrt(d)*x - sqrt(d*x^2 +
c))^2*a*c*d - b*c^3)*a^2)

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maple [B]  time = 0.02, size = 2618, normalized size = 23.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a*d^(3/2)/(a*d-b*c)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(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))-3/4*b/a^2/(-a*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)-1/a^2/c/x*(d*x^2+c)^(3/2)+1/4/a*d^(3/2)/(a*d-b*c)*ln(((x+(-a*b)^(1/2)
/b)*d-(-a*b)^(1/2)/b*d)/d^(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))+3/4*b/a^2/(-a*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)-1/
4/a^2*d^(1/2)/(a*d-b*c)*b*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(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))*c-1/4/a^2*d/(a*d-b*c)*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)*x-1/4/a^2*d^(1/2)/(a*d-b*c)*b*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2
)/b*d)/d^(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))*c+3/4*b/a^2/(
-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))*c-3/4*
b/a^2/(-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)
)*c-1/4/a^2*d/(a*d-b*c)*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)*x+1
/a^2*d/c*x*(d*x^2+c)^(1/2)+1/4/a^2/(a*d-b*c)*b/(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)^(3/2)+1/4/a^2*(-a*b)^(1/2)*d/(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))*c-1/4/a^2*(-a*b)^(1/2)*d/(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)+3/4/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+1/4/a^2/(a*d-b*c)*b/(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)^(3/2)+1/4/a^2*(-a*b)^(1/2)*d/(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)-3/4/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-3/4/a^2*d^(1/2)*ln(((x+(-a*b)^(
1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(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))-3/4/a^2*d^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(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))+1/a^2*d^(1/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/4/a*(-a*b)^(1/2)*
d^2/(a*d-b*c)/b/(-(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
/4/a^2*(-a*b)^(1/2)*d/(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))*c-1/4/a*(-a*b)^(1/2)*d^2/(a*d-b*c)/b/(-(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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + c}}{{\left (b x^{2} + a\right )}^{2} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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