3.8.21 \(\int \frac {\sqrt {c+d x^2}}{x^4 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=147 \[ \frac {b (5 b c-4 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} \sqrt {b c-a d}}+\frac {\sqrt {c+d x^2} (15 b c-2 a d)}{6 a^3 c x}-\frac {5 \sqrt {c+d x^2}}{6 a^2 x^3}+\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \begin {gather*} \frac {\sqrt {c+d x^2} (15 b c-2 a d)}{6 a^3 c x}+\frac {b (5 b c-4 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} \sqrt {b c-a d}}-\frac {5 \sqrt {c+d x^2}}{6 a^2 x^3}+\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^2]/(x^4*(a + b*x^2)^2),x]

[Out]

(-5*Sqrt[c + d*x^2])/(6*a^2*x^3) + ((15*b*c - 2*a*d)*Sqrt[c + d*x^2])/(6*a^3*c*x) + Sqrt[c + d*x^2]/(2*a*x^3*(
a + b*x^2)) + (b*(5*b*c - 4*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2)*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^4 \left (a+b x^2\right )^2} \, dx &=\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )}-\frac {\int \frac {-5 c-4 d x^2}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a}\\ &=-\frac {5 \sqrt {c+d x^2}}{6 a^2 x^3}+\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )}+\frac {\int \frac {-c (15 b c-2 a d)-10 b c d x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^2 c}\\ &=-\frac {5 \sqrt {c+d x^2}}{6 a^2 x^3}+\frac {(15 b c-2 a d) \sqrt {c+d x^2}}{6 a^3 c x}+\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )}-\frac {\int -\frac {3 b c^2 (5 b c-4 a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^3 c^2}\\ &=-\frac {5 \sqrt {c+d x^2}}{6 a^2 x^3}+\frac {(15 b c-2 a d) \sqrt {c+d x^2}}{6 a^3 c x}+\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )}+\frac {(b (5 b c-4 a d)) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a^3}\\ &=-\frac {5 \sqrt {c+d x^2}}{6 a^2 x^3}+\frac {(15 b c-2 a d) \sqrt {c+d x^2}}{6 a^3 c x}+\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )}+\frac {(b (5 b c-4 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^3}\\ &=-\frac {5 \sqrt {c+d x^2}}{6 a^2 x^3}+\frac {(15 b c-2 a d) \sqrt {c+d x^2}}{6 a^3 c x}+\frac {\sqrt {c+d x^2}}{2 a x^3 \left (a+b x^2\right )}+\frac {b (5 b c-4 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} \sqrt {b c-a d}}\\ \end {align*}

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Mathematica [A]  time = 5.14, size = 120, normalized size = 0.82 \begin {gather*} \frac {b (5 b c-4 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} \sqrt {b c-a d}}+\frac {\sqrt {c+d x^2} \left (3 b x^2 \left (\frac {b x^2}{a+b x^2}+4\right )-\frac {2 a \left (c+d x^2\right )}{c}\right )}{6 a^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^2]/(x^4*(a + b*x^2)^2),x]

[Out]

(Sqrt[c + d*x^2]*((-2*a*(c + d*x^2))/c + 3*b*x^2*(4 + (b*x^2)/(a + b*x^2))))/(6*a^3*x^3) + (b*(5*b*c - 4*a*d)*
ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2)*Sqrt[b*c - a*d])

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IntegrateAlgebraic [A]  time = 0.92, size = 170, normalized size = 1.16 \begin {gather*} \frac {\sqrt {b c-a d} \left (5 b^2 c-4 a b d\right ) \tan ^{-1}\left (\frac {a \sqrt {d}-b x \sqrt {c+d x^2}+b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 a^{7/2} (a d-b c)}+\frac {\sqrt {c+d x^2} \left (-2 a^2 c-2 a^2 d x^2+10 a b c x^2-2 a b d x^4+15 b^2 c x^4\right )}{6 a^3 c x^3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[c + d*x^2]/(x^4*(a + b*x^2)^2),x]

[Out]

(Sqrt[c + d*x^2]*(-2*a^2*c + 10*a*b*c*x^2 - 2*a^2*d*x^2 + 15*b^2*c*x^4 - 2*a*b*d*x^4))/(6*a^3*c*x^3*(a + b*x^2
)) + (Sqrt[b*c - a*d]*(5*b^2*c - 4*a*b*d)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^2 - b*x*Sqrt[c + d*x^2])/(Sqrt[a]*Sq
rt[b*c - a*d])])/(2*a^(7/2)*(-(b*c) + a*d))

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fricas [B]  time = 1.32, size = 602, normalized size = 4.10 \begin {gather*} \left [\frac {3 \, {\left ({\left (5 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{5} + {\left (5 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{3}\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^{3} b c^{2} - 2 \, a^{4} c d - {\left (15 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{4} - 2 \, {\left (5 \, a^{2} b^{2} c^{2} - 6 \, a^{3} b c d + a^{4} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left ({\left (a^{4} b^{2} c^{2} - a^{5} b c d\right )} x^{5} + {\left (a^{5} b c^{2} - a^{6} c d\right )} x^{3}\right )}}, \frac {3 \, {\left ({\left (5 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{5} + {\left (5 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{3}\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^{3} b c^{2} - 2 \, a^{4} c d - {\left (15 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{4} - 2 \, {\left (5 \, a^{2} b^{2} c^{2} - 6 \, a^{3} b c d + a^{4} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left ({\left (a^{4} b^{2} c^{2} - a^{5} b c d\right )} x^{5} + {\left (a^{5} b c^{2} - a^{6} c d\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)/x^4/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/24*(3*((5*b^3*c^2 - 4*a*b^2*c*d)*x^5 + (5*a*b^2*c^2 - 4*a^2*b*c*d)*x^3)*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^3*b*c^2 - 2*a^4*c*d - (15*a*b^3*c^2 - 17*
a^2*b^2*c*d + 2*a^3*b*d^2)*x^4 - 2*(5*a^2*b^2*c^2 - 6*a^3*b*c*d + a^4*d^2)*x^2)*sqrt(d*x^2 + c))/((a^4*b^2*c^2
- a^5*b*c*d)*x^5 + (a^5*b*c^2 - a^6*c*d)*x^3), 1/12*(3*((5*b^3*c^2 - 4*a*b^2*c*d)*x^5 + (5*a*b^2*c^2 - 4*a^2*
b*c*d)*x^3)*sqrt(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^3*b*c^2 - 2*a^4*c*d - (15*a*b^3*c^2 - 17*a^2*b^2*c*d +
2*a^3*b*d^2)*x^4 - 2*(5*a^2*b^2*c^2 - 6*a^3*b*c*d + a^4*d^2)*x^2)*sqrt(d*x^2 + c))/((a^4*b^2*c^2 - a^5*b*c*d)*
x^5 + (a^5*b*c^2 - a^6*c*d)*x^3)]

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giac [B]  time = 4.52, size = 361, normalized size = 2.46 \begin {gather*} -\frac {{\left (5 \, b^{2} c \sqrt {d} - 4 \, a b 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^{3}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b d^{\frac {3}{2}} - b^{2} c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} a^{3}} - \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} \sqrt {d} + 6 \, b c^{3} \sqrt {d} - a c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)/x^4/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*(5*b^2*c*sqrt(d) - 4*a*b*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^3) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c*sqrt(d) - 2*(sqrt(d)*x -
sqrt(d*x^2 + c))^2*a*b*d^(3/2) - b^2*c^2*sqrt(d))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d
*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*a^3) - 2/3*(6*(sqrt(d)*x - sqrt(d*x^2 + c))^
4*b*c*sqrt(d) - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*d^(3/2) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^2*sqrt(d)
+ 6*b*c^3*sqrt(d) - a*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a^3)

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maple [B]  time = 0.02, size = 2667, normalized size = 18.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(1/2)/x^4/(b*x^2+a)^2,x)

[Out]

5/4*b/a^3*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))-1/3/a^2/c/x^3*(d*x^2+c)^(3/2)+5/4*b^2/a^3/(-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)+5/4*b/a^3*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))-5/
4*b^2/a^3/(-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)-2*b/a^
3*d^(1/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+5/4*b^2/a^3/(-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-5/4*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))*d+1/4*b^2/a^3*d^(1/2)/(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))
*c+1/4*b/a^3*(-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)+1/4/a^2*(-a*b)^(1/2)*d^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))+1/4*b^2/a^3*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)*x+1/4*b^2/a^3*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)*x+1/4*b^2/a^3*d^(1/2)/(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))*c+5/4*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))*d-5/4*b^2/a^3/(-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-2*b/a^3*d/c*x*(d*x^
2+c)^(1/2)-1/4*b/a^3*(-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)-1/4/a^2*(-a*b)^(1/2)*d^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))-1/4*b^2/a^3/(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)^(3/2)-1/4*b/a^2*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))+2*b/a^3/c/x*
(d*x^2+c)^(3/2)-1/4*b^2/a^3/(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)^(3/2)-1/4*b/a^2*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))-1/4*b/a^3*(-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*b/a^3*(-
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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d x^{2} + c}}{{\left (b x^{2} + a\right )}^{2} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)/x^4/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate(sqrt(d*x^2 + c)/((b*x^2 + a)^2*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d\,x^2+c}}{x^4\,{\left (b\,x^2+a\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)^(1/2)/(x^4*(a + b*x^2)^2),x)

[Out]

int((c + d*x^2)^(1/2)/(x^4*(a + b*x^2)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d x^{2}}}{x^{4} \left (a + b x^{2}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(1/2)/x**4/(b*x**2+a)**2,x)

[Out]

Integral(sqrt(c + d*x**2)/(x**4*(a + b*x**2)**2), x)

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