3.8.57 \(\int \frac {1}{x^4 (a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=277 \[ \frac {b^3 (5 b c-8 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{5/2}}-\frac {\sqrt {c+d x^2} \left (8 a^2 d^2-4 a b c d+5 b^2 c^2\right )}{6 a^2 c^2 x^3 (b c-a d)^2}+\frac {\sqrt {c+d x^2} \left (16 a^3 d^3-8 a^2 b c d^2-14 a b^2 c^2 d+15 b^3 c^3\right )}{6 a^3 c^3 x (b c-a d)^2}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}+\frac {d (2 a d+b c)}{2 a c x^3 \sqrt {c+d x^2} (b c-a d)^2} \]

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Rubi [A]  time = 0.40, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {472, 579, 583, 12, 377, 205} \begin {gather*} \frac {\sqrt {c+d x^2} \left (-8 a^2 b c d^2+16 a^3 d^3-14 a b^2 c^2 d+15 b^3 c^3\right )}{6 a^3 c^3 x (b c-a d)^2}-\frac {\sqrt {c+d x^2} \left (8 a^2 d^2-4 a b c d+5 b^2 c^2\right )}{6 a^2 c^2 x^3 (b c-a d)^2}+\frac {b^3 (5 b c-8 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{5/2}}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}+\frac {d (2 a d+b c)}{2 a c x^3 \sqrt {c+d x^2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(b*c + 2*a*d))/(2*a*c*(b*c - a*d)^2*x^3*Sqrt[c + d*x^2]) + b/(2*a*(b*c - a*d)*x^3*(a + b*x^2)*Sqrt[c + d*x^
2]) - ((5*b^2*c^2 - 4*a*b*c*d + 8*a^2*d^2)*Sqrt[c + d*x^2])/(6*a^2*c^2*(b*c - a*d)^2*x^3) + ((15*b^3*c^3 - 14*
a*b^2*c^2*d - 8*a^2*b*c*d^2 + 16*a^3*d^3)*Sqrt[c + d*x^2])/(6*a^3*c^3*(b*c - a*d)^2*x) + (b^3*(5*b*c - 8*a*d)*
ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2)*(b*c - a*d)^(5/2))

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 472

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

Rule 579

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

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 {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx &=\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\int \frac {-5 b c+2 a d-6 b d x^2}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{2 a (b c-a d)}\\ &=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 x^3 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\int \frac {-5 b^2 c^2+4 a b c d-8 a^2 d^2-4 b d (b c+2 a d) x^2}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a c (b c-a d)^2}\\ &=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 x^3 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\left (5 b^2 c^2-4 a b c d+8 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^2 c^2 (b c-a d)^2 x^3}+\frac {\int \frac {-15 b^3 c^3+14 a b^2 c^2 d+8 a^2 b c d^2-16 a^3 d^3-2 b d \left (5 b^2 c^2-4 a b c d+8 a^2 d^2\right ) x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^2 c^2 (b c-a d)^2}\\ &=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 x^3 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\left (5 b^2 c^2-4 a b c d+8 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^2 c^2 (b c-a d)^2 x^3}+\frac {\left (15 b^3 c^3-14 a b^2 c^2 d-8 a^2 b c d^2+16 a^3 d^3\right ) \sqrt {c+d x^2}}{6 a^3 c^3 (b c-a d)^2 x}-\frac {\int -\frac {3 b^3 c^3 (5 b c-8 a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 a^3 c^3 (b c-a d)^2}\\ &=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 x^3 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\left (5 b^2 c^2-4 a b c d+8 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^2 c^2 (b c-a d)^2 x^3}+\frac {\left (15 b^3 c^3-14 a b^2 c^2 d-8 a^2 b c d^2+16 a^3 d^3\right ) \sqrt {c+d x^2}}{6 a^3 c^3 (b c-a d)^2 x}+\frac {\left (b^3 (5 b c-8 a d)\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a^3 (b c-a d)^2}\\ &=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 x^3 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\left (5 b^2 c^2-4 a b c d+8 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^2 c^2 (b c-a d)^2 x^3}+\frac {\left (15 b^3 c^3-14 a b^2 c^2 d-8 a^2 b c d^2+16 a^3 d^3\right ) \sqrt {c+d x^2}}{6 a^3 c^3 (b c-a d)^2 x}+\frac {\left (b^3 (5 b c-8 a d)\right ) \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 (b c-a d)^2}\\ &=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 x^3 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\left (5 b^2 c^2-4 a b c d+8 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^2 c^2 (b c-a d)^2 x^3}+\frac {\left (15 b^3 c^3-14 a b^2 c^2 d-8 a^2 b c d^2+16 a^3 d^3\right ) \sqrt {c+d x^2}}{6 a^3 c^3 (b c-a d)^2 x}+\frac {b^3 (5 b c-8 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 5.49, size = 167, normalized size = 0.60 \begin {gather*} \frac {b^3 (5 b c-8 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{5/2}}+\sqrt {c+d x^2} \left (\frac {\frac {b^4 x}{2 \left (a+b x^2\right ) (b c-a d)^2}+\frac {2 b}{c^2 x}}{a^3}-\frac {c-5 d x^2}{3 a^2 c^3 x^3}+\frac {d^4 x}{c^3 \left (c+d x^2\right ) (b c-a d)^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^2]*(-1/3*(c - 5*d*x^2)/(a^2*c^3*x^3) + (d^4*x)/(c^3*(b*c - a*d)^2*(c + d*x^2)) + ((2*b)/(c^2*x) +
(b^4*x)/(2*(b*c - a*d)^2*(a + b*x^2)))/a^3) + (b^3*(5*b*c - 8*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c
+ d*x^2])])/(2*a^(7/2)*(b*c - a*d)^(5/2))

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IntegrateAlgebraic [A]  time = 1.93, size = 324, normalized size = 1.17 \begin {gather*} \frac {\left (8 a b^3 d-5 b^4 c\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} (b c-a d)^{5/2}}+\frac {-2 a^4 c^2 d^2+8 a^4 c d^3 x^2+16 a^4 d^4 x^4+4 a^3 b c^3 d-6 a^3 b c^2 d^2 x^2+16 a^3 b d^4 x^6-2 a^2 b^2 c^4-12 a^2 b^2 c^3 d x^2-18 a^2 b^2 c^2 d^2 x^4-8 a^2 b^2 c d^3 x^6+10 a b^3 c^4 x^2-4 a b^3 c^3 d x^4-14 a b^3 c^2 d^2 x^6+15 b^4 c^4 x^4+15 b^4 c^3 d x^6}{6 a^3 c^3 x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (a d-b c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*b^2*c^4 + 4*a^3*b*c^3*d - 2*a^4*c^2*d^2 + 10*a*b^3*c^4*x^2 - 12*a^2*b^2*c^3*d*x^2 - 6*a^3*b*c^2*d^2*x^
2 + 8*a^4*c*d^3*x^2 + 15*b^4*c^4*x^4 - 4*a*b^3*c^3*d*x^4 - 18*a^2*b^2*c^2*d^2*x^4 + 16*a^4*d^4*x^4 + 15*b^4*c^
3*d*x^6 - 14*a*b^3*c^2*d^2*x^6 - 8*a^2*b^2*c*d^3*x^6 + 16*a^3*b*d^4*x^6)/(6*a^3*c^3*(-(b*c) + a*d)^2*x^3*(a +
b*x^2)*Sqrt[c + d*x^2]) + ((-5*b^4*c + 8*a*b^3*d)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^2 - b*x*Sqrt[c + d*x^2])/(Sq
rt[a]*Sqrt[b*c - a*d])])/(2*a^(7/2)*(b*c - a*d)^(5/2))

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fricas [B]  time = 3.02, size = 1252, normalized size = 4.52

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/24*(3*((5*b^5*c^4*d - 8*a*b^4*c^3*d^2)*x^7 + (5*b^5*c^5 - 3*a*b^4*c^4*d - 8*a^2*b^3*c^3*d^2)*x^5 + (5*a*b^4
*c^5 - 8*a^2*b^3*c^4*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^3*c^5 - 6*a^4*b^2*c^4*d + 6*a^5*b*c^3*d^2 - 2*a^6*c^2*d^3 - (15*a*b^5*c^4*d - 29*a^
2*b^4*c^3*d^2 + 6*a^3*b^3*c^2*d^3 + 24*a^4*b^2*c*d^4 - 16*a^5*b*d^5)*x^6 - (15*a*b^5*c^5 - 19*a^2*b^4*c^4*d -
14*a^3*b^3*c^3*d^2 + 18*a^4*b^2*c^2*d^3 + 16*a^5*b*c*d^4 - 16*a^6*d^5)*x^4 - 2*(5*a^2*b^4*c^5 - 11*a^3*b^3*c^4
*d + 3*a^4*b^2*c^3*d^2 + 7*a^5*b*c^2*d^3 - 4*a^6*c*d^4)*x^2)*sqrt(d*x^2 + c))/((a^4*b^4*c^6*d - 3*a^5*b^3*c^5*
d^2 + 3*a^6*b^2*c^4*d^3 - a^7*b*c^3*d^4)*x^7 + (a^4*b^4*c^7 - 2*a^5*b^3*c^6*d + 2*a^7*b*c^4*d^3 - a^8*c^3*d^4)
*x^5 + (a^5*b^3*c^7 - 3*a^6*b^2*c^6*d + 3*a^7*b*c^5*d^2 - a^8*c^4*d^3)*x^3), 1/12*(3*((5*b^5*c^4*d - 8*a*b^4*c
^3*d^2)*x^7 + (5*b^5*c^5 - 3*a*b^4*c^4*d - 8*a^2*b^3*c^3*d^2)*x^5 + (5*a*b^4*c^5 - 8*a^2*b^3*c^4*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^3*c^5 - 6*a^4*b^2*c^4*d + 6*a^5*b*c^3*d^2 - 2*a^6*c^2*d^3 - (15*a*b^
5*c^4*d - 29*a^2*b^4*c^3*d^2 + 6*a^3*b^3*c^2*d^3 + 24*a^4*b^2*c*d^4 - 16*a^5*b*d^5)*x^6 - (15*a*b^5*c^5 - 19*a
^2*b^4*c^4*d - 14*a^3*b^3*c^3*d^2 + 18*a^4*b^2*c^2*d^3 + 16*a^5*b*c*d^4 - 16*a^6*d^5)*x^4 - 2*(5*a^2*b^4*c^5 -
11*a^3*b^3*c^4*d + 3*a^4*b^2*c^3*d^2 + 7*a^5*b*c^2*d^3 - 4*a^6*c*d^4)*x^2)*sqrt(d*x^2 + c))/((a^4*b^4*c^6*d -
3*a^5*b^3*c^5*d^2 + 3*a^6*b^2*c^4*d^3 - a^7*b*c^3*d^4)*x^7 + (a^4*b^4*c^7 - 2*a^5*b^3*c^6*d + 2*a^7*b*c^4*d^3
- a^8*c^3*d^4)*x^5 + (a^5*b^3*c^7 - 3*a^6*b^2*c^6*d + 3*a^7*b*c^5*d^2 - a^8*c^4*d^3)*x^3)]

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giac [A]  time = 5.81, size = 486, normalized size = 1.75 \begin {gather*} \frac {d^{4} x}{{\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt {d x^{2} + c}} - \frac {{\left (5 \, b^{4} c \sqrt {d} - 8 \, a b^{3} 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 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{4} c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{3} d^{\frac {3}{2}} - b^{4} c^{2} \sqrt {d}}{{\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} {\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 )}} - \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} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c d^{\frac {3}{2}} + 6 \, b c^{3} \sqrt {d} + 5 \, 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} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

d^4*x/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2)*sqrt(d*x^2 + c)) - 1/2*(5*b^4*c*sqrt(d) - 8*a*b^3*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))/((a^3*b^2*c^2 - 2*a^4*b*c*d +
a^5*d^2)*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^4*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2
+ c))^2*a*b^3*d^(3/2) - b^4*c^2*sqrt(d))/((a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)*((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)) - 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) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*d^(3/2) + 6*b*c^3*sqrt(d) + 5*a*c^2*d^(3/2)
)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a^3*c^2)

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maple [B]  time = 0.02, size = 1608, normalized size = 5.81

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x)

[Out]

5/4*b^3/a^3/(-a*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)+3/4*b^2/a^3/(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-5/4*b^3/a^3/(-a*b)^(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))-1/4*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)^(1/2)-3/4*b^2/a^3*(-a*b)^(1/2)*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*b^2/a^2*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*b^2/a^3*(-a*b)^(1/2)*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))-5/4*b^3/a^3/(-a*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)+3/4*b^2/a^3/(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+5/4*b^3/a^3/(-a*b)^(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))-1/4*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)^(1/2)+3/4*b^
2/a^3*(-a*b)^(1/2)*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*b^2/a^2*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*b^2/a^3*(-a*b)^(1/2)*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))+2*b/a^3/c/x/(d*x^2+c)^(1/2)+4*b/a^3*d/c^2*x/(d*x^2+c)^(1/2)-1/3/a^2/c/x^3/(d*x^2+c
)^(1/2)+4/3/a^2*d/c^2/x/(d*x^2+c)^(1/2)+8/3/a^2*d^2/c^3*x/(d*x^2+c)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^2*(d*x^2 + c)^(3/2)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^2)^2*(c + d*x^2)^(3/2)),x)

[Out]

int(1/(x^4*(a + b*x^2)^2*(c + d*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Integral(1/(x**4*(a + b*x**2)**2*(c + d*x**2)**(3/2)), x)

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