3.8.10 \(\int \frac {1}{x^2 (a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=178 \[ -\frac {b^3 \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}}-\frac {\sqrt {c+d x^2} (b c-4 a d) (3 b c-2 a d)}{3 a c^3 x (b c-a d)^2}-\frac {d (7 b c-4 a d)}{3 c^2 x \sqrt {c+d x^2} (b c-a d)^2}-\frac {d}{3 c x \left (c+d x^2\right )^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.24, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {b^3 \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}}-\frac {\sqrt {c+d x^2} (b c-4 a d) (3 b c-2 a d)}{3 a c^3 x (b c-a d)^2}-\frac {d (7 b c-4 a d)}{3 c^2 x \sqrt {c+d x^2} (b c-a d)^2}-\frac {d}{3 c x \left (c+d x^2\right )^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

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Mathematica [A]  time = 5.27, size = 143, normalized size = 0.80 \begin {gather*} \frac {\sqrt {c+d x^2} \left (\frac {d^2 x^2 (8 b c-5 a d)}{\left (c+d x^2\right ) (b c-a d)^2}+\frac {c d^2 x^2}{\left (c+d x^2\right )^2 (b c-a d)}-\frac {3}{a}\right )}{3 c^3 x}-\frac {b^3 \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.61, size = 241, normalized size = 1.35 \begin {gather*} \frac {b^3 \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} (b c-a d)^{5/2}}+\frac {-3 a^2 c^2 d^2-12 a^2 c d^3 x^2-8 a^2 d^4 x^4+6 a b c^3 d+21 a b c^2 d^2 x^2+14 a b c d^3 x^4-3 b^2 c^4-6 b^2 c^3 d x^2-3 b^2 c^2 d^2 x^4}{3 a c^3 x \left (c+d x^2\right )^{3/2} (a d-b c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

(-3*b^2*c^4 + 6*a*b*c^3*d - 3*a^2*c^2*d^2 - 6*b^2*c^3*d*x^2 + 21*a*b*c^2*d^2*x^2 - 12*a^2*c*d^3*x^2 - 3*b^2*c^
2*d^2*x^4 + 14*a*b*c*d^3*x^4 - 8*a^2*d^4*x^4)/(3*a*c^3*(-(b*c) + a*d)^2*x*(c + d*x^2)^(3/2)) + (b^3*ArcTan[(Sq
rt[a]*Sqrt[d])/Sqrt[b*c - a*d] + (b*Sqrt[d]*x^2)/(Sqrt[a]*Sqrt[b*c - a*d]) - (b*x*Sqrt[c + d*x^2])/(Sqrt[a]*Sq
rt[b*c - a*d])])/(a^(3/2)*(b*c - a*d)^(5/2))

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fricas [B]  time = 2.17, size = 934, normalized size = 5.25 \begin {gather*} \left [-\frac {3 \, {\left (b^{3} c^{3} d^{2} x^{5} + 2 \, b^{3} c^{4} d x^{3} + b^{3} c^{5} 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 (3 \, a b^{3} c^{5} - 9 \, a^{2} b^{2} c^{4} d + 9 \, a^{3} b c^{3} d^{2} - 3 \, a^{4} c^{2} d^{3} + {\left (3 \, a b^{3} c^{3} d^{2} - 17 \, a^{2} b^{2} c^{2} d^{3} + 22 \, a^{3} b c d^{4} - 8 \, a^{4} d^{5}\right )} x^{4} + 3 \, {\left (2 \, a b^{3} c^{4} d - 9 \, a^{2} b^{2} c^{3} d^{2} + 11 \, a^{3} b c^{2} d^{3} - 4 \, a^{4} c d^{4}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left ({\left (a^{2} b^{3} c^{6} d^{2} - 3 \, a^{3} b^{2} c^{5} d^{3} + 3 \, a^{4} b c^{4} d^{4} - a^{5} c^{3} d^{5}\right )} x^{5} + 2 \, {\left (a^{2} b^{3} c^{7} d - 3 \, a^{3} b^{2} c^{6} d^{2} + 3 \, a^{4} b c^{5} d^{3} - a^{5} c^{4} d^{4}\right )} x^{3} + {\left (a^{2} b^{3} c^{8} - 3 \, a^{3} b^{2} c^{7} d + 3 \, a^{4} b c^{6} d^{2} - a^{5} c^{5} d^{3}\right )} x\right )}}, -\frac {3 \, {\left (b^{3} c^{3} d^{2} x^{5} + 2 \, b^{3} c^{4} d x^{3} + b^{3} c^{5} 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 (3 \, a b^{3} c^{5} - 9 \, a^{2} b^{2} c^{4} d + 9 \, a^{3} b c^{3} d^{2} - 3 \, a^{4} c^{2} d^{3} + {\left (3 \, a b^{3} c^{3} d^{2} - 17 \, a^{2} b^{2} c^{2} d^{3} + 22 \, a^{3} b c d^{4} - 8 \, a^{4} d^{5}\right )} x^{4} + 3 \, {\left (2 \, a b^{3} c^{4} d - 9 \, a^{2} b^{2} c^{3} d^{2} + 11 \, a^{3} b c^{2} d^{3} - 4 \, a^{4} c d^{4}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left ({\left (a^{2} b^{3} c^{6} d^{2} - 3 \, a^{3} b^{2} c^{5} d^{3} + 3 \, a^{4} b c^{4} d^{4} - a^{5} c^{3} d^{5}\right )} x^{5} + 2 \, {\left (a^{2} b^{3} c^{7} d - 3 \, a^{3} b^{2} c^{6} d^{2} + 3 \, a^{4} b c^{5} d^{3} - a^{5} c^{4} d^{4}\right )} x^{3} + {\left (a^{2} b^{3} c^{8} - 3 \, a^{3} b^{2} c^{7} d + 3 \, a^{4} b c^{6} d^{2} - a^{5} c^{5} d^{3}\right )} x\right )}}\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)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(b^3*c^3*d^2*x^5 + 2*b^3*c^4*d*x^3 + b^3*c^5*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)*sqr
t(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(3*a*b^3*c^5 - 9*a^2*b^2*c^4*d + 9*a^3*b*c^3*d^2 - 3*a^4*c^2*d^
3 + (3*a*b^3*c^3*d^2 - 17*a^2*b^2*c^2*d^3 + 22*a^3*b*c*d^4 - 8*a^4*d^5)*x^4 + 3*(2*a*b^3*c^4*d - 9*a^2*b^2*c^3
*d^2 + 11*a^3*b*c^2*d^3 - 4*a^4*c*d^4)*x^2)*sqrt(d*x^2 + c))/((a^2*b^3*c^6*d^2 - 3*a^3*b^2*c^5*d^3 + 3*a^4*b*c
^4*d^4 - a^5*c^3*d^5)*x^5 + 2*(a^2*b^3*c^7*d - 3*a^3*b^2*c^6*d^2 + 3*a^4*b*c^5*d^3 - a^5*c^4*d^4)*x^3 + (a^2*b
^3*c^8 - 3*a^3*b^2*c^7*d + 3*a^4*b*c^6*d^2 - a^5*c^5*d^3)*x), -1/6*(3*(b^3*c^3*d^2*x^5 + 2*b^3*c^4*d*x^3 + b^3
*c^5*x)*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*(3*a*b^3*c^5 - 9*a^2*b^2*c^4*d + 9*a^3*b*c^3*d^2 - 3*a^4*c^2*d^3
+ (3*a*b^3*c^3*d^2 - 17*a^2*b^2*c^2*d^3 + 22*a^3*b*c*d^4 - 8*a^4*d^5)*x^4 + 3*(2*a*b^3*c^4*d - 9*a^2*b^2*c^3*
d^2 + 11*a^3*b*c^2*d^3 - 4*a^4*c*d^4)*x^2)*sqrt(d*x^2 + c))/((a^2*b^3*c^6*d^2 - 3*a^3*b^2*c^5*d^3 + 3*a^4*b*c^
4*d^4 - a^5*c^3*d^5)*x^5 + 2*(a^2*b^3*c^7*d - 3*a^3*b^2*c^6*d^2 + 3*a^4*b*c^5*d^3 - a^5*c^4*d^4)*x^3 + (a^2*b^
3*c^8 - 3*a^3*b^2*c^7*d + 3*a^4*b*c^6*d^2 - a^5*c^5*d^3)*x)]

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giac [B]  time = 3.80, size = 366, normalized size = 2.06 \begin {gather*} \frac {b^{3} \sqrt {d} \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 )}{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {{\left (\frac {{\left (8 \, b^{3} c^{5} d^{4} - 21 \, a b^{2} c^{4} d^{5} + 18 \, a^{2} b c^{3} d^{6} - 5 \, a^{3} c^{2} d^{7}\right )} x^{2}}{b^{4} c^{9} d - 4 \, a b^{3} c^{8} d^{2} + 6 \, a^{2} b^{2} c^{7} d^{3} - 4 \, a^{3} b c^{6} d^{4} + a^{4} c^{5} d^{5}} + \frac {3 \, {\left (3 \, b^{3} c^{6} d^{3} - 8 \, a b^{2} c^{5} d^{4} + 7 \, a^{2} b c^{4} d^{5} - 2 \, a^{3} c^{3} d^{6}\right )}}{b^{4} c^{9} d - 4 \, a b^{3} c^{8} d^{2} + 6 \, a^{2} b^{2} c^{7} d^{3} - 4 \, a^{3} b c^{6} d^{4} + a^{4} c^{5} d^{5}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {2 \, \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )} a c^{2}} \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)^(5/2),x, algorithm="giac")

[Out]

b^3*sqrt(d)*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*b^2*c^2
- 2*a^2*b*c*d + a^3*d^2)*sqrt(a*b*c*d - a^2*d^2)) + 1/3*((8*b^3*c^5*d^4 - 21*a*b^2*c^4*d^5 + 18*a^2*b*c^3*d^6
- 5*a^3*c^2*d^7)*x^2/(b^4*c^9*d - 4*a*b^3*c^8*d^2 + 6*a^2*b^2*c^7*d^3 - 4*a^3*b*c^6*d^4 + a^4*c^5*d^5) + 3*(3*
b^3*c^6*d^3 - 8*a*b^2*c^5*d^4 + 7*a^2*b*c^4*d^5 - 2*a^3*c^3*d^6)/(b^4*c^9*d - 4*a*b^3*c^8*d^2 + 6*a^2*b^2*c^7*
d^3 - 4*a^3*b*c^6*d^4 + a^4*c^5*d^5))*x/(d*x^2 + c)^(3/2) + 2*sqrt(d)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)*a
*c^2)

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maple [B]  time = 0.02, size = 1192, normalized size = 6.70 \begin {gather*} \frac {b^{3} \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}\, \left (a d -b c \right )^{2} \sqrt {-\frac {a d -b c}{b}}\, a}-\frac {b^{3} \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}\, \left (a d -b c \right )^{2} \sqrt {-\frac {a d -b c}{b}}\, a}+\frac {b^{3}}{2 \sqrt {-a b}\, \left (a d -b c \right )^{2} \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}}\, a}-\frac {b^{3}}{2 \sqrt {-a b}\, \left (a d -b c \right )^{2} \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}}\, a}+\frac {b^{2} d x}{2 \left (a d -b c \right )^{2} \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}}\, a c}+\frac {b^{2} d x}{2 \left (a d -b c \right )^{2} \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}}\, a c}-\frac {b^{2}}{6 \sqrt {-a b}\, \left (a d -b c \right ) \left (\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}\right )^{\frac {3}{2}} a}+\frac {b^{2}}{6 \sqrt {-a b}\, \left (a d -b c \right ) \left (\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}\right )^{\frac {3}{2}} a}-\frac {b d x}{6 \left (a d -b c \right ) \left (\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}\right )^{\frac {3}{2}} a c}-\frac {b d x}{6 \left (a d -b c \right ) \left (\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}\right )^{\frac {3}{2}} a c}-\frac {b d x}{3 \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}}\, a \,c^{2}}-\frac {b d x}{3 \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}}\, a \,c^{2}}-\frac {4 d x}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a \,c^{2}}-\frac {8 d x}{3 \sqrt {d \,x^{2}+c}\, a \,c^{3}}-\frac {1}{\left (d \,x^{2}+c \right )^{\frac {3}{2}} 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)^(5/2),x)

[Out]

-1/6*b^2/a/(-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)^(
3/2)-1/6*b/a*d/(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)^(3/2)*x-
1/3*b/a*d/(a*d-b*c)/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)*x+1/2
*b^3/a/(-a*b)^(1/2)/(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)+1/2*b^2/a/(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)*d*
x-1/2*b^3/a/(-a*b)^(1/2)/(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))+1/6*b^2/a/(-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)^(3/2)-1/6*b/a*d/(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)^(3/2)*x-1/3*b/a*d/(a*d-b*c)/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)*x-1/2*b^3/a/(-a*b)^(1/2)/(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)+1/2*b^2/a/(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)*d*x+1/2*b^3/a/(-a*b)^(1/2)/(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))-1/a/c/x/(d*x^2+c)^(3/2)-4/3/a*d/c^2*x/(d*x^2+c)^(3/2)-8/3/a*d/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 )} {\left (d x^{2} + c\right )}^{\frac {5}{2}} 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)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)*(d*x^2 + c)^(5/2)*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,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)^(5/2)),x)

[Out]

int(1/(x^2*(a + b*x^2)*(c + d*x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{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)**(5/2),x)

[Out]

Integral(1/(x**2*(a + b*x**2)*(c + d*x**2)**(5/2)), x)

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