3.774 \(\int \frac {x^9}{(a+b x^4) (c+d x^4)} \, dx\)

Optimal. Leaf size=92 \[ \frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{3/2} (b c-a d)}-\frac {c^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 d^{3/2} (b c-a d)}+\frac {x^2}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {465, 479, 522, 205} \[ \frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{3/2} (b c-a d)}-\frac {c^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 d^{3/2} (b c-a d)}+\frac {x^2}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^9/((a + b*x^4)*(c + d*x^4)),x]

[Out]

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

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 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 479

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rubi steps

\begin {align*} \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx,x,x^2\right )\\ &=\frac {x^2}{2 b d}-\frac {\operatorname {Subst}\left (\int \frac {a c+(b c+a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx,x,x^2\right )}{2 b d}\\ &=\frac {x^2}{2 b d}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^2\right )}{2 b (b c-a d)}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{c+d x^2} \, dx,x,x^2\right )}{2 d (b c-a d)}\\ &=\frac {x^2}{2 b d}+\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{3/2} (b c-a d)}-\frac {c^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 d^{3/2} (b c-a d)}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 82, normalized size = 0.89 \[ \frac {\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{b^{3/2}}+x^2 \left (\frac {c}{d}-\frac {a}{b}\right )-\frac {c^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{d^{3/2}}}{2 b c-2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/((a + b*x^4)*(c + d*x^4)),x]

[Out]

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

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fricas [A]  time = 0.78, size = 416, normalized size = 4.52 \[ \left [-\frac {a d \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} - 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) + b c \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} + 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right ) - 2 \, {\left (b c - a d\right )} x^{2}}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, \frac {2 \, a d \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) - b c \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} + 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right ) + 2 \, {\left (b c - a d\right )} x^{2}}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, -\frac {2 \, b c \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right ) + a d \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} - 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) - 2 \, {\left (b c - a d\right )} x^{2}}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, \frac {a d \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) - b c \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right ) + {\left (b c - a d\right )} x^{2}}{2 \, {\left (b^{2} c d - a b d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 80, normalized size = 0.87 \[ \frac {a^{2} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{2} c - a b d\right )} \sqrt {a b}} - \frac {c^{2} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d - a d^{2}\right )} \sqrt {c d}} + \frac {x^{2}}{2 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 81, normalized size = 0.88 \[ -\frac {a^{2} \arctan \left (\frac {b \,x^{2}}{\sqrt {a b}}\right )}{2 \left (a d -b c \right ) \sqrt {a b}\, b}+\frac {c^{2} \arctan \left (\frac {d \,x^{2}}{\sqrt {c d}}\right )}{2 \left (a d -b c \right ) \sqrt {c d}\, d}+\frac {x^{2}}{2 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)/(d*x^4+c),x)

[Out]

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

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maxima [A]  time = 1.19, size = 80, normalized size = 0.87 \[ \frac {a^{2} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{2} c - a b d\right )} \sqrt {a b}} - \frac {c^{2} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d - a d^{2}\right )} \sqrt {c d}} + \frac {x^{2}}{2 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.72, size = 518, normalized size = 5.63 \[ \frac {\ln \left (b^9\,c^6\,\sqrt {-a^3\,b^3}-a^3\,d^6\,{\left (-a^3\,b^3\right )}^{3/2}+a\,b^{11}\,c^6\,x^2+a^7\,b^5\,d^6\,x^2+2\,b^3\,c^3\,d^3\,{\left (-a^3\,b^3\right )}^{3/2}-2\,a^4\,b^8\,c^3\,d^3\,x^2\right )\,\sqrt {-a^3\,b^3}}{4\,b^4\,c-4\,a\,b^3\,d}-\frac {\ln \left (a^3\,d^6\,{\left (-a^3\,b^3\right )}^{3/2}-b^9\,c^6\,\sqrt {-a^3\,b^3}+a\,b^{11}\,c^6\,x^2+a^7\,b^5\,d^6\,x^2-2\,b^3\,c^3\,d^3\,{\left (-a^3\,b^3\right )}^{3/2}-2\,a^4\,b^8\,c^3\,d^3\,x^2\right )\,\sqrt {-a^3\,b^3}}{4\,\left (b^4\,c-a\,b^3\,d\right )}-\frac {\ln \left (b^6\,c^3\,{\left (-c^3\,d^3\right )}^{3/2}-a^6\,d^9\,\sqrt {-c^3\,d^3}+a^6\,c\,d^{11}\,x^2+b^6\,c^7\,d^5\,x^2-2\,a^3\,b^3\,d^3\,{\left (-c^3\,d^3\right )}^{3/2}-2\,a^3\,b^3\,c^4\,d^8\,x^2\right )\,\sqrt {-c^3\,d^3}}{4\,\left (a\,d^4-b\,c\,d^3\right )}+\frac {\ln \left (a^6\,d^9\,\sqrt {-c^3\,d^3}-b^6\,c^3\,{\left (-c^3\,d^3\right )}^{3/2}+a^6\,c\,d^{11}\,x^2+b^6\,c^7\,d^5\,x^2+2\,a^3\,b^3\,d^3\,{\left (-c^3\,d^3\right )}^{3/2}-2\,a^3\,b^3\,c^4\,d^8\,x^2\right )\,\sqrt {-c^3\,d^3}}{4\,a\,d^4-4\,b\,c\,d^3}+\frac {x^2}{2\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/((a + b*x^4)*(c + d*x^4)),x)

[Out]

(log(b^9*c^6*(-a^3*b^3)^(1/2) - a^3*d^6*(-a^3*b^3)^(3/2) + a*b^11*c^6*x^2 + a^7*b^5*d^6*x^2 + 2*b^3*c^3*d^3*(-
a^3*b^3)^(3/2) - 2*a^4*b^8*c^3*d^3*x^2)*(-a^3*b^3)^(1/2))/(4*b^4*c - 4*a*b^3*d) - (log(a^3*d^6*(-a^3*b^3)^(3/2
) - b^9*c^6*(-a^3*b^3)^(1/2) + a*b^11*c^6*x^2 + a^7*b^5*d^6*x^2 - 2*b^3*c^3*d^3*(-a^3*b^3)^(3/2) - 2*a^4*b^8*c
^3*d^3*x^2)*(-a^3*b^3)^(1/2))/(4*(b^4*c - a*b^3*d)) - (log(b^6*c^3*(-c^3*d^3)^(3/2) - a^6*d^9*(-c^3*d^3)^(1/2)
+ a^6*c*d^11*x^2 + b^6*c^7*d^5*x^2 - 2*a^3*b^3*d^3*(-c^3*d^3)^(3/2) - 2*a^3*b^3*c^4*d^8*x^2)*(-c^3*d^3)^(1/2)
)/(4*(a*d^4 - b*c*d^3)) + (log(a^6*d^9*(-c^3*d^3)^(1/2) - b^6*c^3*(-c^3*d^3)^(3/2) + a^6*c*d^11*x^2 + b^6*c^7*
d^5*x^2 + 2*a^3*b^3*d^3*(-c^3*d^3)^(3/2) - 2*a^3*b^3*c^4*d^8*x^2)*(-c^3*d^3)^(1/2))/(4*a*d^4 - 4*b*c*d^3) + x^
2/(2*b*d)

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sympy [B]  time = 29.70, size = 932, normalized size = 10.13 \[ - \frac {\sqrt {- \frac {a^{3}}{b^{3}}} \log {\left (x^{2} + \frac {- \frac {a^{4} d^{4} \sqrt {- \frac {a^{3}}{b^{3}}}}{a d - b c} - \frac {a^{3} b^{3} d^{6} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {a^{2} b^{4} c d^{5} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {a b^{5} c^{2} d^{4} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {b^{6} c^{3} d^{3} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {b^{4} c^{4} \sqrt {- \frac {a^{3}}{b^{3}}}}{a d - b c}}{a^{3} c d^{2} + a^{2} b c^{2} d + a b^{2} c^{3}} \right )}}{4 \left (a d - b c\right )} + \frac {\sqrt {- \frac {a^{3}}{b^{3}}} \log {\left (x^{2} + \frac {\frac {a^{4} d^{4} \sqrt {- \frac {a^{3}}{b^{3}}}}{a d - b c} + \frac {a^{3} b^{3} d^{6} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {a^{2} b^{4} c d^{5} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {a b^{5} c^{2} d^{4} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {b^{6} c^{3} d^{3} \left (- \frac {a^{3}}{b^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {b^{4} c^{4} \sqrt {- \frac {a^{3}}{b^{3}}}}{a d - b c}}{a^{3} c d^{2} + a^{2} b c^{2} d + a b^{2} c^{3}} \right )}}{4 \left (a d - b c\right )} - \frac {\sqrt {- \frac {c^{3}}{d^{3}}} \log {\left (x^{2} + \frac {- \frac {a^{4} d^{4} \sqrt {- \frac {c^{3}}{d^{3}}}}{a d - b c} - \frac {a^{3} b^{3} d^{6} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {a^{2} b^{4} c d^{5} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {a b^{5} c^{2} d^{4} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {b^{6} c^{3} d^{3} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {b^{4} c^{4} \sqrt {- \frac {c^{3}}{d^{3}}}}{a d - b c}}{a^{3} c d^{2} + a^{2} b c^{2} d + a b^{2} c^{3}} \right )}}{4 \left (a d - b c\right )} + \frac {\sqrt {- \frac {c^{3}}{d^{3}}} \log {\left (x^{2} + \frac {\frac {a^{4} d^{4} \sqrt {- \frac {c^{3}}{d^{3}}}}{a d - b c} + \frac {a^{3} b^{3} d^{6} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {a^{2} b^{4} c d^{5} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} - \frac {a b^{5} c^{2} d^{4} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {b^{6} c^{3} d^{3} \left (- \frac {c^{3}}{d^{3}}\right )^{\frac {3}{2}}}{\left (a d - b c\right )^{3}} + \frac {b^{4} c^{4} \sqrt {- \frac {c^{3}}{d^{3}}}}{a d - b c}}{a^{3} c d^{2} + a^{2} b c^{2} d + a b^{2} c^{3}} \right )}}{4 \left (a d - b c\right )} + \frac {x^{2}}{2 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**4+a)/(d*x**4+c),x)

[Out]

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

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