3.3.82 \(\int \frac {x^2 (c+d x^2)^3}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=147 \[ \frac {d x \left (105 a^2 d^2-190 a b c d+81 b^2 c^2\right )}{30 b^4}+\frac {(b c-7 a d) (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{9/2}}+\frac {d x \left (c+d x^2\right ) (33 b c-35 a d)}{30 b^3}-\frac {x \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {7 d x \left (c+d x^2\right )^2}{10 b^2} \]

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Rubi [A]  time = 0.19, antiderivative size = 147, 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 = {467, 528, 388, 205} \begin {gather*} \frac {d x \left (105 a^2 d^2-190 a b c d+81 b^2 c^2\right )}{30 b^4}+\frac {d x \left (c+d x^2\right ) (33 b c-35 a d)}{30 b^3}+\frac {(b c-7 a d) (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{9/2}}-\frac {x \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {7 d x \left (c+d x^2\right )^2}{10 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(81*b^2*c^2 - 190*a*b*c*d + 105*a^2*d^2)*x)/(30*b^4) + (d*(33*b*c - 35*a*d)*x*(c + d*x^2))/(30*b^3) + (7*d*
x*(c + d*x^2)^2)/(10*b^2) - (x*(c + d*x^2)^3)/(2*b*(a + b*x^2)) + ((b*c - 7*a*d)*(b*c - a*d)^2*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(2*Sqrt[a]*b^(9/2))

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 467

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

Rule 528

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx &=-\frac {x \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\int \frac {\left (c+d x^2\right )^2 \left (c+7 d x^2\right )}{a+b x^2} \, dx}{2 b}\\ &=\frac {7 d x \left (c+d x^2\right )^2}{10 b^2}-\frac {x \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\int \frac {\left (c+d x^2\right ) \left (c (5 b c-7 a d)+d (33 b c-35 a d) x^2\right )}{a+b x^2} \, dx}{10 b^2}\\ &=\frac {d (33 b c-35 a d) x \left (c+d x^2\right )}{30 b^3}+\frac {7 d x \left (c+d x^2\right )^2}{10 b^2}-\frac {x \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\int \frac {c \left (15 b^2 c^2-54 a b c d+35 a^2 d^2\right )+d \left (81 b^2 c^2-190 a b c d+105 a^2 d^2\right ) x^2}{a+b x^2} \, dx}{30 b^3}\\ &=\frac {d \left (81 b^2 c^2-190 a b c d+105 a^2 d^2\right ) x}{30 b^4}+\frac {d (33 b c-35 a d) x \left (c+d x^2\right )}{30 b^3}+\frac {7 d x \left (c+d x^2\right )^2}{10 b^2}-\frac {x \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\left ((b c-7 a d) (b c-a d)^2\right ) \int \frac {1}{a+b x^2} \, dx}{2 b^4}\\ &=\frac {d \left (81 b^2 c^2-190 a b c d+105 a^2 d^2\right ) x}{30 b^4}+\frac {d (33 b c-35 a d) x \left (c+d x^2\right )}{30 b^3}+\frac {7 d x \left (c+d x^2\right )^2}{10 b^2}-\frac {x \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {(b c-7 a d) (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 125, normalized size = 0.85 \begin {gather*} \frac {(b c-7 a d) (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{9/2}}-\frac {x (b c-a d)^3}{2 b^4 \left (a+b x^2\right )}+\frac {3 d x (b c-a d)^2}{b^4}+\frac {d^2 x^3 (3 b c-2 a d)}{3 b^3}+\frac {d^3 x^5}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*(b*c - a*d)^2*x)/b^4 + (d^2*(3*b*c - 2*a*d)*x^3)/(3*b^3) + (d^3*x^5)/(5*b^2) - ((b*c - a*d)^3*x)/(2*b^4*(
a + b*x^2)) + ((b*c - 7*a*d)*(b*c - a*d)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^(9/2))

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.96, size = 508, normalized size = 3.46 \begin {gather*} \left [\frac {12 \, a b^{4} d^{3} x^{7} + 4 \, {\left (15 \, a b^{4} c d^{2} - 7 \, a^{2} b^{3} d^{3}\right )} x^{5} + 20 \, {\left (9 \, a b^{4} c^{2} d - 15 \, a^{2} b^{3} c d^{2} + 7 \, a^{3} b^{2} d^{3}\right )} x^{3} + 15 \, {\left (a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 15 \, a^{3} b c d^{2} - 7 \, a^{4} d^{3} + {\left (b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 15 \, a^{2} b^{2} c d^{2} - 7 \, a^{3} b d^{3}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 30 \, {\left (a b^{4} c^{3} - 9 \, a^{2} b^{3} c^{2} d + 15 \, a^{3} b^{2} c d^{2} - 7 \, a^{4} b d^{3}\right )} x}{60 \, {\left (a b^{6} x^{2} + a^{2} b^{5}\right )}}, \frac {6 \, a b^{4} d^{3} x^{7} + 2 \, {\left (15 \, a b^{4} c d^{2} - 7 \, a^{2} b^{3} d^{3}\right )} x^{5} + 10 \, {\left (9 \, a b^{4} c^{2} d - 15 \, a^{2} b^{3} c d^{2} + 7 \, a^{3} b^{2} d^{3}\right )} x^{3} + 15 \, {\left (a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 15 \, a^{3} b c d^{2} - 7 \, a^{4} d^{3} + {\left (b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 15 \, a^{2} b^{2} c d^{2} - 7 \, a^{3} b d^{3}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 15 \, {\left (a b^{4} c^{3} - 9 \, a^{2} b^{3} c^{2} d + 15 \, a^{3} b^{2} c d^{2} - 7 \, a^{4} b d^{3}\right )} x}{30 \, {\left (a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(12*a*b^4*d^3*x^7 + 4*(15*a*b^4*c*d^2 - 7*a^2*b^3*d^3)*x^5 + 20*(9*a*b^4*c^2*d - 15*a^2*b^3*c*d^2 + 7*a^
3*b^2*d^3)*x^3 + 15*(a*b^3*c^3 - 9*a^2*b^2*c^2*d + 15*a^3*b*c*d^2 - 7*a^4*d^3 + (b^4*c^3 - 9*a*b^3*c^2*d + 15*
a^2*b^2*c*d^2 - 7*a^3*b*d^3)*x^2)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 30*(a*b^4*c^3 - 9
*a^2*b^3*c^2*d + 15*a^3*b^2*c*d^2 - 7*a^4*b*d^3)*x)/(a*b^6*x^2 + a^2*b^5), 1/30*(6*a*b^4*d^3*x^7 + 2*(15*a*b^4
*c*d^2 - 7*a^2*b^3*d^3)*x^5 + 10*(9*a*b^4*c^2*d - 15*a^2*b^3*c*d^2 + 7*a^3*b^2*d^3)*x^3 + 15*(a*b^3*c^3 - 9*a^
2*b^2*c^2*d + 15*a^3*b*c*d^2 - 7*a^4*d^3 + (b^4*c^3 - 9*a*b^3*c^2*d + 15*a^2*b^2*c*d^2 - 7*a^3*b*d^3)*x^2)*sqr
t(a*b)*arctan(sqrt(a*b)*x/a) - 15*(a*b^4*c^3 - 9*a^2*b^3*c^2*d + 15*a^3*b^2*c*d^2 - 7*a^4*b*d^3)*x)/(a*b^6*x^2
+ a^2*b^5)]

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giac [A]  time = 0.36, size = 184, normalized size = 1.25 \begin {gather*} \frac {{\left (b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} - 7 \, a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} - \frac {b^{3} c^{3} x - 3 \, a b^{2} c^{2} d x + 3 \, a^{2} b c d^{2} x - a^{3} d^{3} x}{2 \, {\left (b x^{2} + a\right )} b^{4}} + \frac {3 \, b^{8} d^{3} x^{5} + 15 \, b^{8} c d^{2} x^{3} - 10 \, a b^{7} d^{3} x^{3} + 45 \, b^{8} c^{2} d x - 90 \, a b^{7} c d^{2} x + 45 \, a^{2} b^{6} d^{3} x}{15 \, b^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*c^3 - 9*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 7*a^3*d^3)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/2*(b^3*c^
3*x - 3*a*b^2*c^2*d*x + 3*a^2*b*c*d^2*x - a^3*d^3*x)/((b*x^2 + a)*b^4) + 1/15*(3*b^8*d^3*x^5 + 15*b^8*c*d^2*x^
3 - 10*a*b^7*d^3*x^3 + 45*b^8*c^2*d*x - 90*a*b^7*c*d^2*x + 45*a^2*b^6*d^3*x)/b^10

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maple [A]  time = 0.01, size = 247, normalized size = 1.68 \begin {gather*} \frac {d^{3} x^{5}}{5 b^{2}}-\frac {2 a \,d^{3} x^{3}}{3 b^{3}}+\frac {c \,d^{2} x^{3}}{b^{2}}+\frac {a^{3} d^{3} x}{2 \left (b \,x^{2}+a \right ) b^{4}}-\frac {7 a^{3} d^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{4}}-\frac {3 a^{2} c \,d^{2} x}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {15 a^{2} c \,d^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{3}}+\frac {3 a \,c^{2} d x}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {9 a \,c^{2} d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{2}}-\frac {c^{3} x}{2 \left (b \,x^{2}+a \right ) b}+\frac {c^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b}+\frac {3 a^{2} d^{3} x}{b^{4}}-\frac {6 a c \,d^{2} x}{b^{3}}+\frac {3 c^{2} d x}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*d^3/b^2*x^5-2/3*d^3/b^3*x^3*a+d^2/b^2*x^3*c+3*d^3/b^4*a^2*x-6*d^2/b^3*a*c*x+3*d/b^2*c^2*x+1/2/b^4*x/(b*x^2
+a)*a^3*d^3-3/2/b^3*x/(b*x^2+a)*a^2*c*d^2+3/2/b^2*x/(b*x^2+a)*a*c^2*d-1/2/b*x/(b*x^2+a)*c^3-7/2/b^4/(a*b)^(1/2
)*arctan(1/(a*b)^(1/2)*b*x)*a^3*d^3+15/2/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*a^2*c*d^2-9/2/b^2/(a*b)^(1/
2)*arctan(1/(a*b)^(1/2)*b*x)*a*c^2*d+1/2/b/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*c^3

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maxima [A]  time = 2.29, size = 176, normalized size = 1.20 \begin {gather*} -\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {{\left (b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} - 7 \, a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} + \frac {3 \, b^{2} d^{3} x^{5} + 5 \, {\left (3 \, b^{2} c d^{2} - 2 \, a b d^{3}\right )} x^{3} + 45 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x}{15 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x/(b^5*x^2 + a*b^4) + 1/2*(b^3*c^3 - 9*a*b^2*c^2*d +
15*a^2*b*c*d^2 - 7*a^3*d^3)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/15*(3*b^2*d^3*x^5 + 5*(3*b^2*c*d^2 - 2*a
*b*d^3)*x^3 + 45*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)/b^4

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mupad [B]  time = 0.07, size = 232, normalized size = 1.58 \begin {gather*} x\,\left (\frac {3\,c^2\,d}{b^2}+\frac {2\,a\,\left (\frac {2\,a\,d^3}{b^3}-\frac {3\,c\,d^2}{b^2}\right )}{b}-\frac {a^2\,d^3}{b^4}\right )-x^3\,\left (\frac {2\,a\,d^3}{3\,b^3}-\frac {c\,d^2}{b^2}\right )+\frac {x\,\left (\frac {a^3\,d^3}{2}-\frac {3\,a^2\,b\,c\,d^2}{2}+\frac {3\,a\,b^2\,c^2\,d}{2}-\frac {b^3\,c^3}{2}\right )}{b^5\,x^2+a\,b^4}+\frac {d^3\,x^5}{5\,b^2}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x\,{\left (a\,d-b\,c\right )}^2\,\left (7\,a\,d-b\,c\right )}{\sqrt {a}\,\left (7\,a^3\,d^3-15\,a^2\,b\,c\,d^2+9\,a\,b^2\,c^2\,d-b^3\,c^3\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (7\,a\,d-b\,c\right )}{2\,\sqrt {a}\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((3*c^2*d)/b^2 + (2*a*((2*a*d^3)/b^3 - (3*c*d^2)/b^2))/b - (a^2*d^3)/b^4) - x^3*((2*a*d^3)/(3*b^3) - (c*d^2)
/b^2) + (x*((a^3*d^3)/2 - (b^3*c^3)/2 + (3*a*b^2*c^2*d)/2 - (3*a^2*b*c*d^2)/2))/(a*b^4 + b^5*x^2) + (d^3*x^5)/
(5*b^2) - (atan((b^(1/2)*x*(a*d - b*c)^2*(7*a*d - b*c))/(a^(1/2)*(7*a^3*d^3 - b^3*c^3 + 9*a*b^2*c^2*d - 15*a^2
*b*c*d^2)))*(a*d - b*c)^2*(7*a*d - b*c))/(2*a^(1/2)*b^(9/2))

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sympy [B]  time = 1.22, size = 338, normalized size = 2.30 \begin {gather*} x^{3} \left (- \frac {2 a d^{3}}{3 b^{3}} + \frac {c d^{2}}{b^{2}}\right ) + x \left (\frac {3 a^{2} d^{3}}{b^{4}} - \frac {6 a c d^{2}}{b^{3}} + \frac {3 c^{2} d}{b^{2}}\right ) + \frac {x \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}\right )}{2 a b^{4} + 2 b^{5} x^{2}} + \frac {\sqrt {- \frac {1}{a b^{9}}} \left (a d - b c\right )^{2} \left (7 a d - b c\right ) \log {\left (- \frac {a b^{4} \sqrt {- \frac {1}{a b^{9}}} \left (a d - b c\right )^{2} \left (7 a d - b c\right )}{7 a^{3} d^{3} - 15 a^{2} b c d^{2} + 9 a b^{2} c^{2} d - b^{3} c^{3}} + x \right )}}{4} - \frac {\sqrt {- \frac {1}{a b^{9}}} \left (a d - b c\right )^{2} \left (7 a d - b c\right ) \log {\left (\frac {a b^{4} \sqrt {- \frac {1}{a b^{9}}} \left (a d - b c\right )^{2} \left (7 a d - b c\right )}{7 a^{3} d^{3} - 15 a^{2} b c d^{2} + 9 a b^{2} c^{2} d - b^{3} c^{3}} + x \right )}}{4} + \frac {d^{3} x^{5}}{5 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*(-2*a*d**3/(3*b**3) + c*d**2/b**2) + x*(3*a**2*d**3/b**4 - 6*a*c*d**2/b**3 + 3*c**2*d/b**2) + x*(a**3*d**
3 - 3*a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3)/(2*a*b**4 + 2*b**5*x**2) + sqrt(-1/(a*b**9))*(a*d - b*c)**2
*(7*a*d - b*c)*log(-a*b**4*sqrt(-1/(a*b**9))*(a*d - b*c)**2*(7*a*d - b*c)/(7*a**3*d**3 - 15*a**2*b*c*d**2 + 9*
a*b**2*c**2*d - b**3*c**3) + x)/4 - sqrt(-1/(a*b**9))*(a*d - b*c)**2*(7*a*d - b*c)*log(a*b**4*sqrt(-1/(a*b**9)
)*(a*d - b*c)**2*(7*a*d - b*c)/(7*a**3*d**3 - 15*a**2*b*c*d**2 + 9*a*b**2*c**2*d - b**3*c**3) + x)/4 + d**3*x*
*5/(5*b**2)

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