### 3.1311 $$\int \frac{(b d+2 c d x)^{9/2}}{(a+b x+c x^2)^3} \, dx$$

Optimal. Leaf size=170 $\frac{21 c^2 d^{9/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{21 c^2 d^{9/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}$

[Out]

-(d*(b*d + 2*c*d*x)^(7/2))/(2*(a + b*x + c*x^2)^2) - (7*c*d^3*(b*d + 2*c*d*x)^(3/2))/(2*(a + b*x + c*x^2)) + (
21*c^2*d^(9/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (21*c^2*d^(9/2
)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4)

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Rubi [A]  time = 0.123375, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.231, Rules used = {686, 694, 329, 298, 203, 206} $\frac{21 c^2 d^{9/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{21 c^2 d^{9/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^3,x]

[Out]

-(d*(b*d + 2*c*d*x)^(7/2))/(2*(a + b*x + c*x^2)^2) - (7*c*d^3*(b*d + 2*c*d*x)^(3/2))/(2*(a + b*x + c*x^2)) + (
21*c^2*d^(9/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (21*c^2*d^(9/2
)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4)

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2
*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{1}{2} \left (7 c d^2\right ) \int \frac{(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (21 c^2 d^4\right ) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{4} \left (21 c d^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )\\ &=-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (21 c d^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{a-\frac{b^2}{4 c}+\frac{x^4}{4 c d^2}} \, dx,x,\sqrt{d (b+2 c x)}\right )\\ &=-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}-\left (21 c^2 d^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d-x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )+\left (21 c^2 d^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d+x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )\\ &=-\frac{d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac{21 c^2 d^{9/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{21 c^2 d^{9/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\sqrt [4]{b^2-4 a c}}\\ \end{align*}

Mathematica [C]  time = 0.154552, size = 119, normalized size = 0.7 $\frac{4 (d (b+2 c x))^{9/2} \left (-112 c^2 (a+x (b+c x))^2 \, _2F_1\left (\frac{3}{4},3;\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-5 \left (b^2-4 a c\right ) (b+2 c x)^2+7 \left (b^2-4 a c\right )^2\right )}{5 \left (b^2-4 a c\right ) (b+2 c x)^3 (a+x (b+c x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^3,x]

[Out]

(4*(d*(b + 2*c*x))^(9/2)*(7*(b^2 - 4*a*c)^2 - 5*(b^2 - 4*a*c)*(b + 2*c*x)^2 - 112*c^2*(a + x*(b + c*x))^2*Hype
rgeometric2F1[3/4, 3, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(5*(b^2 - 4*a*c)*(b + 2*c*x)^3*(a + x*(b + c*x))^2)

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Maple [B]  time = 0.199, size = 435, normalized size = 2.6 \begin{align*} -22\,{\frac{{c}^{2}{d}^{5} \left ( 2\,cdx+bd \right ) ^{7/2}}{ \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) ^{2}}}-56\,{\frac{{c}^{3}{d}^{7} \left ( 2\,cdx+bd \right ) ^{3/2}a}{ \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) ^{2}}}+14\,{\frac{{c}^{2}{d}^{7} \left ( 2\,cdx+bd \right ) ^{3/2}{b}^{2}}{ \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) ^{2}}}+{\frac{21\,{c}^{2}{d}^{5}\sqrt{2}}{4}\ln \left ({ \left ( 2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) \left ( 2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+{\frac{21\,{c}^{2}{d}^{5}\sqrt{2}}{2}\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}-{\frac{21\,{c}^{2}{d}^{5}\sqrt{2}}{2}\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^3,x)

[Out]

-22*c^2*d^5/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(7/2)-56*c^3*d^7/(4*c^2*d^2*x^2+4*b*c*d^2*x+
4*a*c*d^2)^2*(2*c*d*x+b*d)^(3/2)*a+14*c^2*d^7/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(3/2)*b^2+
21/4*c^2*d^5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*ln((2*c*d*x+b*d-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2
^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^
2-b^2*d^2)^(1/2)))+21/2*c^2*d^5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*
c*d*x+b*d)^(1/2)+1)-21/2*c^2*d^5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(
2*c*d*x+b*d)^(1/2)+1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.80276, size = 1138, normalized size = 6.69 \begin{align*} -\frac{84 \, \left (\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \arctan \left (-\frac{\left (\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}} \sqrt{2 \, c d x + b d} c^{6} d^{13} - \sqrt{2 \, c^{13} d^{27} x + b c^{12} d^{27} + \sqrt{\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}}{\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{18}} \left (\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}}{c^{8} d^{18}}\right ) + 21 \, \left (\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (9261 \, \sqrt{2 \, c d x + b d} c^{6} d^{13} + 9261 \, \left (\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac{3}{4}}{\left (b^{2} - 4 \, a c\right )}\right ) - 21 \, \left (\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (9261 \, \sqrt{2 \, c d x + b d} c^{6} d^{13} - 9261 \, \left (\frac{c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac{3}{4}}{\left (b^{2} - 4 \, a c\right )}\right ) +{\left (22 \, c^{3} d^{4} x^{3} + 33 \, b c^{2} d^{4} x^{2} +{\left (13 \, b^{2} c + 14 \, a c^{2}\right )} d^{4} x +{\left (b^{3} + 7 \, a b c\right )} d^{4}\right )} \sqrt{2 \, c d x + b d}}{2 \,{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(84*(c^8*d^18/(b^2 - 4*a*c))^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*arctan(-((c^
8*d^18/(b^2 - 4*a*c))^(1/4)*sqrt(2*c*d*x + b*d)*c^6*d^13 - sqrt(2*c^13*d^27*x + b*c^12*d^27 + sqrt(c^8*d^18/(b
^2 - 4*a*c))*(b^2*c^8 - 4*a*c^9)*d^18)*(c^8*d^18/(b^2 - 4*a*c))^(1/4))/(c^8*d^18)) + 21*(c^8*d^18/(b^2 - 4*a*c
))^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*log(9261*sqrt(2*c*d*x + b*d)*c^6*d^13 + 926
1*(c^8*d^18/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) - 21*(c^8*d^18/(b^2 - 4*a*c))^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a
*b*x + (b^2 + 2*a*c)*x^2 + a^2)*log(9261*sqrt(2*c*d*x + b*d)*c^6*d^13 - 9261*(c^8*d^18/(b^2 - 4*a*c))^(3/4)*(b
^2 - 4*a*c)) + (22*c^3*d^4*x^3 + 33*b*c^2*d^4*x^2 + (13*b^2*c + 14*a*c^2)*d^4*x + (b^3 + 7*a*b*c)*d^4)*sqrt(2*
c*d*x + b*d))/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(9/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.269, size = 689, normalized size = 4.05 \begin{align*} -\frac{21 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{3} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} + 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{2} b^{2} - 4 \, \sqrt{2} a c} - \frac{21 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{3} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} - 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{2} b^{2} - 4 \, \sqrt{2} a c} + \frac{21 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d + \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \,{\left (\sqrt{2} b^{2} - 4 \, \sqrt{2} a c\right )}} - \frac{21 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d - \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \,{\left (\sqrt{2} b^{2} - 4 \, \sqrt{2} a c\right )}} + \frac{2 \,{\left (7 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} c^{2} d^{7} - 28 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a c^{3} d^{7} - 11 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} c^{2} d^{5}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-21*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c
*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 21*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2
*d^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)
^(1/4))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) + 21/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^3*log(2*c*d*x + b*d + sqrt(2)*
(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) -
21/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^3*log(2*c*d*x + b*d - sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d
*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) + 2*(7*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^7
- 28*(2*c*d*x + b*d)^(3/2)*a*c^3*d^7 - 11*(2*c*d*x + b*d)^(7/2)*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*
d)^2)^2