### 3.1298 $$\int \frac{(b d+2 c d x)^{9/2}}{(a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=152 $14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac{28}{3} c d^3 (b d+2 c d x)^{3/2}$

[Out]

(28*c*d^3*(b*d + 2*c*d*x)^(3/2))/3 - (d*(b*d + 2*c*d*x)^(7/2))/(a + b*x + c*x^2) + 14*c*(b^2 - 4*a*c)^(3/4)*d^
(9/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 14*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)*ArcTanh[Sqr
t[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

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Rubi [A]  time = 0.12081, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.269, Rules used = {686, 692, 694, 329, 298, 203, 206} $14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac{28}{3} c d^3 (b d+2 c d x)^{3/2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(28*c*d^3*(b*d + 2*c*d*x)^(3/2))/3 - (d*(b*d + 2*c*d*x)^(7/2))/(a + b*x + c*x^2) + 14*c*(b^2 - 4*a*c)^(3/4)*d^
(9/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 14*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)*ArcTanh[Sqr
t[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

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 692

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

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 )^2} \, dx &=-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 c d^2\right ) \int \frac{(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=\frac{28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 c \left (b^2-4 a c\right ) d^4\right ) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=\frac{28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac{1}{2} \left (7 \left (b^2-4 a c\right ) 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{28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 \left (b^2-4 a c\right ) 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{28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}-\left (14 c \left (b^2-4 a c\right ) 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 (14 c \left (b^2-4 a c\right ) 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{28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0910897, size = 92, normalized size = 0.61 $-\frac{8 d^3 (d (b+2 c x))^{3/2} \left (14 c (a+x (b+c x)) \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-2 c \left (7 a+c x^2\right )+3 b^2-2 b c x\right )}{3 (a+x (b+c x))}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-8*d^3*(d*(b + 2*c*x))^(3/2)*(3*b^2 - 2*b*c*x - 2*c*(7*a + c*x^2) + 14*c*(a + x*(b + c*x))*Hypergeometric2F1[
3/4, 2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*(a + x*(b + c*x)))

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Maple [B]  time = 0.2, size = 693, normalized size = 4.6 \begin{align*}{\frac{16\,c{d}^{3}}{3} \left ( 2\,cdx+bd \right ) ^{{\frac{3}{2}}}}+16\,{\frac{{c}^{2}{d}^{5} \left ( 2\,cdx+bd \right ) ^{3/2}a}{4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2}}}-4\,{\frac{c{d}^{5} \left ( 2\,cdx+bd \right ) ^{3/2}{b}^{2}}{4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2}}}-14\,{\frac{{c}^{2}{d}^{5}\sqrt{2}a}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}\ln \left ({\frac{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}}}{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 ) }-28\,{\frac{{c}^{2}{d}^{5}\sqrt{2}a}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }+28\,{\frac{{c}^{2}{d}^{5}\sqrt{2}a}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }+{\frac{7\,c{d}^{5}\sqrt{2}{b}^{2}}{2}\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}}}}}+7\,{\frac{c{d}^{5}\sqrt{2}{b}^{2}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }-7\,{\frac{c{d}^{5}\sqrt{2}{b}^{2}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) } \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)^2,x)

[Out]

16/3*c*d^3*(2*c*d*x+b*d)^(3/2)+16*c^2*d^5*(2*c*d*x+b*d)^(3/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*a-4*c*d^5*
(2*c*d*x+b*d)^(3/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*b^2-14*c^2*d^5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*l
n((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)))-28*c^2*d^5*2^(1/2)/(4*a*c*d^2
-b^2*d^2)^(1/4)*a*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)+28*c^2*d^5*2^(1/2)/(4*a*c*d^
2-b^2*d^2)^(1/4)*a*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)+7/2*c*d^5*2^(1/2)/(4*a*c*d
^2-b^2*d^2)^(1/4)*b^2*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)))+7*c*d^
5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*b^2*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-7*c*d^
5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*b^2*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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.06604, size = 1721, normalized size = 11.32 \begin{align*} \text{result too large to display} \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)^2,x, algorithm="fricas")

[Out]

1/3*(84*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x + a)*arctan((((b^6*c^
4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x
+ b*d)*d^13 - sqrt(2*(b^8*c^7 - 16*a*b^6*c^8 + 96*a^2*b^4*c^9 - 256*a^3*b^2*c^10 + 256*a^4*c^11)*d^27*x + (b^9
*c^6 - 16*a*b^7*c^7 + 96*a^2*b^5*c^8 - 256*a^3*b^3*c^9 + 256*a^4*b*c^10)*d^27 + sqrt((b^6*c^4 - 12*a*b^4*c^5 +
48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)*((b^6*c^4 - 1
2*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4))/((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7
)*d^18)) - 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x + a)*log(343*(b
^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 + 343*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 -
64*a^3*c^7)*d^18)^(3/4)) + 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x
+ a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 - 343*((b^6*c^4 - 12*a*b^4*c^5 + 4
8*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4)) + (32*c^3*d^4*x^3 + 48*b*c^2*d^4*x^2 + 2*(5*b^2*c + 28*a*c^2)*d^4*x -
(3*b^3 - 28*a*b*c)*d^4)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.24153, size = 595, normalized size = 3.91 \begin{align*} -7 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c 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 ) - 7 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c 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 ) + \frac{7}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c 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 ) - \frac{7}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c 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 ) + \frac{16}{3} \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c d^{3} + \frac{4 \,{\left ({\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} c d^{5} - 4 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\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)^2,x, algorithm="giac")

[Out]

-7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqr
t(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - 7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*arctan(-1/2*sqr
t(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)) + 7/2*sqrt(2
)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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)) - 7/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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)) + 16/3*(2*c*d*x + b*d)^(3/2)*c
*d^3 + 4*((2*c*d*x + b*d)^(3/2)*b^2*c*d^5 - 4*(2*c*d*x + b*d)^(3/2)*a*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x
+ b*d)^2)