### 3.1297 $$\int \frac{(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=179 $36 c d^5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x}-18 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-18 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}$

[Out]

36*c*(b^2 - 4*a*c)*d^5*Sqrt[b*d + 2*c*d*x] + (36*c*d^3*(b*d + 2*c*d*x)^(5/2))/5 - (d*(b*d + 2*c*d*x)^(9/2))/(a
+ b*x + c*x^2) - 18*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]
- 18*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

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Rubi [A]  time = 0.150475, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.269, Rules used = {686, 692, 694, 329, 212, 206, 203} $36 c d^5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x}-18 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-18 c d^{11/2} \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x]

[Out]

36*c*(b^2 - 4*a*c)*d^5*Sqrt[b*d + 2*c*d*x] + (36*c*d^3*(b*d + 2*c*d*x)^(5/2))/5 - (d*(b*d + 2*c*d*x)^(9/2))/(a
+ b*x + c*x^2) - 18*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]
- 18*c*(b^2 - 4*a*c)^(5/4)*d^(11/2)*ArcTanh[Sqrt[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 212

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

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])

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 c d^2\right ) \int \frac{(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx\\ &=\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 c \left (b^2-4 a c\right ) d^4\right ) \int \frac{(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=36 c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x}+\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=36 c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x}+\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\frac{1}{2} \left (9 \left (b^2-4 a c\right )^2 d^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )\\ &=36 c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x}+\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}+\left (9 \left (b^2-4 a c\right )^2 d^5\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b^2}{4 c}+\frac{x^4}{4 c d^2}} \, dx,x,\sqrt{d (b+2 c x)}\right )\\ &=36 c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x}+\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}-\left (18 c \left (b^2-4 a c\right )^{3/2} d^6\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 (18 c \left (b^2-4 a c\right )^{3/2} d^6\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 )\\ &=36 c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x}+\frac{36}{5} c d^3 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{9/2}}{a+b x+c x^2}-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-18 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\\ \end{align*}

Mathematica [A]  time = 0.364707, size = 167, normalized size = 0.93 $-\frac{d (d (b+2 c x))^{9/2} \left (-3 \left (b^2-4 a c\right ) \left (-30 \left (b^2-4 a c\right ) \sqrt{b+2 c x}-60 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \left (\tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )+24 (b+2 c x)^{5/2}\right )-8 (b+2 c x)^{9/2}\right )}{10 (b+2 c x)^{9/2} (a+x (b+c x))}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^2,x]

[Out]

-(d*(d*(b + 2*c*x))^(9/2)*(-8*(b + 2*c*x)^(9/2) - 3*(b^2 - 4*a*c)*(-30*(b^2 - 4*a*c)*Sqrt[b + 2*c*x] + 24*(b +
2*c*x)^(5/2) - 60*c*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))*(ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + ArcT
anh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)]))))/(10*(b + 2*c*x)^(9/2)*(a + x*(b + c*x)))

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Maple [B]  time = 0.201, size = 1090, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2,x)

[Out]

16/5*c*d^3*(2*c*d*x+b*d)^(5/2)-128*c^2*d^5*a*(2*c*d*x+b*d)^(1/2)+32*c*d^5*b^2*(2*c*d*x+b*d)^(1/2)-64*c^3*d^7*(
2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*a^2+32*c^2*d^7*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b
*c*d^2*x+4*a*c*d^2)*a*b^2-4*c*d^7*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*b^4+144*c^3*d^7/(4
*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a^2-72*c^2*d^7
/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a*b^2+9*c*d
^7/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*b^4-144*c
^3*d^7/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*a^2+
72*c^2*d^7/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)*
a*b^2-9*c*d^7/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+
1)*b^4+72*c^3*d^7/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*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)))*a^2-36*c^2*d^7/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*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)))*a*b^2+9/2*c*d^7/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*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)))*b^4

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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)^(11/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.17817, size = 2026, 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)^(11/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

1/5*(180*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22
)^(1/4)*(c*x^2 + b*x + a)*arctan(-(((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^
2*c^8 - 1024*a^5*c^9)*d^22)^(3/4)*(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 + ((b^10*c^4 - 20*a*b^8*c^5 + 160*
a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(3/4)*sqrt(2*(b^4*c^3 - 8*a*b^2*c^4 + 1
6*a^2*c^5)*d^11*x + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^11 + sqrt((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*
c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)))/((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 -
640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)) + 45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 -
640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)*(c*x^2 + b*x + a)*log(-9*(b^2*c - 4*a*c^2)*sqrt
(2*c*d*x + b*d)*d^5 + 9*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 102
4*a^5*c^9)*d^22)^(1/4)) - 45*((b^10*c^4 - 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8
- 1024*a^5*c^9)*d^22)^(1/4)*(c*x^2 + b*x + a)*log(-9*(b^2*c - 4*a*c^2)*sqrt(2*c*d*x + b*d)*d^5 - 9*((b^10*c^4
- 20*a*b^8*c^5 + 160*a^2*b^6*c^6 - 640*a^3*b^4*c^7 + 1280*a^4*b^2*c^8 - 1024*a^5*c^9)*d^22)^(1/4)) + (64*c^4*d
^5*x^4 + 128*b*c^3*d^5*x^3 + 48*(5*b^2*c^2 - 12*a*c^3)*d^5*x^2 + 16*(11*b^3*c - 36*a*b*c^2)*d^5*x - (5*b^4 - 2
16*a*b^2*c + 720*a^2*c^2)*d^5)*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)**(11/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.26147, size = 767, normalized size = 4.28 \begin{align*} 32 \, \sqrt{2 \, c d x + b d} b^{2} c d^{5} - 128 \, \sqrt{2 \, c d x + b d} a c^{2} d^{5} + \frac{16}{5} \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} c d^{3} - \frac{9}{2} \, \sqrt{2}{\left (b^{2} c d^{5} - 4 \, a c^{2} d^{5}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \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{9}{2} \, \sqrt{2}{\left (b^{2} c d^{5} - 4 \, a c^{2} d^{5}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \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 ) - 9 \,{\left (\sqrt{2} b^{2} c d^{5} - 4 \, \sqrt{2} a c^{2} d^{5}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \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 ) - 9 \,{\left (\sqrt{2} b^{2} c d^{5} - 4 \, \sqrt{2} a c^{2} d^{5}\right )}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \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{4 \,{\left (\sqrt{2 \, c d x + b d} b^{4} c d^{7} - 8 \, \sqrt{2 \, c d x + b d} a b^{2} c^{2} d^{7} + 16 \, \sqrt{2 \, c d x + b d} a^{2} c^{3} d^{7}\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)^(11/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

32*sqrt(2*c*d*x + b*d)*b^2*c*d^5 - 128*sqrt(2*c*d*x + b*d)*a*c^2*d^5 + 16/5*(2*c*d*x + b*d)^(5/2)*c*d^3 - 9/2*
sqrt(2)*(b^2*c*d^5 - 4*a*c^2*d^5)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) + 9/2*sqrt(2)*(b^2*c*d^5 - 4*a*c^2*d^5)*(-b^2*d^2
+ 4*a*c*d^2)^(1/4)*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)) - 9*(sqrt(2)*b^2*c*d^5 - 4*sqrt(2)*a*c^2*d^5)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) - 9*(sqrt(2)*b^2*
c*d^5 - 4*sqrt(2)*a*c^2*d^5)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) + 4*(sqrt(2*c*d*x + b*d)*b^4*c*d^7 - 8*sqrt(2*c*d
*x + b*d)*a*b^2*c^2*d^7 + 16*sqrt(2*c*d*x + b*d)*a^2*c^3*d^7)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)