### 3.1285 $$\int \frac{(b d+2 c d x)^{11/2}}{a+b x+c x^2} \, dx$$

Optimal. Leaf size=175 $4 d^5 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}+\frac{4}{5} d^3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}-2 d^{11/2} \left (b^2-4 a c\right )^{9/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{11/2} \left (b^2-4 a c\right )^{9/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )+\frac{4}{9} d (b d+2 c d x)^{9/2}$

[Out]

4*(b^2 - 4*a*c)^2*d^5*Sqrt[b*d + 2*c*d*x] + (4*(b^2 - 4*a*c)*d^3*(b*d + 2*c*d*x)^(5/2))/5 + (4*d*(b*d + 2*c*d*
x)^(9/2))/9 - 2*(b^2 - 4*a*c)^(9/4)*d^(11/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 2*(b^
2 - 4*a*c)^(9/4)*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 = 0.214603, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.231, Rules used = {692, 694, 329, 212, 206, 203} $4 d^5 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}+\frac{4}{5} d^3 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}-2 d^{11/2} \left (b^2-4 a c\right )^{9/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{11/2} \left (b^2-4 a c\right )^{9/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )+\frac{4}{9} d (b d+2 c d x)^{9/2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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}}{a+b x+c x^2} \, dx &=\frac{4}{9} d (b d+2 c d x)^{9/2}+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac{(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx\\ &=\frac{4}{5} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{5/2}+\frac{4}{9} d (b d+2 c d x)^{9/2}+\left (\left (b^2-4 a c\right )^2 d^4\right ) \int \frac{(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=4 \left (b^2-4 a c\right )^2 d^5 \sqrt{b d+2 c d x}+\frac{4}{5} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{5/2}+\frac{4}{9} d (b d+2 c d x)^{9/2}+\left (\left (b^2-4 a c\right )^3 d^6\right ) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=4 \left (b^2-4 a c\right )^2 d^5 \sqrt{b d+2 c d x}+\frac{4}{5} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{5/2}+\frac{4}{9} d (b d+2 c d x)^{9/2}+\frac{\left (\left (b^2-4 a c\right )^3 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 )}{2 c}\\ &=4 \left (b^2-4 a c\right )^2 d^5 \sqrt{b d+2 c d x}+\frac{4}{5} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{5/2}+\frac{4}{9} d (b d+2 c d x)^{9/2}+\frac{\left (\left (b^2-4 a c\right )^3 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 )}{c}\\ &=4 \left (b^2-4 a c\right )^2 d^5 \sqrt{b d+2 c d x}+\frac{4}{5} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{5/2}+\frac{4}{9} d (b d+2 c d x)^{9/2}-\left (2 \left (b^2-4 a c\right )^{5/2} 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 (2 \left (b^2-4 a c\right )^{5/2} 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 )\\ &=4 \left (b^2-4 a c\right )^2 d^5 \sqrt{b d+2 c d x}+\frac{4}{5} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{5/2}+\frac{4}{9} d (b d+2 c d x)^{9/2}-2 \left (b^2-4 a c\right )^{9/4} d^{11/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-2 \left (b^2-4 a c\right )^{9/4} 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 = 0.196021, size = 157, normalized size = 0.9 $\frac{4 d (d (b+2 c x))^{9/2} \left (\frac{1}{9} (b+2 c x)^{9/2}-\left (4 a c-b^2\right ) \left (\frac{1}{5} (b+2 c x)^{5/2}-\frac{1}{2} \left (4 a c-b^2\right ) \left (2 \sqrt{b+2 c x}-\sqrt [4]{b^2-4 a c} \left (\tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )\right )\right )\right )}{(b+2 c x)^{9/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*d*(d*(b + 2*c*x))^(9/2)*((b + 2*c*x)^(9/2)/9 - (-b^2 + 4*a*c)*((b + 2*c*x)^(5/2)/5 - ((-b^2 + 4*a*c)*(2*Sqr
t[b + 2*c*x] - (b^2 - 4*a*c)^(1/4)*(ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + ArcTanh[Sqrt[b + 2*c*x]/(b^2
- 4*a*c)^(1/4)])))/2)))/(b + 2*c*x)^(9/2)

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Maple [B]  time = 0.213, size = 1287, normalized size = 7.4 \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),x)

[Out]

4/9*d*(2*c*d*x+b*d)^(9/2)-16/5*(2*c*d*x+b*d)^(5/2)*a*c*d^3+4/5*(2*c*d*x+b*d)^(5/2)*b^2*d^3+64*a^2*c^2*d^5*(2*c
*d*x+b*d)^(1/2)-32*a*b^2*c*d^5*(2*c*d*x+b*d)^(1/2)+4*b^4*d^5*(2*c*d*x+b*d)^(1/2)-64*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^3*c^3+48*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*b^2*c^2-12*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^4*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^6+64*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^3*c^3-48*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*b^2*c^2+12*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^4*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^6-32*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^3*c^3+24*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*b^2*c^2-6*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^4*c+1/2*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)))*b^6

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.51011, size = 3090, normalized size = 17.66 \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),x, algorithm="fricas")

[Out]

4/45*(80*c^4*d^5*x^4 + 160*b*c^3*d^5*x^3 + 12*(13*b^2*c^2 - 12*a*c^3)*d^5*x^2 + 4*(19*b^3*c - 36*a*b*c^2)*d^5*
x + (59*b^4 - 396*a*b^2*c + 720*a^2*c^2)*d^5)*sqrt(2*c*d*x + b*d) - 4*((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2
- 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 5898
24*a^8*b^2*c^8 - 262144*a^9*c^9)*d^22)^(1/4)*arctan((((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c
^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 -
262144*a^9*c^9)*d^22)^(3/4)*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*d*x + b*d)*d^5 - ((b^18 - 36*a*b^16*c + 57
6*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7
*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^22)^(3/4)*sqrt(2*(b^8*c - 16*a*b^6*c^2 + 96*a^2*b^4*c^3 - 25
6*a^3*b^2*c^4 + 256*a^4*c^5)*d^11*x + (b^9 - 16*a*b^7*c + 96*a^2*b^5*c^2 - 256*a^3*b^3*c^3 + 256*a^4*b*c^4)*d^
11 + sqrt((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5
+ 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^22)))/((b^18 - 36*a*b^16*c
+ 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 58982
4*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^22)) - ((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a
^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b
^2*c^8 - 262144*a^9*c^9)*d^22)^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*d*x + b*d)*d^5 + ((b^18 - 36*
a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6
- 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^22)^(1/4)) + ((b^18 - 36*a*b^16*c + 576*a^2*b^1
4*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7
+ 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^22)^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*d*x + b*d)*d^5
- ((b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 3440
64*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^22)^(1/4))

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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),x)

[Out]

Timed out

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Giac [B]  time = 1.26744, size = 779, normalized size = 4.45 \begin{align*} 4 \, \sqrt{2 \, c d x + b d} b^{4} d^{5} - 32 \, \sqrt{2 \, c d x + b d} a b^{2} c d^{5} + 64 \, \sqrt{2 \, c d x + b d} a^{2} c^{2} d^{5} + \frac{4}{5} \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b^{2} d^{3} - \frac{16}{5} \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} a c d^{3} + \frac{4}{9} \,{\left (2 \, c d x + b d\right )}^{\frac{9}{2}} d - \frac{1}{2} \, \sqrt{2}{\left (b^{4} d^{5} - 8 \, a b^{2} c d^{5} + 16 \, a^{2} 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{1}{2} \, \sqrt{2}{\left (b^{4} d^{5} - 8 \, a b^{2} c d^{5} + 16 \, a^{2} 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 ) -{\left (\sqrt{2} b^{4} d^{5} - 8 \, \sqrt{2} a b^{2} c d^{5} + 16 \, \sqrt{2} a^{2} 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 ) -{\left (\sqrt{2} b^{4} d^{5} - 8 \, \sqrt{2} a b^{2} c d^{5} + 16 \, \sqrt{2} a^{2} 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 ) \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),x, algorithm="giac")

[Out]

4*sqrt(2*c*d*x + b*d)*b^4*d^5 - 32*sqrt(2*c*d*x + b*d)*a*b^2*c*d^5 + 64*sqrt(2*c*d*x + b*d)*a^2*c^2*d^5 + 4/5*
(2*c*d*x + b*d)^(5/2)*b^2*d^3 - 16/5*(2*c*d*x + b*d)^(5/2)*a*c*d^3 + 4/9*(2*c*d*x + b*d)^(9/2)*d - 1/2*sqrt(2)
*(b^4*d^5 - 8*a*b^2*c*d^5 + 16*a^2*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)) + 1/2*sqrt(2)*(b^4*d^5 - 8*a*b^2*c*d^5 +
16*a^2*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)) - (sqrt(2)*b^4*d^5 - 8*sqrt(2)*a*b^2*c*d^5 + 16*sqrt(2)*a^2*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)) - (sqrt(2)*b^4*d^5 - 8*sqrt(2)*a*b^2*c*d^5 + 16*sqrt(2)*a^2*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))