### 3.1309 $$\int \frac{(b d+2 c d x)^{13/2}}{(a+b x+c x^2)^3} \, dx$$

Optimal. Leaf size=193 $77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{154}{3} c^2 d^5 (b d+2 c d x)^{3/2}$

[Out]

(154*c^2*d^5*(b*d + 2*c*d*x)^(3/2))/3 - (d*(b*d + 2*c*d*x)^(11/2))/(2*(a + b*x + c*x^2)^2) - (11*c*d^3*(b*d +
2*c*d*x)^(7/2))/(2*(a + b*x + c*x^2)) + 77*c^2*(b^2 - 4*a*c)^(3/4)*d^(13/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 -
4*a*c)^(1/4)*Sqrt[d])] - 77*c^2*(b^2 - 4*a*c)^(3/4)*d^(13/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)
*Sqrt[d])]

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Rubi [A]  time = 0.148014, antiderivative size = 193, 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, 298, 203, 206} $77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{154}{3} c^2 d^5 (b d+2 c d x)^{3/2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^3,x]

[Out]

(154*c^2*d^5*(b*d + 2*c*d*x)^(3/2))/3 - (d*(b*d + 2*c*d*x)^(11/2))/(2*(a + b*x + c*x^2)^2) - (11*c*d^3*(b*d +
2*c*d*x)^(7/2))/(2*(a + b*x + c*x^2)) + 77*c^2*(b^2 - 4*a*c)^(3/4)*d^(13/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 -
4*a*c)^(1/4)*Sqrt[d])] - 77*c^2*(b^2 - 4*a*c)^(3/4)*d^(13/2)*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 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)^{13/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{1}{2} \left (11 c d^2\right ) \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)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (77 c^2 d^4\right ) \int \frac{(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=\frac{154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (77 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=\frac{154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{4} \left (77 c \left (b^2-4 a c\right ) d^5\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{154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (77 c \left (b^2-4 a c\right ) d^5\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{154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}-\left (77 c^2 \left (b^2-4 a c\right ) d^7\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 (77 c^2 \left (b^2-4 a c\right ) d^7\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{154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac{d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+77 c^2 \left (b^2-4 a c\right )^{3/4} d^{13/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-77 c^2 \left (b^2-4 a c\right )^{3/4} d^{13/2} \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 = 0.160206, size = 145, normalized size = 0.75 $-\frac{4 d^5 (d (b+2 c x))^{3/2} \left (-16 c^2 \left (77 a^2+55 a c x^2+5 c^2 x^4\right )+1232 c^2 (a+x (b+c x))^2 \, _2F_1\left (\frac{3}{4},3;\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )+4 b^2 c \left (99 a+25 c x^2\right )-80 b c^2 x \left (11 a+2 c x^2\right )+180 b^3 c x-27 b^4\right )}{15 (a+x (b+c x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^3,x]

[Out]

(-4*d^5*(d*(b + 2*c*x))^(3/2)*(-27*b^4 + 180*b^3*c*x - 80*b*c^2*x*(11*a + 2*c*x^2) + 4*b^2*c*(99*a + 25*c*x^2)
- 16*c^2*(77*a^2 + 55*a*c*x^2 + 5*c^2*x^4) + 1232*c^2*(a + x*(b + c*x))^2*Hypergeometric2F1[3/4, 3, 7/4, (b +
2*c*x)^2/(b^2 - 4*a*c)]))/(15*(a + x*(b + c*x))^2)

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Maple [B]  time = 0.203, size = 857, normalized size = 4.4 \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)^(13/2)/(c*x^2+b*x+a)^3,x)

[Out]

64/3*c^2*d^5*(2*c*d*x+b*d)^(3/2)+152*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)^(7/2)*a-38*
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)^(7/2)*b^2+480*c^4*d^9/(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^2-240*c^3*d^9/(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*b^2+30*c^2*d^9/(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^4-77*c^3*d^7*2^(1/2)/(4*a*c*d^2
-b^2*d^2)^(1/4)*a*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)))-154*c^3*d^
7*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)+154*c^3*
d^7*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)+77/4*
c^2*d^7*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)))+77/2*c^2*d^7*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)-77/2*c^2*d^7*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)^(13/2)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.90145, size = 2148, normalized size = 11.13 \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)^(13/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

1/6*(924*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(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((((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(1/4)*(b^4*
c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*sqrt(2*c*d*x + b*d)*d^19 - sqrt(2*(b^8*c^13 - 16*a*b^6*c^14 + 96*a^2*b^4*c^15
- 256*a^3*b^2*c^16 + 256*a^4*c^17)*d^39*x + (b^9*c^12 - 16*a*b^7*c^13 + 96*a^2*b^5*c^14 - 256*a^3*b^3*c^15 + 2
56*a^4*b*c^16)*d^39 + sqrt((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)*(b^6*c^8 - 12*a*b^4*
c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(1/
4))/((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)) - 231*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b
^2*c^10 - 64*a^3*c^11)*d^26)^(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(456533*(b^4*c
^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*sqrt(2*c*d*x + b*d)*d^19 + 456533*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 -
64*a^3*c^11)*d^26)^(3/4)) + 231*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(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(456533*(b^4*c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*sqrt(2*c*d*x
+ b*d)*d^19 - 456533*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(3/4)) + (256*c^5*d^6*x^
5 + 640*b*c^4*d^6*x^4 + 2*(199*b^2*c^3 + 484*a*c^4)*d^6*x^3 - (43*b^3*c^2 - 1452*a*b*c^3)*d^6*x^2 - (63*b^4*c
- 418*a*b^2*c^2 - 616*a^2*c^3)*d^6*x - (3*b^5 + 33*a*b^3*c - 308*a^2*b*c^2)*d^6)*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)**(13/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.27027, size = 703, normalized size = 3.64 \begin{align*} -\frac{77}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{5} \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{77}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{5} \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{77}{4} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{5} \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{77}{4} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} d^{5} \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{64}{3} \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c^{2} d^{5} + \frac{2 \,{\left (15 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{4} c^{2} d^{9} - 120 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a b^{2} c^{3} d^{9} + 240 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a^{2} c^{4} d^{9} - 19 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} b^{2} c^{2} d^{7} + 76 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} a c^{3} d^{7}\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)^(13/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-77/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*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)) - 77/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*arcta
n(-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)) +
77/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqr
t(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2)) - 77/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*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)) + 64/3*(2*c*d*
x + b*d)^(3/2)*c^2*d^5 + 2*(15*(2*c*d*x + b*d)^(3/2)*b^4*c^2*d^9 - 120*(2*c*d*x + b*d)^(3/2)*a*b^2*c^3*d^9 + 2
40*(2*c*d*x + b*d)^(3/2)*a^2*c^4*d^9 - 19*(2*c*d*x + b*d)^(7/2)*b^2*c^2*d^7 + 76*(2*c*d*x + b*d)^(7/2)*a*c^3*d
^7)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2