### 3.1315 $$\int \frac{\sqrt{b d+2 c d x}}{(a+b x+c x^2)^3} \, dx$$

Optimal. Leaf size=192 $\frac{5 c^2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac{5 c^2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}+\frac{5 c (b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{(b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}$

[Out]

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

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Rubi [A]  time = 0.129688, antiderivative size = 192, 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 = {687, 694, 329, 298, 203, 206} $\frac{5 c^2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac{5 c^2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}+\frac{5 c (b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{(b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && RationalQ[m] && 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{\sqrt{b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac{(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}-\frac{(5 c) \int \frac{\sqrt{b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac{(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac{\left (5 c^2\right ) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx}{2 \left (b^2-4 a c\right )^2}\\ &=-\frac{(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac{(5 c) \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 )}{4 \left (b^2-4 a c\right )^2 d}\\ &=-\frac{(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac{(5 c) \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 )}{2 \left (b^2-4 a c\right )^2 d}\\ &=-\frac{(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac{\left (5 c^2 d\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 (b^2-4 a c\right )^2}+\frac{\left (5 c^2 d\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 (b^2-4 a c\right )^2}\\ &=-\frac{(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac{5 c^2 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac{5 c^2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\left (b^2-4 a c\right )^{9/4}}\\ \end{align*}

Mathematica [C]  time = 0.045127, size = 59, normalized size = 0.31 $-\frac{64 c^2 (d (b+2 c x))^{3/2} \, _2F_1\left (\frac{3}{4},3;\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-64*c^2*(d*(b + 2*c*x))^(3/2)*Hypergeometric2F1[3/4, 3, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(3*(b^2 - 4*a*c)^3
*d)

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Maple [B]  time = 0.194, size = 419, normalized size = 2.2 \begin{align*} 8\,{\frac{{c}^{2}{d}^{5} \left ( 2\,cdx+bd \right ) ^{3/2}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) ^{2}}}+10\,{\frac{{c}^{2}{d}^{5} \left ( 2\,cdx+bd \right ) ^{3/2}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{2} \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) }}+{\frac{5\,{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{9}{4}}}}+{\frac{5\,{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{9}{4}}}}-{\frac{5\,{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{9}{4}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x)

[Out]

8*c^2*d^5*(2*c*d*x+b*d)^(3/2)/(4*a*c*d^2-b^2*d^2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2+10*c^2*d^5/(4*a*c*d^
2-b^2*d^2)^2*(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)+5/4*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(9/4)*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)))+5/2*c^2*d^5/(4*a*c*d^
2-b^2*d^2)^(9/4)*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)-5/2*c^2*d^5/(4*a*c*d^
2-b^2*d^2)^(9/4)*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)

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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)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.11057, size = 4201, normalized size = 21.88 \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)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(20*(c^8*d^2/(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))^(1/4)*(a^2*b^4 - 8*
a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3
+ (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*arctan(-((b^4*c^6 - 8*a*b^2*
c^7 + 16*a^2*c^8)*(c^8*d^2/(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 - 1
29024*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))^(1/4)*sqrt
(2*c*d*x + b*d)*d - sqrt(2*c^13*d^3*x + b*c^12*d^3 + (b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4
*c^11 + 1280*a^4*b^2*c^12 - 1024*a^5*c^13)*sqrt(c^8*d^2/(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^2)*(c^8*d^2/(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))^(1/4
)*(b^4 - 8*a*b^2*c + 16*a^2*c^2))/(c^8*d^2)) + 5*(c^8*d^2/(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))^(1/4)*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*
(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^
3*b*c^2)*x)*log(125*sqrt(2*c*d*x + b*d)*c^6*d + 125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3
+ 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*(c^8*d^2/(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))^(3/4)) - 5*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5
376*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))^(1/4)*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4
)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3
*c + 16*a^3*b*c^2)*x)*log(125*sqrt(2*c*d*x + b*d)*c^6*d - 125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^
3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*(c^8*d^2/(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 - 5
89824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(3/4)) - (10*c^3*x^3 + 15*b*c^2*x^2 - b^3 + 9*a*b*c
+ 3*(b^2*c + 6*a*c^2)*x)*sqrt(2*c*d*x + b*d))/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 1
6*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 -
8*a^2*b^3*c + 16*a^3*b*c^2)*x)

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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)**(1/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.27762, size = 871, normalized size = 4.54 \begin{align*} -\frac{5 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} \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^{6} d - 12 \, \sqrt{2} a b^{4} c d + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d - 64 \, \sqrt{2} a^{3} c^{3} d} - \frac{5 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} \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^{6} d - 12 \, \sqrt{2} a b^{4} c d + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d - 64 \, \sqrt{2} a^{3} c^{3} d} + \frac{5 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} \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^{6} d - 12 \, \sqrt{2} a b^{4} c d + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d - 64 \, \sqrt{2} a^{3} c^{3} d\right )}} - \frac{5 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c^{2} \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^{6} d - 12 \, \sqrt{2} a b^{4} c d + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d - 64 \, \sqrt{2} a^{3} c^{3} d\right )}} - \frac{2 \,{\left (9 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} c^{2} d^{3} - 36 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a c^{3} d^{3} - 5 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} c^{2} d\right )}}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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^6*d - 12*sqrt(2)*a*b^4*c*d + 48*sqrt(2)*a^2*b^2*c^2*d - 64*sq
rt(2)*a^3*c^3*d) - 5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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^6*d - 12*sqrt(2)*a*b^4*c*d + 48*sqrt(2)*a^
2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) + 5/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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^6*d - 12*sqrt(2)*a*b^4*c*d
+ 48*sqrt(2)*a^2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) - 5/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*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^6*d - 12*sqr
t(2)*a*b^4*c*d + 48*sqrt(2)*a^2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) - 2*(9*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^3 - 3
6*(2*c*d*x + b*d)^(3/2)*a*c^3*d^3 - 5*(2*c*d*x + b*d)^(7/2)*c^2*d)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(b^2*d^2 -
4*a*c*d^2 - (2*c*d*x + b*d)^2)^2)