### 3.1304 $$\int \frac{1}{(b d+2 c d x)^{3/2} (a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=172 $-\frac{10 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}+\frac{10 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}-\frac{1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt{b d+2 c d x}}-\frac{20 c}{d \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}}$

[Out]

(-20*c)/((b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]) - 1/((b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)) -
(10*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(3/2)) + (10*c*ArcTanh
[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(3/2))

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Rubi [A]  time = 0.123109, antiderivative size = 172, 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 = {687, 693, 694, 329, 298, 203, 206} $-\frac{10 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}+\frac{10 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{9/4}}-\frac{1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \sqrt{b d+2 c d x}}-\frac{20 c}{d \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-20*c)/((b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]) - 1/((b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)) -
(10*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(3/2)) + (10*c*ArcTanh
[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(9/4)*d^(3/2))

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 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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{1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx &=-\frac{1}{\left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}-\frac{(5 c) \int \frac{1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac{20 c}{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}-\frac{(5 c) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{20 c}{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}-\frac{5 \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 )}{2 \left (b^2-4 a c\right )^2 d^3}\\ &=-\frac{20 c}{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}-\frac{5 \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 )}{\left (b^2-4 a c\right )^2 d^3}\\ &=-\frac{20 c}{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}+\frac{(10 c) \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 d}-\frac{(10 c) \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 d}\\ &=-\frac{20 c}{\left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}-\frac{10 c \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} d^{3/2}}+\frac{10 c \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} d^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0513546, size = 55, normalized size = 0.32 $-\frac{16 c \, _2F_1\left (-\frac{1}{4},2;\frac{3}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right )^2 \sqrt{d (b+2 c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-16*c*Hypergeometric2F1[-1/4, 2, 3/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4*a*c)^2*d*Sqrt[d*(b + 2*c*x)])

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Maple [B]  time = 0.227, size = 404, normalized size = 2.4 \begin{align*} -16\,{\frac{c}{d \left ( 4\,ac-{b}^{2} \right ) ^{2}\sqrt{2\,cdx+bd}}}-4\,{\frac{c \left ( 2\,cdx+bd \right ) ^{3/2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) }}-{\frac{5\,c\sqrt{2}}{2\,d \left ( 4\,ac-{b}^{2} \right ) ^{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}}}}}-5\,{\frac{c\sqrt{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{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 ) }+5\,{\frac{c\sqrt{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{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(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x)

[Out]

-16*c/d/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(1/2)-4*c/d/(4*a*c-b^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)-5/2*c/d/(4*a*c-b^2)^2*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*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*c/d/(4*a*c-b^2)^2*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*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*c/d/(4*a*c-b^2)^2*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*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(1/(2*c*d*x+b*d)^(3/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.33261, size = 4028, normalized size = 23.42 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(20*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 -
6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4)*arctan((sqrt(2*c^7*d*x + b*c^6*d + (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^4*sqrt(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6)))*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d*(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(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*(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4
))/c^4) + 5*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 +
(b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4)*log(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)*d^5*(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(3/4) + 125*sqrt(2*c*d*x + b*d)*c^3) -
5*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 - 6
*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)*(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(1/4)*log(-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)*d^5*(c^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 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^6))^(3/4) + 125*sqrt(2*c*d*x + b*d)*c^3) - (20*c^2
*x^2 + 20*b*c*x + b^2 + 16*a*c)*sqrt(2*c*d*x + b*d))/(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*
c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^2*x^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3
*b*c^2)*d^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(1/(2*c*d*x+b*d)**(3/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.20927, size = 878, normalized size = 5.1 \begin{align*} \frac{5 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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 )}{b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} + \frac{5 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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 )}{b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} - \frac{5 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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 )}{\sqrt{2} b^{6} d^{3} - 12 \, \sqrt{2} a b^{4} c d^{3} + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d^{3} - 64 \, \sqrt{2} a^{3} c^{3} d^{3}} + \frac{5 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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 )}{\sqrt{2} b^{6} d^{3} - 12 \, \sqrt{2} a b^{4} c d^{3} + 48 \, \sqrt{2} a^{2} b^{2} c^{2} d^{3} - 64 \, \sqrt{2} a^{3} c^{3} d^{3}} - \frac{4 \,{\left (4 \, b^{2} c d^{2} - 16 \, a c^{2} d^{2} - 5 \,{\left (2 \, c d x + b d\right )}^{2} c\right )}}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )}{\left (\sqrt{2 \, c d x + b d} b^{2} d^{2} - 4 \, \sqrt{2 \, c d x + b d} a c d^{2} -{\left (2 \, c d x + b d\right )}^{\frac{5}{2}}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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