3.1317 $$\int \frac{1}{(b d+2 c d x)^{3/2} (a+b x+c x^2)^3} \, dx$$

Optimal. Leaf size=223 $\frac{45 c^2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{13/4}}-\frac{45 c^2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{13/4}}+\frac{90 c^2}{d \left (b^2-4 a c\right )^3 \sqrt{b d+2 c d x}}+\frac{9 c}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right ) \sqrt{b d+2 c d x}}-\frac{1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt{b d+2 c d x}}$

[Out]

(90*c^2)/((b^2 - 4*a*c)^3*d*Sqrt[b*d + 2*c*d*x]) - 1/(2*(b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^
2) + (9*c)/(2*(b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)) + (45*c^2*ArcTan[Sqrt[d*(b + 2*c*x)]/((
b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(3/2)) - (45*c^2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a
*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(3/2))

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Rubi [A]  time = 0.168262, antiderivative size = 223, 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 = {687, 693, 694, 329, 298, 203, 206} $\frac{45 c^2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{13/4}}-\frac{45 c^2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{3/2} \left (b^2-4 a c\right )^{13/4}}+\frac{90 c^2}{d \left (b^2-4 a c\right )^3 \sqrt{b d+2 c d x}}+\frac{9 c}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right ) \sqrt{b d+2 c d x}}-\frac{1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 \sqrt{b d+2 c d x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(90*c^2)/((b^2 - 4*a*c)^3*d*Sqrt[b*d + 2*c*d*x]) - 1/(2*(b^2 - 4*a*c)*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^
2) + (9*c)/(2*(b^2 - 4*a*c)^2*d*Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)) + (45*c^2*ArcTan[Sqrt[d*(b + 2*c*x)]/((
b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(3/2)) - (45*c^2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a
*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*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 )^3} \, dx &=-\frac{1}{2 \left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2}-\frac{(9 c) \int \frac{1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac{1}{2 \left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac{9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}+\frac{\left (45 c^2\right ) \int \frac{1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{2 \left (b^2-4 a c\right )^2}\\ &=\frac{90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt{b d+2 c d x}}-\frac{1}{2 \left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac{9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}+\frac{\left (45 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 )^3 d^2}\\ &=\frac{90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt{b d+2 c d x}}-\frac{1}{2 \left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac{9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}+\frac{(45 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 )^3 d^3}\\ &=\frac{90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt{b d+2 c d x}}-\frac{1}{2 \left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac{9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}+\frac{(45 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 )^3 d^3}\\ &=\frac{90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt{b d+2 c d x}}-\frac{1}{2 \left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac{9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}-\frac{\left (45 c^2\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 )^3 d}+\frac{\left (45 c^2\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 )^3 d}\\ &=\frac{90 c^2}{\left (b^2-4 a c\right )^3 d \sqrt{b d+2 c d x}}-\frac{1}{2 \left (b^2-4 a c\right ) d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2}+\frac{9 c}{2 \left (b^2-4 a c\right )^2 d \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )}+\frac{45 c^2 \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 )^{13/4} d^{3/2}}-\frac{45 c^2 \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 )^{13/4} d^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0689023, size = 57, normalized size = 0.26 $\frac{64 c^2 \, _2F_1\left (-\frac{1}{4},3;\frac{3}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right )^3 \sqrt{d (b+2 c x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.209, size = 534, normalized size = 2.4 \begin{align*} -64\,{\frac{{c}^{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{3}\sqrt{2\,cdx+bd}}}-26\,{\frac{{c}^{2} \left ( 2\,cdx+bd \right ) ^{7/2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) ^{2}}}-136\,{\frac{{c}^{3}d \left ( 2\,cdx+bd \right ) ^{3/2}a}{ \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) ^{2}}}+34\,{\frac{{c}^{2}d \left ( 2\,cdx+bd \right ) ^{3/2}{b}^{2}}{ \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) ^{2}}}-{\frac{45\,{c}^{2}\sqrt{2}}{4\,d \left ( 4\,ac-{b}^{2} \right ) ^{3}}\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}}}}}-{\frac{45\,{c}^{2}\sqrt{2}}{2\,d \left ( 4\,ac-{b}^{2} \right ) ^{3}}\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+{\frac{45\,{c}^{2}\sqrt{2}}{2\,d \left ( 4\,ac-{b}^{2} \right ) ^{3}}\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}} \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)^3,x)

[Out]

-64*c^2/d/(4*a*c-b^2)^3/(2*c*d*x+b*d)^(1/2)-26*c^2/d/(4*a*c-b^2)^3/(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)-136*c^3*d/(4*a*c-b^2)^3/(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+34*c^2*
d/(4*a*c-b^2)^3/(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^2-45/4*c^2/d/(4*a*c-b^2)^3*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)))
-45/2*c^2/d/(4*a*c-b^2)^3*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)+45/2*c^2/d/(4*a*c-b^2)^3*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.47029, size = 7175, normalized size = 32.17 \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)^3,x, algorithm="fricas")

[Out]

1/2*(180*(2*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^
2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*d^
2*x^3 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x^2 + 2*(a*b^8 - 11*a^2*b^6*c
+ 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c
^3)*d^2)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^
5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 29
9892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*ar
ctan(-(sqrt(2*c^13*d*x + b*c^12*d + (b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a
^4*b^6*c^12 - 21504*a^5*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*sqrt(c^8/((b^26 - 52*a*b^24*c + 1248*a
^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 2811494
4*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*
c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))*d^4)*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^
3)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16
*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 29989273
6*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*d - (b^6
*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2*c^8 - 64*a^3*c^9)*sqrt(2*c*d*x + b*d)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^
22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7
*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11
+ 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*d)/c^8) - 45*(2*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b
^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*(b^8*c -
11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 22
4*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x^2 + 2*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^
4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*
b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a
^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^1
1 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*log(91125*(b^20 - 40*a*b^18*c + 720*a^2*b^16*c^2
- 7680*a^3*b^14*c^3 + 53760*a^4*b^12*c^4 - 258048*a^5*b^10*c^5 + 860160*a^6*b^8*c^6 - 1966080*a^7*b^6*c^7 + 2
949120*a^8*b^4*c^8 - 2621440*a^9*b^2*c^9 + 1048576*a^10*c^10)*(c^8/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 -
18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 - 28114944*a^7*b^12*c^7
+ 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^11*b^4*c^11 + 2181038
08*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(3/4)*d^5 + 91125*sqrt(2*c*d*x + b*d)*c^6) + 45*(2*(b^6*c^3 - 12*
a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*
d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + (b^9 - 6*a*b^7*c -
24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x^2 + 2*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4
*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*d^2)*(c^8/((b^26 - 52*
a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b
^14*c^6 - 28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 32
7155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(1/4)*log(-91125*(b^20 - 40*a*b^18
*c + 720*a^2*b^16*c^2 - 7680*a^3*b^14*c^3 + 53760*a^4*b^12*c^4 - 258048*a^5*b^10*c^5 + 860160*a^6*b^8*c^6 - 19
66080*a^7*b^6*c^7 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*c^9 + 1048576*a^10*c^10)*(c^8/((b^26 - 52*a*b^24*c +
1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 183040*a^4*b^18*c^4 - 1317888*a^5*b^16*c^5 + 7028736*a^6*b^14*c^6 -
28114944*a^7*b^12*c^7 + 84344832*a^8*b^10*c^8 - 187432960*a^9*b^8*c^9 + 299892736*a^10*b^6*c^10 - 327155712*a^
11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^6))^(3/4)*d^5 + 91125*sqrt(2*c*d*x + b*d)*c^6) +
(180*c^4*x^4 + 360*b*c^3*x^3 - b^4 + 17*a*b^2*c + 128*a^2*c^2 + 27*(7*b^2*c^2 + 12*a*c^3)*x^2 + 9*(b^3*c + 36
*a*b*c^2)*x)*sqrt(2*c*d*x + b*d))/(2*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^2*x^5 + 5*(b^7*c
^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^2*x^4 + 4*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3
*b^2*c^4 - 64*a^4*c^5)*d^2*x^3 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*d^2*x^2
+ 2*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*d^2*x + (a^2*b^7 - 12*a^3*b^5*c + 48
*a^4*b^3*c^2 - 64*a^5*b*c^3)*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)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.2384, size = 1098, normalized size = 4.92 \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)^3,x, algorithm="giac")

[Out]

-45*(-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^8*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3
- 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4*d^3) - 45*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*arctan(-1/2*sqr
t(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^8
*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4
*d^3) + 45/2*(-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^8*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*
d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4*d^3) - 45/2*(-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^8
*d^3 - 16*sqrt(2)*a*b^6*c*d^3 + 96*sqrt(2)*a^2*b^4*c^2*d^3 - 256*sqrt(2)*a^3*b^2*c^3*d^3 + 256*sqrt(2)*a^4*c^4
*d^3) + 64*c^2/((b^6*d - 12*a*b^4*c*d + 48*a^2*b^2*c^2*d - 64*a^3*c^3*d)*sqrt(2*c*d*x + b*d)) - 2*(17*(2*c*d*x
+ b*d)^(3/2)*b^2*c^2*d^2 - 68*(2*c*d*x + b*d)^(3/2)*a*c^3*d^2 - 13*(2*c*d*x + b*d)^(7/2)*c^2)/((b^6*d - 12*a*
b^4*c*d + 48*a^2*b^2*c^2*d - 64*a^3*c^3*d)*(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2)