### 3.1318 $$\int \frac{1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)^3} \, dx$$

Optimal. Leaf size=225 $-\frac{77 c^2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{15/4}}-\frac{77 c^2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{15/4}}+\frac{154 c^2}{3 d \left (b^2-4 a c\right )^3 (b d+2 c d x)^{3/2}}+\frac{11 c}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}-\frac{1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}$

[Out]

(154*c^2)/(3*(b^2 - 4*a*c)^3*d*(b*d + 2*c*d*x)^(3/2)) - 1/(2*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x +
c*x^2)^2) + (11*c)/(2*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)) - (77*c^2*ArcTan[Sqrt[d*(b +
2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(15/4)*d^(5/2)) - (77*c^2*ArcTanh[Sqrt[d*(b + 2*c*x)]/(
(b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(15/4)*d^(5/2))

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Rubi [A]  time = 0.191198, antiderivative size = 225, 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, 212, 206, 203} $-\frac{77 c^2 \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{15/4}}-\frac{77 c^2 \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{15/4}}+\frac{154 c^2}{3 d \left (b^2-4 a c\right )^3 (b d+2 c d x)^{3/2}}+\frac{11 c}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}-\frac{1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^3),x]

[Out]

(154*c^2)/(3*(b^2 - 4*a*c)^3*d*(b*d + 2*c*d*x)^(3/2)) - 1/(2*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x +
c*x^2)^2) + (11*c)/(2*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)) - (77*c^2*ArcTan[Sqrt[d*(b +
2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(15/4)*d^(5/2)) - (77*c^2*ArcTanh[Sqrt[d*(b + 2*c*x)]/(
(b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(15/4)*d^(5/2))

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 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{1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^3} \, dx &=-\frac{1}{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}-\frac{(11 c) \int \frac{1}{(b d+2 c d x)^{5/2} \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 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}+\frac{11 c}{2 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac{\left (77 c^2\right ) \int \frac{1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx}{2 \left (b^2-4 a c\right )^2}\\ &=\frac{154 c^2}{3 \left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2}}-\frac{1}{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}+\frac{11 c}{2 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac{\left (77 c^2\right ) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{2 \left (b^2-4 a c\right )^3 d^2}\\ &=\frac{154 c^2}{3 \left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2}}-\frac{1}{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}+\frac{11 c}{2 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac{(77 c) \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 )}{4 \left (b^2-4 a c\right )^3 d^3}\\ &=\frac{154 c^2}{3 \left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2}}-\frac{1}{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}+\frac{11 c}{2 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac{(77 c) \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 )}{2 \left (b^2-4 a c\right )^3 d^3}\\ &=\frac{154 c^2}{3 \left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2}}-\frac{1}{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}+\frac{11 c}{2 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac{\left (77 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 )^{7/2} d^2}-\frac{\left (77 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 )^{7/2} d^2}\\ &=\frac{154 c^2}{3 \left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2}}-\frac{1}{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2}+\frac{11 c}{2 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac{77 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 )^{15/4} d^{5/2}}-\frac{77 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 )^{15/4} d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0837658, size = 59, normalized size = 0.26 $\frac{64 c^2 \, _2F_1\left (-\frac{3}{4},3;\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^3 (d (b+2 c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^3),x]

[Out]

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

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Maple [B]  time = 0.206, size = 534, normalized size = 2.4 \begin{align*} -{\frac{64\,{c}^{2}}{3\,d \left ( 4\,ac-{b}^{2} \right ) ^{3}} \left ( 2\,cdx+bd \right ) ^{-{\frac{3}{2}}}}-30\,{\frac{{c}^{2} \left ( 2\,cdx+bd \right ) ^{5/2}}{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}}}-152\,{\frac{{c}^{3}d\sqrt{2\,cdx+bd}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}}}+38\,{\frac{{c}^{2}d\sqrt{2\,cdx+bd}{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{77\,{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}-{\frac{77\,{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}+{\frac{77\,{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^3,x)

[Out]

-64/3*c^2/d/(4*a*c-b^2)^3/(2*c*d*x+b*d)^(3/2)-30*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)^(5/2)-152*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)^(1/2)*a+38*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)^(1/2)*b^2-77/4*c^2/d/(4*a*c-b^2)^3/(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)
))-77/2*c^2/d/(4*a*c-b^2)^3/(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)+77/2*c^2/d/(4*a*c-b^2)^3/(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)

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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)^(5/2)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.45367, size = 8675, normalized size = 38.56 \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)^(5/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/6*(924*(4*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^3*x^6 + 12*(b^7*c^3 - 12*a*b^5*c^4 + 48*
a^2*b^3*c^5 - 64*a^3*b*c^6)*d^3*x^5 + (13*b^8*c^2 - 148*a*b^6*c^3 + 528*a^2*b^4*c^4 - 448*a^3*b^2*c^5 - 512*a^
4*c^6)*d^3*x^4 + 2*(3*b^9*c - 28*a*b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^3*b^3*c^4 - 512*a^4*b*c^5)*d^3*x^3 + (b^10
- 2*a*b^8*c - 68*a^2*b^6*c^2 + 368*a^3*b^4*c^3 - 448*a^4*b^2*c^4 - 256*a^5*c^5)*d^3*x^2 + 2*(a*b^9 - 10*a^2*b
^7*c + 24*a^3*b^5*c^2 + 32*a^4*b^3*c^3 - 128*a^5*b*c^4)*d^3*x + (a^2*b^8 - 12*a^3*b^6*c + 48*a^4*b^4*c^2 - 64*
a^5*b^2*c^3)*d^3)*(c^8/((b^30 - 60*a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3*b^24*c^3 + 349440*a^4*b^22*c^4 - 3
075072*a^5*b^20*c^5 + 20500480*a^6*b^18*c^6 - 105431040*a^7*b^16*c^7 + 421724160*a^8*b^14*c^8 - 1312030720*a^9
*b^12*c^9 + 3148873728*a^10*b^10*c^10 - 5725224960*a^11*b^8*c^11 + 7633633280*a^12*b^6*c^12 - 7046430720*a^13*
b^4*c^13 + 4026531840*a^14*b^2*c^14 - 1073741824*a^15*c^15)*d^10))^(1/4)*arctan(-((b^22 - 44*a*b^20*c + 880*a^
2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^
7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*sqrt((b^
16 - 32*a*b^14*c + 448*a^2*b^12*c^2 - 3584*a^3*b^10*c^3 + 17920*a^4*b^8*c^4 - 57344*a^5*b^6*c^5 + 114688*a^6*b
^4*c^6 - 131072*a^7*b^2*c^7 + 65536*a^8*c^8)*d^6*sqrt(c^8/((b^30 - 60*a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3
*b^24*c^3 + 349440*a^4*b^22*c^4 - 3075072*a^5*b^20*c^5 + 20500480*a^6*b^18*c^6 - 105431040*a^7*b^16*c^7 + 4217
24160*a^8*b^14*c^8 - 1312030720*a^9*b^12*c^9 + 3148873728*a^10*b^10*c^10 - 5725224960*a^11*b^8*c^11 + 76336332
80*a^12*b^6*c^12 - 7046430720*a^13*b^4*c^13 + 4026531840*a^14*b^2*c^14 - 1073741824*a^15*c^15)*d^10)) + 2*c^5*
d*x + b*c^4*d)*d^7*(c^8/((b^30 - 60*a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3*b^24*c^3 + 349440*a^4*b^22*c^4 -
3075072*a^5*b^20*c^5 + 20500480*a^6*b^18*c^6 - 105431040*a^7*b^16*c^7 + 421724160*a^8*b^14*c^8 - 1312030720*a^
9*b^12*c^9 + 3148873728*a^10*b^10*c^10 - 5725224960*a^11*b^8*c^11 + 7633633280*a^12*b^6*c^12 - 7046430720*a^13
*b^4*c^13 + 4026531840*a^14*b^2*c^14 - 1073741824*a^15*c^15)*d^10))^(3/4) - (b^22*c^2 - 44*a*b^20*c^3 + 880*a^
2*b^18*c^4 - 10560*a^3*b^16*c^5 + 84480*a^4*b^14*c^6 - 473088*a^5*b^12*c^7 + 1892352*a^6*b^10*c^8 - 5406720*a^
7*b^8*c^9 + 10813440*a^8*b^6*c^10 - 14417920*a^9*b^4*c^11 + 11534336*a^10*b^2*c^12 - 4194304*a^11*c^13)*sqrt(2
*c*d*x + b*d)*d^7*(c^8/((b^30 - 60*a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3*b^24*c^3 + 349440*a^4*b^22*c^4 - 3
075072*a^5*b^20*c^5 + 20500480*a^6*b^18*c^6 - 105431040*a^7*b^16*c^7 + 421724160*a^8*b^14*c^8 - 1312030720*a^9
*b^12*c^9 + 3148873728*a^10*b^10*c^10 - 5725224960*a^11*b^8*c^11 + 7633633280*a^12*b^6*c^12 - 7046430720*a^13*
b^4*c^13 + 4026531840*a^14*b^2*c^14 - 1073741824*a^15*c^15)*d^10))^(3/4))/c^8) + 231*(4*(b^6*c^4 - 12*a*b^4*c^
5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^3*x^6 + 12*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^3*x^5
+ (13*b^8*c^2 - 148*a*b^6*c^3 + 528*a^2*b^4*c^4 - 448*a^3*b^2*c^5 - 512*a^4*c^6)*d^3*x^4 + 2*(3*b^9*c - 28*a*
b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^3*b^3*c^4 - 512*a^4*b*c^5)*d^3*x^3 + (b^10 - 2*a*b^8*c - 68*a^2*b^6*c^2 + 368
*a^3*b^4*c^3 - 448*a^4*b^2*c^4 - 256*a^5*c^5)*d^3*x^2 + 2*(a*b^9 - 10*a^2*b^7*c + 24*a^3*b^5*c^2 + 32*a^4*b^3*
c^3 - 128*a^5*b*c^4)*d^3*x + (a^2*b^8 - 12*a^3*b^6*c + 48*a^4*b^4*c^2 - 64*a^5*b^2*c^3)*d^3)*(c^8/((b^30 - 60*
a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3*b^24*c^3 + 349440*a^4*b^22*c^4 - 3075072*a^5*b^20*c^5 + 20500480*a^6*
b^18*c^6 - 105431040*a^7*b^16*c^7 + 421724160*a^8*b^14*c^8 - 1312030720*a^9*b^12*c^9 + 3148873728*a^10*b^10*c^
10 - 5725224960*a^11*b^8*c^11 + 7633633280*a^12*b^6*c^12 - 7046430720*a^13*b^4*c^13 + 4026531840*a^14*b^2*c^14
- 1073741824*a^15*c^15)*d^10))^(1/4)*log(77*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^
4)*d^3*(c^8/((b^30 - 60*a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3*b^24*c^3 + 349440*a^4*b^22*c^4 - 3075072*a^5*
b^20*c^5 + 20500480*a^6*b^18*c^6 - 105431040*a^7*b^16*c^7 + 421724160*a^8*b^14*c^8 - 1312030720*a^9*b^12*c^9 +
3148873728*a^10*b^10*c^10 - 5725224960*a^11*b^8*c^11 + 7633633280*a^12*b^6*c^12 - 7046430720*a^13*b^4*c^13 +
4026531840*a^14*b^2*c^14 - 1073741824*a^15*c^15)*d^10))^(1/4) + 77*sqrt(2*c*d*x + b*d)*c^2) - 231*(4*(b^6*c^4
- 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^3*x^6 + 12*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b
*c^6)*d^3*x^5 + (13*b^8*c^2 - 148*a*b^6*c^3 + 528*a^2*b^4*c^4 - 448*a^3*b^2*c^5 - 512*a^4*c^6)*d^3*x^4 + 2*(3*
b^9*c - 28*a*b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^3*b^3*c^4 - 512*a^4*b*c^5)*d^3*x^3 + (b^10 - 2*a*b^8*c - 68*a^2*
b^6*c^2 + 368*a^3*b^4*c^3 - 448*a^4*b^2*c^4 - 256*a^5*c^5)*d^3*x^2 + 2*(a*b^9 - 10*a^2*b^7*c + 24*a^3*b^5*c^2
+ 32*a^4*b^3*c^3 - 128*a^5*b*c^4)*d^3*x + (a^2*b^8 - 12*a^3*b^6*c + 48*a^4*b^4*c^2 - 64*a^5*b^2*c^3)*d^3)*(c^8
/((b^30 - 60*a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3*b^24*c^3 + 349440*a^4*b^22*c^4 - 3075072*a^5*b^20*c^5 +
20500480*a^6*b^18*c^6 - 105431040*a^7*b^16*c^7 + 421724160*a^8*b^14*c^8 - 1312030720*a^9*b^12*c^9 + 3148873728
*a^10*b^10*c^10 - 5725224960*a^11*b^8*c^11 + 7633633280*a^12*b^6*c^12 - 7046430720*a^13*b^4*c^13 + 4026531840*
a^14*b^2*c^14 - 1073741824*a^15*c^15)*d^10))^(1/4)*log(-77*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^
3 + 256*a^4*c^4)*d^3*(c^8/((b^30 - 60*a*b^28*c + 1680*a^2*b^26*c^2 - 29120*a^3*b^24*c^3 + 349440*a^4*b^22*c^4
- 3075072*a^5*b^20*c^5 + 20500480*a^6*b^18*c^6 - 105431040*a^7*b^16*c^7 + 421724160*a^8*b^14*c^8 - 1312030720*
a^9*b^12*c^9 + 3148873728*a^10*b^10*c^10 - 5725224960*a^11*b^8*c^11 + 7633633280*a^12*b^6*c^12 - 7046430720*a^
13*b^4*c^13 + 4026531840*a^14*b^2*c^14 - 1073741824*a^15*c^15)*d^10))^(1/4) + 77*sqrt(2*c*d*x + b*d)*c^2) - (3
08*c^4*x^4 + 616*b*c^3*x^3 - 3*b^4 + 57*a*b^2*c + 128*a^2*c^2 + 11*(31*b^2*c^2 + 44*a*c^3)*x^2 + 11*(3*b^3*c +
44*a*b*c^2)*x)*sqrt(2*c*d*x + b*d))/(4*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^3*x^6 + 12*(b
^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^3*x^5 + (13*b^8*c^2 - 148*a*b^6*c^3 + 528*a^2*b^4*c^4
- 448*a^3*b^2*c^5 - 512*a^4*c^6)*d^3*x^4 + 2*(3*b^9*c - 28*a*b^7*c^2 + 48*a^2*b^5*c^3 + 192*a^3*b^3*c^4 - 512
*a^4*b*c^5)*d^3*x^3 + (b^10 - 2*a*b^8*c - 68*a^2*b^6*c^2 + 368*a^3*b^4*c^3 - 448*a^4*b^2*c^4 - 256*a^5*c^5)*d^
3*x^2 + 2*(a*b^9 - 10*a^2*b^7*c + 24*a^3*b^5*c^2 + 32*a^4*b^3*c^3 - 128*a^5*b*c^4)*d^3*x + (a^2*b^8 - 12*a^3*b
^6*c + 48*a^4*b^4*c^2 - 64*a^5*b^2*c^3)*d^3)

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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)**(5/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.27741, size = 1098, normalized size = 4.88 \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)^(5/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-77*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) - 77*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) - 77/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) + 77/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) + 64/3*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)*(2*c*d*x + b*d)^(3/2)) - 2*(19*sqrt
(2*c*d*x + b*d)*b^2*c^2*d^2 - 76*sqrt(2*c*d*x + b*d)*a*c^3*d^2 - 15*(2*c*d*x + b*d)^(5/2)*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)