3.1305 \(\int \frac{1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=174 \[ \frac{14 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}+\frac{14 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}-\frac{1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}-\frac{28 c}{3 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}} \]

[Out]

(-28*c)/(3*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(3/2)) - 1/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^
2)) + (14*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*d^(5/2)) + (14*c*
ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*d^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.141952, antiderivative size = 174, 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, 212, 206, 203} \[ \frac{14 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}+\frac{14 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{5/2} \left (b^2-4 a c\right )^{11/4}}-\frac{1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{3/2}}-\frac{28 c}{3 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-28*c)/(3*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(3/2)) - 1/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^
2)) + (14*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*d^(5/2)) + (14*c*
ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*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 )^2} \, dx &=-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac{(7 c) \int \frac{1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac{28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac{(7 c) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac{7 \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 )}{2 \left (b^2-4 a c\right )^2 d^3}\\ &=-\frac{28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}-\frac{7 \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 )}{\left (b^2-4 a c\right )^2 d^3}\\ &=-\frac{28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac{(14 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 )^{5/2} d^2}+\frac{(14 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 )^{5/2} d^2}\\ &=-\frac{28 c}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )}+\frac{14 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 )^{11/4} d^{5/2}}+\frac{14 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 )^{11/4} d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0657688, size = 57, normalized size = 0.33 \[ -\frac{16 c \, _2F_1\left (-\frac{3}{4},2;\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^2 (d (b+2 c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.203, size = 404, normalized size = 2.3 \begin{align*} -{\frac{16\,c}{3\,d \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( 2\,cdx+bd \right ) ^{-{\frac{3}{2}}}}-4\,{\frac{c\sqrt{2\,cdx+bd}}{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{7\,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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{3}{4}}}}-7\,{\frac{c\sqrt{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }+7\,{\frac{c\sqrt{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{3/4}}\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)^(5/2)/(c*x^2+b*x+a)^2,x)

[Out]

-16/3*c/d/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(3/2)-4*c/d/(4*a*c-b^2)^2*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x
+4*a*c*d^2)-7/2*c/d/(4*a*c-b^2)^2/(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)))-7*c/d/(4*a*c-b^2)^2/(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)+7*c/d/(4*a*c-b^2)^2/(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)

________________________________________________________________________________________

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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.36246, size = 5273, normalized size = 30.3 \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)^2,x, algorithm="fricas")

[Out]

-1/3*(84*(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 +
(5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*
c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)*(c^4/((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 + 1081344
0*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)*d^10))^(1/4)*arctan(((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)*sqrt((b^12 - 24*a*b^10*c + 240*a^2*b^8*c^2 - 1280*a^3*b^6*c^3 + 3840
*a^4*b^4*c^4 - 6144*a^5*b^2*c^5 + 4096*a^6*c^6)*d^6*sqrt(c^4/((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)*d^10)) + 2*c^3*d*x + b*c^2*d
)*d^7*(c^4/((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^1
0*b^2*c^10 - 4194304*a^11*c^11)*d^10))^(3/4) - (b^16*c - 32*a*b^14*c^2 + 448*a^2*b^12*c^3 - 3584*a^3*b^10*c^4
+ 17920*a^4*b^8*c^5 - 57344*a^5*b^6*c^6 + 114688*a^6*b^4*c^7 - 131072*a^7*b^2*c^8 + 65536*a^8*c^9)*sqrt(2*c*d*
x + b*d)*d^7*(c^4/((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 + 11534
336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^10))^(3/4))/c^4) - 21*(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4
+ 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d
^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)
*(c^4/((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)*d^10))^(1/4)*log(7*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^3*(c^4/((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 - 4194
304*a^11*c^11)*d^10))^(1/4) + 7*sqrt(2*c*d*x + b*d)*c) + 21*(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 +
8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*
x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)*(c
^4/((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 + 1
892352*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)*d^10))^(1/4)*log(-7*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^3*(c^4/((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 - 419430
4*a^11*c^11)*d^10))^(1/4) + 7*sqrt(2*c*d*x + b*d)*c) + (28*c^2*x^2 + 28*b*c*x + 3*b^2 + 16*a*c)*sqrt(2*c*d*x +
b*d))/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5
*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^
3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.20674, size = 872, normalized size = 5.01 \begin{align*} \frac{7 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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{7 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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{7 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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{7 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{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 \, \sqrt{2 \, c d x + b d} c}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )}{\left (b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}\right )}} - \frac{16 \, c}{3 \,{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )}{\left (2 \, c d x + b d\right )}^{\frac{3}{2}}} \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)^2,x, algorithm="giac")

[Out]

7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) +
7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) +
7*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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) - 7*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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*sqrt(2*c*d*x + b*d)*c/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*(b^2*d^2
- 4*a*c*d^2 - (2*c*d*x + b*d)^2)) - 16/3*c/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*(2*c*d*x + b*d)^(3/2))