3.1306 \(\int \frac{1}{(b d+2 c d x)^{7/2} (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=203 \[ -\frac{36 c}{d^3 \left (b^2-4 a c\right )^3 \sqrt{b d+2 c d x}}-\frac{18 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}+\frac{18 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}-\frac{1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{5/2}}-\frac{36 c}{5 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}} \]

[Out]

(-36*c)/(5*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(5/2)) - (36*c)/((b^2 - 4*a*c)^3*d^3*Sqrt[b*d + 2*c*d*x]) - 1/((b
^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)) - (18*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)
*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(7/2)) + (18*c*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])
/((b^2 - 4*a*c)^(13/4)*d^(7/2))

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Rubi [A]  time = 0.171861, antiderivative size = 203, 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} \[ -\frac{36 c}{d^3 \left (b^2-4 a c\right )^3 \sqrt{b d+2 c d x}}-\frac{18 c \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}+\frac{18 c \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{13/4}}-\frac{1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) (b d+2 c d x)^{5/2}}-\frac{36 c}{5 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-36*c)/(5*(b^2 - 4*a*c)^2*d*(b*d + 2*c*d*x)^(5/2)) - (36*c)/((b^2 - 4*a*c)^3*d^3*Sqrt[b*d + 2*c*d*x]) - 1/((b
^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)) - (18*c*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)
*Sqrt[d])])/((b^2 - 4*a*c)^(13/4)*d^(7/2)) + (18*c*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])
/((b^2 - 4*a*c)^(13/4)*d^(7/2))

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)^{7/2} \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)^{5/2} \left (a+b x+c x^2\right )}-\frac{(9 c) \int \frac{1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac{36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac{(9 c) \int \frac{1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac{36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac{(9 c) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3 d^4}\\ &=-\frac{36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac{36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac{9 \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 )^3 d^5}\\ &=-\frac{36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac{36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac{9 \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 )^3 d^5}\\ &=-\frac{36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac{36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}+\frac{(18 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 )^3 d^3}-\frac{(18 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 )^3 d^3}\\ &=-\frac{36 c}{5 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{5/2}}-\frac{36 c}{\left (b^2-4 a c\right )^3 d^3 \sqrt{b d+2 c d x}}-\frac{1}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )}-\frac{18 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 )^{13/4} d^{7/2}}+\frac{18 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 )^{13/4} d^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.081183, size = 57, normalized size = 0.28 \[ -\frac{16 c \, _2F_1\left (-\frac{5}{4},2;-\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{5 d \left (b^2-4 a c\right )^2 (d (b+2 c x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^2),x]

[Out]

(-16*c*Hypergeometric2F1[-5/4, 2, -1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(5*(b^2 - 4*a*c)^2*d*(d*(b + 2*c*x))^(5/
2))

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Maple [B]  time = 0.202, size = 433, normalized size = 2.1 \begin{align*} -{\frac{16\,c}{5\,d \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( 2\,cdx+bd \right ) ^{-{\frac{5}{2}}}}+32\,{\frac{c}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3}\sqrt{2\,cdx+bd}}}+4\,{\frac{c \left ( 2\,cdx+bd \right ) ^{3/2}}{{d}^{3} \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 ) }}+{\frac{9\,c\sqrt{2}}{2\,{d}^{3} \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}}}}}+9\,{\frac{c\sqrt{2}}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3}\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 ) }-9\,{\frac{c\sqrt{2}}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3}\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)^(7/2)/(c*x^2+b*x+a)^2,x)

[Out]

-16/5*c/d/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(5/2)+32*c/d^3/(4*a*c-b^2)^3/(2*c*d*x+b*d)^(1/2)+4*c/d^3/(4*a*c-b^2)^3*(
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)+9/2*c/d^3/(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)))+9*c/d^3/(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)-9*c/d^3/(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)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.10764, size = 7275, normalized size = 35.84 \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)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-1/5*(180*(8*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*
a^2*b^3*c^5 - 64*a^3*b*c^6)*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a
^4*c^6)*d^4*x^3 + (7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10
- 6*a*b^8*c - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*
c^2 - 64*a^4*b^3*c^3)*d^4)*(c^4/((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^1
8*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 - 18743296
0*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^14))^(1/4)*arctan(-(sqrt((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*
c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^8*sqrt(c^4/((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^14)) + 2*c^7*d*x + b*c^6*d)*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c
^2 - 64*a^3*c^3)*d^3*(c^4/((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^
14))^(1/4) - (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*sqrt(2*c*d*x + b*d)*d^3*(c^4/((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 - 327
155712*a^11*b^4*c^11 + 218103808*a^12*b^2*c^12 - 67108864*a^13*c^13)*d^14))^(1/4))/c^4) - 45*(8*(b^6*c^4 - 12*
a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)
*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*x^3 + (7*b^9*c
- 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10 - 6*a*b^8*c - 24*a^2*b^6*c
^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)*
(c^4/((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^14))^(1/4)*log(729*(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 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*c^9 + 1048576*a^10*c^10)*d^11*(c^4/((
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 + 702
8736*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^14))^(3/4) + 729*sqrt(2*c*d*
x + b*d)*c^3) + 45*(8*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*
c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^
5 - 256*a^4*c^6)*d^4*x^3 + (7*b^9*c - 72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^
2 + (b^10 - 6*a*b^8*c - 24*a^2*b^6*c^2 + 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48
*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)*(c^4/((b^26 - 52*a*b^24*c + 1248*a^2*b^22*c^2 - 18304*a^3*b^20*c^3 + 18304
0*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 - 6710886
4*a^13*c^13)*d^14))^(1/4)*log(-729*(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 + 2949120*a^8*b^4*c^8 - 2621440*a^9*b^2*
c^9 + 1048576*a^10*c^10)*d^11*(c^4/((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 - 18743
2960*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^14))^(3/4) + 729*sqrt(2*c*d*x + b*d)*c^3) + (720*c^4*x^4 + 1440*b*c^3*x^3 + 5*b^4 + 176*a*b^2*c - 64*
a^2*c^2 + 72*(13*b^2*c^2 + 8*a*c^3)*x^2 + 72*(3*b^3*c + 8*a*b*c^2)*x)*sqrt(2*c*d*x + b*d))/(8*(b^6*c^4 - 12*a*
b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^4*x^5 + 20*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*d
^4*x^4 + 2*(9*b^8*c^2 - 104*a*b^6*c^3 + 384*a^2*b^4*c^4 - 384*a^3*b^2*c^5 - 256*a^4*c^6)*d^4*x^3 + (7*b^9*c -
72*a*b^7*c^2 + 192*a^2*b^5*c^3 + 128*a^3*b^3*c^4 - 768*a^4*b*c^5)*d^4*x^2 + (b^10 - 6*a*b^8*c - 24*a^2*b^6*c^2
+ 224*a^3*b^4*c^3 - 384*a^4*b^2*c^4)*d^4*x + (a*b^9 - 12*a^2*b^7*c + 48*a^3*b^5*c^2 - 64*a^4*b^3*c^3)*d^4)

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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)**(7/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.21528, size = 1052, normalized size = 5.18 \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)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

9*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^8*d^5 - 16*a*b^6*c*d^5 + 96*a^2*b^4*c^2*d^5 - 256*a^3*b^2*c^3*d^
5 + 256*a^4*c^4*d^5) + 9*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^8*d^5 - 16*a*b^6*c*d^5 + 96*a^2*b^4*c^2*
d^5 - 256*a^3*b^2*c^3*d^5 + 256*a^4*c^4*d^5) - 9*(-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^8*d^5 - 16*sqrt(2)*a*b
^6*c*d^5 + 96*sqrt(2)*a^2*b^4*c^2*d^5 - 256*sqrt(2)*a^3*b^2*c^3*d^5 + 256*sqrt(2)*a^4*c^4*d^5) + 9*(-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^8*d^5 - 16*sqrt(2)*a*b^6*c*d^5 + 96*sqrt(2)*a^2*b^4*c^2*d^5 - 256*sqrt(2)*a^3*b^2*
c^3*d^5 + 256*sqrt(2)*a^4*c^4*d^5) + 4*(2*c*d*x + b*d)^(3/2)*c/((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)*(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)) - 16/5*(b^2*c*d^2 - 4*a*c^2*d^2 + 10*(2*c*d*x +
b*d)^2*c)/((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)*(2*c*d*x + b*d)^(5/2))