3.1307 \(\int \frac{(b d+2 c d x)^{17/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=222 \[ 110 c^2 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}+165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2} \]

[Out]

110*c^2*(b^2 - 4*a*c)*d^7*(b*d + 2*c*d*x)^(3/2) + (330*c^2*d^5*(b*d + 2*c*d*x)^(7/2))/7 - (d*(b*d + 2*c*d*x)^(
15/2))/(2*(a + b*x + c*x^2)^2) - (15*c*d^3*(b*d + 2*c*d*x)^(11/2))/(2*(a + b*x + c*x^2)) + 165*c^2*(b^2 - 4*a*
c)^(7/4)*d^(17/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 165*c^2*(b^2 - 4*a*c)^(7/4)*d^(1
7/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

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Rubi [A]  time = 0.19136, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {686, 692, 694, 329, 298, 203, 206} \[ 110 c^2 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}+165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(17/2)/(a + b*x + c*x^2)^3,x]

[Out]

110*c^2*(b^2 - 4*a*c)*d^7*(b*d + 2*c*d*x)^(3/2) + (330*c^2*d^5*(b*d + 2*c*d*x)^(7/2))/7 - (d*(b*d + 2*c*d*x)^(
15/2))/(2*(a + b*x + c*x^2)^2) - (15*c*d^3*(b*d + 2*c*d*x)^(11/2))/(2*(a + b*x + c*x^2)) + 165*c^2*(b^2 - 4*a*
c)^(7/4)*d^(17/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 165*c^2*(b^2 - 4*a*c)^(7/4)*d^(1
7/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && IntegerQ[2*p]

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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{(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{1}{2} \left (15 c d^2\right ) \int \frac{(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (165 c^2 d^4\right ) \int \frac{(b d+2 c d x)^{9/2}}{a+b x+c x^2} \, dx\\ &=\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (165 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac{(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (165 c^2 \left (b^2-4 a c\right )^2 d^8\right ) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{4} \left (165 c \left (b^2-4 a c\right )^2 d^7\right ) \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 )\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (165 c \left (b^2-4 a c\right )^2 d^7\right ) \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 )\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}-\left (165 c^2 \left (b^2-4 a c\right )^2 d^9\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 (165 c^2 \left (b^2-4 a c\right )^2 d^9\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 )\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac{330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac{d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+165 c^2 \left (b^2-4 a c\right )^{7/4} d^{17/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-165 c^2 \left (b^2-4 a c\right )^{7/4} d^{17/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\\ \end{align*}

Mathematica [C]  time = 0.324156, size = 138, normalized size = 0.62 \[ \frac{4 (d (b+2 c x))^{17/2} \left (\left (b^2-4 a c\right ) \left (77 \left (\left (b^2-4 a c\right )^2-16 c^2 (a+x (b+c x))^2 \, _2F_1\left (\frac{3}{4},3;\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )\right )-55 \left (b^2-4 a c\right ) (b+2 c x)^2+5 (b+2 c x)^4\right )+(b+2 c x)^6\right )}{7 (b+2 c x)^7 (a+x (b+c x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(17/2)/(a + b*x + c*x^2)^3,x]

[Out]

(4*(d*(b + 2*c*x))^(17/2)*((b + 2*c*x)^6 + (b^2 - 4*a*c)*(-55*(b^2 - 4*a*c)*(b + 2*c*x)^2 + 5*(b + 2*c*x)^4 +
77*((b^2 - 4*a*c)^2 - 16*c^2*(a + x*(b + c*x))^2*Hypergeometric2F1[3/4, 3, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))
))/(7*(b + 2*c*x)^7*(a + x*(b + c*x))^2)

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Maple [B]  time = 0.219, size = 1310, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(17/2)/(c*x^2+b*x+a)^3,x)

[Out]

64/7*c^2*d^5*(2*c*d*x+b*d)^(7/2)-256*c^3*d^7*(2*c*d*x+b*d)^(3/2)*a+64*c^2*d^7*(2*c*d*x+b*d)^(3/2)*b^2-864*c^4*
d^9/(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)*a^2+432*c^3*d^9/(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)*a*b^2-54*c^2*d^9/(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)*b^4
-2944*c^5*d^11/(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^3+2208*c^4*d^11/(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^2*b^2-552*c^3*d^11/(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*b^4+46*c^2*d^11/(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^6+660*c^4*d^9*2^
(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a^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)))+1320*c^4*d^9*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a^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)-1320*c^4*d^9*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a^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)-330*c^3*d^9*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*b^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)))-660*c^3*d^9*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*b^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)+660*c^3*d^9*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*b^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)+165/4*c^2*d^9*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*b^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)))+165/2*c^2*d^9*2^(1/2)/(4*a*c*d
^2-b^2*d^2)^(1/4)*b^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)-165/2*c^2*d^9*2^(1/2)/(4
*a*c*d^2-b^2*d^2)^(1/4)*b^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((2*c*d*x+b*d)^(17/2)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.7111, size = 4231, normalized size = 19.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(17/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/14*(4620*((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)*d^34)^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^
2 + a^2)*arctan(-(((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 - 215
04*a^5*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34)^(1/4)*(b^10*c^6 - 20*a*b^8*c^7 + 160*a^2*b^6*c^8
- 640*a^3*b^4*c^9 + 1280*a^4*b^2*c^10 - 1024*a^5*c^11)*sqrt(2*c*d*x + b*d)*d^25 + sqrt(2*(b^20*c^13 - 40*a*b^1
8*c^14 + 720*a^2*b^16*c^15 - 7680*a^3*b^14*c^16 + 53760*a^4*b^12*c^17 - 258048*a^5*b^10*c^18 + 860160*a^6*b^8*
c^19 - 1966080*a^7*b^6*c^20 + 2949120*a^8*b^4*c^21 - 2621440*a^9*b^2*c^22 + 1048576*a^10*c^23)*d^51*x + (b^21*
c^12 - 40*a*b^19*c^13 + 720*a^2*b^17*c^14 - 7680*a^3*b^15*c^15 + 53760*a^4*b^13*c^16 - 258048*a^5*b^11*c^17 +
860160*a^6*b^9*c^18 - 1966080*a^7*b^7*c^19 + 2949120*a^8*b^5*c^20 - 2621440*a^9*b^3*c^21 + 1048576*a^10*b*c^22
)*d^51 + sqrt((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)*d^34)*(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)*d^34)*((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)*d^34)^(1/4))/((b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 896
0*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)*d^34)) - 1155*((b^14*c^8 - 28*a*b^1
2*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)*d^34)^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*log(-4492125*(b^10*c^6 -
20*a*b^8*c^7 + 160*a^2*b^6*c^8 - 640*a^3*b^4*c^9 + 1280*a^4*b^2*c^10 - 1024*a^5*c^11)*sqrt(2*c*d*x + b*d)*d^2
5 + 4492125*((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)*d^34)^(3/4)) + 1155*((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)*d^34
)^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*log(-4492125*(b^10*c^6 - 20*a*b^8*c^7 + 160*
a^2*b^6*c^8 - 640*a^3*b^4*c^9 + 1280*a^4*b^2*c^10 - 1024*a^5*c^11)*sqrt(2*c*d*x + b*d)*d^25 - 4492125*((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)*d^34)^(3/4)) - (1024*c^7*d^8*x^7 + 3584*b*c^6*d^8*x^6 + 512*(13*b^2*c^5 - 10*a*c^
6)*d^8*x^5 + 2560*(3*b^3*c^4 - 5*a*b*c^5)*d^8*x^4 + 10*(423*b^4*c^3 - 312*a*b^2*c^4 - 1936*a^2*c^5)*d^8*x^3 +
(457*b^5*c^2 + 8120*a*b^3*c^3 - 29040*a^2*b*c^4)*d^8*x^2 - (203*b^6*c - 3350*a*b^4*c^2 + 5280*a^2*b^2*c^3 + 12
320*a^3*c^4)*d^8*x - (7*b^7 + 105*a*b^5*c - 2200*a^2*b^3*c^2 + 6160*a^3*b*c^3)*d^8)*sqrt(2*c*d*x + b*d))/(c^2*
x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^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((2*c*d*x+b*d)**(17/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.40116, size = 903, normalized size = 4.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(17/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

64*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^7 - 256*(2*c*d*x + b*d)^(3/2)*a*c^3*d^7 + 64/7*(2*c*d*x + b*d)^(7/2)*c^2*d^
5 - 165/2*sqrt(2)*(b^2*c^2*d^7 - 4*a*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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)) - 165/2*sqrt(2)*(b^2*c^2*d^7 - 4*
a*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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)) + 165/4*sqrt(2)*(b^2*c^2*d^7 - 4*a*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^
(3/4)*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
)) - 165/4*sqrt(2)*(b^2*c^2*d^7 - 4*a*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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)) + 2*(23*(2*c*d*x + b*d)^(3/2)*b^6*c^2
*d^11 - 276*(2*c*d*x + b*d)^(3/2)*a*b^4*c^3*d^11 + 1104*(2*c*d*x + b*d)^(3/2)*a^2*b^2*c^4*d^11 - 1472*(2*c*d*x
+ b*d)^(3/2)*a^3*c^5*d^11 - 27*(2*c*d*x + b*d)^(7/2)*b^4*c^2*d^9 + 216*(2*c*d*x + b*d)^(7/2)*a*b^2*c^3*d^9 -
432*(2*c*d*x + b*d)^(7/2)*a^2*c^4*d^9)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2