3.1295 \(\int \frac{(b d+2 c d x)^{15/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=210 \[ 52 c d^7 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}+\frac{52}{5} c d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}-26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2} \]

[Out]

52*c*(b^2 - 4*a*c)^2*d^7*Sqrt[b*d + 2*c*d*x] + (52*c*(b^2 - 4*a*c)*d^5*(b*d + 2*c*d*x)^(5/2))/5 + (52*c*d^3*(b
*d + 2*c*d*x)^(9/2))/9 - (d*(b*d + 2*c*d*x)^(13/2))/(a + b*x + c*x^2) - 26*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*ArcT
an[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 26*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*ArcTanh[Sqrt[d*(b +
2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

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Rubi [A]  time = 0.179978, antiderivative size = 210, 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, 212, 206, 203} \[ 52 c d^7 \left (b^2-4 a c\right )^2 \sqrt{b d+2 c d x}+\frac{52}{5} c d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}-26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x]

[Out]

52*c*(b^2 - 4*a*c)^2*d^7*Sqrt[b*d + 2*c*d*x] + (52*c*(b^2 - 4*a*c)*d^5*(b*d + 2*c*d*x)^(5/2))/5 + (52*c*d^3*(b
*d + 2*c*d*x)^(9/2))/9 - (d*(b*d + 2*c*d*x)^(13/2))/(a + b*x + c*x^2) - 26*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*ArcT
an[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 26*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*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 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{(b d+2 c d x)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c d^2\right ) \int \frac{(b d+2 c d x)^{11/2}}{a+b x+c x^2} \, dx\\ &=\frac{52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c \left (b^2-4 a c\right ) d^4\right ) \int \frac{(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx\\ &=\frac{52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac{(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt{b d+2 c d x}+\frac{52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c \left (b^2-4 a c\right )^3 d^8\right ) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt{b d+2 c d x}+\frac{52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\frac{1}{2} \left (13 \left (b^2-4 a c\right )^3 d^7\right ) \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 )\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt{b d+2 c d x}+\frac{52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 \left (b^2-4 a c\right )^3 d^7\right ) \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 )\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt{b d+2 c d x}+\frac{52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}-\left (26 c \left (b^2-4 a c\right )^{5/2} d^8\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 (26 c \left (b^2-4 a c\right )^{5/2} d^8\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 )\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt{b d+2 c d x}+\frac{52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac{52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac{d (b d+2 c d x)^{13/2}}{a+b x+c x^2}-26 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-26 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\\ \end{align*}

Mathematica [A]  time = 0.601822, size = 189, normalized size = 0.9 \[ -\frac{(d (b+2 c x))^{15/2} \left (-13 \left (b^2-4 a c\right ) \left (3 \left (b^2-4 a c\right ) \left (-30 \left (b^2-4 a c\right ) \sqrt{b+2 c x}-60 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \left (\tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )+24 (b+2 c x)^{5/2}\right )+8 (b+2 c x)^{9/2}\right )-40 (b+2 c x)^{13/2}\right )}{90 (b+2 c x)^{15/2} (a+x (b+c x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x]

[Out]

-((d*(b + 2*c*x))^(15/2)*(-40*(b + 2*c*x)^(13/2) - 13*(b^2 - 4*a*c)*(8*(b + 2*c*x)^(9/2) + 3*(b^2 - 4*a*c)*(-3
0*(b^2 - 4*a*c)*Sqrt[b + 2*c*x] + 24*(b + 2*c*x)^(5/2) - 60*c*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))*(ArcTan[Sq
rt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + ArcTanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)])))))/(90*(b + 2*c*x)^(15/2)*
(a + x*(b + c*x)))

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Maple [B]  time = 0.201, size = 1512, normalized size = 7.2 \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)^(15/2)/(c*x^2+b*x+a)^2,x)

[Out]

16/9*c*d^3*(2*c*d*x+b*d)^(9/2)-128/5*c^2*d^5*(2*c*d*x+b*d)^(5/2)*a+32/5*c*d^5*(2*c*d*x+b*d)^(5/2)*b^2+768*c^3*
d^7*a^2*(2*c*d*x+b*d)^(1/2)-384*c^2*d^7*a*b^2*(2*c*d*x+b*d)^(1/2)+48*c*d^7*b^4*(2*c*d*x+b*d)^(1/2)+256*c^4*d^9
*(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)*a^3-192*c^3*d^9*(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)*a^2*b^2+48*c^2*d^9*(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)*a*b^4-4*c*
d^9*(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)*b^6-832*c^4*d^9/(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)*a^3+624*c^3*d^9/(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)*a^2*b^2-156*c^2*d^9/(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)*a*b^4+13*c*d^9/(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)*b^6+832*c^4*d^9/(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)*a^3-624*c^3*d^9/(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)*a^2*b^2+156*c^2*d^9
/(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)*a*b^4-13*c
*d^9/(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)*b^6-41
6*c^4*d^9/(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)))*a^3+312*c^3*d^9/(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)))*a^2*b^2-78*c^2*d^9/(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)))*a*b^4+13/2*c*d^9/(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)))*b^6

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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)^(15/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.28474, size = 3564, normalized size = 16.97 \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)^(15/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-1/45*(2340*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^
5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)*(c*
x^2 + b*x + a)*arctan((((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8
- 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30
)^(3/4)*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d^7 - ((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14
*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11
+ 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(3/4)*sqrt(2*(b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^
3*b^2*c^6 + 256*a^4*c^7)*d^15*x + (b^9*c^2 - 16*a*b^7*c^3 + 96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*
d^15 + sqrt((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5
*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)))/((b^18*c
^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a
^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)) + 585*((b^18*c^4 - 36*a*b^16
*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 -
589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)*(c*x^2 + b*x + a)*log(13*(b^4*c - 8*a
*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d^7 + 13*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b
^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2
*c^12 - 262144*a^9*c^13)*d^30)^(1/4)) - 585*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7
+ 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 -
262144*a^9*c^13)*d^30)^(1/4)*(c*x^2 + b*x + a)*log(13*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d
^7 - 13*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^
8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)) - (128
0*c^6*d^7*x^6 + 3840*b*c^5*d^7*x^5 + 256*(22*b^2*c^4 - 13*a*c^5)*d^7*x^4 + 256*(19*b^3*c^3 - 26*a*b*c^4)*d^7*x
^3 + 96*(45*b^4*c^2 - 208*a*b^2*c^3 + 312*a^2*c^4)*d^7*x^2 + 32*(79*b^5*c - 520*a*b^3*c^2 + 936*a^2*b*c^3)*d^7
*x - (45*b^6 - 3068*a*b^4*c + 20592*a^2*b^2*c^2 - 37440*a^3*c^3)*d^7)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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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)**(15/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.26822, size = 972, normalized size = 4.63 \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)^(15/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

48*sqrt(2*c*d*x + b*d)*b^4*c*d^7 - 384*sqrt(2*c*d*x + b*d)*a*b^2*c^2*d^7 + 768*sqrt(2*c*d*x + b*d)*a^2*c^3*d^7
+ 32/5*(2*c*d*x + b*d)^(5/2)*b^2*c*d^5 - 128/5*(2*c*d*x + b*d)^(5/2)*a*c^2*d^5 + 16/9*(2*c*d*x + b*d)^(9/2)*c
*d^3 - 13/2*sqrt(2)*(b^4*c*d^7 - 8*a*b^2*c^2*d^7 + 16*a^2*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) + 13/2*sqrt(2)*(b
^4*c*d^7 - 8*a*b^2*c^2*d^7 + 16*a^2*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) - 13*(sqrt(2)*b^4*c*d^7 - 8*sqrt(2)*a*b
^2*c^2*d^7 + 16*sqrt(2)*a^2*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) - 13*(sqrt(2)*b^4*c*d^7 - 8*sqrt(2)*a*b^2*
c^2*d^7 + 16*sqrt(2)*a^2*c^3*d^7)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*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)) + 4*(sqrt(2*c*d*x + b*d)*b^6*c*d^9 - 12*sqrt
(2*c*d*x + b*d)*a*b^4*c^2*d^9 + 48*sqrt(2*c*d*x + b*d)*a^2*b^2*c^3*d^9 - 64*sqrt(2*c*d*x + b*d)*a^3*c^4*d^9)/(
b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)