3.1308 \(\int \frac{(b d+2 c d x)^{15/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=222 \[ 234 c^2 d^7 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x}-117 c^2 d^{15/2} \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-117 c^2 d^{15/2} \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2} \]

[Out]

234*c^2*(b^2 - 4*a*c)*d^7*Sqrt[b*d + 2*c*d*x] + (234*c^2*d^5*(b*d + 2*c*d*x)^(5/2))/5 - (d*(b*d + 2*c*d*x)^(13
/2))/(2*(a + b*x + c*x^2)^2) - (13*c*d^3*(b*d + 2*c*d*x)^(9/2))/(2*(a + b*x + c*x^2)) - 117*c^2*(b^2 - 4*a*c)^
(5/4)*d^(15/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 117*c^2*(b^2 - 4*a*c)^(5/4)*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 = 0.18192, 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, 212, 206, 203} \[ 234 c^2 d^7 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x}-117 c^2 d^{15/2} \left (b^2-4 a c\right )^{5/4} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-117 c^2 d^{15/2} \left (b^2-4 a c\right )^{5/4} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^3,x]

[Out]

234*c^2*(b^2 - 4*a*c)*d^7*Sqrt[b*d + 2*c*d*x] + (234*c^2*d^5*(b*d + 2*c*d*x)^(5/2))/5 - (d*(b*d + 2*c*d*x)^(13
/2))/(2*(a + b*x + c*x^2)^2) - (13*c*d^3*(b*d + 2*c*d*x)^(9/2))/(2*(a + b*x + c*x^2)) - 117*c^2*(b^2 - 4*a*c)^
(5/4)*d^(15/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 117*c^2*(b^2 - 4*a*c)^(5/4)*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 )^3} \, dx &=-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{1}{2} \left (13 c d^2\right ) \int \frac{(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (117 c^2 d^4\right ) \int \frac{(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx\\ &=\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (117 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac{(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=234 c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x}+\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (117 c^2 \left (b^2-4 a c\right )^2 d^8\right ) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=234 c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x}+\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{4} \left (117 c \left (b^2-4 a c\right )^2 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 )\\ &=234 c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x}+\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (117 c \left (b^2-4 a c\right )^2 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 )\\ &=234 c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x}+\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}-\left (117 c^2 \left (b^2-4 a c\right )^{3/2} 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 (117 c^2 \left (b^2-4 a c\right )^{3/2} 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 )\\ &=234 c^2 \left (b^2-4 a c\right ) d^7 \sqrt{b d+2 c d x}+\frac{234}{5} c^2 d^5 (b d+2 c d x)^{5/2}-\frac{d (b d+2 c d x)^{13/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{13 c d^3 (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )}-117 c^2 \left (b^2-4 a c\right )^{5/4} d^{15/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-117 c^2 \left (b^2-4 a c\right )^{5/4} 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 = 0.75566, size = 225, normalized size = 1.01 \[ \frac{(d (b+2 c x))^{15/2} \left (91 \left (b^2-4 a c\right ) \left (-192 \left (b^2-4 a c\right ) (b+2 c x)^{5/2}+120 \left (b^2-4 a c\right )^2 \sqrt{b+2 c x}+60 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \left (2 \left (b^2-4 a c\right )^{3/4} \sqrt{b+2 c x}-12 c (a+x (b+c x)) \left (\tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )\right )+64 (b+2 c x)^{9/2}\right )+448 (b+2 c x)^{13/2}\right )}{560 (b+2 c x)^{15/2} (a+x (b+c x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^3,x]

[Out]

((d*(b + 2*c*x))^(15/2)*(448*(b + 2*c*x)^(13/2) + 91*(b^2 - 4*a*c)*(120*(b^2 - 4*a*c)^2*Sqrt[b + 2*c*x] - 192*
(b^2 - 4*a*c)*(b + 2*c*x)^(5/2) + 64*(b + 2*c*x)^(9/2) + 60*c*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))*(2*(b^2 -
4*a*c)^(3/4)*Sqrt[b + 2*c*x] - 12*c*(a + x*(b + c*x))*(ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + ArcTanh[S
qrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)])))))/(560*(b + 2*c*x)^(15/2)*(a + x*(b + c*x))^2)

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Maple [B]  time = 0.211, 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)^(15/2)/(c*x^2+b*x+a)^3,x)

[Out]

64/5*c^2*d^5*(2*c*d*x+b*d)^(5/2)-768*c^3*d^7*a*(2*c*d*x+b*d)^(1/2)+192*c^2*d^7*b^2*(2*c*d*x+b*d)^(1/2)-800*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)^(5/2)*a^2+400*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)^(5/2)*a*b^2-50*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)^(5/2)*b^
4-2688*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)^(1/2)*a^3+2016*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)^(1/2)*a^2*b^2-504*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)^(1/2)*a*b^4+42*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)^(1/2)*b^6+936*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^2-468*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*b^2+117
/2*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)*b
^4-936*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^2+468*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*b^2-117/2*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)*b^4+468*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^2-234*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*b^2+117/4*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)))*b^4

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.84988, size = 2502, normalized size = 11.27 \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)^3,x, algorithm="fricas")

[Out]

1/10*(2340*((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*a^4*b^2*c^12 - 1024*a^5*c^13
)*d^30)^(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^10*c^8 - 20*a*b^8*c^9 + 1
60*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*a^4*b^2*c^12 - 1024*a^5*c^13)*d^30)^(3/4)*(b^2*c^2 - 4*a*c^3)*sqrt(2
*c*d*x + b*d)*d^7 + ((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*a^4*b^2*c^12 - 1024
*a^5*c^13)*d^30)^(3/4)*sqrt(2*(b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*d^15*x + (b^5*c^4 - 8*a*b^3*c^5 + 16*a^2*b*
c^6)*d^15 + sqrt((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*a^4*b^2*c^12 - 1024*a^5
*c^13)*d^30)))/((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*a^4*b^2*c^12 - 1024*a^5*
c^13)*d^30)) + 585*((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*a^4*b^2*c^12 - 1024*
a^5*c^13)*d^30)^(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(-117*(b^2*c^2 - 4*a*c^3)*s
qrt(2*c*d*x + b*d)*d^7 + 117*((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*a^4*b^2*c^
12 - 1024*a^5*c^13)*d^30)^(1/4)) - 585*((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^11 + 1280*
a^4*b^2*c^12 - 1024*a^5*c^13)*d^30)^(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(-117*(
b^2*c^2 - 4*a*c^3)*sqrt(2*c*d*x + b*d)*d^7 - 117*((b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4*c^
11 + 1280*a^4*b^2*c^12 - 1024*a^5*c^13)*d^30)^(1/4)) + (512*c^6*d^7*x^6 + 1536*b*c^5*d^7*x^5 + 512*(7*b^2*c^4
- 13*a*c^5)*d^7*x^4 + 512*(9*b^3*c^3 - 26*a*b*c^4)*d^7*x^3 + 3*(641*b^4*c^2 - 520*a*b^2*c^3 - 5616*a^2*c^4)*d^
7*x^2 - (125*b^5*c - 5096*a*b^3*c^2 + 16848*a^2*b*c^3)*d^7*x - (5*b^6 + 65*a*b^4*c - 2808*a^2*b^2*c^2 + 9360*a
^3*c^3)*d^7)*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)**(15/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.29784, 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)^(15/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

192*sqrt(2*c*d*x + b*d)*b^2*c^2*d^7 - 768*sqrt(2*c*d*x + b*d)*a*c^3*d^7 + 64/5*(2*c*d*x + b*d)^(5/2)*c^2*d^5 -
117/2*sqrt(2)*(b^2*c^2*d^7 - 4*a*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)) - 117/2*sqrt(2)*(b^2*c^2*d^7 - 4*a*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)) - 117/4*sqrt(2)*(b^2*c^2*d^7 - 4*a*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))
+ 117/4*sqrt(2)*(b^2*c^2*d^7 - 4*a*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)) + 2*(21*sqrt(2*c*d*x + b*d)*b^6*c^2*d^11
- 252*sqrt(2*c*d*x + b*d)*a*b^4*c^3*d^11 + 1008*sqrt(2*c*d*x + b*d)*a^2*b^2*c^4*d^11 - 1344*sqrt(2*c*d*x + b*
d)*a^3*c^5*d^11 - 25*(2*c*d*x + b*d)^(5/2)*b^4*c^2*d^9 + 200*(2*c*d*x + b*d)^(5/2)*a*b^2*c^3*d^9 - 400*(2*c*d*
x + b*d)^(5/2)*a^2*c^4*d^9)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2