### 3.1395 $$\int \frac{(b d+2 c d x)^{13/2}}{(a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=299 $-\frac{616 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{5 \sqrt{a+b x+c x^2}}+\frac{616 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

[Out]

(-2*d*(b*d + 2*c*d*x)^(11/2))/(3*(a + b*x + c*x^2)^(3/2)) - (44*c*d^3*(b*d + 2*c*d*x)^(7/2))/(3*Sqrt[a + b*x +
c*x^2]) + (1232*c^2*d^5*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/15 + (616*c*(b^2 - 4*a*c)^(7/4)*d^(13/2)
*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d
])], -1])/(5*Sqrt[a + b*x + c*x^2]) - (616*c*(b^2 - 4*a*c)^(7/4)*d^(13/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 -
4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.286027, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {686, 692, 691, 690, 307, 221, 1199, 424} $-\frac{616 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 \sqrt{a+b x+c x^2}}+\frac{616 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(11/2))/(3*(a + b*x + c*x^2)^(3/2)) - (44*c*d^3*(b*d + 2*c*d*x)^(7/2))/(3*Sqrt[a + b*x +
c*x^2]) + (1232*c^2*d^5*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/15 + (616*c*(b^2 - 4*a*c)^(7/4)*d^(13/2)
*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d
])], -1])/(5*Sqrt[a + b*x + c*x^2]) - (616*c*(b^2 - 4*a*c)^(7/4)*d^(13/2)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 -
4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(5*Sqrt[a + b*x + c*x^2])

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 691

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[-((c*(a + b*x + c
*x^2))/(b^2 - 4*a*c))]/Sqrt[a + b*x + c*x^2], Int[(d + e*x)^m/Sqrt[-((a*c)/(b^2 - 4*a*c)) - (b*c*x)/(b^2 - 4*a
*c) - (c^2*x^2)/(b^2 - 4*a*c)], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && EqQ[m^2, 1/4]

Rule 690

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 - 4*a*
c))])/e, Subst[Int[x^2/Sqrt[Simp[1 - (b^2*x^4)/(d^2*(b^2 - 4*a*c)), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a
, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^{13/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac{1}{3} \left (22 c d^2\right ) \int \frac{(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1}{3} \left (308 c^2 d^4\right ) \int \frac{(b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{1}{5} \left (308 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{\left (308 c^2 \left (b^2-4 a c\right ) d^6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{-\frac{a c}{b^2-4 a c}-\frac{b c x}{b^2-4 a c}-\frac{c^2 x^2}{b^2-4 a c}}} \, dx}{5 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{\left (616 c \left (b^2-4 a c\right ) d^5 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}-\frac{\left (616 c \left (b^2-4 a c\right )^{3/2} d^6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 \sqrt{a+b x+c x^2}}+\frac{\left (616 c \left (b^2-4 a c\right )^{3/2} d^6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}-\frac{616 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 \sqrt{a+b x+c x^2}}+\frac{\left (616 c \left (b^2-4 a c\right )^{3/2} d^6 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}}{\sqrt{1-\frac{x^2}{\sqrt{b^2-4 a c} d}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{5 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{11/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{44 c d^3 (b d+2 c d x)^{7/2}}{3 \sqrt{a+b x+c x^2}}+\frac{1232}{15} c^2 d^5 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}+\frac{616 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 \sqrt{a+b x+c x^2}}-\frac{616 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{5 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.228671, size = 202, normalized size = 0.68 $\frac{4 d^5 (d (b+2 c x))^{3/2} \left (616 c \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \left (4 a^2 c+a \left (-b^2+4 b c x+4 c^2 x^2\right )-b^2 x (b+c x)\right ) \, _2F_1\left (\frac{3}{4},\frac{5}{2};\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )+16 c^2 \left (-77 a^2-33 a c x^2+3 c^2 x^4\right )+4 b^2 c \left (121 a+51 c x^2\right )+48 b c^2 x \left (2 c x^2-11 a\right )+156 b^3 c x-41 b^4\right )}{15 (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(4*d^5*(d*(b + 2*c*x))^(3/2)*(-41*b^4 + 156*b^3*c*x + 48*b*c^2*x*(-11*a + 2*c*x^2) + 4*b^2*c*(121*a + 51*c*x^2
) + 16*c^2*(-77*a^2 - 33*a*c*x^2 + 3*c^2*x^4) + 616*c*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*(4*a^2*c - b^
2*x*(b + c*x) + a*(-b^2 + 4*b*c*x + 4*c^2*x^2))*Hypergeometric2F1[3/4, 5/2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])
)/(15*(a + x*(b + c*x))^(3/2))

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Maple [B]  time = 0.292, size = 1328, normalized size = 4.4 \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)^(13/2)/(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/15*(d*(2*c*x+b))^(1/2)*(7392*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),
2^(1/2))*x^2*a^2*c^4*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(
1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)-3696*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))
/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^2*a*b^2*c^3*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1
/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)+462*Ellipti
cE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^2*b^4*c^2*((b+2*c*x+(-4*a*c+
b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4
*a*c+b^2)^(1/2))^(1/2)-384*x^6*c^6+7392*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*
2^(1/2),2^(1/2))*x*a^2*b*c^3*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^
(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)-3696*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2
)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x*a*b^3*c^2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/
2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)+462*E
llipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x*b^5*c*((b+2*c*x+(-4*a*
c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(
-4*a*c+b^2)^(1/2))^(1/2)-1152*x^5*b*c^5+7392*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+
b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*((b+2*c*x+
(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a^3*c^3-3696*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*
a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))
^(1/2)*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a^2*b^2*c^2+462*
((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a
*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)
*2^(1/2),2^(1/2))*a*b^4*c-3168*x^4*a*c^5-648*x^4*b^2*c^4-6336*x^3*a*b*c^4+624*x^3*b^3*c^3-2464*x^2*a^2*c^4-352
0*x^2*a*b^2*c^3+674*x^2*b^4*c^2-2464*x*a^2*b*c^3-352*x*a*b^3*c^2+170*x*b^5*c-616*a^2*b^2*c^2+110*a*b^4*c+5*b^6
)*d^6/(2*c*x+b)/(c*x^2+b*x+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{13}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(13/2)/(c*x^2 + b*x + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (64 \, c^{6} d^{6} x^{6} + 192 \, b c^{5} d^{6} x^{5} + 240 \, b^{2} c^{4} d^{6} x^{4} + 160 \, b^{3} c^{3} d^{6} x^{3} + 60 \, b^{4} c^{2} d^{6} x^{2} + 12 \, b^{5} c d^{6} x + b^{6} d^{6}\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((64*c^6*d^6*x^6 + 192*b*c^5*d^6*x^5 + 240*b^2*c^4*d^6*x^4 + 160*b^3*c^3*d^6*x^3 + 60*b^4*c^2*d^6*x^2
+ 12*b^5*c*d^6*x + b^6*d^6)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^
2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)

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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)**(13/2)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{\frac{13}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(13/2)/(c*x^2 + b*x + a)^(5/2), x)