### 3.1357 $$\int (b d+2 c d x)^{5/2} (a+b x+c x^2)^{5/2} \, dx$$

Optimal. Leaf size=373 $\frac{d^{5/2} \left (b^2-4 a c\right )^{19/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{884 c^4 \sqrt{a+b x+c x^2}}-\frac{d^{5/2} \left (b^2-4 a c\right )^{19/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{884 c^4 \sqrt{a+b x+c x^2}}-\frac{d \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{7/2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{7/2}}{442 c^2 d}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{7/2}}{17 c d}$

[Out]

-((b^2 - 4*a*c)^3*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(1326*c^3) + (5*(b^2 - 4*a*c)^2*(b*d + 2*c*d*
x)^(7/2)*Sqrt[a + b*x + c*x^2])/(2652*c^3*d) - (5*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^(3/2))
/(442*c^2*d) + ((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^(5/2))/(17*c*d) - ((b^2 - 4*a*c)^(19/4)*d^(5/2)*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
])/(884*c^4*Sqrt[a + b*x + c*x^2]) + ((b^2 - 4*a*c)^(19/4)*d^(5/2)*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])/(884*c^4*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.372943, antiderivative size = 373, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {685, 692, 691, 690, 307, 221, 1199, 424} $\frac{d^{5/2} \left (b^2-4 a c\right )^{19/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{884 c^4 \sqrt{a+b x+c x^2}}-\frac{d^{5/2} \left (b^2-4 a c\right )^{19/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{884 c^4 \sqrt{a+b x+c x^2}}-\frac{d \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{7/2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{7/2}}{442 c^2 d}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{7/2}}{17 c d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((b^2 - 4*a*c)^3*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(1326*c^3) + (5*(b^2 - 4*a*c)^2*(b*d + 2*c*d*
x)^(7/2)*Sqrt[a + b*x + c*x^2])/(2652*c^3*d) - (5*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^(3/2))
/(442*c^2*d) + ((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)^(5/2))/(17*c*d) - ((b^2 - 4*a*c)^(19/4)*d^(5/2)*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
])/(884*c^4*Sqrt[a + b*x + c*x^2]) + ((b^2 - 4*a*c)^(19/4)*d^(5/2)*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])/(884*c^4*Sqrt[a + b*x + c*x^2])

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), 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] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && 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 (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2} \, dx}{34 c}\\ &=-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}+\frac{\left (15 \left (b^2-4 a c\right )^2\right ) \int (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2} \, dx}{884 c^2}\\ &=\frac{5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}-\frac{\left (5 \left (b^2-4 a c\right )^3\right ) \int \frac{(b d+2 c d x)^{5/2}}{\sqrt{a+b x+c x^2}} \, dx}{5304 c^3}\\ &=-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}-\frac{\left (\left (b^2-4 a c\right )^4 d^2\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{1768 c^3}\\ &=-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}-\frac{\left (\left (b^2-4 a c\right )^4 d^2 \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}{1768 c^3 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}-\frac{\left (\left (b^2-4 a c\right )^4 d \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 )}{884 c^4 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}+\frac{\left (\left (b^2-4 a c\right )^{9/2} d^2 \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 )}{884 c^4 \sqrt{a+b x+c x^2}}-\frac{\left (\left (b^2-4 a c\right )^{9/2} d^2 \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 )}{884 c^4 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}+\frac{\left (b^2-4 a c\right )^{19/4} d^{5/2} \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 )}{884 c^4 \sqrt{a+b x+c x^2}}-\frac{\left (\left (b^2-4 a c\right )^{9/2} d^2 \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 )}{884 c^4 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right )^3 d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{1326 c^3}+\frac{5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2}}{2652 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{3/2}}{442 c^2 d}+\frac{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )^{5/2}}{17 c d}-\frac{\left (b^2-4 a c\right )^{19/4} d^{5/2} \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 )}{884 c^4 \sqrt{a+b x+c x^2}}+\frac{\left (b^2-4 a c\right )^{19/4} d^{5/2} \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 )}{884 c^4 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.185608, size = 117, normalized size = 0.31 $\frac{2}{17} d \sqrt{a+x (b+c x)} (d (b+2 c x))^{3/2} \left (\frac{\left (b^2-4 a c\right )^3 \, _2F_1\left (-\frac{5}{2},\frac{3}{4};\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{64 c^3 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}+2 (a+x (b+c x))^3\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*d*(d*(b + 2*c*x))^(3/2)*Sqrt[a + x*(b + c*x)]*(2*(a + x*(b + c*x))^3 + ((b^2 - 4*a*c)^3*Hypergeometric2F1[-
5/2, 3/4, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(64*c^3*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])))/17

________________________________________________________________________________________

Maple [B]  time = 0.227, size = 1190, normalized size = 3.2 \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)^(5/2)*(c*x^2+b*x+a)^(5/2),x)

[Out]

-1/5304*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^2*(-11112*x^5*b^5*c^5-10*x^2*b^8*c^2-24960*x^9*b*c^9-1516*x^
4*b^6*c^4-6*x*b^9*c-34168*x^6*b^4*c^6-56064*x^7*b^3*c^7-51456*x^8*b^2*c^8-18816*x^8*a*c^9-25216*x^6*a^2*c^8-12
416*x^4*a^3*c^7-1024*x^2*a^4*c^6-256*a^4*b^2*c^4-520*a^3*b^4*c^3+92*a^2*b^6*c^2-6*a*b^8*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)*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))*
b^10+3072*((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^5*c^5-75648*x^5*a^2*b*c^7-93888*x^5*a*b^3*c^6-85248*x^4*a^2*b^2*c^6-37368*x^4*a*
b^4*c^5-24832*x^3*a^3*b*c^6-44416*x^3*a^2*b^3*c^5-6064*x^3*a*b^5*c^4-17600*x^2*a^3*b^2*c^5-10056*x^2*a^2*b^4*c
^4+160*x^2*a*b^6*c^3-1024*x*a^4*b*c^5-5184*x*a^3*b^3*c^4-456*x*a^2*b^5*c^3+88*x*a*b^7*c^2-75264*x^7*a*b*c^8-11
9104*x^6*a*b^2*c^7+1920*((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*b^4*c^3-480*((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^6*c^2+60*((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^8*c-3840*((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^4*b^2*c^4-4992*x^10*c^10)/c^4/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (4 \, c^{4} d^{2} x^{6} + 12 \, b c^{3} d^{2} x^{5} +{\left (13 \, b^{2} c^{2} + 8 \, a c^{3}\right )} d^{2} x^{4} + a^{2} b^{2} d^{2} + 2 \,{\left (3 \, b^{3} c + 8 \, a b c^{2}\right )} d^{2} x^{3} +{\left (b^{4} + 10 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d^{2} x^{2} + 2 \,{\left (a b^{3} + 2 \, a^{2} b c\right )} d^{2} x\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((4*c^4*d^2*x^6 + 12*b*c^3*d^2*x^5 + (13*b^2*c^2 + 8*a*c^3)*d^2*x^4 + a^2*b^2*d^2 + 2*(3*b^3*c + 8*a*b
*c^2)*d^2*x^3 + (b^4 + 10*a*b^2*c + 4*a^2*c^2)*d^2*x^2 + 2*(a*b^3 + 2*a^2*b*c)*d^2*x)*sqrt(2*c*d*x + b*d)*sqrt
(c*x^2 + b*x + a), x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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