3.1358 \(\int \sqrt{b d+2 c d x} (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=328 \[ \frac{\sqrt{d} \left (b^2-4 a c\right )^{15/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{156 c^4 \sqrt{a+b x+c x^2}}+\frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{3/2}}{234 c^2 d}-\frac{\sqrt{d} \left (b^2-4 a c\right )^{15/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{156 c^4 \sqrt{a+b x+c x^2}}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{3/2}}{13 c d} \]

[Out]

((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(156*c^3*d) - (5*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(
3/2)*(a + b*x + c*x^2)^(3/2))/(234*c^2*d) + ((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(5/2))/(13*c*d) - ((b^2 -
4*a*c)^(15/4)*Sqrt[d]*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])/(156*c^4*Sqrt[a + b*x + c*x^2]) + ((b^2 - 4*a*c)^(15/4)*Sqrt[d]*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])/(156*
c^4*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.298775, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {685, 691, 690, 307, 221, 1199, 424} \[ \frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{3/2}}{234 c^2 d}+\frac{\sqrt{d} \left (b^2-4 a c\right )^{15/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{156 c^4 \sqrt{a+b x+c x^2}}-\frac{\sqrt{d} \left (b^2-4 a c\right )^{15/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{156 c^4 \sqrt{a+b x+c x^2}}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{3/2}}{13 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(156*c^3*d) - (5*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(
3/2)*(a + b*x + c*x^2)^(3/2))/(234*c^2*d) + ((b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(5/2))/(13*c*d) - ((b^2 -
4*a*c)^(15/4)*Sqrt[d]*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])/(156*c^4*Sqrt[a + b*x + c*x^2]) + ((b^2 - 4*a*c)^(15/4)*Sqrt[d]*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])/(156*
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 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 \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2} \, dx}{26 c}\\ &=-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d}+\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}+\frac{\left (5 \left (b^2-4 a c\right )^2\right ) \int \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2} \, dx}{156 c^2}\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d}+\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}-\frac{\left (b^2-4 a c\right )^3 \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{312 c^3}\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d}+\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}-\frac{\left (\left (b^2-4 a c\right )^3 \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}{312 c^3 \sqrt{a+b x+c x^2}}\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d}+\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}-\frac{\left (\left (b^2-4 a c\right )^3 \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 )}{156 c^4 d \sqrt{a+b x+c x^2}}\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d}+\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}+\frac{\left (\left (b^2-4 a c\right )^{7/2} \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 )}{156 c^4 \sqrt{a+b x+c x^2}}-\frac{\left (\left (b^2-4 a c\right )^{7/2} \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 )}{156 c^4 \sqrt{a+b x+c x^2}}\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d}+\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}+\frac{\left (b^2-4 a c\right )^{15/4} \sqrt{d} \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 )}{156 c^4 \sqrt{a+b x+c x^2}}-\frac{\left (\left (b^2-4 a c\right )^{7/2} \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 )}{156 c^4 \sqrt{a+b x+c x^2}}\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{156 c^3 d}-\frac{5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{234 c^2 d}+\frac{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{5/2}}{13 c d}-\frac{\left (b^2-4 a c\right )^{15/4} \sqrt{d} \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 )}{156 c^4 \sqrt{a+b x+c x^2}}+\frac{\left (b^2-4 a c\right )^{15/4} \sqrt{d} \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 )}{156 c^4 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.067064, size = 101, normalized size = 0.31 \[ \frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} (d (b+2 c x))^{3/2} \, _2F_1\left (-\frac{5}{2},\frac{3}{4};\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{96 c^3 d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)^2*(d*(b + 2*c*x))^(3/2)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, 3/4, 7/4, (b + 2*c*x)^2/(
b^2 - 4*a*c)])/(96*c^3*d*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.219, size = 924, normalized size = 2.8 \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)^(1/2)*(c*x^2+b*x+a)^(5/2),x)

[Out]

1/936*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*(288*x^8*c^8+6*x*b^7*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^8+268*x^4*b^4*c^4+1
128*x^5*b^3*c^5+1184*x^6*a*c^7+1888*x^4*a^2*c^6+992*x^2*a^3*c^5+1152*x^7*b*c^7+10*x^2*b^6*c^2+1720*x^6*b^2*c^6
+248*a^3*b^2*c^3-68*a^2*b^4*c^2+6*a*b^6*c-64*x*a*b^5*c^2+3552*x^5*a*b*c^6+3496*x^4*a*b^2*c^5+3776*x^3*a^2*b*c^
5+1072*x^3*a*b^3*c^4+2088*x^2*a^2*b^2*c^4-120*x^2*a*b^4*c^3+992*x*a^3*b*c^4+200*x*a^2*b^3*c^3+768*((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*c^4-768*((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^2*c^3+288*((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^4*c^2-48*((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^6*c
)/c^4/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{2 \, c d x + b d}{\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)^(1/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(2*c*d*x + b*d)*(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 ({\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\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)^(1/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a),
x)

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Sympy [A]  time = 47.0422, size = 539, normalized size = 1.64 \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)**(1/2)*(c*x**2+b*x+a)**(5/2),x)

[Out]

a**2*(b*d + 2*c*d*x)**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), (b*d + 2*c*d*x)**2*exp_polar(I*pi)/(4*c*d**2
*polar_lift(a - b**2/(4*c))))*sqrt(polar_lift(a - b**2/(4*c)))/(4*c*d*gamma(7/4)) - a*b**2*(b*d + 2*c*d*x)**(3
/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), (b*d + 2*c*d*x)**2*exp_polar(I*pi)/(4*c*d**2*polar_lift(a - b**2/(4*
c))))*sqrt(polar_lift(a - b**2/(4*c)))/(8*c**2*d*gamma(7/4)) + a*(b*d + 2*c*d*x)**(7/2)*gamma(7/4)*hyper((-1/2
, 7/4), (11/4,), (b*d + 2*c*d*x)**2*exp_polar(I*pi)/(4*c*d**2*polar_lift(a - b**2/(4*c))))*sqrt(polar_lift(a -
b**2/(4*c)))/(8*c**2*d**3*gamma(11/4)) + b**4*(b*d + 2*c*d*x)**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), (b
*d + 2*c*d*x)**2*exp_polar(I*pi)/(4*c*d**2*polar_lift(a - b**2/(4*c))))*sqrt(polar_lift(a - b**2/(4*c)))/(64*c
**3*d*gamma(7/4)) - b**2*(b*d + 2*c*d*x)**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), (b*d + 2*c*d*x)**2*exp_
polar(I*pi)/(4*c*d**2*polar_lift(a - b**2/(4*c))))*sqrt(polar_lift(a - b**2/(4*c)))/(32*c**3*d**3*gamma(11/4))
+ (b*d + 2*c*d*x)**(11/2)*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), (b*d + 2*c*d*x)**2*exp_polar(I*pi)/(4*c*d*
*2*polar_lift(a - b**2/(4*c))))*sqrt(polar_lift(a - b**2/(4*c)))/(64*c**3*d**5*gamma(15/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{2 \, c d x + b d}{\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)^(1/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

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