### 3.1359 $$\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{3/2}} \, dx$$

Optimal. Leaf size=316 $-\frac{\left (b^2-4 a c\right )^{11/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{12 c^4 d^{3/2} \sqrt{a+b x+c x^2}}-\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{12 c^3 d^3}+\frac{\left (b^2-4 a c\right )^{11/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt{a+b x+c x^2}}+\frac{5 \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}$

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(12*c^3*d^3) + (5*(b*d + 2*c*d*x)^(3/2)*(a + b*x
+ c*x^2)^(3/2))/(18*c^2*d^3) - (a + b*x + c*x^2)^(5/2)/(c*d*Sqrt[b*d + 2*c*d*x]) + ((b^2 - 4*a*c)^(11/4)*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])/(12*c^4*d^(3/2)*Sqrt[a + b*x + c*x^2]) - ((b^2 - 4*a*c)^(11/4)*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])/(12*c^4*d^(3/2)*Sqrt[a + b*x + c*x
^2])

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Rubi [A]  time = 0.292052, antiderivative size = 316, 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 = {684, 685, 691, 690, 307, 221, 1199, 424} $-\frac{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{12 c^3 d^3}-\frac{\left (b^2-4 a c\right )^{11/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt{a+b x+c x^2}}+\frac{\left (b^2-4 a c\right )^{11/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt{a+b x+c x^2}}+\frac{5 \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(3/2),x]

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(12*c^3*d^3) + (5*(b*d + 2*c*d*x)^(3/2)*(a + b*x
+ c*x^2)^(3/2))/(18*c^2*d^3) - (a + b*x + c*x^2)^(5/2)/(c*d*Sqrt[b*d + 2*c*d*x]) + ((b^2 - 4*a*c)^(11/4)*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])/(12*c^4*d^(3/2)*Sqrt[a + b*x + c*x^2]) - ((b^2 - 4*a*c)^(11/4)*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])/(12*c^4*d^(3/2)*Sqrt[a + b*x + c*x
^2])

Rule 684

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 + 1)), x] - Dist[(b*p)/(d*e*(m + 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] &&
GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && IntegerQ[2*p]

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 \frac{\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{3/2}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}+\frac{5 \int \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2} \, dx}{2 c d^2}\\ &=\frac{5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2} \, dx}{12 c^2 d^2}\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{12 c^3 d^3}+\frac{5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}+\frac{\left (b^2-4 a c\right )^2 \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{24 c^3 d^2}\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{12 c^3 d^3}+\frac{5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}+\frac{\left (\left (b^2-4 a c\right )^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}{24 c^3 d^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{12 c^3 d^3}+\frac{5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}+\frac{\left (\left (b^2-4 a c\right )^2 \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 )}{12 c^4 d^3 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{12 c^3 d^3}+\frac{5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}-\frac{\left (\left (b^2-4 a c\right )^{5/2} \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 )}{12 c^4 d^2 \sqrt{a+b x+c x^2}}+\frac{\left (\left (b^2-4 a c\right )^{5/2} \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 )}{12 c^4 d^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{12 c^3 d^3}+\frac{5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}-\frac{\left (b^2-4 a c\right )^{11/4} \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt{a+b x+c x^2}}+\frac{\left (\left (b^2-4 a c\right )^{5/2} \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 )}{12 c^4 d^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{12 c^3 d^3}+\frac{5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt{b d+2 c d x}}+\frac{\left (b^2-4 a c\right )^{11/4} \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt{a+b x+c x^2}}-\frac{\left (b^2-4 a c\right )^{11/4} \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0536455, size = 101, normalized size = 0.32 $-\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \, _2F_1\left (-\frac{5}{2},-\frac{1}{4};\frac{3}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{32 c^3 d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \sqrt{d (b+2 c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(3/2),x]

[Out]

-((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -1/4, 3/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(32*c
^3*d*Sqrt[d*(b + 2*c*x)]*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.227, size = 700, normalized size = 2.2 \begin{align*}{\frac{1}{72\,{c}^{4}{d}^{2} \left ( 2\,{c}^{2}{x}^{3}+3\,bc{x}^{2}+2\,acx+{b}^{2}x+ab \right ) }\sqrt{c{x}^{2}+bx+a}\sqrt{d \left ( 2\,cx+b \right ) } \left ( 8\,{x}^{6}{c}^{6}+24\,{x}^{5}b{c}^{5}+192\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticE} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ){a}^{3}{c}^{3}-144\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticE} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ){a}^{2}{b}^{2}{c}^{2}+36\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticE} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ) a{b}^{4}c-3\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticE} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ){b}^{6}+40\,{x}^{4}a{c}^{5}+20\,{x}^{4}{b}^{2}{c}^{4}+80\,{x}^{3}ab{c}^{4}-40\,{x}^{2}{a}^{2}{c}^{4}+80\,{x}^{2}a{b}^{2}{c}^{3}-10\,{x}^{2}{b}^{4}{c}^{2}-40\,x{a}^{2}b{c}^{3}+40\,xa{b}^{3}{c}^{2}-6\,x{b}^{5}c-72\,{a}^{3}{c}^{3}+44\,{a}^{2}{b}^{2}{c}^{2}-6\,a{b}^{4}c \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),x)

[Out]

1/72*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(8*x^6*c^6+24*x^5*b*c^5+192*((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^3-144*((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+36*((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-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^6+40*x^4*a*c^5+20*x^4*b^2*c^4+80*x^3
*a*b*c^4-40*x^2*a^2*c^4+80*x^2*a*b^2*c^3-10*x^2*b^4*c^2-40*x*a^2*b*c^3+40*x*a*b^3*c^2-6*x*b^5*c-72*a^3*c^3+44*
a^2*b^2*c^2-6*a*b^4*c)/d^2/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 \frac{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\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}}{4 \, c^{2} d^{2} x^{2} + 4 \, b c d^{2} x + b^{2} d^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),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)/(
4*c^2*d^2*x^2 + 4*b*c*d^2*x + b^2*d^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x + c x^{2}\right )^{\frac{5}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(3/2),x)

[Out]

Integral((a + b*x + c*x**2)**(5/2)/(d*(b + 2*c*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(3/2), x)