### 3.1360 $$\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{7/2}} \, dx$$

Optimal. Leaf size=310 $\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}+\frac{3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}$

[Out]

(3*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(20*c^3*d^5) - (a + b*x + c*x^2)^(3/2)/(2*c^2*d^3*Sqrt[b*d + 2
*c*d*x]) - (a + b*x + c*x^2)^(5/2)/(5*c*d*(b*d + 2*c*d*x)^(5/2)) - (3*(b^2 - 4*a*c)^(7/4)*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])/(20*c^4*d^(
7/2)*Sqrt[a + b*x + c*x^2]) + (3*(b^2 - 4*a*c)^(7/4)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[Ar
cSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(20*c^4*d^(7/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.285641, antiderivative size = 310, 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} $\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}+\frac{3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(7/2),x]

[Out]

(3*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(20*c^3*d^5) - (a + b*x + c*x^2)^(3/2)/(2*c^2*d^3*Sqrt[b*d + 2
*c*d*x]) - (a + b*x + c*x^2)^(5/2)/(5*c*d*(b*d + 2*c*d*x)^(5/2)) - (3*(b^2 - 4*a*c)^(7/4)*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])/(20*c^4*d^(
7/2)*Sqrt[a + b*x + c*x^2]) + (3*(b^2 - 4*a*c)^(7/4)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[Ar
cSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(20*c^4*d^(7/2)*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)^{7/2}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{\int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{3/2}} \, dx}{2 c d^2}\\ &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{3 \int \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2} \, dx}{4 c^2 d^4}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{40 c^3 d^4}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{\left (3 \left (b^2-4 a c\right ) \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}{40 c^3 d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{\left (3 \left (b^2-4 a c\right ) \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 )}{20 c^4 d^5 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{\left (3 \left (b^2-4 a c\right )^{3/2} \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 )}{20 c^4 d^4 \sqrt{a+b x+c x^2}}-\frac{\left (3 \left (b^2-4 a c\right )^{3/2} \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 )}{20 c^4 d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{3 \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{\left (3 \left (b^2-4 a c\right )^{3/2} \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 )}{20 c^4 d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{3 \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}+\frac{3 \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 d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0647175, size = 101, normalized size = 0.33 $-\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \, _2F_1\left (-\frac{5}{2},-\frac{5}{4};-\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{160 c^3 d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(7/2),x]

[Out]

-((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -5/4, -1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(160
*c^3*d*(d*(b + 2*c*x))^(5/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.228, size = 1362, 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((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(7/2),x)

[Out]

1/40*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(192*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)-96*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)+12*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*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)+8*x^6*c^6+192*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)-96*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)+12*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*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)+24*x^5*b*c^5+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^2*b^2*c^2-24*((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^6-88*x^4*a*c^5+52*x^4*b^2*c^4-176*x^3*a*b*c^4+64*x^3*b^3*c^3-104*x^2*a^2*c^4-80*x^2*a*b^2
*c^3+34*x^2*b^4*c^2-104*x*a^2*b*c^3+8*x*a*b^3*c^2+6*x*b^5*c-8*a^3*c^3-20*a^2*b^2*c^2+6*a*b^4*c)/d^4/(2*c^2*x^3
+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/(2*c*x+b)^2/c^4

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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{7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(7/2), 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}}{16 \, c^{4} d^{4} x^{4} + 32 \, b c^{3} d^{4} x^{3} + 24 \, b^{2} c^{2} d^{4} x^{2} + 8 \, b^{3} c d^{4} x + b^{4} d^{4}}, 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)^(7/2),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)/(
16*c^4*d^4*x^4 + 32*b*c^3*d^4*x^3 + 24*b^2*c^2*d^4*x^2 + 8*b^3*c*d^4*x + b^4*d^4), 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{7}{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)**(7/2),x)

[Out]

Integral((a + b*x + c*x**2)**(5/2)/(d*(b + 2*c*x))**(7/2), 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{7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(7/2), x)