3.1397 \(\int \frac{(b d+2 c d x)^{5/2}}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=264 \[ -\frac{8 c d^{5/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt{a+b x+c x^2}}+\frac{8 c d^{5/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt{a+b x+c x^2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

(-2*d*(b*d + 2*c*d*x)^(3/2))/(3*(a + b*x + c*x^2)^(3/2)) - (4*c*d*(b*d + 2*c*d*x)^(3/2))/((b^2 - 4*a*c)*Sqrt[a
+ b*x + c*x^2]) + (8*c*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])/((b^2 - 4*a*c)^(1/4)*Sqrt[a + b*x + c*x^2]) - (8*c*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])/(
(b^2 - 4*a*c)^(1/4)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.237248, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {686, 687, 691, 690, 307, 221, 1199, 424} \[ -\frac{8 c d^{5/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt{a+b x+c x^2}}+\frac{8 c d^{5/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt{a+b x+c x^2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(b*d + 2*c*d*x)^(3/2))/(3*(a + b*x + c*x^2)^(3/2)) - (4*c*d*(b*d + 2*c*d*x)^(3/2))/((b^2 - 4*a*c)*Sqrt[a
+ b*x + c*x^2]) + (8*c*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])/((b^2 - 4*a*c)^(1/4)*Sqrt[a + b*x + c*x^2]) - (8*c*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])/(
(b^2 - 4*a*c)^(1/4)*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 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), 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] && LtQ[p, -1] &&  !GtQ[m, 1] && 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{(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}}+\left (2 c d^2\right ) \int \frac{\sqrt{b d+2 c d x}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{\left (4 c^2 d^2\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{\left (4 c^2 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}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{\left (8 c 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 )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{\left (8 c 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 )}{\sqrt{b^2-4 a c} \sqrt{a+b x+c x^2}}+\frac{\left (8 c 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 )}{\sqrt{b^2-4 a c} \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{8 c 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 )}{\sqrt [4]{b^2-4 a c} \sqrt{a+b x+c x^2}}+\frac{\left (8 c 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 )}{\sqrt{b^2-4 a c} \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{3/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 c d (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{8 c 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 )}{\sqrt [4]{b^2-4 a c} \sqrt{a+b x+c x^2}}-\frac{8 c 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 )}{\sqrt [4]{b^2-4 a c} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.116073, size = 116, normalized size = 0.44 \[ -\frac{4 d (d (b+2 c x))^{3/2} \left (8 c (a+x (b+c x)) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{3}{4},\frac{5}{2};\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-4 a c+b^2\right )}{3 \left (b^2-4 a c\right ) (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d*(d*(b + 2*c*x))^(3/2)*(b^2 - 4*a*c + 8*c*(a + x*(b + c*x))*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hy
pergeometric2F1[3/4, 5/2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(3/2))

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Maple [B]  time = 0.227, size = 871, normalized size = 3.3 \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]

-2/3*(d*(2*c*x+b))^(1/2)*(24*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*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)-6*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^2*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)+24*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*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)-6*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^3*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*((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*c^2-6*((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^2*c-24*c^4*x^4-48*b
*c^3*x^3-8*x^2*a*c^3-34*x^2*b^2*c^2-8*b*a*c^2*x-10*b^3*c*x-2*a*c*b^2-b^4)*d^2/(2*c*x+b)/(4*a*c-b^2)/(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{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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (4 \, c^{2} d^{2} x^{2} + 4 \, b c d^{2} x + b^{2} d^{2}\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)^(5/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((4*c^2*d^2*x^2 + 4*b*c*d^2*x + b^2*d^2)*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)**(5/2)/(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{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)