3.1396 \(\int \frac{(b d+2 c d x)^{9/2}}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=258 \[ -\frac{56 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{\sqrt{a+b x+c x^2}}+\frac{56 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{\sqrt{a+b x+c x^2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

(-2*d*(b*d + 2*c*d*x)^(7/2))/(3*(a + b*x + c*x^2)^(3/2)) - (28*c*d^3*(b*d + 2*c*d*x)^(3/2))/(3*Sqrt[a + b*x +
c*x^2]) + (56*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)*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])/Sqrt[a + b*x + c*x^2] - (56*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)
*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])/Sqrt[a + b*x + c*x^2]

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Rubi [A]  time = 0.237905, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {686, 691, 690, 307, 221, 1199, 424} \[ -\frac{56 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{\sqrt{a+b x+c x^2}}+\frac{56 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{\sqrt{a+b x+c x^2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(7/2))/(3*(a + b*x + c*x^2)^(3/2)) - (28*c*d^3*(b*d + 2*c*d*x)^(3/2))/(3*Sqrt[a + b*x +
c*x^2]) + (56*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)*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])/Sqrt[a + b*x + c*x^2] - (56*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)
*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])/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 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)^{9/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac{1}{3} \left (14 c d^2\right ) \int \frac{(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}+\left (28 c^2 d^4\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}+\frac{\left (28 c^2 d^4 \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}{\sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}+\frac{\left (56 c d^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 )}{\sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}-\frac{\left (56 c \sqrt{b^2-4 a c} d^4 \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{a+b x+c x^2}}+\frac{\left (56 c \sqrt{b^2-4 a c} d^4 \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{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}-\frac{56 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \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{a+b x+c x^2}}+\frac{\left (56 c \sqrt{b^2-4 a c} d^4 \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{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{7/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{28 c d^3 (b d+2 c d x)^{3/2}}{3 \sqrt{a+b x+c x^2}}+\frac{56 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \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{a+b x+c x^2}}-\frac{56 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \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{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.132319, size = 122, normalized size = 0.47 \[ -\frac{16 d^3 (d (b+2 c x))^{3/2} \left (14 c (a+x (b+c x)) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{3}{4},\frac{5}{2};\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-c \left (7 a+3 c x^2\right )+b^2-3 b c x\right )}{3 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-16*d^3*(d*(b + 2*c*x))^(3/2)*(b^2 - 3*b*c*x - c*(7*a + 3*c*x^2) + 14*c*(a + x*(b + c*x))*Sqrt[(c*(a + x*(b +
c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[3/4, 5/2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*(a + x*(b + c*x))^(
3/2))

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Maple [B]  time = 0.229, size = 859, 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)^(9/2)/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/3*(d*(2*c*x+b))^(1/2)*(168*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)-42*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)+168*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)-42*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)+168*((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^2-42*((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-72*c^4*x^4
-144*b*c^3*x^3-56*x^2*a*c^3-94*x^2*b^2*c^2-56*b*a*c^2*x-22*b^3*c*x-14*a*c*b^2-b^4)*d^4/(2*c*x+b)/(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{9}{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)^(9/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(9/2)/(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 (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}\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)^(9/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((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)*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)**(9/2)/(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{9}{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)^(9/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(9/2)/(c*x^2 + b*x + a)^(5/2), x)