3.1352 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{5/2}} \, dx\)

Optimal. Leaf size=219 \[ \frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{84 c^4 d^{5/2} \sqrt{a+b x+c x^2}}-\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{84 c^3 d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}} \]

[Out]

(-5*(b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(84*c^3*d^3) + (5*Sqrt[b*d + 2*c*d*x]*(a + b*x +
c*x^2)^(3/2))/(42*c^2*d^3) - (a + b*x + c*x^2)^(5/2)/(3*c*d*(b*d + 2*c*d*x)^(3/2)) + (5*(b^2 - 4*a*c)^(9/4)*Sq
rt[-((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])/(84*c^4*d^(5/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.183386, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {684, 685, 691, 689, 221} \[ -\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{84 c^3 d^3}+\frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{84 c^4 d^{5/2} \sqrt{a+b x+c x^2}}+\frac{5 \left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(84*c^3*d^3) + (5*Sqrt[b*d + 2*c*d*x]*(a + b*x +
c*x^2)^(3/2))/(42*c^2*d^3) - (a + b*x + c*x^2)^(5/2)/(3*c*d*(b*d + 2*c*d*x)^(3/2)) + (5*(b^2 - 4*a*c)^(9/4)*Sq
rt[-((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])/(84*c^4*d^(5/2)*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 689

Int[1/(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[1/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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{5/2}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}}+\frac{5 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{\sqrt{b d+2 c d x}} \, dx}{6 c d^2}\\ &=\frac{5 \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int \frac{\sqrt{a+b x+c x^2}}{\sqrt{b d+2 c d x}} \, dx}{28 c^2 d^2}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{84 c^3 d^3}+\frac{5 \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}}+\frac{\left (5 \left (b^2-4 a c\right )^2\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{168 c^3 d^2}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{84 c^3 d^3}+\frac{5 \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}}+\frac{\left (5 \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{1}{\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}{168 c^3 d^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{84 c^3 d^3}+\frac{5 \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}}+\frac{\left (5 \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{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 )}{84 c^4 d^3 \sqrt{a+b x+c x^2}}\\ &=-\frac{5 \left (b^2-4 a c\right ) \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{84 c^3 d^3}+\frac{5 \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}}+\frac{5 \left (b^2-4 a c\right )^{9/4} \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 )}{84 c^4 d^{5/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0628769, size = 101, normalized size = 0.46 \[ -\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \, _2F_1\left (-\frac{5}{2},-\frac{3}{4};\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{96 c^3 d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -3/4, 1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(96*c
^3*d*(d*(b + 2*c*x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.245, size = 1000, normalized size = 4.6 \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)^(5/2),x)

[Out]

1/168*(24*x^6*c^6+160*((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)*EllipticF(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))*(-4*a*c+b^2)^(1/2)*x*a^2*c^3-80*((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)*EllipticF(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))*(-4*a*c+b^2)^(1/2)*x
*a*b^2*c^2+10*(-4*a*c+b^2)^(1/2)*((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)*EllipticF(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^4*c+72*x^5*b*c^5+80*((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)*EllipticF(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))*(-4*a*c+b^2)^(1/2
)*a^2*b*c^2-40*((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)*EllipticF(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))*(-4*a*c+b^2)^(1/2)*a*b^3*c+5*(-4*a*c+b^2)^(1/2)*((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)*EllipticF(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^5+152*x^4
*a*c^5+52*x^4*b^2*c^4+304*x^3*a*b*c^4-16*x^3*b^3*c^3+72*x^2*a^2*c^4+192*x^2*a*b^2*c^3-30*x^2*b^4*c^2+72*x*a^2*
b*c^3+40*x*a*b^3*c^2-10*x*b^5*c-56*a^3*c^3+60*a^2*b^2*c^2-10*a*b^4*c)/d^3*(d*(2*c*x+b))^(1/2)/(c*x^2+b*x+a)^(1
/2)/(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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(5/2), 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}}{8 \, c^{3} d^{3} x^{3} + 12 \, b c^{2} d^{3} x^{2} + 6 \, b^{2} c d^{3} x + b^{3} d^{3}}, 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)^(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)/(
8*c^3*d^3*x^3 + 12*b*c^2*d^3*x^2 + 6*b^2*c*d^3*x + b^3*d^3), 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{5}{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)**(5/2),x)

[Out]

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

[Out]

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