3.1355 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{17/2}} \, dx\)

Optimal. Leaf size=258 \[ \frac{\sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{924 c^4 d^{17/2} \left (b^2-4 a c\right )^{3/4} \sqrt{a+b x+c x^2}}+\frac{\sqrt{a+b x+c x^2}}{462 c^3 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}} \]

[Out]

-Sqrt[a + b*x + c*x^2]/(308*c^3*d^5*(b*d + 2*c*d*x)^(7/2)) + Sqrt[a + b*x + c*x^2]/(462*c^3*(b^2 - 4*a*c)*d^7*
(b*d + 2*c*d*x)^(3/2)) - (a + b*x + c*x^2)^(3/2)/(66*c^2*d^3*(b*d + 2*c*d*x)^(11/2)) - (a + b*x + c*x^2)^(5/2)
/(15*c*d*(b*d + 2*c*d*x)^(15/2)) + (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])/(924*c^4*(b^2 - 4*a*c)^(3/4)*d^(17/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.213809, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {684, 693, 691, 689, 221} \[ \frac{\sqrt{a+b x+c x^2}}{462 c^3 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}+\frac{\sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{924 c^4 d^{17/2} \left (b^2-4 a c\right )^{3/4} \sqrt{a+b x+c x^2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(17/2),x]

[Out]

-Sqrt[a + b*x + c*x^2]/(308*c^3*d^5*(b*d + 2*c*d*x)^(7/2)) + Sqrt[a + b*x + c*x^2]/(462*c^3*(b^2 - 4*a*c)*d^7*
(b*d + 2*c*d*x)^(3/2)) - (a + b*x + c*x^2)^(3/2)/(66*c^2*d^3*(b*d + 2*c*d*x)^(11/2)) - (a + b*x + c*x^2)^(5/2)
/(15*c*d*(b*d + 2*c*d*x)^(15/2)) + (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])/(924*c^4*(b^2 - 4*a*c)^(3/4)*d^(17/2)*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 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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)^{17/2}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac{\int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{13/2}} \, dx}{6 c d^2}\\ &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac{\int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^{9/2}} \, dx}{44 c^2 d^4}\\ &=-\frac{\sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac{\int \frac{1}{(b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}} \, dx}{616 c^3 d^6}\\ &=-\frac{\sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac{\sqrt{a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac{\int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{1848 c^3 \left (b^2-4 a c\right ) d^8}\\ &=-\frac{\sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac{\sqrt{a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac{\sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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}{1848 c^3 \left (b^2-4 a c\right ) d^8 \sqrt{a+b x+c x^2}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac{\sqrt{a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac{\sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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 )}{924 c^4 \left (b^2-4 a c\right ) d^9 \sqrt{a+b x+c x^2}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{308 c^3 d^5 (b d+2 c d x)^{7/2}}+\frac{\sqrt{a+b x+c x^2}}{462 c^3 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{66 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac{\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 )}{924 c^4 \left (b^2-4 a c\right )^{3/4} d^{17/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0950798, size = 109, normalized size = 0.42 \[ -\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \sqrt{d (b+2 c x)} \, _2F_1\left (-\frac{15}{4},-\frac{5}{2};-\frac{11}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{480 c^3 d^9 (b+2 c x)^8 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(17/2),x]

[Out]

-((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-15/4, -5/2, -11/4, (b + 2*c*x)^
2/(b^2 - 4*a*c)])/(480*c^3*d^9*(b + 2*c*x)^8*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.234, size = 1431, normalized size = 5.6 \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)^(17/2),x)

[Out]

-1/9240*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(2240*((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^6*b*c^6+3360*((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^5*b^2*c^5+1280*x^8*c^8-10*x*b^7*c+70*((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
*b^6*c+420*((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^2*b^5*c^2+2800*((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)*Elli
pticF(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^4*b^3*
c^4+1400*((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^3*b^4*c^3+640*((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)*Ellipti
cF(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^7*c^7-948
*x^4*b^4*c^4+3032*x^5*b^3*c^5+7904*x^6*a*c^7+13792*x^4*a^2*c^6+9632*x^2*a^3*c^5+5120*x^7*b*c^7-976*x^3*b^5*c^3
-150*x^2*b^6*c^2+6984*x^6*b^2*c^6+2464*a^4*c^4-56*a^3*b^2*c^3-20*a^2*b^4*c^2-10*a*b^6*c-160*x*a*b^5*c^2+5*((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)*b^7+23712*x^5*a*b*c^6+22744*x^4*a*b^2*c^5+27584*x^3*a^2*b*c^5+5968*x^3*a*b^3*
c^4+13464*x^2*a^2*b^2*c^4-1128*x^2*a*b^4*c^3+9632*x*a^3*b*c^4-328*x*a^2*b^3*c^3)/d^9/(2*c^2*x^3+3*b*c*x^2+2*a*
c*x+b^2*x+a*b)/(2*c*x+b)^7/(4*a*c-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{17}{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)^(17/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(17/2), 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}}{512 \, c^{9} d^{9} x^{9} + 2304 \, b c^{8} d^{9} x^{8} + 4608 \, b^{2} c^{7} d^{9} x^{7} + 5376 \, b^{3} c^{6} d^{9} x^{6} + 4032 \, b^{4} c^{5} d^{9} x^{5} + 2016 \, b^{5} c^{4} d^{9} x^{4} + 672 \, b^{6} c^{3} d^{9} x^{3} + 144 \, b^{7} c^{2} d^{9} x^{2} + 18 \, b^{8} c d^{9} x + b^{9} d^{9}}, 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)^(17/2),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)/(
512*c^9*d^9*x^9 + 2304*b*c^8*d^9*x^8 + 4608*b^2*c^7*d^9*x^7 + 5376*b^3*c^6*d^9*x^6 + 4032*b^4*c^5*d^9*x^5 + 20
16*b^5*c^4*d^9*x^4 + 672*b^6*c^3*d^9*x^3 + 144*b^7*c^2*d^9*x^2 + 18*b^8*c*d^9*x + b^9*d^9), 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((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(17/2),x)

[Out]

Timed out

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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{17}{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)^(17/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(17/2), x)