3.1350 \(\int (b d+2 c d x)^{3/2} (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=274 \[ -\frac{d^{3/2} \left (b^2-4 a c\right )^{17/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{924 c^4 \sqrt{a+b x+c x^2}}-\frac{d \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{462 c^3}+\frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{308 c^3 d}-\frac{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{66 c^2 d}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d} \]

[Out]

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

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Rubi [A]  time = 0.22119, antiderivative size = 274, 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 = {685, 692, 691, 689, 221} \[ -\frac{d^{3/2} \left (b^2-4 a c\right )^{17/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{924 c^4 \sqrt{a+b x+c x^2}}-\frac{d \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{462 c^3}+\frac{\left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{308 c^3 d}-\frac{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{5/2}}{66 c^2 d}+\frac{\left (a+b x+c x^2\right )^{5/2} (b d+2 c d x)^{5/2}}{15 c d} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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

Mathematica [C]  time = 0.119294, size = 117, normalized size = 0.43 \[ \frac{2}{15} d \sqrt{a+x (b+c x)} \sqrt{d (b+2 c x)} \left (\frac{\left (b^2-4 a c\right )^3 \, _2F_1\left (-\frac{5}{2},\frac{1}{4};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{64 c^3 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}+2 (a+x (b+c x))^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2)^(5/2),x]

[Out]

(2*d*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*(2*(a + x*(b + c*x))^3 + ((b^2 - 4*a*c)^3*Hypergeometric2F1[-5/
2, 1/4, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(64*c^3*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])))/15

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Maple [B]  time = 0.222, size = 1055, normalized size = 3.9 \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)^(3/2)*(c*x^2+b*x+a)^(5/2),x)

[Out]

-1/9240*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d*(-22176*x^8*b*c^8-19264*x^7*a*c^8-39536*x^7*b^2*c^7-34888*x^
6*b^3*c^6-27584*x^5*a^2*c^7-15248*x^5*b^4*c^5-2644*x^4*b^5*c^4-15808*x^3*a^3*c^6+4*x^3*b^6*c^3-10*x^2*b^7*c^2-
2560*x*a^4*c^5-10*x*b^8*c-1280*(-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))*a^3*b^2*c^3-1280*a^4*b*c^4-696*a^3*b^3*c^3
+140*a^2*b^5*c^2-10*a*b^7*c-67424*x^6*a*b*c^7-87344*x^5*a*b^2*c^6-68960*x^4*a^2*b*c^6-49800*x^4*a*b^3*c^5-5710
4*x^3*a^2*b^2*c^5-10624*x^3*a*b^4*c^4-23712*x^2*a^3*b*c^5-16696*x^2*a^2*b^3*c^4+152*x^2*a*b^5*c^3-9296*x*a^3*b
^2*c^4-688*x*a^2*b^4*c^3+140*x*a*b^6*c^2+480*(-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)*Ellip
ticF(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^4*c^2-80*(-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))*a*b^6*c-4928*x^9*c^9+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)*Ell
ipticF(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^8+1280*(-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))*a^4*c^4)/c^4/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, c d x + b d\right )}^{\frac{3}{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)^(3/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(3/2)*(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 ({\left (2 \, c^{3} d x^{5} + 5 \, b c^{2} d x^{4} + 4 \,{\left (b^{2} c + a c^{2}\right )} d x^{3} + a^{2} b d +{\left (b^{3} + 6 \, a b c\right )} d x^{2} + 2 \,{\left (a b^{2} + a^{2} c\right )} d x\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((2*c^3*d*x^5 + 5*b*c^2*d*x^4 + 4*(b^2*c + a*c^2)*d*x^3 + a^2*b*d + (b^3 + 6*a*b*c)*d*x^2 + 2*(a*b^2 +
a^2*c)*d*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a), 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)**(3/2)*(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{\left (2 \, c d x + b d\right )}^{\frac{3}{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)^(3/2)*(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(3/2)*(c*x^2 + b*x + a)^(5/2), x)