3.1327 \(\int (b d+2 c d x)^{7/2} \sqrt{a+b x+c x^2} \, dx\)

Optimal. Leaf size=227 \[ -\frac{5 d^{7/2} \left (b^2-4 a c\right )^{13/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{231 c^2 \sqrt{a+b x+c x^2}}-\frac{10 d^3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{231 c}-\frac{2 d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{77 c}+\frac{\sqrt{a+b x+c x^2} (b d+2 c d x)^{9/2}}{11 c d} \]

[Out]

(-10*(b^2 - 4*a*c)^2*d^3*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(231*c) - (2*(b^2 - 4*a*c)*d*(b*d + 2*c*d*
x)^(5/2)*Sqrt[a + b*x + c*x^2])/(77*c) + ((b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2])/(11*c*d) - (5*(b^2 - 4*
a*c)^(13/4)*d^(7/2)*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])/(231*c^2*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.207217, antiderivative size = 227, 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 = {685, 692, 691, 689, 221} \[ -\frac{5 d^{7/2} \left (b^2-4 a c\right )^{13/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{231 c^2 \sqrt{a+b x+c x^2}}-\frac{10 d^3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{231 c}-\frac{2 d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}}{77 c}+\frac{\sqrt{a+b x+c x^2} (b d+2 c d x)^{9/2}}{11 c d} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(7/2)*Sqrt[a + b*x + c*x^2],x]

[Out]

(-10*(b^2 - 4*a*c)^2*d^3*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(231*c) - (2*(b^2 - 4*a*c)*d*(b*d + 2*c*d*
x)^(5/2)*Sqrt[a + b*x + c*x^2])/(77*c) + ((b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2])/(11*c*d) - (5*(b^2 - 4*
a*c)^(13/4)*d^(7/2)*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])/(231*c^2*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)^{7/2} \sqrt{a+b x+c x^2} \, dx &=\frac{(b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{11 c d}-\frac{\left (b^2-4 a c\right ) \int \frac{(b d+2 c d x)^{7/2}}{\sqrt{a+b x+c x^2}} \, dx}{22 c}\\ &=-\frac{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{77 c}+\frac{(b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )^2 d^2\right ) \int \frac{(b d+2 c d x)^{3/2}}{\sqrt{a+b x+c x^2}} \, dx}{154 c}\\ &=-\frac{10 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{231 c}-\frac{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{77 c}+\frac{(b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )^3 d^4\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{462 c}\\ &=-\frac{10 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{231 c}-\frac{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{77 c}+\frac{(b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )^3 d^4 \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}{462 c \sqrt{a+b x+c x^2}}\\ &=-\frac{10 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{231 c}-\frac{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{77 c}+\frac{(b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{11 c d}-\frac{\left (5 \left (b^2-4 a c\right )^3 d^3 \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 )}{231 c^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{10 \left (b^2-4 a c\right )^2 d^3 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}}{231 c}-\frac{2 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}}{77 c}+\frac{(b d+2 c d x)^{9/2} \sqrt{a+b x+c x^2}}{11 c d}-\frac{5 \left (b^2-4 a c\right )^{13/4} d^{7/2} \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 )}{231 c^2 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.40669, size = 155, normalized size = 0.68 \[ \frac{4 \sqrt{a+x (b+c x)} (d (b+2 c x))^{7/2} \left (7 (b+2 c x)^2 (a+x (b+c x))-10 c \left (a-\frac{b^2}{4 c}\right ) \left (\frac{\left (b^2-4 a c\right ) \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{4 c \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}+2 (a+x (b+c x))\right )\right )}{77 (b+2 c x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(7/2)*Sqrt[a + b*x + c*x^2],x]

[Out]

(4*(d*(b + 2*c*x))^(7/2)*Sqrt[a + x*(b + c*x)]*(7*(b + 2*c*x)^2*(a + x*(b + c*x)) - 10*(a - b^2/(4*c))*c*(2*(a
+ x*(b + c*x)) + ((b^2 - 4*a*c)*Hypergeometric2F1[-1/2, 1/4, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(4*c*Sqrt[(c*
(a + x*(b + c*x)))/(-b^2 + 4*a*c)]))))/(77*(b + 2*c*x)^3)

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Maple [B]  time = 0.381, size = 798, normalized size = 3.5 \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)^(7/2)*(c*x^2+b*x+a)^(1/2),x)

[Out]

1/462*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^3*(1344*x^7*c^7+4704*x^6*b*c^6+1728*x^5*a*c^6+6624*x^5*b^2*c^5
+320*(-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*c^3-240*(-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^2*b^2*c^2+60*(-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^4*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)*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))*b^6+4320*x^4*a*b*c^5+4800*x^4*
b^3*c^4-256*x^3*a^2*c^5+4448*x^3*a*b^2*c^4+1844*x^3*b^4*c^3-384*x^2*a^2*b*c^4+2352*x^2*a*b^3*c^3+318*x^2*b^5*c
^2-640*x*a^3*c^4+288*x*a^2*b^2*c^3+516*x*a*b^4*c^2+10*x*b^6*c-320*a^3*b*c^3+208*a^2*b^3*c^2+10*a*b^5*c)/c^2/(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{7}{2}} \sqrt{c x^{2} + b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(7/2)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(7/2)*sqrt(c*x^2 + b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (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}\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)^(7/2)*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral((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)*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)**(7/2)*(c*x**2+b*x+a)**(1/2),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{7}{2}} \sqrt{c x^{2} + b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(7/2)*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(7/2)*sqrt(c*x^2 + b*x + a), x)