3.1379 \(\int \frac{(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=205 \[ \frac{60 d^{11/2} \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{7 \sqrt{a+b x+c x^2}}+\frac{120}{7} c d^5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}+\frac{72}{7} c d^3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}} \]

[Out]

(-2*d*(b*d + 2*c*d*x)^(9/2))/Sqrt[a + b*x + c*x^2] + (120*c*(b^2 - 4*a*c)*d^5*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x
+ c*x^2])/7 + (72*c*d^3*(b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/7 + (60*(b^2 - 4*a*c)^(9/4)*d^(11/2)*Sqr
t[-((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])/(7*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.175973, antiderivative size = 205, 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 = {686, 692, 691, 689, 221} \[ \frac{120}{7} c d^5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}+\frac{60 d^{11/2} \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{7 \sqrt{a+b x+c x^2}}+\frac{72}{7} c d^3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{5/2}-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(9/2))/Sqrt[a + b*x + c*x^2] + (120*c*(b^2 - 4*a*c)*d^5*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x
+ c*x^2])/7 + (72*c*d^3*(b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/7 + (60*(b^2 - 4*a*c)^(9/4)*d^(11/2)*Sqr
t[-((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])/(7*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 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 \frac{(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}}+\left (18 c d^2\right ) \int \frac{(b d+2 c d x)^{7/2}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}}+\frac{72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{1}{7} \left (90 c \left (b^2-4 a c\right ) d^4\right ) \int \frac{(b d+2 c d x)^{3/2}}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}}+\frac{120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{1}{7} \left (30 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}}+\frac{120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{\left (30 c \left (b^2-4 a c\right )^2 d^6 \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}{7 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}}+\frac{120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{\left (60 \left (b^2-4 a c\right )^2 d^5 \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 )}{7 \sqrt{a+b x+c x^2}}\\ &=-\frac{2 d (b d+2 c d x)^{9/2}}{\sqrt{a+b x+c x^2}}+\frac{120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}+\frac{72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}+\frac{60 \left (b^2-4 a c\right )^{9/4} d^{11/2} \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 )}{7 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.197787, size = 172, normalized size = 0.84 \[ \frac{2 d^5 \sqrt{d (b+2 c x)} \left (16 c^2 \left (-15 a^2-6 a c x^2+2 c^2 x^4\right )+30 \left (b^2-4 a c\right )^2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )+24 b^2 c \left (4 a+3 c x^2\right )+32 b c^2 x \left (2 c x^2-3 a\right )+40 b^3 c x-7 b^4\right )}{7 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*d^5*Sqrt[d*(b + 2*c*x)]*(-7*b^4 + 40*b^3*c*x + 32*b*c^2*x*(-3*a + 2*c*x^2) + 24*b^2*c*(4*a + 3*c*x^2) + 16*
c^2*(-15*a^2 - 6*a*c*x^2 + 2*c^2*x^4) + 30*(b^2 - 4*a*c)^2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeo
metric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(7*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.343, size = 569, normalized size = 2.8 \begin{align*}{\frac{2\,{d}^{5}}{14\,{c}^{2}{x}^{3}+21\,bc{x}^{2}+14\,acx+7\,{b}^{2}x+7\,ab}\sqrt{d \left ( 2\,cx+b \right ) }\sqrt{c{x}^{2}+bx+a} \left ( 64\,{x}^{5}{c}^{5}+240\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticF} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ) \sqrt{-4\,ac+{b}^{2}}{a}^{2}{c}^{2}-120\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticF} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ) \sqrt{-4\,ac+{b}^{2}}a{b}^{2}c+15\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{-{\frac{2\,cx+b}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{{\frac{-b-2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}{\it EllipticF} \left ( 1/2\,\sqrt{{\frac{b+2\,cx+\sqrt{-4\,ac+{b}^{2}}}{\sqrt{-4\,ac+{b}^{2}}}}}\sqrt{2},\sqrt{2} \right ) \sqrt{-4\,ac+{b}^{2}}{b}^{4}+160\,{x}^{4}b{c}^{4}-192\,{x}^{3}a{c}^{4}+208\,{x}^{3}{b}^{2}{c}^{3}-288\,{x}^{2}ab{c}^{3}+152\,{x}^{2}{b}^{3}{c}^{2}-480\,x{a}^{2}{c}^{3}+96\,xa{b}^{2}{c}^{2}+26\,x{b}^{4}c-240\,{a}^{2}b{c}^{2}+96\,a{b}^{3}c-7\,{b}^{5} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x)

[Out]

2/7*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^5*(64*x^5*c^5+240*((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))*(-4*a*c+b^2)^(1/2)*a^2*c^2-1
20*((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^2*c+15*((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^4+160*x^4*b*c^4-192*
x^3*a*c^4+208*x^3*b^2*c^3-288*x^2*a*b*c^3+152*x^2*b^3*c^2-480*x*a^2*c^3+96*x*a*b^2*c^2+26*x*b^4*c-240*a^2*b*c^
2+96*a*b^3*c-7*b^5)/(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 \frac{{\left (2 \, c d x + b d\right )}^{\frac{11}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(11/2)/(c*x^2 + b*x + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (32 \, c^{5} d^{5} x^{5} + 80 \, b c^{4} d^{5} x^{4} + 80 \, b^{2} c^{3} d^{5} x^{3} + 40 \, b^{3} c^{2} d^{5} x^{2} + 10 \, b^{4} c d^{5} x + b^{5} d^{5}\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral((32*c^5*d^5*x^5 + 80*b*c^4*d^5*x^4 + 80*b^2*c^3*d^5*x^3 + 40*b^3*c^2*d^5*x^2 + 10*b^4*c*d^5*x + b^5*d
^5)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^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)**(11/2)/(c*x**2+b*x+a)**(3/2),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{11}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(11/2)/(c*x^2 + b*x + a)^(3/2), x)