3.1383 \(\int \frac{1}{(b d+2 c d x)^{5/2} (a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=184 \[ -\frac{20 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{3 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt{a+b x+c x^2}}-\frac{40 c \sqrt{a+b x+c x^2}}{3 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}-\frac{2}{d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}} \]

[Out]

-2/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2]) - (40*c*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 4*a*
c)^2*d*(b*d + 2*c*d*x)^(3/2)) - (20*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])/(3*(b^2 - 4*a*c)^(7/4)*d^(5/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.143852, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {687, 693, 691, 689, 221} \[ -\frac{20 \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{3 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt{a+b x+c x^2}}-\frac{40 c \sqrt{a+b x+c x^2}}{3 d \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}-\frac{2}{d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

-2/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2]) - (40*c*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 4*a*
c)^2*d*(b*d + 2*c*d*x)^(3/2)) - (20*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])/(3*(b^2 - 4*a*c)^(7/4)*d^(5/2)*Sqrt[a + b*x + c*x^2])

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), 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] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && 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{1}{(b d+2 c d x)^{5/2} \left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}-\frac{(10 c) \int \frac{1}{(b d+2 c d x)^{5/2} \sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}-\frac{40 c \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{(10 c) \int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}-\frac{40 c \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{\left (10 c \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}{3 \left (b^2-4 a c\right )^2 d^2 \sqrt{a+b x+c x^2}}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}-\frac{40 c \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{\left (20 \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 )}{3 \left (b^2-4 a c\right )^2 d^3 \sqrt{a+b x+c x^2}}\\ &=-\frac{2}{\left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}-\frac{40 c \sqrt{a+b x+c x^2}}{3 \left (b^2-4 a c\right )^2 d (b d+2 c d x)^{3/2}}-\frac{20 \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 )}{3 \left (b^2-4 a c\right )^{7/4} d^{5/2} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0619869, size = 98, normalized size = 0.53 \[ \frac{8 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (-\frac{3}{4},\frac{3}{2};\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} (d (b+2 c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.23, size = 365, normalized size = 2. \begin{align*} -{\frac{2}{3\,{d}^{3} \left ( 2\,{c}^{2}{x}^{3}+3\,bc{x}^{2}+2\,acx+{b}^{2}x+ab \right ) \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( 2\,cx+b \right ) }\sqrt{d \left ( 2\,cx+b \right ) }\sqrt{c{x}^{2}+bx+a} \left ( 10\,\sqrt{-4\,ac+{b}^{2}}\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 ) xc+5\,\sqrt{-4\,ac+{b}^{2}}\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 ) b+20\,{c}^{2}{x}^{2}+20\,bcx+8\,ac+3\,{b}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(3/2),x)

[Out]

-2/3*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*(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)*E
llipticF(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*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+20*c^2*x^2+20*b*c*x+8*a*c+3*b^2)/d^3/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/(4*a*c-b^2)^
2/(2*c*x+b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (2 \, c d x + b d\right )}^{\frac{5}{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(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((2*c*d*x + b*d)^(5/2)*(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{\sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{8 \, c^{5} d^{3} x^{7} + 28 \, b c^{4} d^{3} x^{6} + 2 \,{\left (19 \, b^{2} c^{3} + 8 \, a c^{4}\right )} d^{3} x^{5} + a^{2} b^{3} d^{3} + 5 \,{\left (5 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} d^{3} x^{4} + 4 \,{\left (2 \, b^{4} c + 9 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} d^{3} x^{3} +{\left (b^{5} + 14 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} d^{3} x^{2} + 2 \,{\left (a b^{4} + 3 \, a^{2} b^{2} c\right )} d^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(8*c^5*d^3*x^7 + 28*b*c^4*d^3*x^6 + 2*(19*b^2*c^3 + 8*a*c^4
)*d^3*x^5 + a^2*b^3*d^3 + 5*(5*b^3*c^2 + 8*a*b*c^3)*d^3*x^4 + 4*(2*b^4*c + 9*a*b^2*c^2 + 2*a^2*c^3)*d^3*x^3 +
(b^5 + 14*a*b^3*c + 12*a^2*b*c^2)*d^3*x^2 + 2*(a*b^4 + 3*a^2*b^2*c)*d^3*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \left (b + 2 c x\right )\right )^{\frac{5}{2}} \left (a + b x + c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/((d*(b + 2*c*x))**(5/2)*(a + b*x + c*x**2)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (2 \, c d x + b d\right )}^{\frac{5}{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(1/(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((2*c*d*x + b*d)^(5/2)*(c*x^2 + b*x + a)^(3/2)), x)