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

Optimal. Leaf size=135 \[ \frac {x \left (a^2 d^4+2 a c d^2+b^2 d^2+c^2\right )+2 b d^2 \left (a+\frac {c}{d^2}\right )}{d^4 \sqrt {1-d^2 x^2}}-\frac {\sin ^{-1}(d x) \left (c \left (4 a+\frac {3 c}{d^2}\right )+2 b^2\right )}{2 d^3}+\frac {2 b c \sqrt {1-d^2 x^2}}{d^4}+\frac {c^2 x \sqrt {1-d^2 x^2}}{2 d^4} \]

[Out]

-1/2*(2*b^2+c*(4*a+3*c/d^2))*arcsin(d*x)/d^3+(2*b*(a+c/d^2)*d^2+(a^2*d^4+2*a*c*d^2+b^2*d^2+c^2)*x)/d^4/(-d^2*x
^2+1)^(1/2)+2*b*c*(-d^2*x^2+1)^(1/2)/d^4+1/2*c^2*x*(-d^2*x^2+1)^(1/2)/d^4

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Rubi [A]  time = 0.19, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {899, 1814, 1815, 641, 216} \[ \frac {x \left (a^2 d^4+2 a c d^2+b^2 d^2+c^2\right )+2 b d^2 \left (a+\frac {c}{d^2}\right )}{d^4 \sqrt {1-d^2 x^2}}-\frac {\sin ^{-1}(d x) \left (c \left (4 a+\frac {3 c}{d^2}\right )+2 b^2\right )}{2 d^3}+\frac {2 b c \sqrt {1-d^2 x^2}}{d^4}+\frac {c^2 x \sqrt {1-d^2 x^2}}{2 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 899

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Int[(d*f + e*g*x^2)^m*(a + b*x + c*x^2)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m - n, 0] &&
EqQ[e*f + d*g, 0] && (IntegerQ[m] || (GtQ[d, 0] && GtQ[f, 0]))

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^2}{(1-d x)^{3/2} (1+d x)^{3/2}} \, dx &=\int \frac {\left (a+b x+c x^2\right )^2}{\left (1-d^2 x^2\right )^{3/2}} \, dx\\ &=\frac {2 b \left (a+\frac {c}{d^2}\right ) d^2+\left (c^2+b^2 d^2+2 a c d^2+a^2 d^4\right ) x}{d^4 \sqrt {1-d^2 x^2}}-\int \frac {\frac {c^2+b^2 d^2+2 a c d^2}{d^4}+\frac {2 b c x}{d^2}+\frac {c^2 x^2}{d^2}}{\sqrt {1-d^2 x^2}} \, dx\\ &=\frac {2 b \left (a+\frac {c}{d^2}\right ) d^2+\left (c^2+b^2 d^2+2 a c d^2+a^2 d^4\right ) x}{d^4 \sqrt {1-d^2 x^2}}+\frac {c^2 x \sqrt {1-d^2 x^2}}{2 d^4}+\frac {\int \frac {-2 b^2-c \left (4 a+\frac {3 c}{d^2}\right )-4 b c x}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2}\\ &=\frac {2 b \left (a+\frac {c}{d^2}\right ) d^2+\left (c^2+b^2 d^2+2 a c d^2+a^2 d^4\right ) x}{d^4 \sqrt {1-d^2 x^2}}+\frac {2 b c \sqrt {1-d^2 x^2}}{d^4}+\frac {c^2 x \sqrt {1-d^2 x^2}}{2 d^4}-\frac {\left (2 b^2+c \left (4 a+\frac {3 c}{d^2}\right )\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2}\\ &=\frac {2 b \left (a+\frac {c}{d^2}\right ) d^2+\left (c^2+b^2 d^2+2 a c d^2+a^2 d^4\right ) x}{d^4 \sqrt {1-d^2 x^2}}+\frac {2 b c \sqrt {1-d^2 x^2}}{d^4}+\frac {c^2 x \sqrt {1-d^2 x^2}}{2 d^4}-\frac {\left (2 b^2+c \left (4 a+\frac {3 c}{d^2}\right )\right ) \sin ^{-1}(d x)}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 127, normalized size = 0.94 \[ \frac {d x \left (2 a^2 d^4+4 a c d^2+c^2 \left (3-d^2 x^2\right )\right )-\sqrt {1-d^2 x^2} \sin ^{-1}(d x) \left (4 a c d^2+2 b^2 d^2+3 c^2\right )+4 b d \left (a d^2+c \left (2-d^2 x^2\right )\right )+2 b^2 d^3 x}{2 d^5 \sqrt {1-d^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 204, normalized size = 1.51 \[ -\frac {4 \, a b d^{3} + 8 \, b c d - 4 \, {\left (a b d^{5} + 2 \, b c d^{3}\right )} x^{2} - {\left (c^{2} d^{3} x^{3} + 4 \, b c d^{3} x^{2} - 4 \, a b d^{3} - 8 \, b c d - {\left (2 \, a^{2} d^{5} + 2 \, {\left (b^{2} + 2 \, a c\right )} d^{3} + 3 \, c^{2} d\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 2 \, {\left (2 \, {\left (b^{2} + 2 \, a c\right )} d^{2} - {\left (2 \, {\left (b^{2} + 2 \, a c\right )} d^{4} + 3 \, c^{2} d^{2}\right )} x^{2} + 3 \, c^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{2 \, {\left (d^{7} x^{2} - d^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*a*b*d^3 + 8*b*c*d - 4*(a*b*d^5 + 2*b*c*d^3)*x^2 - (c^2*d^3*x^3 + 4*b*c*d^3*x^2 - 4*a*b*d^3 - 8*b*c*d -
(2*a^2*d^5 + 2*(b^2 + 2*a*c)*d^3 + 3*c^2*d)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*(2*(b^2 + 2*a*c)*d^2 - (2*(b^
2 + 2*a*c)*d^4 + 3*c^2*d^2)*x^2 + 3*c^2)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^7*x^2 - d^5)

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giac [B]  time = 0.39, size = 387, normalized size = 2.87 \[ \frac {\sqrt {d x + 1} \sqrt {-d x + 1} {\left ({\left (d x + 1\right )} {\left (\frac {{\left (d x + 1\right )} c^{2}}{d^{5}} + \frac {4 \, b c d^{16} - 3 \, c^{2} d^{15}}{d^{20}}\right )} - \frac {a^{2} d^{19} + 2 \, a b d^{18} + b^{2} d^{17} + 2 \, a c d^{17} + 10 \, b c d^{16} - c^{2} d^{15}}{d^{20}}\right )}}{2 \, {\left (d x - 1\right )}} - \frac {{\left (2 \, b^{2} d^{2} + 4 \, a c d^{2} + 3 \, c^{2}\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{5}} + \frac {\frac {a^{2} d^{4} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}} - \frac {2 \, a b d^{3} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}} + \frac {b^{2} d^{2} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}} + \frac {2 \, a c d^{2} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}} - \frac {2 \, b c d {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}} + \frac {c^{2} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}}}{4 \, d^{5}} - \frac {{\left (a^{2} d^{4} - 2 \, a b d^{3} + b^{2} d^{2} + 2 \, a c d^{2} - 2 \, b c d + c^{2}\right )} \sqrt {d x + 1}}{4 \, d^{5} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*((d*x + 1)*c^2/d^5 + (4*b*c*d^16 - 3*c^2*d^15)/d^20) - (a^2*d^19 +
2*a*b*d^18 + b^2*d^17 + 2*a*c*d^17 + 10*b*c*d^16 - c^2*d^15)/d^20)/(d*x - 1) - (2*b^2*d^2 + 4*a*c*d^2 + 3*c^2
)*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^5 + 1/4*(a^2*d^4*(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - 2*a*b*d^3*(s
qrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) + b^2*d^2*(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) + 2*a*c*d^2*(sqrt(2)
- sqrt(-d*x + 1))/sqrt(d*x + 1) - 2*b*c*d*(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) + c^2*(sqrt(2) - sqrt(-d*x
+ 1))/sqrt(d*x + 1))/d^5 - 1/4*(a^2*d^4 - 2*a*b*d^3 + b^2*d^2 + 2*a*c*d^2 - 2*b*c*d + c^2)*sqrt(d*x + 1)/(d^5
*(sqrt(2) - sqrt(-d*x + 1)))

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maple [C]  time = 0.03, size = 380, normalized size = 2.81 \[ \frac {\sqrt {-d x +1}\, \left (-2 \sqrt {-d^{2} x^{2}+1}\, a^{2} d^{5} x \,\mathrm {csgn}\relax (d )-4 a c \,d^{4} x^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-2 b^{2} d^{4} x^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+\sqrt {-d^{2} x^{2}+1}\, c^{2} d^{3} x^{3} \mathrm {csgn}\relax (d )+4 \sqrt {-d^{2} x^{2}+1}\, b c \,d^{3} x^{2} \mathrm {csgn}\relax (d )-4 \sqrt {-d^{2} x^{2}+1}\, a c \,d^{3} x \,\mathrm {csgn}\relax (d )-2 \sqrt {-d^{2} x^{2}+1}\, b^{2} d^{3} x \,\mathrm {csgn}\relax (d )-3 c^{2} d^{2} x^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-4 \sqrt {-d^{2} x^{2}+1}\, a b \,d^{3} \mathrm {csgn}\relax (d )+4 a c \,d^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+2 b^{2} d^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-3 \sqrt {-d^{2} x^{2}+1}\, c^{2} d x \,\mathrm {csgn}\relax (d )-8 \sqrt {-d^{2} x^{2}+1}\, b c d \,\mathrm {csgn}\relax (d )+3 c^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )\right ) \mathrm {csgn}\relax (d )}{2 \left (d x -1\right ) \sqrt {-d^{2} x^{2}+1}\, \sqrt {d x +1}\, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-d*x+1)^(1/2)*((-d^2*x^2+1)^(1/2)*c^2*d^3*x^3*csgn(d)-2*csgn(d)*d^5*(-d^2*x^2+1)^(1/2)*x*a^2+4*(-d^2*x^2+
1)^(1/2)*b*c*d^3*x^2*csgn(d)-4*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))*x^2*a*c*d^4-2*arctan(1/(-d^2*x^2+1)^(1
/2)*d*x*csgn(d))*x^2*b^2*d^4-4*(-d^2*x^2+1)^(1/2)*a*c*d^3*x*csgn(d)-2*(-d^2*x^2+1)^(1/2)*b^2*d^3*x*csgn(d)-4*(
-d^2*x^2+1)^(1/2)*a*b*d^3*csgn(d)-3*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))*x^2*c^2*d^2-3*(-d^2*x^2+1)^(1/2)*
c^2*d*x*csgn(d)-8*(-d^2*x^2+1)^(1/2)*b*c*d*csgn(d)+4*a*c*d^2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))+2*b^2*d^
2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))+3*c^2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d)))*csgn(d)/(d*x-1)/(-d
^2*x^2+1)^(1/2)/d^5/(d*x+1)^(1/2)

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maxima [A]  time = 0.97, size = 176, normalized size = 1.30 \[ \frac {a^{2} x}{\sqrt {-d^{2} x^{2} + 1}} - \frac {c^{2} x^{3}}{2 \, \sqrt {-d^{2} x^{2} + 1} d^{2}} - \frac {2 \, b c x^{2}}{\sqrt {-d^{2} x^{2} + 1} d^{2}} + \frac {2 \, a b}{\sqrt {-d^{2} x^{2} + 1} d^{2}} + \frac {{\left (b^{2} + 2 \, a c\right )} x}{\sqrt {-d^{2} x^{2} + 1} d^{2}} - \frac {{\left (b^{2} + 2 \, a c\right )} \arcsin \left (d x\right )}{d^{3}} + \frac {3 \, c^{2} x}{2 \, \sqrt {-d^{2} x^{2} + 1} d^{4}} - \frac {3 \, c^{2} \arcsin \left (d x\right )}{2 \, d^{5}} + \frac {4 \, b c}{\sqrt {-d^{2} x^{2} + 1} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x/sqrt(-d^2*x^2 + 1) - 1/2*c^2*x^3/(sqrt(-d^2*x^2 + 1)*d^2) - 2*b*c*x^2/(sqrt(-d^2*x^2 + 1)*d^2) + 2*a*b/(
sqrt(-d^2*x^2 + 1)*d^2) + (b^2 + 2*a*c)*x/(sqrt(-d^2*x^2 + 1)*d^2) - (b^2 + 2*a*c)*arcsin(d*x)/d^3 + 3/2*c^2*x
/(sqrt(-d^2*x^2 + 1)*d^4) - 3/2*c^2*arcsin(d*x)/d^5 + 4*b*c/(sqrt(-d^2*x^2 + 1)*d^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^2}{{\left (1-d\,x\right )}^{3/2}\,{\left (d\,x+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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