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

Optimal. Leaf size=40 \[ \frac {x \left (a d^2+c\right )+b}{d^2 \sqrt {1-d^2 x^2}}-\frac {c \sin ^{-1}(d x)}{d^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {899, 1814, 12, 216} \begin {gather*} \frac {x \left (a d^2+c\right )+b}{d^2 \sqrt {1-d^2 x^2}}-\frac {c \sin ^{-1}(d x)}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 39, normalized size = 0.98 \begin {gather*} \frac {\frac {d \left (x \left (a d^2+c\right )+b\right )}{\sqrt {1-d^2 x^2}}-c \sin ^{-1}(d x)}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.13, size = 115, normalized size = 2.88 \begin {gather*} \frac {\sqrt {d x+1} \left (-\frac {a d^2 (1-d x)}{d x+1}+a d^2+\frac {b d (1-d x)}{d x+1}+b d-\frac {c (1-d x)}{d x+1}+c\right )}{2 d^3 \sqrt {1-d x}}+\frac {2 c \tan ^{-1}\left (\frac {\sqrt {1-d x}}{\sqrt {d x+1}}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 101, normalized size = 2.52 \begin {gather*} \frac {b d^{3} x^{2} - {\left (b d + {\left (a d^{3} + c d\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - b d + 2 \, {\left (c d^{2} x^{2} - c\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{d^{5} x^{2} - d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.30, size = 182, normalized size = 4.55 \begin {gather*} -\frac {2 \, c \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{3}} + \frac {\frac {a d^{2} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}} - \frac {b d {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}} + \frac {c {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}}{\sqrt {d x + 1}}}{4 \, d^{3}} - \frac {{\left (a d^{2} - b d + c\right )} \sqrt {d x + 1}}{4 \, d^{3} {\left (\sqrt {2} - \sqrt {-d x + 1}\right )}} - \frac {{\left (a d^{5} + b d^{4} + c d^{3}\right )} \sqrt {d x + 1} \sqrt {-d x + 1}}{2 \, {\left (d x - 1\right )} d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^3 + 1/4*(a*d^2*(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - b*d*(sqrt(2
) - sqrt(-d*x + 1))/sqrt(d*x + 1) + c*(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1))/d^3 - 1/4*(a*d^2 - b*d + c)*sq
rt(d*x + 1)/(d^3*(sqrt(2) - sqrt(-d*x + 1))) - 1/2*(a*d^5 + b*d^4 + c*d^3)*sqrt(d*x + 1)*sqrt(-d*x + 1)/((d*x
- 1)*d^6)

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maple [C]  time = 0.03, size = 151, normalized size = 3.78 \begin {gather*} \frac {\left (-\sqrt {-d^{2} x^{2}+1}\, a \,d^{3} x \,\mathrm {csgn}\relax (d )-c \,d^{2} x^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-\left (d x -1\right ) \left (d x +1\right )}}\right )-\sqrt {-d^{2} x^{2}+1}\, c d x \,\mathrm {csgn}\relax (d )-\sqrt {-d^{2} x^{2}+1}\, b d \,\mathrm {csgn}\relax (d )+c \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-\left (d x -1\right ) \left (d x +1\right )}}\right )\right ) \sqrt {-d x +1}\, \mathrm {csgn}\relax (d )}{\left (d x -1\right ) \sqrt {-d^{2} x^{2}+1}\, \sqrt {d x +1}\, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 61, normalized size = 1.52 \begin {gather*} \frac {a x}{\sqrt {-d^{2} x^{2} + 1}} + \frac {c x}{\sqrt {-d^{2} x^{2} + 1} d^{2}} - \frac {c \arcsin \left (d x\right )}{d^{3}} + \frac {b}{\sqrt {-d^{2} x^{2} + 1} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {c\,x^2+b\,x+a}{{\left (1-d\,x\right )}^{3/2}\,{\left (d\,x+1\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a + b*x + c*x^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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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