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

Optimal. Leaf size=107 \[ \frac {(a c-d)^2 \tan ^{-1}\left (\frac {a^2 c x+d}{\sqrt {1-a^2 x^2} \sqrt {a^2 c^2-d^2}}\right )}{d^2 \sqrt {a^2 c^2-d^2}}-\frac {\sqrt {1-a^2 x^2}}{d}-\frac {(a c-2 d) \sin ^{-1}(a x)}{d^2} \]

[Out]

-(a*c-2*d)*arcsin(a*x)/d^2+(a*c-d)^2*arctan((a^2*c*x+d)/(a^2*c^2-d^2)^(1/2)/(-a^2*x^2+1)^(1/2))/d^2/(a^2*c^2-d
^2)^(1/2)-(-a^2*x^2+1)^(1/2)/d

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Rubi [A]  time = 0.24, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {853, 1654, 844, 216, 725, 204} \[ \frac {(a c-d)^2 \tan ^{-1}\left (\frac {a^2 c x+d}{\sqrt {1-a^2 x^2} \sqrt {a^2 c^2-d^2}}\right )}{d^2 \sqrt {a^2 c^2-d^2}}-\frac {\sqrt {1-a^2 x^2}}{d}-\frac {(a c-2 d) \sin ^{-1}(a x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 853

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

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rubi steps

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

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Mathematica [C]  time = 0.30, size = 148, normalized size = 1.38 \[ -\frac {\frac {i (d-a c)^2 \log \left (\frac {2 d^3 \left (\sqrt {1-a^2 x^2} \sqrt {a^2 c^2-d^2}+i a^2 c x+i d\right )}{(d-a c)^2 \sqrt {a^2 c^2-d^2} (c+d x)}\right )}{\sqrt {a^2 c^2-d^2}}+d \sqrt {1-a^2 x^2}+(a c-2 d) \sin ^{-1}(a x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.47, size = 318, normalized size = 2.97 \[ \left [-\frac {{\left (a c - d\right )} \sqrt {-\frac {a c - d}{a c + d}} \log \left (\frac {a^{2} c d x + d^{2} - {\left (a^{2} c^{2} - d^{2}\right )} \sqrt {-a^{2} x^{2} + 1} - {\left (a c d + d^{2} + {\left (a^{3} c^{2} + a^{2} c d\right )} x + \sqrt {-a^{2} x^{2} + 1} {\left (a c d + d^{2}\right )}\right )} \sqrt {-\frac {a c - d}{a c + d}}}{d x + c}\right ) - 2 \, {\left (a c - 2 \, d\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} d}{d^{2}}, \frac {2 \, {\left (a c - d\right )} \sqrt {\frac {a c - d}{a c + d}} \arctan \left (\frac {{\left (d x - \sqrt {-a^{2} x^{2} + 1} c + c\right )} \sqrt {\frac {a c - d}{a c + d}}}{{\left (a c - d\right )} x}\right ) + 2 \, {\left (a c - 2 \, d\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - \sqrt {-a^{2} x^{2} + 1} d}{d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((a*c - d)*sqrt(-(a*c - d)/(a*c + d))*log((a^2*c*d*x + d^2 - (a^2*c^2 - d^2)*sqrt(-a^2*x^2 + 1) - (a*c*d + d
^2 + (a^3*c^2 + a^2*c*d)*x + sqrt(-a^2*x^2 + 1)*(a*c*d + d^2))*sqrt(-(a*c - d)/(a*c + d)))/(d*x + c)) - 2*(a*c
- 2*d)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*d)/d^2, (2*(a*c - d)*sqrt((a*c - d)/(a*c +
d))*arctan((d*x - sqrt(-a^2*x^2 + 1)*c + c)*sqrt((a*c - d)/(a*c + d))/((a*c - d)*x)) + 2*(a*c - 2*d)*arctan((
sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - sqrt(-a^2*x^2 + 1)*d)/d^2]

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giac [B]  time = 0.59, size = 208, normalized size = 1.94 \[ -{\left (\frac {{\left (a x - 1\right )} \sqrt {-\frac {2}{a x - 1} - 1} \mathrm {sgn}\left (\frac {1}{a x - 1}\right ) \mathrm {sgn}\relax (a)}{a d} - \frac {2 \, {\left (a c \mathrm {sgn}\left (\frac {1}{a x - 1}\right ) \mathrm {sgn}\relax (a) - 2 \, d \mathrm {sgn}\left (\frac {1}{a x - 1}\right ) \mathrm {sgn}\relax (a)\right )} \arctan \left (\sqrt {-\frac {2}{a x - 1} - 1}\right )}{a d^{2}} + \frac {2 \, {\left (a^{2} c^{2} \mathrm {sgn}\left (\frac {1}{a x - 1}\right ) \mathrm {sgn}\relax (a) - 2 \, a c d \mathrm {sgn}\left (\frac {1}{a x - 1}\right ) \mathrm {sgn}\relax (a) + d^{2} \mathrm {sgn}\left (\frac {1}{a x - 1}\right ) \mathrm {sgn}\relax (a)\right )} \arctan \left (\frac {a c \sqrt {-\frac {2}{a x - 1} - 1} + d \sqrt {-\frac {2}{a x - 1} - 1}}{\sqrt {a^{2} c^{2} - d^{2}}}\right )}{\sqrt {a^{2} c^{2} - d^{2}} a d^{2}}\right )} {\left | a \right |} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*x - 1)*sqrt(-2/(a*x - 1) - 1)*sgn(1/(a*x - 1))*sgn(a)/(a*d) - 2*(a*c*sgn(1/(a*x - 1))*sgn(a) - 2*d*sgn(1/
(a*x - 1))*sgn(a))*arctan(sqrt(-2/(a*x - 1) - 1))/(a*d^2) + 2*(a^2*c^2*sgn(1/(a*x - 1))*sgn(a) - 2*a*c*d*sgn(1
/(a*x - 1))*sgn(a) + d^2*sgn(1/(a*x - 1))*sgn(a))*arctan((a*c*sqrt(-2/(a*x - 1) - 1) + d*sqrt(-2/(a*x - 1) - 1
))/sqrt(a^2*c^2 - d^2))/(sqrt(a^2*c^2 - d^2)*a*d^2))*abs(a)

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maple [B]  time = 0.04, size = 1178, normalized size = 11.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x - 1\right )}^{2} {\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.29, size = 148, normalized size = 1.38 \[ -\frac {\sqrt {1-a^2\,x^2}}{d}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\left (2\,a\,\sqrt {-a^2}-\frac {a^2\,c\,\sqrt {-a^2}}{d}\right )}{a^2\,d}-\frac {\left (\ln \left (\sqrt {1-\frac {a^2\,c^2}{d^2}}\,\sqrt {1-a^2\,x^2}+\frac {a^2\,c\,x}{d}+1\right )-\ln \left (c+d\,x\right )\right )\,\left (a^2\,c^2-2\,a\,c\,d+d^2\right )}{d^3\,\sqrt {1-\frac {a^2\,c^2}{d^2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (1 - a^2*x^2)^(1/2)/d - (asinh(x*(-a^2)^(1/2))*(2*a*(-a^2)^(1/2) - (a^2*c*(-a^2)^(1/2))/d))/(a^2*d) - ((log(
(1 - (a^2*c^2)/d^2)^(1/2)*(1 - a^2*x^2)^(1/2) + (a^2*c*x)/d + 1) - log(c + d*x))*(d^2 + a^2*c^2 - 2*a*c*d))/(d
^3*(1 - (a^2*c^2)/d^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (c + d x\right ) \left (a x - 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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