3.621 \(\int \frac{(1+a x)^2}{(c+d x) \sqrt{1-a^2 x^2}} \, dx\)

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]

-(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])

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Rubi [A]  time = 0.179428, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {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*x)^2/((c + d*x)*Sqrt[1 - a^2*x^2]),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 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]))

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 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 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])

Rubi steps

\begin{align*} \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*}

Mathematica [A]  time = 0.108653, size = 120, normalized size = 1.12 \[ \frac{(a c-d) \sqrt{a^2 c^2-d^2} \tan ^{-1}\left (\frac{a^2 c x+d}{\sqrt{1-a^2 x^2} \sqrt{a^2 c^2-d^2}}\right )}{d^2 (a c+d)}-\frac{\sqrt{1-a^2 x^2}}{d}-\frac{(a c-d) \sin ^{-1}(a x)}{d^2}+\frac{\sin ^{-1}(a x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.23, size = 524, normalized size = 4.9 \begin{align*} -{\frac{1}{d}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}c}{{d}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+2\,{\frac{a}{d\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }-{\frac{{a}^{2}{c}^{2}}{{d}^{3}}\ln \left ({ \left ( -2\,{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}+2\,{\frac{{a}^{2}c}{d} \left ( x+{\frac{c}{d}} \right ) }+2\,\sqrt{-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}}\sqrt{- \left ( x+{\frac{c}{d}} \right ) ^{2}{a}^{2}+2\,{\frac{{a}^{2}c}{d} \left ( x+{\frac{c}{d}} \right ) }-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}} \right ) \left ( x+{\frac{c}{d}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}}}}}+2\,{\frac{ac}{{d}^{2}}\ln \left ({ \left ( -2\,{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}+2\,{\frac{{a}^{2}c}{d} \left ( x+{\frac{c}{d}} \right ) }+2\,\sqrt{-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}}\sqrt{- \left ( x+{\frac{c}{d}} \right ) ^{2}{a}^{2}+2\,{\frac{{a}^{2}c}{d} \left ( x+{\frac{c}{d}} \right ) }-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}} \right ) \left ( x+{\frac{c}{d}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}}}}}-{\frac{1}{d}\ln \left ({ \left ( -2\,{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}+2\,{\frac{{a}^{2}c}{d} \left ( x+{\frac{c}{d}} \right ) }+2\,\sqrt{-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}}\sqrt{- \left ( x+{\frac{c}{d}} \right ) ^{2}{a}^{2}+2\,{\frac{{a}^{2}c}{d} \left ( x+{\frac{c}{d}} \right ) }-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}} \right ) \left ( x+{\frac{c}{d}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{{a}^{2}{c}^{2}-{d}^{2}}{{d}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a^2*x^2+1)^(1/2)/d-a^2/d^2*c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2*a/d/(a^2)^(1/2)*arctan(
(a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-1/d^3/(-(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+2/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*c-1/d/(-(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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.19296, size = 679, normalized size = 6.35 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34627, size = 177, normalized size = 1.65 \begin{align*} -\frac{{\left (a^{2} c - 2 \, a d\right )} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{d^{2}{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{d} - \frac{2 \,{\left (a^{3} c^{2} - 2 \, a^{2} c d + a d^{2}\right )} \arctan \left (\frac{d + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c}{a x}}{\sqrt{a^{2} c^{2} - d^{2}}}\right )}{\sqrt{a^{2} c^{2} - d^{2}} d^{2}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*c - 2*a*d)*arcsin(a*x)*sgn(a)/(d^2*abs(a)) - sqrt(-a^2*x^2 + 1)/d - 2*(a^3*c^2 - 2*a^2*c*d + a*d^2)*arct
an((d + (sqrt(-a^2*x^2 + 1)*abs(a) + a)*c/(a*x))/sqrt(a^2*c^2 - d^2))/(sqrt(a^2*c^2 - d^2)*d^2*abs(a))