\(\int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx\) [621]

Optimal result

Integrand size = 29, antiderivative size = 107 \[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{d}-\frac {(a c-2 d) \arcsin (a x)}{d^2}+\frac {(a c-d)^2 \arctan \left (\frac {d+a^2 c x}{\sqrt {a^2 c^2-d^2} \sqrt {1-a^2 x^2}}\right )}{d^2 \sqrt {a^2 c^2-d^2}} \]

[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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1668, 858, 222, 739, 210} \[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=\frac {(a c-d)^2 \arctan \left (\frac {a^2 c x+d}{\sqrt {1-a^2 x^2} \sqrt {a^2 c^2-d^2}}\right )}{d^2 \sqrt {a^2 c^2-d^2}}-\frac {\sqrt {1-a^2 x^2}}{d}-\frac {\arcsin (a x) (a c-2 d)}{d^2} \]

[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 210

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

Rule 222

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 739

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 858

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 1668

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*} \text {integral}& = -\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 \text {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] (verified)

Time = 0.39 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=\frac {-d \sqrt {1-a^2 x^2}+(-2 a c+4 d) \arctan \left (\frac {a x}{-1+\sqrt {1-a^2 x^2}}\right )-\frac {2 (a c-d) \sqrt {a^2 c^2-d^2} \arctan \left (\frac {\sqrt {a^2 c^2-d^2} x}{c+d x-c \sqrt {1-a^2 x^2}}\right )}{a c+d}}{d^2} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(232\) vs. \(2(99)=198\).

Time = 0.47 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.18

method	result	size
risch	\(\frac {a^{2} x^{2}-1}{d \sqrt {-a^{2} x^{2}+1}}-\frac {\frac {a \left (a c -2 d \right ) \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{d \sqrt {a^{2}}}-\frac {\left (-a^{2} c^{2}+2 a c d -d^{2}\right ) \ln \left (\frac {-\frac {2 \left (a^{2} c^{2}-d^{2}\right )}{d^{2}}+\frac {2 a^{2} c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {a^{2} c^{2}-d^{2}}{d^{2}}}\, \sqrt {-a^{2} \left (x +\frac {c}{d}\right )^{2}+\frac {2 a^{2} c \left (x +\frac {c}{d}\right )}{d}-\frac {a^{2} c^{2}-d^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {-\frac {a^{2} c^{2}-d^{2}}{d^{2}}}}}{d}\)	\(233\)
default	\(-\frac {a \left (-\frac {2 d \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+\frac {a c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+\frac {d \sqrt {-a^{2} x^{2}+1}}{a}\right )}{d^{2}}-\frac {\left (a^{2} c^{2}-2 a c d +d^{2}\right ) \ln \left (\frac {-\frac {2 \left (a^{2} c^{2}-d^{2}\right )}{d^{2}}+\frac {2 a^{2} c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {a^{2} c^{2}-d^{2}}{d^{2}}}\, \sqrt {-a^{2} \left (x +\frac {c}{d}\right )^{2}+\frac {2 a^{2} c \left (x +\frac {c}{d}\right )}{d}-\frac {a^{2} c^{2}-d^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{3} \sqrt {-\frac {a^{2} c^{2}-d^{2}}{d^{2}}}}\)	\(242\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.97 \[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=\left [-\frac {{\left (a c - d\right )} \sqrt {-\frac {a c - d}{a c + d}} \log \left (\frac {a^{2} c d x + d^{2} - {\left (a^{2} c^{2} - d^{2}\right )} \sqrt {-a^{2} x^{2} + 1} - {\left (a c d + d^{2} + {\left (a^{3} c^{2} + a^{2} c d\right )} x + \sqrt {-a^{2} x^{2} + 1} {\left (a c d + d^{2}\right )}\right )} \sqrt {-\frac {a c - d}{a c + d}}}{d x + c}\right ) - 2 \, {\left (a c - 2 \, d\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} d}{d^{2}}, \frac {2 \, {\left (a c - d\right )} \sqrt {\frac {a c - d}{a c + d}} \arctan \left (\frac {{\left (d x - \sqrt {-a^{2} x^{2} + 1} c + c\right )} \sqrt {\frac {a c - d}{a c + d}}}{{\left (a c - d\right )} x}\right ) + 2 \, {\left (a c - 2 \, d\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - \sqrt {-a^{2} x^{2} + 1} d}{d^{2}}\right ] \]

[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]

Sympy [F]

\[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=\int \frac {\left (a x + 1\right )^{2}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (c + d x\right )}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d-a*c>0)', see `assume?` for m
ore details)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22 \[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=-\frac {{\left (a^{2} c - 2 \, a d\right )} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{d^{2} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{d} - \frac {2 \, {\left (a^{3} c^{2} - 2 \, a^{2} c d + a d^{2}\right )} \arctan \left (\frac {d + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c}{a x}}{\sqrt {a^{2} c^{2} - d^{2}}}\right )}{\sqrt {a^{2} c^{2} - d^{2}} d^{2} {\left | a \right |}} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.38 \[ \int \frac {(1+a x)^2}{(c+d x) \sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {1-a^2\,x^2}}{d}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\left (2\,a\,\sqrt {-a^2}-\frac {a^2\,c\,\sqrt {-a^2}}{d}\right )}{a^2\,d}-\frac {\left (\ln \left (\sqrt {1-\frac {a^2\,c^2}{d^2}}\,\sqrt {1-a^2\,x^2}+\frac {a^2\,c\,x}{d}+1\right )-\ln \left (c+d\,x\right )\right )\,\left (a^2\,c^2-2\,a\,c\,d+d^2\right )}{d^3\,\sqrt {1-\frac {a^2\,c^2}{d^2}}} \]

[In]

int((a*x + 1)^2/((1 - a^2*x^2)^(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))