\(\int (a+b \arccos (1+d x^2)) \, dx\) [76]

Optimal result

Integrand size = 12, antiderivative size = 43 \[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx=a x-\frac {2 b \sqrt {-2 d x^2-d^2 x^4}}{d x}+b x \arccos \left (1+d x^2\right ) \]

[Out]

a*x+b*x*arccos(d*x^2+1)-2*b*(-d^2*x^4-2*d*x^2)^(1/2)/d/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4925, 12, 1602} \[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx=a x+b x \arccos \left (d x^2+1\right )-\frac {2 b \sqrt {-d^2 x^4-2 d x^2}}{d x} \]

[In]

Int[a + b*ArcCos[1 + d*x^2],x]

[Out]

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

Rule 12

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

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 4925

Int[ArcCos[u_], x_Symbol] :> Simp[x*ArcCos[u], x] + Int[SimplifyIntegrand[x*(D[u, x]/Sqrt[1 - u^2]), x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps \begin{align*} \text {integral}& = a x+b \int \arccos \left (1+d x^2\right ) \, dx \\ & = a x+b x \arccos \left (1+d x^2\right )+b \int \frac {2 d x^2}{\sqrt {-2 d x^2-d^2 x^4}} \, dx \\ & = a x+b x \arccos \left (1+d x^2\right )+(2 b d) \int \frac {x^2}{\sqrt {-2 d x^2-d^2 x^4}} \, dx \\ & = a x-\frac {2 b \sqrt {-2 d x^2-d^2 x^4}}{d x}+b x \arccos \left (1+d x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx=a x-\frac {2 b \sqrt {-d x^2 \left (2+d x^2\right )}}{d x}+b x \arccos \left (1+d x^2\right ) \]

[In]

Integrate[a + b*ArcCos[1 + d*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05

method	result	size
default	\(a x +b \left (x \arccos \left (d \,x^{2}+1\right )+\frac {2 x \left (d \,x^{2}+2\right )}{\sqrt {-d^{2} x^{4}-2 d \,x^{2}}}\right )\)	\(45\)
parts	\(a x +b \left (x \arccos \left (d \,x^{2}+1\right )+\frac {2 x \left (d \,x^{2}+2\right )}{\sqrt {-d^{2} x^{4}-2 d \,x^{2}}}\right )\)	\(45\)

[In]

int(a+b*arccos(d*x^2+1),x,method=_RETURNVERBOSE)

[Out]

a*x+b*(x*arccos(d*x^2+1)+2/(-d^2*x^4-2*d*x^2)^(1/2)*x*(d*x^2+2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12 \[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx=\frac {b d x^{2} \arccos \left (d x^{2} + 1\right ) + a d x^{2} - 2 \, \sqrt {-d^{2} x^{4} - 2 \, d x^{2}} b}{d x} \]

[In]

integrate(a+b*arccos(d*x^2+1),x, algorithm="fricas")

[Out]

(b*d*x^2*arccos(d*x^2 + 1) + a*d*x^2 - 2*sqrt(-d^2*x^4 - 2*d*x^2)*b)/(d*x)

Sympy [F]

\[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx=\int \left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )\, dx \]

[In]

integrate(a+b*acos(d*x**2+1),x)

[Out]

Integral(a + b*acos(d*x**2 + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx={\left (x \arccos \left (d x^{2} + 1\right ) + \frac {2 \, {\left (d^{\frac {3}{2}} x^{2} + 2 \, \sqrt {d}\right )}}{\sqrt {-d x^{2} - 2} d}\right )} b + a x \]

[In]

integrate(a+b*arccos(d*x^2+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx={\left (x \arccos \left (d x^{2} + 1\right ) + \frac {2 \, \sqrt {2} \sqrt {-d} \mathrm {sgn}\left (x\right )}{d} - \frac {2 \, \sqrt {-d^{2} x^{2} - 2 \, d}}{d \mathrm {sgn}\left (x\right )}\right )} b + a x \]

[In]

integrate(a+b*arccos(d*x^2+1),x, algorithm="giac")

[Out]

(x*arccos(d*x^2 + 1) + 2*sqrt(2)*sqrt(-d)*sgn(x)/d - 2*sqrt(-d^2*x^2 - 2*d)/(d*sgn(x)))*b + a*x

Mupad [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \left (a+b \arccos \left (1+d x^2\right )\right ) \, dx=a\,x+b\,x\,\mathrm {acos}\left (d\,x^2+1\right )-\frac {2\,b\,\sqrt {1-{\left (d\,x^2+1\right )}^2}}{d\,x} \]

[In]

int(a + b*acos(d*x^2 + 1),x)

[Out]

a*x + b*x*acos(d*x^2 + 1) - (2*b*(1 - (d*x^2 + 1)^2)^(1/2))/(d*x)