\(\int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx\) [102]

Optimal result

Integrand size = 17, antiderivative size = 63 \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=-\frac {3 a \sqrt {1+(a+b x)^2}}{2 b^3}+\frac {x \sqrt {1+(a+b x)^2}}{2 b^2}-\frac {\left (1-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \]

[Out]

-1/2*(-2*a^2+1)*arcsinh(b*x+a)/b^3-3/2*a*(1+(b*x+a)^2)^(1/2)/b^3+1/2*x*(1+(b*x+a)^2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {378, 757, 655, 221} \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=-\frac {\left (1-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3}-\frac {3 a \sqrt {(a+b x)^2+1}}{2 b^3}+\frac {x \sqrt {(a+b x)^2+1}}{2 b^2} \]

[In]

Int[x^2/Sqrt[1 + (a + b*x)^2],x]

[Out]

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

Rule 221

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

Rule 378

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(-a+x)^2}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^3} \\ & = \frac {x \sqrt {1+(a+b x)^2}}{2 b^2}+\frac {\text {Subst}\left (\int \frac {-1+2 a^2-3 a x}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^3} \\ & = -\frac {3 a \sqrt {1+(a+b x)^2}}{2 b^3}+\frac {x \sqrt {1+(a+b x)^2}}{2 b^2}-\frac {\left (1-2 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^3} \\ & = -\frac {3 a \sqrt {1+(a+b x)^2}}{2 b^3}+\frac {x \sqrt {1+(a+b x)^2}}{2 b^2}-\frac {\left (1-2 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.37 \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=\frac {(-3 a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}}{2 b^3}+\frac {\left (1-2 a^2\right ) \text {arctanh}\left (\frac {b x}{\sqrt {1+a^2}-\sqrt {1+a^2+2 a b x+b^2 x^2}}\right )}{b^3} \]

[In]

Integrate[x^2/Sqrt[1 + (a + b*x)^2],x]

[Out]

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

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.38

method	result	size
risch	\(-\frac {\left (-b x +3 a \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{3}}+\frac {\left (2 a^{2}-1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\)	\(87\)
default	\(\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\)	\(155\)

[In]

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

[Out]

-1/2*(-b*x+3*a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)/b^3+1/2/b^2*(2*a^2-1)*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*
x+a^2+1)^(1/2))/(b^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11 \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=-\frac {{\left (2 \, a^{2} - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 3 \, a\right )}}{2 \, b^{3}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (54) = 108\).

Time = 0.64 (sec) , antiderivative size = 228, normalized size of antiderivative = 3.62 \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=\begin {cases} \left (- \frac {3 a}{2 b^{3}} + \frac {x}{2 b^{2}}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} + \frac {\left (\frac {3 a^{2}}{2 b^{2}} - \frac {a^{2} + 1}{2 b^{2}}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} & \text {for}\: b^{2} \neq 0 \\\frac {a^{4} \sqrt {a^{2} + 2 a b x + 1} + 2 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \sqrt {a^{2} + 2 a b x + 1}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3}}{3 \sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2/(1+(b*x+a)**2)**(1/2),x)

[Out]

Piecewise(((-3*a/(2*b**3) + x/(2*b**2))*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + (3*a**2/(2*b**2) - (a**2 + 1)/(
2*b**2))*log(2*a*b + 2*b**2*x + 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*sqrt(b**2))/sqrt(b**2), Ne(b**2, 0)), (
(a**4*sqrt(a**2 + 2*a*b*x + 1) + 2*a**2*sqrt(a**2 + 2*a*b*x + 1) + (-2*a**2 - 2)*(a**2 + 2*a*b*x + 1)**(3/2)/3
+ (a**2 + 2*a*b*x + 1)**(5/2)/5 + sqrt(a**2 + 2*a*b*x + 1))/(4*a**3*b**3), Ne(a*b, 0)), (x**3/(3*sqrt(a**2 +
1)), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (53) = 106\).

Time = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.14 \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=\frac {3 \, a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b^{2}} - \frac {{\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{3}} \]

[In]

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

[Out]

3/2*a^2*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1
)*x/b^2 - 1/2*(a^2 + 1)*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 - 3/2*sqrt(b^2*x^2 + 2
*a*b*x + a^2 + 1)*a/b^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.37 \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=\frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (\frac {x}{b^{2}} - \frac {3 \, a}{b^{3}}\right )} - \frac {{\left (2 \, a^{2} - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b^{2} {\left | b \right |}} \]

[In]

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

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(x/b^2 - 3*a/b^3) - 1/2*(2*a^2 - 1)*log(-a*b - (x*abs(b) - sqrt(b^2*x^2
+ 2*a*b*x + a^2 + 1))*abs(b))/(b^2*abs(b))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {1+(a+b x)^2}} \, dx=\int \frac {x^2}{\sqrt {{\left (a+b\,x\right )}^2+1}} \,d x \]

[In]

int(x^2/((a + b*x)^2 + 1)^(1/2),x)

[Out]

int(x^2/((a + b*x)^2 + 1)^(1/2), x)