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

Optimal. Leaf size=63 \[ -\frac{\left (1-2 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3}-\frac{3 a \sqrt{(a+b x)^2+1}}{2 b^3}+\frac{x \sqrt{(a+b x)^2+1}}{2 b^2} \]

[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)

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Rubi [A]  time = 0.0404596, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {371, 743, 641, 215} \[ -\frac{\left (1-2 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3}-\frac{3 a \sqrt{(a+b x)^2+1}}{2 b^3}+\frac{x \sqrt{(a+b x)^2+1}}{2 b^2} \]

Antiderivative was successfully verified.

[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 371

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 743

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]

Rule 641

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 215

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]

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{1+(a+b x)^2}} \, dx &=\frac{\operatorname{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{\operatorname{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 ) \operatorname{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]  time = 0.0565015, size = 51, normalized size = 0.81 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2+1} (b x-3 a)+\left (2 a^2-1\right ) \sinh ^{-1}(a+b x)}{2 b^3} \]

Antiderivative was successfully verified.

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

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Maple [B]  time = 0.01, size = 146, normalized size = 2.3 \begin{align*}{\frac{x}{2\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,abx+{a}^{2}+1}}-{\frac{3\,a}{2\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,abx+{a}^{2}+1}}+{\frac{{a}^{2}}{{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,abx+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{1}{2\,{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,abx+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74386, size = 165, normalized size = 2.62 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1), x)

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Giac [A]  time = 1.1905, size = 95, normalized size = 1.51 \begin{align*} \frac{1}{2} \, \sqrt{{\left (b x + a\right )}^{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{{\left (b x + a\right )}^{2} + 1}\right )}{\left | b \right |}\right )}{2 \, b^{2}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt((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*x + a)^2 + 1))*ab
s(b))/(b^2*abs(b))