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]

-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

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.85, size = 70, normalized size = 1.11 \[ -\frac {{\left (2 \, a^{2} - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 3 \, a\right )}}{2 \, b^{3}} \]

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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giac [A]  time = 0.42, size = 70, normalized size = 1.11 \[ \frac {1}{2} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (\frac {x}{b^{2}} - \frac {3 \, a}{b^{3}}\right )} - \frac {{\left (2 \, a^{2} - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b^{2} {\left | b \right |}} \]

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))

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maple [B]  time = 0.01, size = 146, normalized size = 2.32 \[ \frac {a^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}\, b^{2}}+\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{2 b^{2}}-\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 \sqrt {b^{2}}\, b^{2}}-\frac {3 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a}{2 b^{3}} \]

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 [B]  time = 0.59, size = 135, normalized size = 2.14 \[ \frac {3 \, a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b^{2}} - \frac {{\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{3}} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2}{\sqrt {{\left (a+b\,x\right )}^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + b*x)^2 + 1)^(1/2),x)

[Out]

int(x^2/((a + b*x)^2 + 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \]

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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