3.6.89 \(\int \frac {(a+b x)^2}{\sqrt {1+x^2}} \, dx\) [589]

Optimal. Leaf size=52 \[ \frac {3}{2} a b \sqrt {1+x^2}+\frac {1}{2} b (a+b x) \sqrt {1+x^2}+\frac {1}{2} \left (2 a^2-b^2\right ) \sinh ^{-1}(x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {757, 655, 221} \begin {gather*} \frac {1}{2} \left (2 a^2-b^2\right ) \sinh ^{-1}(x)+\frac {3}{2} a b \sqrt {x^2+1}+\frac {1}{2} b \sqrt {x^2+1} (a+b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/Sqrt[1 + x^2],x]

[Out]

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

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

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Mathematica [A]
time = 0.16, size = 49, normalized size = 0.94 \begin {gather*} \frac {1}{2} b (4 a+b x) \sqrt {1+x^2}+\frac {1}{2} \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {x}{\sqrt {1+x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.45, size = 38, normalized size = 0.73

method	result	size
risch	\(\frac {b \left (b x +4 a \right ) \sqrt {x^{2}+1}}{2}+\left (a^{2}-\frac {b^{2}}{2}\right ) \arcsinh \left (x \right )\)	\(31\)
default	\(b^{2} \left (\frac {x \sqrt {x^{2}+1}}{2}-\frac {\arcsinh \left (x \right )}{2}\right )+2 a b \sqrt {x^{2}+1}+a^{2} \arcsinh \left (x \right )\)	\(38\)
trager	\(\frac {b \left (b x +4 a \right ) \sqrt {x^{2}+1}}{2}+\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (\sqrt {x^{2}+1}+x \right )}{2}\)	\(42\)
meijerg	\(\frac {b^{2} \left (\sqrt {\pi }\, x \sqrt {x^{2}+1}-\sqrt {\pi }\, \arcsinh \left (x \right )\right )}{2 \sqrt {\pi }}+\frac {a b \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {x^{2}+1}\right )}{\sqrt {\pi }}+a^{2} \arcsinh \left (x \right )\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 38, normalized size = 0.73 \begin {gather*} \frac {1}{2} \, \sqrt {x^{2} + 1} b^{2} x + a^{2} \operatorname {arsinh}\left (x\right ) - \frac {1}{2} \, b^{2} \operatorname {arsinh}\left (x\right ) + 2 \, \sqrt {x^{2} + 1} a b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.66, size = 45, normalized size = 0.87 \begin {gather*} -\frac {1}{2} \, {\left (2 \, a^{2} - b^{2}\right )} \log \left (-x + \sqrt {x^{2} + 1}\right ) + \frac {1}{2} \, {\left (b^{2} x + 4 \, a b\right )} \sqrt {x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.08, size = 42, normalized size = 0.81 \begin {gather*} a^{2} \operatorname {asinh}{\left (x \right )} + 2 a b \sqrt {x^{2} + 1} + \frac {b^{2} x \sqrt {x^{2} + 1}}{2} - \frac {b^{2} \operatorname {asinh}{\left (x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*asinh(x) + 2*a*b*sqrt(x**2 + 1) + b**2*x*sqrt(x**2 + 1)/2 - b**2*asinh(x)/2

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Giac [A]
time = 1.95, size = 45, normalized size = 0.87 \begin {gather*} -\frac {1}{2} \, {\left (2 \, a^{2} - b^{2}\right )} \log \left (-x + \sqrt {x^{2} + 1}\right ) + \frac {1}{2} \, {\left (b^{2} x + 4 \, a b\right )} \sqrt {x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.03, size = 32, normalized size = 0.62 \begin {gather*} \left (\frac {x\,b^2}{2}+2\,a\,b\right )\,\sqrt {x^2+1}+\mathrm {asinh}\left (x\right )\,\left (a^2-\frac {b^2}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(x^2 + 1)^(1/2),x)

[Out]

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

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