3.4.57 \(\int \frac {x (A+B x)}{\sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=56 \[ \frac {\sqrt {a+c x^2} (2 A+B x)}{2 c}-\frac {a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {780, 217, 206} \begin {gather*} \frac {\sqrt {a+c x^2} (2 A+B x)}{2 c}-\frac {a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

((2*A + B*x)*Sqrt[a + c*x^2])/(2*c) - (a*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x (A+B x)}{\sqrt {a+c x^2}} \, dx &=\frac {(2 A+B x) \sqrt {a+c x^2}}{2 c}-\frac {(a B) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 c}\\ &=\frac {(2 A+B x) \sqrt {a+c x^2}}{2 c}-\frac {(a B) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{2 c}\\ &=\frac {(2 A+B x) \sqrt {a+c x^2}}{2 c}-\frac {a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 1.02 \begin {gather*} \frac {\sqrt {c} \sqrt {a+c x^2} (2 A+B x)-a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

(Sqrt[c]*(2*A + B*x)*Sqrt[a + c*x^2] - a*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(3/2))

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IntegrateAlgebraic [A]  time = 0.24, size = 58, normalized size = 1.04 \begin {gather*} \frac {\sqrt {a+c x^2} (2 A+B x)}{2 c}+\frac {a B \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

((2*A + B*x)*Sqrt[a + c*x^2])/(2*c) + (a*B*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(2*c^(3/2))

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fricas [A]  time = 0.49, size = 109, normalized size = 1.95 \begin {gather*} \left [\frac {B a \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{4 \, c^{2}}, \frac {B a \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{2 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(B*a*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(B*c*x + 2*A*c)*sqrt(c*x^2 + a))/c^2, 1/
2*(B*a*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (B*c*x + 2*A*c)*sqrt(c*x^2 + a))/c^2]

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giac [A]  time = 0.20, size = 50, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + a} {\left (\frac {B x}{c} + \frac {2 \, A}{c}\right )} + \frac {B a \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + a)*(B*x/c + 2*A/c) + 1/2*B*a*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)

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maple [A]  time = 0.05, size = 55, normalized size = 0.98 \begin {gather*} -\frac {B a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+a}\, B x}{2 c}+\frac {\sqrt {c \,x^{2}+a}\, A}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

1/2*B*x/c*(c*x^2+a)^(1/2)-1/2*B*a/c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))+A/c*(c*x^2+a)^(1/2)

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maxima [A]  time = 0.49, size = 47, normalized size = 0.84 \begin {gather*} \frac {\sqrt {c x^{2} + a} B x}{2 \, c} - \frac {B a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + a} A}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + a)*B*x/c - 1/2*B*a*arcsinh(c*x/sqrt(a*c))/c^(3/2) + sqrt(c*x^2 + a)*A/c

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mupad [B]  time = 1.43, size = 82, normalized size = 1.46 \begin {gather*} \left \{\begin {array}{cl} \frac {2\,B\,x^3+3\,A\,x^2}{6\,\sqrt {a}} & \text {\ if\ \ }c=0\\ \frac {A\,\sqrt {c\,x^2+a}}{c}-\frac {B\,a\,\ln \left (2\,\sqrt {c}\,x+2\,\sqrt {c\,x^2+a}\right )}{2\,c^{3/2}}+\frac {B\,x\,\sqrt {c\,x^2+a}}{2\,c} & \text {\ if\ \ }c\neq 0 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(A + B*x))/(a + c*x^2)^(1/2),x)

[Out]

piecewise(c == 0, (3*A*x^2 + 2*B*x^3)/(6*a^(1/2)), c ~= 0, (A*(a + c*x^2)^(1/2))/c - (B*a*log(2*c^(1/2)*x + 2*
(a + c*x^2)^(1/2)))/(2*c^(3/2)) + (B*x*(a + c*x^2)^(1/2))/(2*c))

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sympy [A]  time = 4.20, size = 70, normalized size = 1.25 \begin {gather*} A \left (\begin {cases} \frac {x^{2}}{2 \sqrt {a}} & \text {for}\: c = 0 \\\frac {\sqrt {a + c x^{2}}}{c} & \text {otherwise} \end {cases}\right ) + \frac {B \sqrt {a} x \sqrt {1 + \frac {c x^{2}}{a}}}{2 c} - \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

A*Piecewise((x**2/(2*sqrt(a)), Eq(c, 0)), (sqrt(a + c*x**2)/c, True)) + B*sqrt(a)*x*sqrt(1 + c*x**2/a)/(2*c) -
B*a*asinh(sqrt(c)*x/sqrt(a))/(2*c**(3/2))

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