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

Optimal. Leaf size=79 \[ \frac {1}{2} \sqrt {a+c x^2} (2 A+B x)-\sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c}} \]

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Rubi [A]  time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {815, 844, 217, 206, 266, 63, 208} \begin {gather*} \frac {1}{2} \sqrt {a+c x^2} (2 A+B x)-\sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {a B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + c*x^2])/x,x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 208

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 100, normalized size = 1.27 \begin {gather*} \frac {1}{2} \left (\frac {a^{3/2} B \sqrt {\frac {c x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {c} \sqrt {a+c x^2}}+\sqrt {a+c x^2} (2 A+B x)-2 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A + B*x)*Sqrt[a + c*x^2] + (a^(3/2)*B*Sqrt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[c]*Sqrt[a +
c*x^2]) - 2*Sqrt[a]*A*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/2

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IntegrateAlgebraic [A]  time = 0.26, size = 95, normalized size = 1.20 \begin {gather*} \frac {1}{2} \sqrt {a+c x^2} (2 A+B x)+2 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {a B \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 341, normalized size = 4.32 \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 \, A \sqrt {a} c \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{4 \, c}, -\frac {B a \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - A \sqrt {a} c \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{2 \, c}, \frac {4 \, A \sqrt {-a} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 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}, -\frac {B a \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 2 \, A \sqrt {-a} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{2 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.20, size = 78, normalized size = 0.99 \begin {gather*} \frac {2 \, A a \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {B a \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, \sqrt {c}} + \frac {1}{2} \, \sqrt {c x^{2} + a} {\left (B x + 2 \, A\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 78, normalized size = 0.99 \begin {gather*} -A \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )+\frac {B a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}+\frac {\sqrt {c \,x^{2}+a}\, B x}{2}+\sqrt {c \,x^{2}+a}\, A \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 59, normalized size = 0.75 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + a} B x + \frac {B a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c}} - A \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \sqrt {c x^{2} + a} A \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.36, size = 68, normalized size = 0.86 \begin {gather*} A\,\sqrt {c\,x^2+a}-A\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )+\frac {B\,x\,\sqrt {c\,x^2+a}}{2}+\frac {B\,a\,\ln \left (\sqrt {c}\,x+\sqrt {c\,x^2+a}\right )}{2\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(a + c*x^2)^(1/2) - A*a^(1/2)*atanh((a + c*x^2)^(1/2)/a^(1/2)) + (B*x*(a + c*x^2)^(1/2))/2 + (B*a*log(c^(1/2
)*x + (a + c*x^2)^(1/2)))/(2*c^(1/2))

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sympy [A]  time = 6.74, size = 107, normalized size = 1.35 \begin {gather*} - A \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )} + \frac {A a}{\sqrt {c} x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {A \sqrt {c} x}{\sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B \sqrt {a} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} + \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*sqrt(a)*asinh(sqrt(a)/(sqrt(c)*x)) + A*a/(sqrt(c)*x*sqrt(a/(c*x**2) + 1)) + A*sqrt(c)*x/sqrt(a/(c*x**2) + 1
) + B*sqrt(a)*x*sqrt(1 + c*x**2/a)/2 + B*a*asinh(sqrt(c)*x/sqrt(a))/(2*sqrt(c))

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