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

Optimal. Leaf size=92 \[ -\frac {b (b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}-\frac {\sqrt {b x+c x^2} (b B-4 A c)}{4 c}+\frac {B \left (b x+c x^2\right )^{3/2}}{2 c x} \]

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Rubi [A]  time = 0.07, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {794, 664, 620, 206} \begin {gather*} -\frac {b (b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}-\frac {\sqrt {b x+c x^2} (b B-4 A c)}{4 c}+\frac {B \left (b x+c x^2\right )^{3/2}}{2 c x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x,x]

[Out]

-((b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(4*c) + (B*(b*x + c*x^2)^(3/2))/(2*c*x) - (b*(b*B - 4*A*c)*ArcTanh[(Sqrt[c]
*x)/Sqrt[b*x + c*x^2]])/(4*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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx &=\frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}+\frac {\left (b B-A c+\frac {3}{2} (-b B+2 A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x} \, dx}{2 c}\\ &=-\frac {(b B-4 A c) \sqrt {b x+c x^2}}{4 c}+\frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {(b (b B-4 A c)) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{8 c}\\ &=-\frac {(b B-4 A c) \sqrt {b x+c x^2}}{4 c}+\frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {(b (b B-4 A c)) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{4 c}\\ &=-\frac {(b B-4 A c) \sqrt {b x+c x^2}}{4 c}+\frac {B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac {b (b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 89, normalized size = 0.97 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} (4 A c+b B+2 B c x)-\frac {\sqrt {b} (b B-4 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{4 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(b*B + 4*A*c + 2*B*c*x) - (Sqrt[b]*(b*B - 4*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]
])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(4*c^(3/2))

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IntegrateAlgebraic [A]  time = 0.45, size = 85, normalized size = 0.92 \begin {gather*} \frac {\left (b^2 B-4 A b c\right ) \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{8 c^{3/2}}+\frac {\sqrt {b x+c x^2} (4 A c+b B+2 B c x)}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B + 4*A*c + 2*B*c*x)*Sqrt[b*x + c*x^2])/(4*c) + ((b^2*B - 4*A*b*c)*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[b*x
+ c*x^2]])/(8*c^(3/2))

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fricas [A]  time = 0.42, size = 153, normalized size = 1.66 \begin {gather*} \left [-\frac {{\left (B b^{2} - 4 \, A b c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{2} x + B b c + 4 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{8 \, c^{2}}, \frac {{\left (B b^{2} - 4 \, A b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (2 \, B c^{2} x + B b c + 4 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{4 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((B*b^2 - 4*A*b*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(2*B*c^2*x + B*b*c + 4*A*c^2
)*sqrt(c*x^2 + b*x))/c^2, 1/4*((B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (2*B*c^2*
x + B*b*c + 4*A*c^2)*sqrt(c*x^2 + b*x))/c^2]

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giac [A]  time = 0.23, size = 77, normalized size = 0.84 \begin {gather*} \frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (2 \, B x + \frac {B b + 4 \, A c}{c}\right )} + \frac {{\left (B b^{2} - 4 \, A b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x)*(2*B*x + (B*b + 4*A*c)/c) + 1/8*(B*b^2 - 4*A*b*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b
*x))*sqrt(c) - b))/c^(3/2)

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maple [A]  time = 0.05, size = 112, normalized size = 1.22 \begin {gather*} \frac {A b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}-\frac {B \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+b x}\, B x}{2}+\sqrt {c \,x^{2}+b x}\, A +\frac {\sqrt {c \,x^{2}+b x}\, B b}{4 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x*(c*x^2+b*x)^(1/2)+1/4*B/c*(c*x^2+b*x)^(1/2)*b-1/8*B*b^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/
2))+A*(c*x^2+b*x)^(1/2)+1/2*A*b*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))/c^(1/2)

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maxima [A]  time = 0.89, size = 109, normalized size = 1.18 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + b x} B x - \frac {B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {A b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, \sqrt {c}} + \sqrt {c x^{2} + b x} A + \frac {\sqrt {c x^{2} + b x} B b}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + b*x)*B*x - 1/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 1/2*A*b*log(2*c*x
+ b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) + sqrt(c*x^2 + b*x)*A + 1/4*sqrt(c*x^2 + b*x)*B*b/c

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mupad [B]  time = 1.32, size = 101, normalized size = 1.10 \begin {gather*} A\,\sqrt {c\,x^2+b\,x}+B\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {B\,b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {A\,b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{2\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(b*x + c*x^2)^(1/2) + B*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (B*b^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)
^(1/2)))/(8*c^(3/2)) + (A*b*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(2*c^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x, x)

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