3.1.71 \(\int \frac {(A+B x) \sqrt {b x+c x^2}}{x} \, dx\) [71]

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

[Out]

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

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Rubi [A]
time = 0.05, 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 = {808, 678, 634, 212} \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]

-1/4*((b*B - 4*A*c)*Sqrt[b*x + c*x^2])/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 212

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

Rule 634

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 678

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 808

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)) \text {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 = 0.12, size = 87, normalized size = 0.95 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} (b B+4 A c+2 B c x)+\frac {b (b B-4 A c) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\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) + (b*(b*B - 4*A*c)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]
)/(Sqrt[x]*Sqrt[b + c*x])))/(4*c^(3/2))

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Maple [A]
time = 0.53, size = 103, normalized size = 1.12

method	result	size
risch	\(\frac {\left (2 B c x +4 A c +B b \right ) x \left (c x +b \right )}{4 c \sqrt {x \left (c x +b \right )}}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) A}{2 \sqrt {c}}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) B}{8 c^{\frac {3}{2}}}\)	\(99\)
default	\(B \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )+A \left (\sqrt {c \,x^{2}+b x}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}\right )\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, 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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Fricas [A]
time = 3.93, 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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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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Giac [A]
time = 1.26, 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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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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