3.1.73 \(\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^3} \, dx\) [73]

Optimal. Leaf size=73 \[ -\frac {2 B \sqrt {b x+c x^2}}{x}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}+2 B \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 73, 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 = {806, 676, 634, 212} \begin {gather*} -\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}-\frac {2 B \sqrt {b x+c x^2}}{x}+2 B \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^3,x]

[Out]

(-2*B*Sqrt[b*x + c*x^2])/x - (2*A*(b*x + c*x^2)^(3/2))/(3*b*x^3) + 2*B*Sqrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x +
c*x^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 676

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 + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(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] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 91, normalized size = 1.25 \begin {gather*} -\frac {2 \sqrt {x (b+c x)} (A b+3 b B x+A c x)}{3 b x^2}-\frac {2 B \sqrt {c} \sqrt {x (b+c x)} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^3,x]

[Out]

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

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Maple [A]
time = 0.53, size = 92, normalized size = 1.26

method	result	size
risch	\(-\frac {2 \left (c x +b \right ) \left (A c x +3 b B x +A b \right )}{3 x \sqrt {x \left (c x +b \right )}\, b}+B \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )\)	\(66\)
default	\(-\frac {2 A \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 b \,x^{3}}+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{b \,x^{2}}+\frac {2 c \left (\sqrt {c \,x^{2}+b x}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}\right )}{b}\right )\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

-2/3*A*(c*x^2+b*x)^(3/2)/b/x^3+B*(-2/b/x^2*(c*x^2+b*x)^(3/2)+2*c/b*((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 = 0.30, size = 85, normalized size = 1.16 \begin {gather*} {\left (\sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {2 \, \sqrt {c x^{2} + b x}}{x}\right )} B - \frac {2}{3} \, A {\left (\frac {\sqrt {c x^{2} + b x} c}{b x} + \frac {\sqrt {c x^{2} + b x}}{x^{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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.64, size = 141, normalized size = 1.93 \begin {gather*} \left [\frac {3 \, B b \sqrt {c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} {\left (A b + {\left (3 \, B b + A c\right )} x\right )}}{3 \, b x^{2}}, -\frac {2 \, {\left (3 \, B b \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} {\left (A b + {\left (3 \, B b + A c\right )} x\right )}\right )}}{3 \, b x^{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^3,x, algorithm="fricas")

[Out]

[1/3*(3*B*b*sqrt(c)*x^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*sqrt(c*x^2 + b*x)*(A*b + (3*B*b + A*c
)*x))/(b*x^2), -2/3*(3*B*b*sqrt(-c)*x^2*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + sqrt(c*x^2 + b*x)*(A*b + (3
*B*b + A*c)*x))/(b*x^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^{3}}\, 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**3,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (61) = 122\).
time = 1.16, size = 151, normalized size = 2.07 \begin {gather*} -B \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b \sqrt {c} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b c + A b^{2} \sqrt {c}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} \sqrt {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^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^3} \,d x \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^3,x)

[Out]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^3, x)

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