3.200 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=95 \[ \frac{\sqrt{b x+c x^2} (A c+2 b B)}{b \sqrt{x}}-\frac{(A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{\sqrt{b}}-\frac{A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}} \]

[Out]

((2*b*B + A*c)*Sqrt[b*x + c*x^2])/(b*Sqrt[x]) - (A*(b*x + c*x^2)^(3/2))/(b*x^(5/2)) - ((2*b*B + A*c)*ArcTanh[S
qrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/Sqrt[b]

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Rubi [A]  time = 0.0912353, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {792, 664, 660, 207} \[ \frac{\sqrt{b x+c x^2} (A c+2 b B)}{b \sqrt{x}}-\frac{(A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{\sqrt{b}}-\frac{A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^(5/2),x]

[Out]

((2*b*B + A*c)*Sqrt[b*x + c*x^2])/(b*Sqrt[x]) - (A*(b*x + c*x^2)^(3/2))/(b*x^(5/2)) - ((2*b*B + A*c)*ArcTanh[S
qrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/Sqrt[b]

Rule 792

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]

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 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*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]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{b x+c x^2}}{x^{5/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}+\frac{\left (-\frac{5}{2} (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x^{3/2}} \, dx}{b}\\ &=\frac{(2 b B+A c) \sqrt{b x+c x^2}}{b \sqrt{x}}-\frac{A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}+\frac{1}{2} (2 b B+A c) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx\\ &=\frac{(2 b B+A c) \sqrt{b x+c x^2}}{b \sqrt{x}}-\frac{A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}+(2 b B+A c) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )\\ &=\frac{(2 b B+A c) \sqrt{b x+c x^2}}{b \sqrt{x}}-\frac{A \left (b x+c x^2\right )^{3/2}}{b x^{5/2}}-\frac{(2 b B+A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{\sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0493042, size = 80, normalized size = 0.84 \[ -\frac{\sqrt{x (b+c x)} \left (\sqrt{b} (A-2 B x) \sqrt{b+c x}+x (A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b+c x}}{\sqrt{b}}\right )\right )}{\sqrt{b} x^{3/2} \sqrt{b+c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^(5/2),x]

[Out]

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

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Maple [A]  time = 0.017, size = 86, normalized size = 0.9 \begin{align*}{ \left ( -A{\it Artanh} \left ({\sqrt{cx+b}{\frac{1}{\sqrt{b}}}} \right ) xc+2\,B\sqrt{cx+b}x\sqrt{b}-2\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) xb-A\sqrt{cx+b}\sqrt{b} \right ) \sqrt{x \left ( cx+b \right ) }{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{cx+b}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x}{\left (B x + A\right )}}{x^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x)*(B*x + A)/x^(5/2), x)

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Fricas [A]  time = 1.64405, size = 378, normalized size = 3.98 \begin{align*} \left [\frac{{\left (2 \, B b + A c\right )} \sqrt{b} x^{2} \log \left (-\frac{c x^{2} + 2 \, b x - 2 \, \sqrt{c x^{2} + b x} \sqrt{b} \sqrt{x}}{x^{2}}\right ) + 2 \,{\left (2 \, B b x - A b\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{2 \, b x^{2}}, \frac{{\left (2 \, B b + A c\right )} \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (2 \, B b x - A b\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{b x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )} \left (A + B x\right )}{x^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**(5/2),x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**(5/2), x)

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Giac [A]  time = 1.38784, size = 82, normalized size = 0.86 \begin{align*} \frac{2 \, \sqrt{c x + b} B c - \frac{\sqrt{c x + b} A c}{x} + \frac{{\left (2 \, B b c + A c^{2}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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