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

Optimal. Leaf size=88 \[ \frac{\sqrt{x} (3 b B-A c)}{b c^2}-\frac{(3 b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} c^{5/2}}-\frac{x^{3/2} (b B-A c)}{b c (b+c x)} \]

[Out]

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

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Rubi [A]  time = 0.0431694, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {781, 78, 50, 63, 205} \[ \frac{\sqrt{x} (3 b B-A c)}{b c^2}-\frac{(3 b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} c^{5/2}}-\frac{x^{3/2} (b B-A c)}{b c (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 781

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 205

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

Rubi steps

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

Mathematica [A]  time = 0.0446419, size = 69, normalized size = 0.78 \[ \frac{\sqrt{x} (-A c+3 b B+2 B c x)}{c^2 (b+c x)}-\frac{(3 b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 87, normalized size = 1. \begin{align*} 2\,{\frac{B\sqrt{x}}{{c}^{2}}}-{\frac{A}{c \left ( cx+b \right ) }\sqrt{x}}+{\frac{bB}{{c}^{2} \left ( cx+b \right ) }\sqrt{x}}+{\frac{A}{c}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}-3\,{\frac{bB}{{c}^{2}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64546, size = 441, normalized size = 5.01 \begin{align*} \left [\frac{{\left (3 \, B b^{2} - A b c +{\left (3 \, B b c - A c^{2}\right )} x\right )} \sqrt{-b c} \log \left (\frac{c x - b - 2 \, \sqrt{-b c} \sqrt{x}}{c x + b}\right ) + 2 \,{\left (2 \, B b c^{2} x + 3 \, B b^{2} c - A b c^{2}\right )} \sqrt{x}}{2 \,{\left (b c^{4} x + b^{2} c^{3}\right )}}, \frac{{\left (3 \, B b^{2} - A b c +{\left (3 \, B b c - A c^{2}\right )} x\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c}}{c \sqrt{x}}\right ) +{\left (2 \, B b c^{2} x + 3 \, B b^{2} c - A b c^{2}\right )} \sqrt{x}}{b c^{4} x + b^{2} c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((3*B*b^2 - A*b*c + (3*B*b*c - A*c^2)*x)*sqrt(-b*c)*log((c*x - b - 2*sqrt(-b*c)*sqrt(x))/(c*x + b)) + 2*(
2*B*b*c^2*x + 3*B*b^2*c - A*b*c^2)*sqrt(x))/(b*c^4*x + b^2*c^3), ((3*B*b^2 - A*b*c + (3*B*b*c - A*c^2)*x)*sqrt
(b*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (2*B*b*c^2*x + 3*B*b^2*c - A*b*c^2)*sqrt(x))/(b*c^4*x + b^2*c^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.15383, size = 88, normalized size = 1. \begin{align*} \frac{2 \, B \sqrt{x}}{c^{2}} - \frac{{\left (3 \, B b - A c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} c^{2}} + \frac{B b \sqrt{x} - A c \sqrt{x}}{{\left (c x + b\right )} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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