3.189 \(\int \frac{x^{5/2} (A+B x)}{(b x+c x^2)^3} \, dx\)

Optimal. Leaf size=100 \[ \frac{(3 A c+b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{5/2} c^{3/2}}+\frac{\sqrt{x} (3 A c+b B)}{4 b^2 c (b+c x)}-\frac{\sqrt{x} (b B-A c)}{2 b c (b+c x)^2} \]

[Out]

-((b*B - A*c)*Sqrt[x])/(2*b*c*(b + c*x)^2) + ((b*B + 3*A*c)*Sqrt[x])/(4*b^2*c*(b + c*x)) + ((b*B + 3*A*c)*ArcT
an[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(5/2)*c^(3/2))

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Rubi [A]  time = 0.0479337, antiderivative size = 100, 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, 51, 63, 205} \[ \frac{(3 A c+b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{5/2} c^{3/2}}+\frac{\sqrt{x} (3 A c+b B)}{4 b^2 c (b+c x)}-\frac{\sqrt{x} (b B-A c)}{2 b c (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

-((b*B - A*c)*Sqrt[x])/(2*b*c*(b + c*x)^2) + ((b*B + 3*A*c)*Sqrt[x])/(4*b^2*c*(b + c*x)) + ((b*B + 3*A*c)*ArcT
an[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(5/2)*c^(3/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 )^3} \, dx &=\int \frac{A+B x}{\sqrt{x} (b+c x)^3} \, dx\\ &=-\frac{(b B-A c) \sqrt{x}}{2 b c (b+c x)^2}+\frac{(b B+3 A c) \int \frac{1}{\sqrt{x} (b+c x)^2} \, dx}{4 b c}\\ &=-\frac{(b B-A c) \sqrt{x}}{2 b c (b+c x)^2}+\frac{(b B+3 A c) \sqrt{x}}{4 b^2 c (b+c x)}+\frac{(b B+3 A c) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{8 b^2 c}\\ &=-\frac{(b B-A c) \sqrt{x}}{2 b c (b+c x)^2}+\frac{(b B+3 A c) \sqrt{x}}{4 b^2 c (b+c x)}+\frac{(b B+3 A c) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{4 b^2 c}\\ &=-\frac{(b B-A c) \sqrt{x}}{2 b c (b+c x)^2}+\frac{(b B+3 A c) \sqrt{x}}{4 b^2 c (b+c x)}+\frac{(b B+3 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{5/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.104995, size = 91, normalized size = 0.91 \[ \frac{\sqrt{x} \left (\frac{b^2 (A c-b B)}{(b+c x)^2}-\frac{1}{2} (-3 A c-b B) \left (\frac{b}{b+c x}+\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{c} \sqrt{x}}\right )\right )}{2 b^3 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.014, size = 95, normalized size = 1. \begin{align*} 2\,{\frac{1}{ \left ( cx+b \right ) ^{2}} \left ( 1/8\,{\frac{ \left ( 3\,Ac+bB \right ){x}^{3/2}}{{b}^{2}}}+1/8\,{\frac{ \left ( 5\,Ac-bB \right ) \sqrt{x}}{bc}} \right ) }+{\frac{3\,A}{4\,{b}^{2}}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{B}{4\,bc}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \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)^3,x)

[Out]

2*(1/8*(3*A*c+B*b)/b^2*x^(3/2)+1/8*(5*A*c-B*b)/b/c*x^(1/2))/(c*x+b)^2+3/4/b^2/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*
c)^(1/2))*A+1/4/b/c/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*c)^(1/2))*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61963, size = 633, normalized size = 6.33 \begin{align*} \left [-\frac{{\left (B b^{3} + 3 \, A b^{2} c +{\left (B b c^{2} + 3 \, A c^{3}\right )} x^{2} + 2 \,{\left (B b^{2} c + 3 \, A b 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 (B b^{3} c - 5 \, A b^{2} c^{2} -{\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x\right )} \sqrt{x}}{8 \,{\left (b^{3} c^{4} x^{2} + 2 \, b^{4} c^{3} x + b^{5} c^{2}\right )}}, -\frac{{\left (B b^{3} + 3 \, A b^{2} c +{\left (B b c^{2} + 3 \, A c^{3}\right )} x^{2} + 2 \,{\left (B b^{2} c + 3 \, A b c^{2}\right )} x\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c}}{c \sqrt{x}}\right ) +{\left (B b^{3} c - 5 \, A b^{2} c^{2} -{\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x\right )} \sqrt{x}}{4 \,{\left (b^{3} c^{4} x^{2} + 2 \, b^{4} c^{3} x + b^{5} c^{2}\right )}}\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)^3,x, algorithm="fricas")

[Out]

[-1/8*((B*b^3 + 3*A*b^2*c + (B*b*c^2 + 3*A*c^3)*x^2 + 2*(B*b^2*c + 3*A*b*c^2)*x)*sqrt(-b*c)*log((c*x - b - 2*s
qrt(-b*c)*sqrt(x))/(c*x + b)) + 2*(B*b^3*c - 5*A*b^2*c^2 - (B*b^2*c^2 + 3*A*b*c^3)*x)*sqrt(x))/(b^3*c^4*x^2 +
2*b^4*c^3*x + b^5*c^2), -1/4*((B*b^3 + 3*A*b^2*c + (B*b*c^2 + 3*A*c^3)*x^2 + 2*(B*b^2*c + 3*A*b*c^2)*x)*sqrt(b
*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (B*b^3*c - 5*A*b^2*c^2 - (B*b^2*c^2 + 3*A*b*c^3)*x)*sqrt(x))/(b^3*c^4*x^2
+ 2*b^4*c^3*x + b^5*c^2)]

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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.09444, size = 111, normalized size = 1.11 \begin{align*} \frac{{\left (B b + 3 \, A c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{4 \, \sqrt{b c} b^{2} c} + \frac{B b c x^{\frac{3}{2}} + 3 \, A c^{2} x^{\frac{3}{2}} - B b^{2} \sqrt{x} + 5 \, A b c \sqrt{x}}{4 \,{\left (c x + b\right )}^{2} b^{2} c} \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)^3,x, algorithm="giac")

[Out]

1/4*(B*b + 3*A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^2*c) + 1/4*(B*b*c*x^(3/2) + 3*A*c^2*x^(3/2) - B*b^2
*sqrt(x) + 5*A*b*c*sqrt(x))/((c*x + b)^2*b^2*c)