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

Optimal. Leaf size=123 \[ \frac{3 (b B-5 A c)}{4 b^3 c \sqrt{x}}-\frac{b B-5 A c}{4 b^2 c \sqrt{x} (b+c x)}+\frac{3 (b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2} \sqrt{c}}-\frac{b B-A c}{2 b c \sqrt{x} (b+c x)^2} \]

[Out]

(3*(b*B - 5*A*c))/(4*b^3*c*Sqrt[x]) - (b*B - A*c)/(2*b*c*Sqrt[x]*(b + c*x)^2) - (b*B - 5*A*c)/(4*b^2*c*Sqrt[x]
*(b + c*x)) + (3*(b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(7/2)*Sqrt[c])

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Rubi [A]  time = 0.0580332, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{3 (b B-5 A c)}{4 b^3 c \sqrt{x}}-\frac{b B-5 A c}{4 b^2 c \sqrt{x} (b+c x)}+\frac{3 (b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2} \sqrt{c}}-\frac{b B-A c}{2 b c \sqrt{x} (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*B - 5*A*c))/(4*b^3*c*Sqrt[x]) - (b*B - A*c)/(2*b*c*Sqrt[x]*(b + c*x)^2) - (b*B - 5*A*c)/(4*b^2*c*Sqrt[x]
*(b + c*x)) + (3*(b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(7/2)*Sqrt[c])

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^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac{A+B x}{x^{3/2} (b+c x)^3} \, dx\\ &=-\frac{b B-A c}{2 b c \sqrt{x} (b+c x)^2}-\frac{\left (\frac{b B}{2}-\frac{5 A c}{2}\right ) \int \frac{1}{x^{3/2} (b+c x)^2} \, dx}{2 b c}\\ &=-\frac{b B-A c}{2 b c \sqrt{x} (b+c x)^2}-\frac{b B-5 A c}{4 b^2 c \sqrt{x} (b+c x)}-\frac{(3 (b B-5 A c)) \int \frac{1}{x^{3/2} (b+c x)} \, dx}{8 b^2 c}\\ &=\frac{3 (b B-5 A c)}{4 b^3 c \sqrt{x}}-\frac{b B-A c}{2 b c \sqrt{x} (b+c x)^2}-\frac{b B-5 A c}{4 b^2 c \sqrt{x} (b+c x)}+\frac{(3 (b B-5 A c)) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{8 b^3}\\ &=\frac{3 (b B-5 A c)}{4 b^3 c \sqrt{x}}-\frac{b B-A c}{2 b c \sqrt{x} (b+c x)^2}-\frac{b B-5 A c}{4 b^2 c \sqrt{x} (b+c x)}+\frac{(3 (b B-5 A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{4 b^3}\\ &=\frac{3 (b B-5 A c)}{4 b^3 c \sqrt{x}}-\frac{b B-A c}{2 b c \sqrt{x} (b+c x)^2}-\frac{b B-5 A c}{4 b^2 c \sqrt{x} (b+c x)}+\frac{3 (b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2} \sqrt{c}}\\ \end{align*}

Mathematica [C]  time = 0.0224241, size = 59, normalized size = 0.48 \[ \frac{\frac{b^2 (A c-b B)}{(b+c x)^2}+(b B-5 A c) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};-\frac{c x}{b}\right )}{2 b^3 c \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*(-(b*B) + A*c))/(b + c*x)^2 + (b*B - 5*A*c)*Hypergeometric2F1[-1/2, 2, 1/2, -((c*x)/b)])/(2*b^3*c*Sqrt[x
])

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Maple [A]  time = 0.017, size = 125, normalized size = 1. \begin{align*} -2\,{\frac{A}{{b}^{3}\sqrt{x}}}-{\frac{7\,A{c}^{2}}{4\,{b}^{3} \left ( cx+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{3\,Bc}{4\,{b}^{2} \left ( cx+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{9\,Ac}{4\,{b}^{2} \left ( cx+b \right ) ^{2}}\sqrt{x}}+{\frac{5\,B}{4\,b \left ( cx+b \right ) ^{2}}\sqrt{x}}-{\frac{15\,Ac}{4\,{b}^{3}}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{3\,B}{4\,{b}^{2}}\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^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x)

[Out]

-2*A/b^3/x^(1/2)-7/4/b^3/(c*x+b)^2*x^(3/2)*A*c^2+3/4/b^2/(c*x+b)^2*x^(3/2)*B*c-9/4/b^2/(c*x+b)^2*A*x^(1/2)*c+5
/4/b/(c*x+b)^2*B*x^(1/2)-15/4/b^3/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*c)^(1/2))*A*c+3/4/b^2/(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^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72983, size = 718, normalized size = 5.84 \begin{align*} \left [\frac{3 \,{\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{3} + 2 \,{\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{2} +{\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt{-b c} \log \left (\frac{c x - b + 2 \, \sqrt{-b c} \sqrt{x}}{c x + b}\right ) - 2 \,{\left (8 \, A b^{3} c - 3 \,{\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} - 5 \,{\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt{x}}{8 \,{\left (b^{4} c^{3} x^{3} + 2 \, b^{5} c^{2} x^{2} + b^{6} c x\right )}}, -\frac{3 \,{\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{3} + 2 \,{\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{2} +{\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c}}{c \sqrt{x}}\right ) +{\left (8 \, A b^{3} c - 3 \,{\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} - 5 \,{\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt{x}}{4 \,{\left (b^{4} c^{3} x^{3} + 2 \, b^{5} c^{2} x^{2} + b^{6} c x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[1/8*(3*((B*b*c^2 - 5*A*c^3)*x^3 + 2*(B*b^2*c - 5*A*b*c^2)*x^2 + (B*b^3 - 5*A*b^2*c)*x)*sqrt(-b*c)*log((c*x -
b + 2*sqrt(-b*c)*sqrt(x))/(c*x + b)) - 2*(8*A*b^3*c - 3*(B*b^2*c^2 - 5*A*b*c^3)*x^2 - 5*(B*b^3*c - 5*A*b^2*c^2
)*x)*sqrt(x))/(b^4*c^3*x^3 + 2*b^5*c^2*x^2 + b^6*c*x), -1/4*(3*((B*b*c^2 - 5*A*c^3)*x^3 + 2*(B*b^2*c - 5*A*b*c
^2)*x^2 + (B*b^3 - 5*A*b^2*c)*x)*sqrt(b*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (8*A*b^3*c - 3*(B*b^2*c^2 - 5*A*b*c
^3)*x^2 - 5*(B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(x))/(b^4*c^3*x^3 + 2*b^5*c^2*x^2 + b^6*c*x)]

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

[Out]

Timed out

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Giac [A]  time = 1.13637, size = 116, normalized size = 0.94 \begin{align*} \frac{3 \,{\left (B b - 5 \, A c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{4 \, \sqrt{b c} b^{3}} - \frac{2 \, A}{b^{3} \sqrt{x}} + \frac{3 \, B b c x^{\frac{3}{2}} - 7 \, A c^{2} x^{\frac{3}{2}} + 5 \, B b^{2} \sqrt{x} - 9 \, A b c \sqrt{x}}{4 \,{\left (c x + b\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

3/4*(B*b - 5*A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^3) - 2*A/(b^3*sqrt(x)) + 1/4*(3*B*b*c*x^(3/2) - 7*A
*c^2*x^(3/2) + 5*B*b^2*sqrt(x) - 9*A*b*c*sqrt(x))/((c*x + b)^2*b^3)