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

Optimal. Leaf size=156 \[ -\frac{c^2 (7 b B-9 A c)}{b^5 \sqrt{x}}-\frac{c^{5/2} (7 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{11/2}}+\frac{c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac{7 b B-9 A c}{5 b^3 x^{5/2}}+\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{b B-A c}{b c x^{7/2} (b+c x)} \]

[Out]

(7*b*B - 9*A*c)/(7*b^2*c*x^(7/2)) - (7*b*B - 9*A*c)/(5*b^3*x^(5/2)) + (c*(7*b*B - 9*A*c))/(3*b^4*x^(3/2)) - (c
^2*(7*b*B - 9*A*c))/(b^5*Sqrt[x]) - (b*B - A*c)/(b*c*x^(7/2)*(b + c*x)) - (c^(5/2)*(7*b*B - 9*A*c)*ArcTan[(Sqr
t[c]*Sqrt[x])/Sqrt[b]])/b^(11/2)

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Rubi [A]  time = 0.082545, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{c^2 (7 b B-9 A c)}{b^5 \sqrt{x}}-\frac{c^{5/2} (7 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{11/2}}+\frac{c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac{7 b B-9 A c}{5 b^3 x^{5/2}}+\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{b B-A c}{b c x^{7/2} (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b*B - 9*A*c)/(7*b^2*c*x^(7/2)) - (7*b*B - 9*A*c)/(5*b^3*x^(5/2)) + (c*(7*b*B - 9*A*c))/(3*b^4*x^(3/2)) - (c
^2*(7*b*B - 9*A*c))/(b^5*Sqrt[x]) - (b*B - A*c)/(b*c*x^(7/2)*(b + c*x)) - (c^(5/2)*(7*b*B - 9*A*c)*ArcTan[(Sqr
t[c]*Sqrt[x])/Sqrt[b]])/b^(11/2)

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{A+B x}{x^{5/2} \left (b x+c x^2\right )^2} \, dx &=\int \frac{A+B x}{x^{9/2} (b+c x)^2} \, dx\\ &=-\frac{b B-A c}{b c x^{7/2} (b+c x)}-\frac{\left (\frac{7 b B}{2}-\frac{9 A c}{2}\right ) \int \frac{1}{x^{9/2} (b+c x)} \, dx}{b c}\\ &=\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{b B-A c}{b c x^{7/2} (b+c x)}+\frac{(7 b B-9 A c) \int \frac{1}{x^{7/2} (b+c x)} \, dx}{2 b^2}\\ &=\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{7 b B-9 A c}{5 b^3 x^{5/2}}-\frac{b B-A c}{b c x^{7/2} (b+c x)}-\frac{(c (7 b B-9 A c)) \int \frac{1}{x^{5/2} (b+c x)} \, dx}{2 b^3}\\ &=\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{7 b B-9 A c}{5 b^3 x^{5/2}}+\frac{c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac{b B-A c}{b c x^{7/2} (b+c x)}+\frac{\left (c^2 (7 b B-9 A c)\right ) \int \frac{1}{x^{3/2} (b+c x)} \, dx}{2 b^4}\\ &=\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{7 b B-9 A c}{5 b^3 x^{5/2}}+\frac{c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac{c^2 (7 b B-9 A c)}{b^5 \sqrt{x}}-\frac{b B-A c}{b c x^{7/2} (b+c x)}-\frac{\left (c^3 (7 b B-9 A c)\right ) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{2 b^5}\\ &=\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{7 b B-9 A c}{5 b^3 x^{5/2}}+\frac{c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac{c^2 (7 b B-9 A c)}{b^5 \sqrt{x}}-\frac{b B-A c}{b c x^{7/2} (b+c x)}-\frac{\left (c^3 (7 b B-9 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{b^5}\\ &=\frac{7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac{7 b B-9 A c}{5 b^3 x^{5/2}}+\frac{c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac{c^2 (7 b B-9 A c)}{b^5 \sqrt{x}}-\frac{b B-A c}{b c x^{7/2} (b+c x)}-\frac{c^{5/2} (7 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0176821, size = 64, normalized size = 0.41 \[ \frac{(b+c x) (7 b B-9 A c) \, _2F_1\left (-\frac{7}{2},1;-\frac{5}{2};-\frac{c x}{b}\right )+7 b (A c-b B)}{7 b^2 c x^{7/2} (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b*(-(b*B) + A*c) + (7*b*B - 9*A*c)*(b + c*x)*Hypergeometric2F1[-7/2, 1, -5/2, -((c*x)/b)])/(7*b^2*c*x^(7/2)
*(b + c*x))

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Maple [A]  time = 0.017, size = 163, normalized size = 1. \begin{align*} -{\frac{2\,A}{7\,{b}^{2}}{x}^{-{\frac{7}{2}}}}+{\frac{4\,Ac}{5\,{b}^{3}}{x}^{-{\frac{5}{2}}}}-{\frac{2\,B}{5\,{b}^{2}}{x}^{-{\frac{5}{2}}}}-2\,{\frac{A{c}^{2}}{{b}^{4}{x}^{3/2}}}+{\frac{4\,Bc}{3\,{b}^{3}}{x}^{-{\frac{3}{2}}}}+8\,{\frac{A{c}^{3}}{{b}^{5}\sqrt{x}}}-6\,{\frac{B{c}^{2}}{{b}^{4}\sqrt{x}}}+{\frac{{c}^{4}A}{{b}^{5} \left ( cx+b \right ) }\sqrt{x}}-{\frac{{c}^{3}B}{{b}^{4} \left ( cx+b \right ) }\sqrt{x}}+9\,{\frac{{c}^{4}A}{{b}^{5}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }-7\,{\frac{{c}^{3}B}{{b}^{4}\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((B*x+A)/x^(5/2)/(c*x^2+b*x)^2,x)

[Out]

-2/7*A/b^2/x^(7/2)+4/5/b^3/x^(5/2)*A*c-2/5/b^2/x^(5/2)*B-2*c^2/b^4/x^(3/2)*A+4/3*c/b^3/x^(3/2)*B+8*c^3/b^5/x^(
1/2)*A-6*c^2/b^4/x^(1/2)*B+1/b^5*c^4*x^(1/2)/(c*x+b)*A-1/b^4*c^3*x^(1/2)/(c*x+b)*B+9/b^5*c^4/(b*c)^(1/2)*arcta
n(x^(1/2)*c/(b*c)^(1/2))*A-7/b^4*c^3/(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((B*x+A)/x^(5/2)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82785, size = 813, normalized size = 5.21 \begin{align*} \left [-\frac{105 \,{\left ({\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} +{\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x + 2 \, b \sqrt{x} \sqrt{-\frac{c}{b}} - b}{c x + b}\right ) + 2 \,{\left (30 \, A b^{4} + 105 \,{\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} + 70 \,{\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 14 \,{\left (7 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 6 \,{\left (7 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt{x}}{210 \,{\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}, \frac{105 \,{\left ({\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} +{\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4}\right )} \sqrt{\frac{c}{b}} \arctan \left (\frac{b \sqrt{\frac{c}{b}}}{c \sqrt{x}}\right ) -{\left (30 \, A b^{4} + 105 \,{\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} + 70 \,{\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 14 \,{\left (7 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 6 \,{\left (7 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt{x}}{105 \,{\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/210*(105*((7*B*b*c^3 - 9*A*c^4)*x^5 + (7*B*b^2*c^2 - 9*A*b*c^3)*x^4)*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sqr
t(-c/b) - b)/(c*x + b)) + 2*(30*A*b^4 + 105*(7*B*b*c^3 - 9*A*c^4)*x^4 + 70*(7*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 14*
(7*B*b^3*c - 9*A*b^2*c^2)*x^2 + 6*(7*B*b^4 - 9*A*b^3*c)*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4), 1/105*(105*((7*B*b*
c^3 - 9*A*c^4)*x^5 + (7*B*b^2*c^2 - 9*A*b*c^3)*x^4)*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) - (30*A*b^4 + 10
5*(7*B*b*c^3 - 9*A*c^4)*x^4 + 70*(7*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 14*(7*B*b^3*c - 9*A*b^2*c^2)*x^2 + 6*(7*B*b^4
- 9*A*b^3*c)*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4)]

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

[Out]

Timed out

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Giac [A]  time = 1.14354, size = 184, normalized size = 1.18 \begin{align*} -\frac{{\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} b^{5}} - \frac{B b c^{3} \sqrt{x} - A c^{4} \sqrt{x}}{{\left (c x + b\right )} b^{5}} - \frac{2 \,{\left (315 \, B b c^{2} x^{3} - 420 \, A c^{3} x^{3} - 70 \, B b^{2} c x^{2} + 105 \, A b c^{2} x^{2} + 21 \, B b^{3} x - 42 \, A b^{2} c x + 15 \, A b^{3}\right )}}{105 \, b^{5} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(7*B*b*c^3 - 9*A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^5) - (B*b*c^3*sqrt(x) - A*c^4*sqrt(x))/((c*x +
b)*b^5) - 2/105*(315*B*b*c^2*x^3 - 420*A*c^3*x^3 - 70*B*b^2*c*x^2 + 105*A*b*c^2*x^2 + 21*B*b^3*x - 42*A*b^2*c
*x + 15*A*b^3)/(b^5*x^(7/2))