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

Optimal. Leaf size=193 \[ -\frac{9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt{x}}-\frac{9 c^{5/2} (7 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{13/2}}+\frac{3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac{9 (7 b B-11 A c)}{20 b^4 x^{5/2}}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}+\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2} \]

[Out]

(9*(7*b*B - 11*A*c))/(28*b^3*c*x^(7/2)) - (9*(7*b*B - 11*A*c))/(20*b^4*x^(5/2)) + (3*c*(7*b*B - 11*A*c))/(4*b^
5*x^(3/2)) - (9*c^2*(7*b*B - 11*A*c))/(4*b^6*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(7/2)*(b + c*x)^2) - (7*b*B - 11*
A*c)/(4*b^2*c*x^(7/2)*(b + c*x)) - (9*c^(5/2)*(7*b*B - 11*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(13/2))

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Rubi [A]  time = 0.101074, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt{x}}-\frac{9 c^{5/2} (7 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{13/2}}+\frac{3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac{9 (7 b B-11 A c)}{20 b^4 x^{5/2}}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}+\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*(7*b*B - 11*A*c))/(28*b^3*c*x^(7/2)) - (9*(7*b*B - 11*A*c))/(20*b^4*x^(5/2)) + (3*c*(7*b*B - 11*A*c))/(4*b^
5*x^(3/2)) - (9*c^2*(7*b*B - 11*A*c))/(4*b^6*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(7/2)*(b + c*x)^2) - (7*b*B - 11*
A*c)/(4*b^2*c*x^(7/2)*(b + c*x)) - (9*c^(5/2)*(7*b*B - 11*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(13/2))

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^{3/2} \left (b x+c x^2\right )^3} \, dx &=\int \frac{A+B x}{x^{9/2} (b+c x)^3} \, dx\\ &=-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{\left (\frac{7 b B}{2}-\frac{11 A c}{2}\right ) \int \frac{1}{x^{9/2} (b+c x)^2} \, dx}{2 b c}\\ &=-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac{(9 (7 b B-11 A c)) \int \frac{1}{x^{9/2} (b+c x)} \, dx}{8 b^2 c}\\ &=\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}+\frac{(9 (7 b B-11 A c)) \int \frac{1}{x^{7/2} (b+c x)} \, dx}{8 b^3}\\ &=\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{9 (7 b B-11 A c)}{20 b^4 x^{5/2}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac{(9 c (7 b B-11 A c)) \int \frac{1}{x^{5/2} (b+c x)} \, dx}{8 b^4}\\ &=\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac{3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}+\frac{\left (9 c^2 (7 b B-11 A c)\right ) \int \frac{1}{x^{3/2} (b+c x)} \, dx}{8 b^5}\\ &=\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac{3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac{9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt{x}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac{\left (9 c^3 (7 b B-11 A c)\right ) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{8 b^6}\\ &=\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac{3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac{9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt{x}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac{\left (9 c^3 (7 b B-11 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{4 b^6}\\ &=\frac{9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac{9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac{3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac{9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt{x}}-\frac{b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac{7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac{9 c^{5/2} (7 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{13/2}}\\ \end{align*}

Mathematica [C]  time = 0.0283695, size = 61, normalized size = 0.32 \[ \frac{\frac{7 b^2 (A c-b B)}{(b+c x)^2}+(7 b B-11 A c) \, _2F_1\left (-\frac{7}{2},2;-\frac{5}{2};-\frac{c x}{b}\right )}{14 b^3 c x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*b^2*(-(b*B) + A*c))/(b + c*x)^2 + (7*b*B - 11*A*c)*Hypergeometric2F1[-7/2, 2, -5/2, -((c*x)/b)])/(14*b^3*c
*x^(7/2))

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Maple [A]  time = 0.022, size = 202, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{7\,{b}^{3}}{x}^{-{\frac{7}{2}}}}+{\frac{6\,Ac}{5\,{b}^{4}}{x}^{-{\frac{5}{2}}}}-{\frac{2\,B}{5\,{b}^{3}}{x}^{-{\frac{5}{2}}}}-4\,{\frac{A{c}^{2}}{{b}^{5}{x}^{3/2}}}+2\,{\frac{Bc}{{b}^{4}{x}^{3/2}}}+20\,{\frac{A{c}^{3}}{{b}^{6}\sqrt{x}}}-12\,{\frac{B{c}^{2}}{{b}^{5}\sqrt{x}}}+{\frac{19\,{c}^{5}A}{4\,{b}^{6} \left ( cx+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{15\,{c}^{4}B}{4\,{b}^{5} \left ( cx+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{21\,{c}^{4}A}{4\,{b}^{5} \left ( cx+b \right ) ^{2}}\sqrt{x}}-{\frac{17\,{c}^{3}B}{4\,{b}^{4} \left ( cx+b \right ) ^{2}}\sqrt{x}}+{\frac{99\,{c}^{4}A}{4\,{b}^{6}}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}-{\frac{63\,{c}^{3}B}{4\,{b}^{5}}\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((B*x+A)/x^(3/2)/(c*x^2+b*x)^3,x)

[Out]

-2/7*A/b^3/x^(7/2)+6/5/b^4/x^(5/2)*A*c-2/5/b^3/x^(5/2)*B-4*c^2/b^5/x^(3/2)*A+2*c/b^4/x^(3/2)*B+20*c^3/b^6/x^(1
/2)*A-12*c^2/b^5/x^(1/2)*B+19/4/b^6*c^5/(c*x+b)^2*x^(3/2)*A-15/4/b^5*c^4/(c*x+b)^2*x^(3/2)*B+21/4/b^5*c^4/(c*x
+b)^2*A*x^(1/2)-17/4/b^4*c^3/(c*x+b)^2*B*x^(1/2)+99/4/b^6*c^4/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*c)^(1/2))*A-63/4
/b^5*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^(3/2)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73646, size = 1080, normalized size = 5.6 \begin{align*} \left [-\frac{315 \,{\left ({\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{6} + 2 \,{\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{5} +{\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} 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 (40 \, A b^{5} + 315 \,{\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \,{\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \,{\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \,{\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \,{\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x\right )} \sqrt{x}}{280 \,{\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}, \frac{315 \,{\left ({\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{6} + 2 \,{\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{5} +{\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt{\frac{c}{b}} \arctan \left (\frac{b \sqrt{\frac{c}{b}}}{c \sqrt{x}}\right ) -{\left (40 \, A b^{5} + 315 \,{\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \,{\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \,{\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \,{\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \,{\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x\right )} \sqrt{x}}{140 \,{\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/280*(315*((7*B*b*c^4 - 11*A*c^5)*x^6 + 2*(7*B*b^2*c^3 - 11*A*b*c^4)*x^5 + (7*B*b^3*c^2 - 11*A*b^2*c^3)*x^4
)*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5
+ 525*(7*B*b^2*c^3 - 11*A*b*c^4)*x^4 + 168*(7*B*b^3*c^2 - 11*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A*b^3*c^2)*x^
2 + 8*(7*B*b^5 - 11*A*b^4*c)*x)*sqrt(x))/(b^6*c^2*x^6 + 2*b^7*c*x^5 + b^8*x^4), 1/140*(315*((7*B*b*c^4 - 11*A*
c^5)*x^6 + 2*(7*B*b^2*c^3 - 11*A*b*c^4)*x^5 + (7*B*b^3*c^2 - 11*A*b^2*c^3)*x^4)*sqrt(c/b)*arctan(b*sqrt(c/b)/(
c*sqrt(x))) - (40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5 + 525*(7*B*b^2*c^3 - 11*A*b*c^4)*x^4 + 168*(7*B*b^3*c
^2 - 11*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A*b^3*c^2)*x^2 + 8*(7*B*b^5 - 11*A*b^4*c)*x)*sqrt(x))/(b^6*c^2*x^6
+ 2*b^7*c*x^5 + b^8*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**(3/2)/(c*x**2+b*x)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1311, size = 215, normalized size = 1.11 \begin{align*} -\frac{9 \,{\left (7 \, B b c^{3} - 11 \, A c^{4}\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{4 \, \sqrt{b c} b^{6}} - \frac{15 \, B b c^{4} x^{\frac{3}{2}} - 19 \, A c^{5} x^{\frac{3}{2}} + 17 \, B b^{2} c^{3} \sqrt{x} - 21 \, A b c^{4} \sqrt{x}}{4 \,{\left (c x + b\right )}^{2} b^{6}} - \frac{2 \,{\left (210 \, B b c^{2} x^{3} - 350 \, A c^{3} x^{3} - 35 \, B b^{2} c x^{2} + 70 \, A b c^{2} x^{2} + 7 \, B b^{3} x - 21 \, A b^{2} c x + 5 \, A b^{3}\right )}}{35 \, b^{6} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/4*(7*B*b*c^3 - 11*A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^6) - 1/4*(15*B*b*c^4*x^(3/2) - 19*A*c^5*x
^(3/2) + 17*B*b^2*c^3*sqrt(x) - 21*A*b*c^4*sqrt(x))/((c*x + b)^2*b^6) - 2/35*(210*B*b*c^2*x^3 - 350*A*c^3*x^3
- 35*B*b^2*c*x^2 + 70*A*b*c^2*x^2 + 7*B*b^3*x - 21*A*b^2*c*x + 5*A*b^3)/(b^6*x^(7/2))