3.177 \(\int \frac{x^{9/2} (A+B x)}{(b x+c x^2)^2} \, dx\)

Optimal. Leaf size=131 \[ -\frac{b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{9/2}}+\frac{x^{5/2} (7 b B-5 A c)}{5 b c^2}-\frac{x^{3/2} (7 b B-5 A c)}{3 c^3}+\frac{b \sqrt{x} (7 b B-5 A c)}{c^4}-\frac{x^{7/2} (b B-A c)}{b c (b+c x)} \]

[Out]

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

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Rubi [A]  time = 0.0697185, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{9/2}}+\frac{x^{5/2} (7 b B-5 A c)}{5 b c^2}-\frac{x^{3/2} (7 b B-5 A c)}{3 c^3}+\frac{b \sqrt{x} (7 b B-5 A c)}{c^4}-\frac{x^{7/2} (b B-A c)}{b c (b+c x)} \]

Antiderivative was successfully verified.

[In]

Int[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.073685, size = 110, normalized size = 0.84 \[ \frac{\sqrt{x} \left (b^2 (70 B c x-75 A c)-2 b c^2 x (25 A+7 B x)+2 c^3 x^2 (5 A+3 B x)+105 b^3 B\right )}{15 c^4 (b+c x)}-\frac{b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^2,x]

[Out]

(Sqrt[x]*(105*b^3*B + 2*c^3*x^2*(5*A + 3*B*x) - 2*b*c^2*x*(25*A + 7*B*x) + b^2*(-75*A*c + 70*B*c*x)))/(15*c^4*
(b + c*x)) - (b^(3/2)*(7*b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(9/2)

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Maple [A]  time = 0.016, size = 139, normalized size = 1.1 \begin{align*}{\frac{2\,B}{5\,{c}^{2}}{x}^{{\frac{5}{2}}}}+{\frac{2\,A}{3\,{c}^{2}}{x}^{{\frac{3}{2}}}}-{\frac{4\,bB}{3\,{c}^{3}}{x}^{{\frac{3}{2}}}}-4\,{\frac{Ab\sqrt{x}}{{c}^{3}}}+6\,{\frac{{b}^{2}B\sqrt{x}}{{c}^{4}}}-{\frac{A{b}^{2}}{{c}^{3} \left ( cx+b \right ) }\sqrt{x}}+{\frac{{b}^{3}B}{{c}^{4} \left ( cx+b \right ) }\sqrt{x}}+5\,{\frac{A{b}^{2}}{{c}^{3}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }-7\,{\frac{{b}^{3}B}{{c}^{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(x^(9/2)*(B*x+A)/(c*x^2+b*x)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65631, size = 652, normalized size = 4.98 \begin{align*} \left [-\frac{15 \,{\left (7 \, B b^{3} - 5 \, A b^{2} c +{\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x + 2 \, c \sqrt{x} \sqrt{-\frac{b}{c}} - b}{c x + b}\right ) - 2 \,{\left (6 \, B c^{3} x^{3} + 105 \, B b^{3} - 75 \, A b^{2} c - 2 \,{\left (7 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} + 10 \,{\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt{x}}{30 \,{\left (c^{5} x + b c^{4}\right )}}, -\frac{15 \,{\left (7 \, B b^{3} - 5 \, A b^{2} c +{\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c \sqrt{x} \sqrt{\frac{b}{c}}}{b}\right ) -{\left (6 \, B c^{3} x^{3} + 105 \, B b^{3} - 75 \, A b^{2} c - 2 \,{\left (7 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} + 10 \,{\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt{x}}{15 \,{\left (c^{5} x + b c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

[-1/30*(15*(7*B*b^3 - 5*A*b^2*c + (7*B*b^2*c - 5*A*b*c^2)*x)*sqrt(-b/c)*log((c*x + 2*c*sqrt(x)*sqrt(-b/c) - b)
/(c*x + b)) - 2*(6*B*c^3*x^3 + 105*B*b^3 - 75*A*b^2*c - 2*(7*B*b*c^2 - 5*A*c^3)*x^2 + 10*(7*B*b^2*c - 5*A*b*c^
2)*x)*sqrt(x))/(c^5*x + b*c^4), -1/15*(15*(7*B*b^3 - 5*A*b^2*c + (7*B*b^2*c - 5*A*b*c^2)*x)*sqrt(b/c)*arctan(c
*sqrt(x)*sqrt(b/c)/b) - (6*B*c^3*x^3 + 105*B*b^3 - 75*A*b^2*c - 2*(7*B*b*c^2 - 5*A*c^3)*x^2 + 10*(7*B*b^2*c -
5*A*b*c^2)*x)*sqrt(x))/(c^5*x + b*c^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(x**(9/2)*(B*x+A)/(c*x**2+b*x)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.1342, size = 165, normalized size = 1.26 \begin{align*} -\frac{{\left (7 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} c^{4}} + \frac{B b^{3} \sqrt{x} - A b^{2} c \sqrt{x}}{{\left (c x + b\right )} c^{4}} + \frac{2 \,{\left (3 \, B c^{8} x^{\frac{5}{2}} - 10 \, B b c^{7} x^{\frac{3}{2}} + 5 \, A c^{8} x^{\frac{3}{2}} + 45 \, B b^{2} c^{6} \sqrt{x} - 30 \, A b c^{7} \sqrt{x}\right )}}{15 \, c^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

-(7*B*b^3 - 5*A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + (B*b^3*sqrt(x) - A*b^2*c*sqrt(x))/((c*x +
b)*c^4) + 2/15*(3*B*c^8*x^(5/2) - 10*B*b*c^7*x^(3/2) + 5*A*c^8*x^(3/2) + 45*B*b^2*c^6*sqrt(x) - 30*A*b*c^7*sq
rt(x))/c^10