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

Optimal. Leaf size=147 \[ -\frac{3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}+\frac{5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac{5 (3 b B-7 A c)}{4 b^4 \sqrt{x}}-\frac{5 \sqrt{c} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{9/2}}-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2} \]

[Out]

(5*(3*b*B - 7*A*c))/(12*b^3*c*x^(3/2)) - (5*(3*b*B - 7*A*c))/(4*b^4*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(3/2)*(b +
c*x)^2) - (3*b*B - 7*A*c)/(4*b^2*c*x^(3/2)*(b + c*x)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/S
qrt[b]])/(4*b^(9/2))

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Rubi [A]  time = 0.0683295, antiderivative size = 147, 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, 51, 63, 205} \[ -\frac{3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}+\frac{5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac{5 (3 b B-7 A c)}{4 b^4 \sqrt{x}}-\frac{5 \sqrt{c} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{9/2}}-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(3*b*B - 7*A*c))/(12*b^3*c*x^(3/2)) - (5*(3*b*B - 7*A*c))/(4*b^4*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(3/2)*(b +
c*x)^2) - (3*b*B - 7*A*c)/(4*b^2*c*x^(3/2)*(b + c*x)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/S
qrt[b]])/(4*b^(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 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{\sqrt{x} (A+B x)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac{A+B x}{x^{5/2} (b+c x)^3} \, dx\\ &=-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac{\left (\frac{3 b B}{2}-\frac{7 A c}{2}\right ) \int \frac{1}{x^{5/2} (b+c x)^2} \, dx}{2 b c}\\ &=-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac{3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac{(5 (3 b B-7 A c)) \int \frac{1}{x^{5/2} (b+c x)} \, dx}{8 b^2 c}\\ &=\frac{5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac{3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}+\frac{(5 (3 b B-7 A c)) \int \frac{1}{x^{3/2} (b+c x)} \, dx}{8 b^3}\\ &=\frac{5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac{5 (3 b B-7 A c)}{4 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac{3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac{(5 c (3 b B-7 A c)) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{8 b^4}\\ &=\frac{5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac{5 (3 b B-7 A c)}{4 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac{3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac{(5 c (3 b B-7 A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{4 b^4}\\ &=\frac{5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac{5 (3 b B-7 A c)}{4 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac{3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac{5 \sqrt{c} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0263652, size = 61, normalized size = 0.41 \[ \frac{\frac{3 b^2 (A c-b B)}{(b+c x)^2}+(3 b B-7 A c) \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};-\frac{c x}{b}\right )}{6 b^3 c x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b^2*(-(b*B) + A*c))/(b + c*x)^2 + (3*b*B - 7*A*c)*Hypergeometric2F1[-3/2, 2, -1/2, -((c*x)/b)])/(6*b^3*c*x
^(3/2))

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Maple [A]  time = 0.02, size = 152, normalized size = 1. \begin{align*} -{\frac{2\,A}{3\,{b}^{3}}{x}^{-{\frac{3}{2}}}}+6\,{\frac{Ac}{{b}^{4}\sqrt{x}}}-2\,{\frac{B}{{b}^{3}\sqrt{x}}}+{\frac{11\,{c}^{3}A}{4\,{b}^{4} \left ( cx+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{7\,{c}^{2}B}{4\,{b}^{3} \left ( cx+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{13\,A{c}^{2}}{4\,{b}^{3} \left ( cx+b \right ) ^{2}}\sqrt{x}}-{\frac{9\,Bc}{4\,{b}^{2} \left ( cx+b \right ) ^{2}}\sqrt{x}}+{\frac{35\,A{c}^{2}}{4\,{b}^{4}}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}-{\frac{15\,Bc}{4\,{b}^{3}}\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^(1/2)/(c*x^2+b*x)^3,x)

[Out]

-2/3*A/b^3/x^(3/2)+6/b^4/x^(1/2)*A*c-2/b^3/x^(1/2)*B+11/4/b^4*c^3/(c*x+b)^2*x^(3/2)*A-7/4/b^3*c^2/(c*x+b)^2*x^
(3/2)*B+13/4/b^3*c^2/(c*x+b)^2*A*x^(1/2)-9/4/b^2*c/(c*x+b)^2*B*x^(1/2)+35/4/b^4*c^2/(b*c)^(1/2)*arctan(x^(1/2)
*c/(b*c)^(1/2))*A-15/4/b^3*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((B*x+A)*x^(1/2)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64127, size = 821, normalized size = 5.59 \begin{align*} \left [-\frac{15 \,{\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + 2 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{3} +{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2}\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 (8 \, A b^{3} + 15 \,{\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} + 25 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 8 \,{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt{x}}{24 \,{\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}, \frac{15 \,{\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + 2 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{3} +{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2}\right )} \sqrt{\frac{c}{b}} \arctan \left (\frac{b \sqrt{\frac{c}{b}}}{c \sqrt{x}}\right ) -{\left (8 \, A b^{3} + 15 \,{\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} + 25 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 8 \,{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt{x}}{12 \,{\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[-1/24*(15*((3*B*b*c^2 - 7*A*c^3)*x^4 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^3 + (3*B*b^3 - 7*A*b^2*c)*x^2)*sqrt(-c/b)*
log((c*x + 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(8*A*b^3 + 15*(3*B*b*c^2 - 7*A*c^3)*x^3 + 25*(3*B*b^2*c
- 7*A*b*c^2)*x^2 + 8*(3*B*b^3 - 7*A*b^2*c)*x)*sqrt(x))/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2), 1/12*(15*((3*B*b
*c^2 - 7*A*c^3)*x^4 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^3 + (3*B*b^3 - 7*A*b^2*c)*x^2)*sqrt(c/b)*arctan(b*sqrt(c/b)/
(c*sqrt(x))) - (8*A*b^3 + 15*(3*B*b*c^2 - 7*A*c^3)*x^3 + 25*(3*B*b^2*c - 7*A*b*c^2)*x^2 + 8*(3*B*b^3 - 7*A*b^2
*c)*x)*sqrt(x))/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^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((B*x+A)*x**(1/2)/(c*x**2+b*x)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15351, size = 146, normalized size = 0.99 \begin{align*} -\frac{5 \,{\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{4 \, \sqrt{b c} b^{4}} - \frac{2 \,{\left (3 \, B b x - 9 \, A c x + A b\right )}}{3 \, b^{4} x^{\frac{3}{2}}} - \frac{7 \, B b c^{2} x^{\frac{3}{2}} - 11 \, A c^{3} x^{\frac{3}{2}} + 9 \, B b^{2} c \sqrt{x} - 13 \, A b c^{2} \sqrt{x}}{4 \,{\left (c x + b\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

-5/4*(3*B*b*c - 7*A*c^2)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4) - 2/3*(3*B*b*x - 9*A*c*x + A*b)/(b^4*x^(3
/2)) - 1/4*(7*B*b*c^2*x^(3/2) - 11*A*c^3*x^(3/2) + 9*B*b^2*c*sqrt(x) - 13*A*b*c^2*sqrt(x))/((c*x + b)^2*b^4)