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

Optimal. Leaf size=90 \[ \frac{2 c^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}+\frac{2 c (b B-A c)}{b^3 \sqrt{x}}-\frac{2 A}{5 b x^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0534538, antiderivative size = 90, 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} \[ \frac{2 c^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}+\frac{2 c (b B-A c)}{b^3 \sqrt{x}}-\frac{2 A}{5 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0162336, size = 43, normalized size = 0.48 \[ \frac{-10 x (b B-A c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{c x}{b}\right )-6 A b}{15 b^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*A*b - 10*(b*B - A*c)*x*Hypergeometric2F1[-3/2, 1, -1/2, -((c*x)/b)])/(15*b^2*x^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 102, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{5\,b}{x}^{-{\frac{5}{2}}}}+{\frac{2\,Ac}{3\,{b}^{2}}{x}^{-{\frac{3}{2}}}}-{\frac{2\,B}{3\,b}{x}^{-{\frac{3}{2}}}}-2\,{\frac{A{c}^{2}}{{b}^{3}\sqrt{x}}}+2\,{\frac{Bc}{{b}^{2}\sqrt{x}}}-2\,{\frac{A{c}^{3}}{{b}^{3}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }+2\,{\frac{{c}^{2}B}{{b}^{2}\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),x)

[Out]

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

________________________________________________________________________________________

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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.54063, size = 441, normalized size = 4.9 \begin{align*} \left [-\frac{15 \,{\left (B b c - A c^{2}\right )} x^{3} \sqrt{-\frac{c}{b}} \log \left (\frac{c x - 2 \, b \sqrt{x} \sqrt{-\frac{c}{b}} - b}{c x + b}\right ) + 2 \,{\left (3 \, A b^{2} - 15 \,{\left (B b c - A c^{2}\right )} x^{2} + 5 \,{\left (B b^{2} - A b c\right )} x\right )} \sqrt{x}}{15 \, b^{3} x^{3}}, -\frac{2 \,{\left (15 \,{\left (B b c - A c^{2}\right )} x^{3} \sqrt{\frac{c}{b}} \arctan \left (\frac{b \sqrt{\frac{c}{b}}}{c \sqrt{x}}\right ) +{\left (3 \, A b^{2} - 15 \,{\left (B b c - A c^{2}\right )} x^{2} + 5 \,{\left (B b^{2} - A b c\right )} x\right )} \sqrt{x}\right )}}{15 \, b^{3} x^{3}}\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),x, algorithm="fricas")

[Out]

[-1/15*(15*(B*b*c - A*c^2)*x^3*sqrt(-c/b)*log((c*x - 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(3*A*b^2 - 15*
(B*b*c - A*c^2)*x^2 + 5*(B*b^2 - A*b*c)*x)*sqrt(x))/(b^3*x^3), -2/15*(15*(B*b*c - A*c^2)*x^3*sqrt(c/b)*arctan(
b*sqrt(c/b)/(c*sqrt(x))) + (3*A*b^2 - 15*(B*b*c - A*c^2)*x^2 + 5*(B*b^2 - A*b*c)*x)*sqrt(x))/(b^3*x^3)]

________________________________________________________________________________________

Sympy [A]  time = 28.1894, size = 289, normalized size = 3.21 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{2 A}{7 x^{\frac{7}{2}}} - \frac{2 B}{5 x^{\frac{5}{2}}}\right ) & \text{for}\: b = 0 \wedge c = 0 \\\frac{- \frac{2 A}{7 x^{\frac{7}{2}}} - \frac{2 B}{5 x^{\frac{5}{2}}}}{c} & \text{for}\: b = 0 \\\frac{- \frac{2 A}{5 x^{\frac{5}{2}}} - \frac{2 B}{3 x^{\frac{3}{2}}}}{b} & \text{for}\: c = 0 \\- \frac{2 A}{5 b x^{\frac{5}{2}}} + \frac{2 A c}{3 b^{2} x^{\frac{3}{2}}} - \frac{2 A c^{2}}{b^{3} \sqrt{x}} + \frac{i A c^{2} \log{\left (- i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{b^{\frac{7}{2}} \sqrt{\frac{1}{c}}} - \frac{i A c^{2} \log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{b^{\frac{7}{2}} \sqrt{\frac{1}{c}}} - \frac{2 B}{3 b x^{\frac{3}{2}}} + \frac{2 B c}{b^{2} \sqrt{x}} - \frac{i B c \log{\left (- i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{b^{\frac{5}{2}} \sqrt{\frac{1}{c}}} + \frac{i B c \log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{b^{\frac{5}{2}} \sqrt{\frac{1}{c}}} & \text{otherwise} \end{cases} \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),x)

[Out]

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(
5/2)))/c, Eq(b, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2)))/b, Eq(c, 0)), (-2*A/(5*b*x**(5/2)) + 2*A*c/(3*b**
2*x**(3/2)) - 2*A*c**2/(b**3*sqrt(x)) + I*A*c**2*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(7/2)*sqrt(1/c)) - I*
A*c**2*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(7/2)*sqrt(1/c)) - 2*B/(3*b*x**(3/2)) + 2*B*c/(b**2*sqrt(x)) - I
*B*c*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(5/2)*sqrt(1/c)) + I*B*c*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(
5/2)*sqrt(1/c)), True))

________________________________________________________________________________________

Giac [A]  time = 1.12833, size = 108, normalized size = 1.2 \begin{align*} \frac{2 \,{\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} b^{3}} + \frac{2 \,{\left (15 \, B b c x^{2} - 15 \, A c^{2} x^{2} - 5 \, B b^{2} x + 5 \, A b c x - 3 \, A b^{2}\right )}}{15 \, b^{3} x^{\frac{5}{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),x, algorithm="giac")

[Out]

2*(B*b*c^2 - A*c^3)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^3) + 2/15*(15*B*b*c*x^2 - 15*A*c^2*x^2 - 5*B*b^2*
x + 5*A*b*c*x - 3*A*b^2)/(b^3*x^(5/2))