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

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

[Out]

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

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Rubi [A]  time = 0.0484572, 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, 80, 50, 63, 205} \[ -\frac{2 b^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{7/2}}-\frac{2 x^{3/2} (b B-A c)}{3 c^2}+\frac{2 b \sqrt{x} (b B-A c)}{c^3}+\frac{2 B x^{5/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(b*B - A*c)*Sqrt[x])/c^3 - (2*(b*B - A*c)*x^(3/2))/(3*c^2) + (2*B*x^(5/2))/(5*c) - (2*b^(3/2)*(b*B - A*c)
*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A]  time = 0.0610002, size = 81, normalized size = 0.9 \[ \frac{2 \sqrt{x} \left (-5 b c (3 A+B x)+c^2 x (5 A+3 B x)+15 b^2 B\right )}{15 c^3}-\frac{2 b^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(15*b^2*B - 5*b*c*(3*A + B*x) + c^2*x*(5*A + 3*B*x)))/(15*c^3) - (2*b^(3/2)*(b*B - A*c)*ArcTan[(Sqr
t[c]*Sqrt[x])/Sqrt[b]])/c^(7/2)

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Maple [A]  time = 0.009, size = 102, normalized size = 1.1 \begin{align*}{\frac{2\,B}{5\,c}{x}^{{\frac{5}{2}}}}+{\frac{2\,A}{3\,c}{x}^{{\frac{3}{2}}}}-{\frac{2\,bB}{3\,{c}^{2}}{x}^{{\frac{3}{2}}}}-2\,{\frac{Ab\sqrt{x}}{{c}^{2}}}+2\,{\frac{{b}^{2}B\sqrt{x}}{{c}^{3}}}+2\,{\frac{A{b}^{2}}{{c}^{2}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }-2\,{\frac{{b}^{3}B}{{c}^{3}\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^(5/2)*(B*x+A)/(c*x^2+b*x),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.93468, size = 414, normalized size = 4.6 \begin{align*} \left [-\frac{15 \,{\left (B b^{2} - A b c\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 (3 \, B c^{2} x^{2} + 15 \, B b^{2} - 15 \, A b c - 5 \,{\left (B b c - A c^{2}\right )} x\right )} \sqrt{x}}{15 \, c^{3}}, -\frac{2 \,{\left (15 \,{\left (B b^{2} - A b c\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c \sqrt{x} \sqrt{\frac{b}{c}}}{b}\right ) -{\left (3 \, B c^{2} x^{2} + 15 \, B b^{2} - 15 \, A b c - 5 \,{\left (B b c - A c^{2}\right )} x\right )} \sqrt{x}\right )}}{15 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 45.4449, size = 245, normalized size = 2.72 \begin{align*} \begin{cases} - \frac{i A b^{\frac{3}{2}} \log{\left (- i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{c^{3} \sqrt{\frac{1}{c}}} + \frac{i A b^{\frac{3}{2}} \log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{c^{3} \sqrt{\frac{1}{c}}} - \frac{2 A b \sqrt{x}}{c^{2}} + \frac{2 A x^{\frac{3}{2}}}{3 c} + \frac{i B b^{\frac{5}{2}} \log{\left (- i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{c^{4} \sqrt{\frac{1}{c}}} - \frac{i B b^{\frac{5}{2}} \log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{c^{4} \sqrt{\frac{1}{c}}} + \frac{2 B b^{2} \sqrt{x}}{c^{3}} - \frac{2 B b x^{\frac{3}{2}}}{3 c^{2}} + \frac{2 B x^{\frac{5}{2}}}{5 c} & \text{for}\: c \neq 0 \\\frac{\frac{2 A x^{\frac{5}{2}}}{5} + \frac{2 B x^{\frac{7}{2}}}{7}}{b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*A*b**(3/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(c**3*sqrt(1/c)) + I*A*b**(3/2)*log(I*sqrt(b)*sqr
t(1/c) + sqrt(x))/(c**3*sqrt(1/c)) - 2*A*b*sqrt(x)/c**2 + 2*A*x**(3/2)/(3*c) + I*B*b**(5/2)*log(-I*sqrt(b)*sqr
t(1/c) + sqrt(x))/(c**4*sqrt(1/c)) - I*B*b**(5/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(c**4*sqrt(1/c)) + 2*B*b*
*2*sqrt(x)/c**3 - 2*B*b*x**(3/2)/(3*c**2) + 2*B*x**(5/2)/(5*c), Ne(c, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/
b, True))

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Giac [A]  time = 1.10734, size = 123, normalized size = 1.37 \begin{align*} -\frac{2 \,{\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} c^{3}} + \frac{2 \,{\left (3 \, B c^{4} x^{\frac{5}{2}} - 5 \, B b c^{3} x^{\frac{3}{2}} + 5 \, A c^{4} x^{\frac{3}{2}} + 15 \, B b^{2} c^{2} \sqrt{x} - 15 \, A b c^{3} \sqrt{x}\right )}}{15 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*b^3 - A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^3) + 2/15*(3*B*c^4*x^(5/2) - 5*B*b*c^3*x^(3/2) +
5*A*c^4*x^(3/2) + 15*B*b^2*c^2*sqrt(x) - 15*A*b*c^3*sqrt(x))/c^5