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

Optimal. Leaf size=69 \[ -\frac{2 \sqrt{x} (b B-A c)}{c^2}+\frac{2 \sqrt{b} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{5/2}}+\frac{2 B x^{3/2}}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.0385251, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ -\frac{2 \sqrt{x} (b B-A c)}{c^2}+\frac{2 \sqrt{b} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{5/2}}+\frac{2 B x^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0327641, size = 63, normalized size = 0.91 \[ \frac{2 \sqrt{x} (3 A c-3 b B+B c x)}{3 c^2}+\frac{2 \sqrt{b} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 78, normalized size = 1.1 \begin{align*}{\frac{2\,B}{3\,c}{x}^{{\frac{3}{2}}}}+2\,{\frac{A\sqrt{x}}{c}}-2\,{\frac{bB\sqrt{x}}{{c}^{2}}}-2\,{\frac{Ab}{c\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }+2\,{\frac{{b}^{2}B}{{c}^{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(x^(3/2)*(B*x+A)/(c*x^2+b*x),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.18636, size = 305, normalized size = 4.42 \begin{align*} \left [-\frac{3 \,{\left (B b - A 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 (B c x - 3 \, B b + 3 \, A c\right )} \sqrt{x}}{3 \, c^{2}}, \frac{2 \,{\left (3 \,{\left (B b - A c\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c \sqrt{x} \sqrt{\frac{b}{c}}}{b}\right ) +{\left (B c x - 3 \, B b + 3 \, A c\right )} \sqrt{x}\right )}}{3 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(B*x+A)/(c*x**2+b*x),x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.13373, size = 86, normalized size = 1.25 \begin{align*} \frac{2 \,{\left (B b^{2} - A b c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} c^{2}} + \frac{2 \,{\left (B c^{2} x^{\frac{3}{2}} - 3 \, B b c \sqrt{x} + 3 \, A c^{2} \sqrt{x}\right )}}{3 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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