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

Optimal. Leaf size=64 \[ \frac{(A c+b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{3/2} c^{3/2}}-\frac{\sqrt{x} (b B-A c)}{b c (b+c x)} \]

[Out]

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

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Rubi [A]  time = 0.0311665, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {781, 78, 63, 205} \[ \frac{(A c+b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{3/2} c^{3/2}}-\frac{\sqrt{x} (b B-A c)}{b c (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0374522, size = 63, normalized size = 0.98 \[ \frac{(A c+b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{3/2} c^{3/2}}+\frac{\sqrt{x} (A c-b B)}{b c (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 69, normalized size = 1.1 \begin{align*}{\frac{Ac-bB}{bc \left ( cx+b \right ) }\sqrt{x}}+{\frac{A}{b}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{B}{c}\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(x^(3/2)*(B*x+A)/(c*x^2+b*x)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62597, size = 396, normalized size = 6.19 \begin{align*} \left [-\frac{{\left (B b^{2} + A b c +{\left (B b c + A c^{2}\right )} x\right )} \sqrt{-b c} \log \left (\frac{c x - b - 2 \, \sqrt{-b c} \sqrt{x}}{c x + b}\right ) + 2 \,{\left (B b^{2} c - A b c^{2}\right )} \sqrt{x}}{2 \,{\left (b^{2} c^{3} x + b^{3} c^{2}\right )}}, -\frac{{\left (B b^{2} + A b c +{\left (B b c + A c^{2}\right )} x\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c}}{c \sqrt{x}}\right ) +{\left (B b^{2} c - A b c^{2}\right )} \sqrt{x}}{b^{2} c^{3} x + b^{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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 116.044, size = 716, normalized size = 11.19 \begin{align*} \text{result too large to display} \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)**2,x)

[Out]

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

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Giac [A]  time = 1.1372, size = 81, normalized size = 1.27 \begin{align*} \frac{{\left (B b + A c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} b c} - \frac{B b \sqrt{x} - A c \sqrt{x}}{{\left (c x + b\right )} b c} \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)^2,x, algorithm="giac")

[Out]

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