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

Optimal. Leaf size=88 \[ -\frac{3 A b-a B}{a^2 b \sqrt{x}}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{5/2} \sqrt{b}}+\frac{A b-a B}{a b \sqrt{x} (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0374867, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 51, 63, 205} \[ -\frac{3 A b-a B}{a^2 b \sqrt{x}}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{5/2} \sqrt{b}}+\frac{A b-a B}{a b \sqrt{x} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0162266, size = 59, normalized size = 0.67 \[ \frac{(a+b x) (a B-3 A b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{b x}{a}\right )+a (A b-a B)}{a^2 b \sqrt{x} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A*b - a*B) + (-3*A*b + a*B)*(a + b*x)*Hypergeometric2F1[-1/2, 1, 1/2, -((b*x)/a)])/(a^2*b*Sqrt[x]*(a + b*x
))

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Maple [A]  time = 0.015, size = 87, normalized size = 1. \begin{align*} -2\,{\frac{A}{{a}^{2}\sqrt{x}}}-{\frac{Ab}{{a}^{2} \left ( bx+a \right ) }\sqrt{x}}+{\frac{B}{a \left ( bx+a \right ) }\sqrt{x}}-3\,{\frac{Ab}{{a}^{2}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) }+{\frac{B}{a}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(3/2)/(b*x+a)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.36884, size = 475, normalized size = 5.4 \begin{align*} \left [\frac{{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{2} +{\left (B a^{2} - 3 \, A a b\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x - a + 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right ) - 2 \,{\left (2 \, A a^{2} b -{\left (B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{x}}{2 \,{\left (a^{3} b^{2} x^{2} + a^{4} b x\right )}}, -\frac{{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{2} +{\left (B a^{2} - 3 \, A a b\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (2 \, A a^{2} b -{\left (B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{x}}{a^{3} b^{2} x^{2} + a^{4} b x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((B*a*b - 3*A*b^2)*x^2 + (B*a^2 - 3*A*a*b)*x)*sqrt(-a*b)*log((b*x - a + 2*sqrt(-a*b)*sqrt(x))/(b*x + a))
- 2*(2*A*a^2*b - (B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(a^3*b^2*x^2 + a^4*b*x), -(((B*a*b - 3*A*b^2)*x^2 + (B*a^2
- 3*A*a*b)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (2*A*a^2*b - (B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(a^3*b
^2*x^2 + a^4*b*x)]

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Sympy [A]  time = 36.1825, size = 884, normalized size = 10.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(
3/2)))/b**2, Eq(a, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/a**2, Eq(b, 0)), (-4*I*A*a**(3/2)*b*sqrt(1/b)/(2*I*a**(7
/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)) - 6*I*A*sqrt(a)*b**2*x*sqrt(1/b)/(2*I*a**(7/2)
*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)) - 3*A*a*b*sqrt(x)*log(-I*sqrt(a)*sqrt(1/b) + sqrt
(x))/(2*I*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)) + 3*A*a*b*sqrt(x)*log(I*sqrt(a)
*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)) - 3*A*b**2*x**
(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(
1/b)) + 3*A*b**2*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*
b**2*x**(3/2)*sqrt(1/b)) + 2*I*B*a**(3/2)*b*x*sqrt(1/b)/(2*I*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*
x**(3/2)*sqrt(1/b)) + B*a**2*sqrt(x)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2
*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)) - B*a**2*sqrt(x)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(7/2)*b*sqrt(
x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)) + B*a*b*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I
*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)) - B*a*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b)
+ sqrt(x))/(2*I*a**(7/2)*b*sqrt(x)*sqrt(1/b) + 2*I*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)), True))

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Giac [A]  time = 1.17731, size = 81, normalized size = 0.92 \begin{align*} \frac{{\left (B a - 3 \, A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{2}} + \frac{B a x - 3 \, A b x - 2 \, A a}{{\left (b x^{\frac{3}{2}} + a \sqrt{x}\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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