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

Optimal. Leaf size=126 \[ -\frac{3 (5 A b-a B)}{4 a^3 b \sqrt{x}}+\frac{5 A b-a B}{4 a^2 b \sqrt{x} (a+b x)}-\frac{3 (5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2} \sqrt{b}}+\frac{A b-a B}{2 a b \sqrt{x} (a+b x)^2} \]

[Out]

(-3*(5*A*b - a*B))/(4*a^3*b*Sqrt[x]) + (A*b - a*B)/(2*a*b*Sqrt[x]*(a + b*x)^2) + (5*A*b - a*B)/(4*a^2*b*Sqrt[x
]*(a + b*x)) - (3*(5*A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(7/2)*Sqrt[b])

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Rubi [A]  time = 0.0515795, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 51, 63, 205} \[ -\frac{3 (5 A b-a B)}{4 a^3 b \sqrt{x}}+\frac{5 A b-a B}{4 a^2 b \sqrt{x} (a+b x)}-\frac{3 (5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2} \sqrt{b}}+\frac{A b-a B}{2 a b \sqrt{x} (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(5*A*b - a*B))/(4*a^3*b*Sqrt[x]) + (A*b - a*B)/(2*a*b*Sqrt[x]*(a + b*x)^2) + (5*A*b - a*B)/(4*a^2*b*Sqrt[x
]*(a + b*x)) - (3*(5*A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(7/2)*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)^3} \, dx &=\frac{A b-a B}{2 a b \sqrt{x} (a+b x)^2}-\frac{\left (-\frac{5 A b}{2}+\frac{a B}{2}\right ) \int \frac{1}{x^{3/2} (a+b x)^2} \, dx}{2 a b}\\ &=\frac{A b-a B}{2 a b \sqrt{x} (a+b x)^2}+\frac{5 A b-a B}{4 a^2 b \sqrt{x} (a+b x)}+\frac{(3 (5 A b-a B)) \int \frac{1}{x^{3/2} (a+b x)} \, dx}{8 a^2 b}\\ &=-\frac{3 (5 A b-a B)}{4 a^3 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} (a+b x)^2}+\frac{5 A b-a B}{4 a^2 b \sqrt{x} (a+b x)}-\frac{(3 (5 A b-a B)) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{8 a^3}\\ &=-\frac{3 (5 A b-a B)}{4 a^3 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} (a+b x)^2}+\frac{5 A b-a B}{4 a^2 b \sqrt{x} (a+b x)}-\frac{(3 (5 A b-a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{4 a^3}\\ &=-\frac{3 (5 A b-a B)}{4 a^3 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} (a+b x)^2}+\frac{5 A b-a B}{4 a^2 b \sqrt{x} (a+b x)}-\frac{3 (5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2} \sqrt{b}}\\ \end{align*}

Mathematica [C]  time = 0.0239584, size = 59, normalized size = 0.47 \[ \frac{\frac{a^2 (A b-a B)}{(a+b x)^2}+(a B-5 A b) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};-\frac{b x}{a}\right )}{2 a^3 b \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2*(A*b - a*B))/(a + b*x)^2 + (-5*A*b + a*B)*Hypergeometric2F1[-1/2, 2, 1/2, -((b*x)/a)])/(2*a^3*b*Sqrt[x])

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Maple [A]  time = 0.014, size = 125, normalized size = 1. \begin{align*} -2\,{\frac{A}{{a}^{3}\sqrt{x}}}-{\frac{7\,A{b}^{2}}{4\,{a}^{3} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{3\,Bb}{4\,{a}^{2} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{9\,Ab}{4\,{a}^{2} \left ( bx+a \right ) ^{2}}\sqrt{x}}+{\frac{5\,B}{4\,a \left ( bx+a \right ) ^{2}}\sqrt{x}}-{\frac{15\,Ab}{4\,{a}^{3}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,B}{4\,{a}^{2}}\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)^3,x)

[Out]

-2*A/a^3/x^(1/2)-7/4/a^3/(b*x+a)^2*x^(3/2)*A*b^2+3/4/a^2/(b*x+a)^2*x^(3/2)*B*b-9/4/a^2/(b*x+a)^2*A*x^(1/2)*b+5
/4/a/(b*x+a)^2*B*x^(1/2)-15/4/a^3/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))*A*b+3/4/a^2/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.44172, size = 718, normalized size = 5.7 \begin{align*} \left [\frac{3 \,{\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \,{\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} +{\left (B a^{3} - 5 \, A a^{2} 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 (8 \, A a^{3} b - 3 \,{\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \,{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{8 \,{\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}, -\frac{3 \,{\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \,{\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} +{\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (8 \, A a^{3} b - 3 \,{\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \,{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{4 \,{\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*((B*a*b^2 - 5*A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(-a*b)*log((b*x -
a + 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) - 2*(8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)*x^2 - 5*(B*a^3*b - 5*A*a^2*b^2
)*x)*sqrt(x))/(a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x), -1/4*(3*((B*a*b^2 - 5*A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b
^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b
^3)*x^2 - 5*(B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x)]

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Sympy [A]  time = 149.272, size = 1761, normalized size = 13.98 \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)**3,x)

[Out]

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(
5/2)))/b**3, Eq(a, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/a**3, Eq(b, 0)), (-16*I*A*a**(5/2)*b*sqrt(1/b)/(8*I*a**(
11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) - 50
*I*A*a**(3/2)*b**2*x*sqrt(1/b)/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*
I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) - 30*I*A*sqrt(a)*b**3*x**2*sqrt(1/b)/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) +
16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) - 15*A*a**2*b*sqrt(x)*log(-I*sqr
t(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**
(7/2)*b**3*x**(5/2)*sqrt(1/b)) + 15*A*a**2*b*sqrt(x)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(
x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) - 30*A*a*b**2*x**
(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqr
t(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) + 30*A*a*b**2*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*
a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b))
- 15*A*b**3*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b*
*2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) + 15*A*b**3*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + s
qrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2
)*sqrt(1/b)) + 10*I*B*a**(5/2)*b*x*sqrt(1/b)/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*
sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) + 6*I*B*a**(3/2)*b**2*x**2*sqrt(1/b)/(8*I*a**(11/2)*b*sqrt(x
)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) + 3*B*a**3*sqrt(x)
*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b
) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) - 3*B*a**3*sqrt(x)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)
*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) + 6*B*a**
2*b*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3
/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) - 6*B*a**2*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))
/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(
1/b)) + 3*B*a*b**2*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(
9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3*x**(5/2)*sqrt(1/b)) - 3*B*a*b**2*x**(5/2)*log(I*sqrt(a)*sqrt(
1/b) + sqrt(x))/(8*I*a**(11/2)*b*sqrt(x)*sqrt(1/b) + 16*I*a**(9/2)*b**2*x**(3/2)*sqrt(1/b) + 8*I*a**(7/2)*b**3
*x**(5/2)*sqrt(1/b)), True))

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Giac [A]  time = 1.22428, size = 116, normalized size = 0.92 \begin{align*} \frac{3 \,{\left (B a - 5 \, A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{3}} - \frac{2 \, A}{a^{3} \sqrt{x}} + \frac{3 \, B a b x^{\frac{3}{2}} - 7 \, A b^{2} x^{\frac{3}{2}} + 5 \, B a^{2} \sqrt{x} - 9 \, A a b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*(B*a - 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2*A/(a^3*sqrt(x)) + 1/4*(3*B*a*b*x^(3/2) - 7*A
*b^2*x^(3/2) + 5*B*a^2*sqrt(x) - 9*A*a*b*sqrt(x))/((b*x + a)^2*a^3)