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

Optimal. Leaf size=110 \[ \frac{3 b (5 A b-4 a B)}{4 a^3 \sqrt{a+b x}}+\frac{5 A b-4 a B}{4 a^2 x \sqrt{a+b x}}-\frac{3 b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{A}{2 a x^2 \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.0445824, antiderivative size = 112, normalized size of antiderivative = 1.02, 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, 208} \[ \frac{3 \sqrt{a+b x} (5 A b-4 a B)}{4 a^3 x}-\frac{5 A b-4 a B}{2 a^2 x \sqrt{a+b x}}-\frac{3 b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{A}{2 a x^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [C]  time = 0.0159692, size = 56, normalized size = 0.51 \[ \frac{b x^2 (5 A b-4 a B) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x}{a}+1\right )-a^2 A}{2 a^3 x^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2*A) + b*(5*A*b - 4*a*B)*x^2*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (b*x)/a])/(2*a^3*x^2*Sqrt[a + b*x])

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Maple [A]  time = 0.013, size = 101, normalized size = 0.9 \begin{align*} 2\,b \left ({\frac{1}{{a}^{3}} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ( \left ({\frac{7\,Ab}{8}}-1/2\,Ba \right ) \left ( bx+a \right ) ^{3/2}+ \left ( -{\frac{9\,Aba}{8}}+1/2\,B{a}^{2} \right ) \sqrt{bx+a} \right ) }-3/8\,{\frac{5\,Ab-4\,Ba}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-{\frac{-Ab+Ba}{{a}^{3}\sqrt{bx+a}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*(1/a^3*(((7/8*A*b-1/2*B*a)*(b*x+a)^(3/2)+(-9/8*A*b*a+1/2*B*a^2)*(b*x+a)^(1/2))/b^2/x^2-3/8*(5*A*b-4*B*a)/a
^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-1/a^3*(-A*b+B*a)/(b*x+a)^(1/2))

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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/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.55501, size = 606, normalized size = 5.51 \begin{align*} \left [-\frac{3 \,{\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} +{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt{a} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (2 \, A a^{3} + 3 \,{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} +{\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{8 \,{\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, -\frac{3 \,{\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} +{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (2 \, A a^{3} + 3 \,{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} +{\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{4 \,{\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 76.3539, size = 185, normalized size = 1.68 \begin{align*} A \left (- \frac{1}{2 a \sqrt{b} x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{5 \sqrt{b}}{4 a^{2} x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{15 b^{\frac{3}{2}}}{4 a^{3} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} - \frac{15 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4 a^{\frac{7}{2}}}\right ) + B \left (- \frac{1}{a \sqrt{b} x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{3 \sqrt{b}}{a^{2} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} + \frac{3 b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{a^{\frac{5}{2}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-1/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + 5*sqrt(b)/(4*a**2*x**(3/2)*sqrt(a/(b*x) + 1)) + 15*b**(3/2)/(
4*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) - 15*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2))) + B*(-1/(a*sqrt(b)*
x**(3/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(a**2*sqrt(x)*sqrt(a/(b*x) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x))
)/a**(5/2))

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Giac [A]  time = 1.24564, size = 169, normalized size = 1.54 \begin{align*} -\frac{3 \,{\left (4 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a} a^{3}} - \frac{2 \,{\left (B a b - A b^{2}\right )}}{\sqrt{b x + a} a^{3}} - \frac{4 \,{\left (b x + a\right )}^{\frac{3}{2}} B a b - 4 \, \sqrt{b x + a} B a^{2} b - 7 \,{\left (b x + a\right )}^{\frac{3}{2}} A b^{2} + 9 \, \sqrt{b x + a} A a b^{2}}{4 \, a^{3} b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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