3.431 \(\int \frac{A+B x}{x^4 \sqrt{a+b x}} \, dx\)

Optimal. Leaf size=115 \[ \frac{b^2 (5 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{7/2}}+\frac{\sqrt{a+b x} (5 A b-6 a B)}{12 a^2 x^2}-\frac{b \sqrt{a+b x} (5 A b-6 a B)}{8 a^3 x}-\frac{A \sqrt{a+b x}}{3 a x^3} \]

[Out]

-(A*Sqrt[a + b*x])/(3*a*x^3) + ((5*A*b - 6*a*B)*Sqrt[a + b*x])/(12*a^2*x^2) - (b*(5*A*b - 6*a*B)*Sqrt[a + b*x]
)/(8*a^3*x) + (b^2*(5*A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(7/2))

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Rubi [A]  time = 0.04895, antiderivative size = 115, 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, 208} \[ \frac{b^2 (5 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{7/2}}+\frac{\sqrt{a+b x} (5 A b-6 a B)}{12 a^2 x^2}-\frac{b \sqrt{a+b x} (5 A b-6 a B)}{8 a^3 x}-\frac{A \sqrt{a+b x}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^4*Sqrt[a + b*x]),x]

[Out]

-(A*Sqrt[a + b*x])/(3*a*x^3) + ((5*A*b - 6*a*B)*Sqrt[a + b*x])/(12*a^2*x^2) - (b*(5*A*b - 6*a*B)*Sqrt[a + b*x]
)/(8*a^3*x) + (b^2*(5*A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*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^4 \sqrt{a+b x}} \, dx &=-\frac{A \sqrt{a+b x}}{3 a x^3}+\frac{\left (-\frac{5 A b}{2}+3 a B\right ) \int \frac{1}{x^3 \sqrt{a+b x}} \, dx}{3 a}\\ &=-\frac{A \sqrt{a+b x}}{3 a x^3}+\frac{(5 A b-6 a B) \sqrt{a+b x}}{12 a^2 x^2}+\frac{(b (5 A b-6 a B)) \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{8 a^2}\\ &=-\frac{A \sqrt{a+b x}}{3 a x^3}+\frac{(5 A b-6 a B) \sqrt{a+b x}}{12 a^2 x^2}-\frac{b (5 A b-6 a B) \sqrt{a+b x}}{8 a^3 x}-\frac{\left (b^2 (5 A b-6 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{16 a^3}\\ &=-\frac{A \sqrt{a+b x}}{3 a x^3}+\frac{(5 A b-6 a B) \sqrt{a+b x}}{12 a^2 x^2}-\frac{b (5 A b-6 a B) \sqrt{a+b x}}{8 a^3 x}-\frac{(b (5 A b-6 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{8 a^3}\\ &=-\frac{A \sqrt{a+b x}}{3 a x^3}+\frac{(5 A b-6 a B) \sqrt{a+b x}}{12 a^2 x^2}-\frac{b (5 A b-6 a B) \sqrt{a+b x}}{8 a^3 x}+\frac{b^2 (5 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0157858, size = 57, normalized size = 0.5 \[ -\frac{\sqrt{a+b x} \left (a^3 A+b^2 x^3 (6 a B-5 A b) \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{b x}{a}+1\right )\right )}{3 a^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^4*Sqrt[a + b*x]),x]

[Out]

-(Sqrt[a + b*x]*(a^3*A + b^2*(-5*A*b + 6*a*B)*x^3*Hypergeometric2F1[1/2, 3, 3/2, 1 + (b*x)/a]))/(3*a^4*x^3)

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Maple [A]  time = 0.011, size = 104, normalized size = 0.9 \begin{align*} 2\,{b}^{2} \left ({\frac{1}{{b}^{3}{x}^{3}} \left ( -1/16\,{\frac{ \left ( 5\,Ab-6\,Ba \right ) \left ( bx+a \right ) ^{5/2}}{{a}^{3}}}+1/6\,{\frac{ \left ( 5\,Ab-6\,Ba \right ) \left ( bx+a \right ) ^{3/2}}{{a}^{2}}}-1/16\,{\frac{ \left ( 11\,Ab-10\,Ba \right ) \sqrt{bx+a}}{a}} \right ) }+1/16\,{\frac{5\,Ab-6\,Ba}{{a}^{7/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^4/(b*x+a)^(1/2),x)

[Out]

2*b^2*((-1/16*(5*A*b-6*B*a)/a^3*(b*x+a)^(5/2)+1/6/a^2*(5*A*b-6*B*a)*(b*x+a)^(3/2)-1/16*(11*A*b-10*B*a)/a*(b*x+
a)^(1/2))/b^3/x^3+1/16*(5*A*b-6*B*a)/a^(7/2)*arctanh((b*x+a)^(1/2)/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^4/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.59494, size = 494, normalized size = 4.3 \begin{align*} \left [-\frac{3 \,{\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt{a} x^{3} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (8 \, A a^{3} - 3 \,{\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + 2 \,{\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{48 \, a^{4} x^{3}}, \frac{3 \,{\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt{-a} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (8 \, A a^{3} - 3 \,{\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + 2 \,{\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{24 \, a^{4} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(6*B*a*b^2 - 5*A*b^3)*sqrt(a)*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*A*a^3 - 3*(6*B
*a^2*b - 5*A*a*b^2)*x^2 + 2*(6*B*a^3 - 5*A*a^2*b)*x)*sqrt(b*x + a))/(a^4*x^3), 1/24*(3*(6*B*a*b^2 - 5*A*b^3)*s
qrt(-a)*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (8*A*a^3 - 3*(6*B*a^2*b - 5*A*a*b^2)*x^2 + 2*(6*B*a^3 - 5*A*a^2
*b)*x)*sqrt(b*x + a))/(a^4*x^3)]

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Sympy [B]  time = 62.4809, size = 245, normalized size = 2.13 \begin{align*} - \frac{A}{3 \sqrt{b} x^{\frac{7}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{A \sqrt{b}}{12 a x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{5 A b^{\frac{3}{2}}}{24 a^{2} x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{5 A b^{\frac{5}{2}}}{8 a^{3} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} + \frac{5 A b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{8 a^{\frac{7}{2}}} - \frac{B}{2 \sqrt{b} x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{B \sqrt{b}}{4 a x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{3 B b^{\frac{3}{2}}}{4 a^{2} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} - \frac{3 B b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**4/(b*x+a)**(1/2),x)

[Out]

-A/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) + A*sqrt(b)/(12*a*x**(5/2)*sqrt(a/(b*x) + 1)) - 5*A*b**(3/2)/(24*a**
2*x**(3/2)*sqrt(a/(b*x) + 1)) - 5*A*b**(5/2)/(8*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) + 5*A*b**3*asinh(sqrt(a)/(sqrt
(b)*sqrt(x)))/(8*a**(7/2)) - B/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + B*sqrt(b)/(4*a*x**(3/2)*sqrt(a/(b*x) +
1)) + 3*B*b**(3/2)/(4*a**2*sqrt(x)*sqrt(a/(b*x) + 1)) - 3*B*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(5/2)
)

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Giac [A]  time = 1.25252, size = 194, normalized size = 1.69 \begin{align*} \frac{\frac{3 \,{\left (6 \, B a b^{3} - 5 \, A b^{4}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{18 \,{\left (b x + a\right )}^{\frac{5}{2}} B a b^{3} - 48 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{2} b^{3} + 30 \, \sqrt{b x + a} B a^{3} b^{3} - 15 \,{\left (b x + a\right )}^{\frac{5}{2}} A b^{4} + 40 \,{\left (b x + a\right )}^{\frac{3}{2}} A a b^{4} - 33 \, \sqrt{b x + a} A a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/24*(3*(6*B*a*b^3 - 5*A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (18*(b*x + a)^(5/2)*B*a*b^3 - 48
*(b*x + a)^(3/2)*B*a^2*b^3 + 30*sqrt(b*x + a)*B*a^3*b^3 - 15*(b*x + a)^(5/2)*A*b^4 + 40*(b*x + a)^(3/2)*A*a*b^
4 - 33*sqrt(b*x + a)*A*a^2*b^4)/(a^3*b^3*x^3))/b