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

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

[Out]

1/8*b^2*(5*A*b-6*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(7/2)-1/3*A*(b*x+a)^(1/2)/x^3/a+1/12*(5*A*b-6*B*a)*(b*x
+a)^(1/2)/a^2/x^2-1/8*b*(5*A*b-6*B*a)*(b*x+a)^(1/2)/a^3/x

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Rubi [A]  time = 0.05, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.50 \[ -\frac {\sqrt {a+b x} \left (a^3 A+b^2 x^3 (6 a B-5 A b) \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b x}{a}+1\right )\right )}{3 a^4 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(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]))/(a^4*x^3)

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fricas [A]  time = 0.68, size = 212, normalized size = 1.84 \[ \left [-\frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt {a} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, A a^{3} - 3 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{48 \, a^{4} x^{3}}, \frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (8 \, A a^{3} - 3 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{24 \, a^{4} x^{3}}\right ] \]

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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giac [A]  time = 1.38, size = 144, normalized size = 1.25 \[ \frac {\frac {3 \, {\left (6 \, B a b^{3} - 5 \, A b^{4}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {18 \, {\left (b x + a\right )}^{\frac {5}{2}} B a b^{3} - 48 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{2} b^{3} + 30 \, \sqrt {b x + a} B a^{3} b^{3} - 15 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{4} + 40 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{4} - 33 \, \sqrt {b x + a} A a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \]

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

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maple [A]  time = 0.01, size = 104, normalized size = 0.90 \[ 2 \left (\frac {\left (5 A b -6 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {7}{2}}}+\frac {-\frac {\left (11 A b -10 B a \right ) \sqrt {b x +a}}{16 a}+\frac {\left (5 A b -6 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{6 a^{2}}-\frac {\left (5 A b -6 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{16 a^{3}}}{b^{3} x^{3}}\right ) b^{2} \]

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))/x^3/b^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 [A]  time = 1.95, size = 161, normalized size = 1.40 \[ \frac {1}{48} \, b^{3} {\left (\frac {2 \, {\left (3 \, {\left (6 \, B a - 5 \, A b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 8 \, {\left (6 \, B a^{2} - 5 \, A a b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 3 \, {\left (10 \, B a^{3} - 11 \, A a^{2} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{3} a^{3} b - 3 \, {\left (b x + a\right )}^{2} a^{4} b + 3 \, {\left (b x + a\right )} a^{5} b - a^{6} b} + \frac {3 \, {\left (6 \, B a - 5 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}} b}\right )} \]

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]

1/48*b^3*(2*(3*(6*B*a - 5*A*b)*(b*x + a)^(5/2) - 8*(6*B*a^2 - 5*A*a*b)*(b*x + a)^(3/2) + 3*(10*B*a^3 - 11*A*a^
2*b)*sqrt(b*x + a))/((b*x + a)^3*a^3*b - 3*(b*x + a)^2*a^4*b + 3*(b*x + a)*a^5*b - a^6*b) + 3*(6*B*a - 5*A*b)*
log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(7/2)*b))

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mupad [B]  time = 0.12, size = 145, normalized size = 1.26 \[ \frac {\frac {\left (5\,A\,b^3-6\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^{5/2}}{8\,a^3}-\frac {\left (5\,A\,b^3-6\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^{3/2}}{3\,a^2}+\frac {\left (11\,A\,b^3-10\,B\,a\,b^2\right )\,\sqrt {a+b\,x}}{8\,a}}{3\,a\,{\left (a+b\,x\right )}^2-3\,a^2\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^3+a^3}+\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (5\,A\,b-6\,B\,a\right )}{8\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((5*A*b^3 - 6*B*a*b^2)*(a + b*x)^(5/2))/(8*a^3) - ((5*A*b^3 - 6*B*a*b^2)*(a + b*x)^(3/2))/(3*a^2) + ((11*A*b^
3 - 10*B*a*b^2)*(a + b*x)^(1/2))/(8*a))/(3*a*(a + b*x)^2 - 3*a^2*(a + b*x) - (a + b*x)^3 + a^3) + (b^2*atanh((
a + b*x)^(1/2)/a^(1/2))*(5*A*b - 6*B*a))/(8*a^(7/2))

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sympy [B]  time = 108.13, size = 245, normalized size = 2.13 \[ - \frac {A}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A \sqrt {b}}{12 a x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A b^{\frac {3}{2}}}{24 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A b^{\frac {5}{2}}}{8 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 A b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {7}{2}}} - \frac {B}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {B \sqrt {b}}{4 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {3 B b^{\frac {3}{2}}}{4 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {3 B b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {5}{2}}} \]

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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