3.5.7 \(\int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx\)

Optimal. Leaf size=146 \[ -\frac {5 b^3 (7 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{9/2}}+\frac {5 b^2 \sqrt {a+b x} (7 A b-8 a B)}{64 a^4 x}-\frac {5 b \sqrt {a+b x} (7 A b-8 a B)}{96 a^3 x^2}+\frac {\sqrt {a+b x} (7 A b-8 a B)}{24 a^2 x^3}-\frac {A \sqrt {a+b x}}{4 a x^4} \]

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Rubi [A]  time = 0.06, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 51, 63, 208} \begin {gather*} \frac {5 b^2 \sqrt {a+b x} (7 A b-8 a B)}{64 a^4 x}-\frac {5 b^3 (7 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{9/2}}-\frac {5 b \sqrt {a+b x} (7 A b-8 a B)}{96 a^3 x^2}+\frac {\sqrt {a+b x} (7 A b-8 a B)}{24 a^2 x^3}-\frac {A \sqrt {a+b x}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^5*Sqrt[a + b*x]),x]

[Out]

-(A*Sqrt[a + b*x])/(4*a*x^4) + ((7*A*b - 8*a*B)*Sqrt[a + b*x])/(24*a^2*x^3) - (5*b*(7*A*b - 8*a*B)*Sqrt[a + b*
x])/(96*a^3*x^2) + (5*b^2*(7*A*b - 8*a*B)*Sqrt[a + b*x])/(64*a^4*x) - (5*b^3*(7*A*b - 8*a*B)*ArcTanh[Sqrt[a +
b*x]/Sqrt[a]])/(64*a^(9/2))

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^5 \sqrt {a+b x}} \, dx &=-\frac {A \sqrt {a+b x}}{4 a x^4}+\frac {\left (-\frac {7 A b}{2}+4 a B\right ) \int \frac {1}{x^4 \sqrt {a+b x}} \, dx}{4 a}\\ &=-\frac {A \sqrt {a+b x}}{4 a x^4}+\frac {(7 A b-8 a B) \sqrt {a+b x}}{24 a^2 x^3}+\frac {(5 b (7 A b-8 a B)) \int \frac {1}{x^3 \sqrt {a+b x}} \, dx}{48 a^2}\\ &=-\frac {A \sqrt {a+b x}}{4 a x^4}+\frac {(7 A b-8 a B) \sqrt {a+b x}}{24 a^2 x^3}-\frac {5 b (7 A b-8 a B) \sqrt {a+b x}}{96 a^3 x^2}-\frac {\left (5 b^2 (7 A b-8 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{64 a^3}\\ &=-\frac {A \sqrt {a+b x}}{4 a x^4}+\frac {(7 A b-8 a B) \sqrt {a+b x}}{24 a^2 x^3}-\frac {5 b (7 A b-8 a B) \sqrt {a+b x}}{96 a^3 x^2}+\frac {5 b^2 (7 A b-8 a B) \sqrt {a+b x}}{64 a^4 x}+\frac {\left (5 b^3 (7 A b-8 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a^4}\\ &=-\frac {A \sqrt {a+b x}}{4 a x^4}+\frac {(7 A b-8 a B) \sqrt {a+b x}}{24 a^2 x^3}-\frac {5 b (7 A b-8 a B) \sqrt {a+b x}}{96 a^3 x^2}+\frac {5 b^2 (7 A b-8 a B) \sqrt {a+b x}}{64 a^4 x}+\frac {\left (5 b^2 (7 A b-8 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a^4}\\ &=-\frac {A \sqrt {a+b x}}{4 a x^4}+\frac {(7 A b-8 a B) \sqrt {a+b x}}{24 a^2 x^3}-\frac {5 b (7 A b-8 a B) \sqrt {a+b x}}{96 a^3 x^2}+\frac {5 b^2 (7 A b-8 a B) \sqrt {a+b x}}{64 a^4 x}-\frac {5 b^3 (7 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 57, normalized size = 0.39 \begin {gather*} -\frac {\sqrt {a+b x} \left (a^4 A+b^3 x^4 (7 A b-8 a B) \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};\frac {b x}{a}+1\right )\right )}{4 a^5 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^5*Sqrt[a + b*x]),x]

[Out]

-1/4*(Sqrt[a + b*x]*(a^4*A + b^3*(7*A*b - 8*a*B)*x^4*Hypergeometric2F1[1/2, 4, 3/2, 1 + (b*x)/a]))/(a^5*x^4)

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IntegrateAlgebraic [A]  time = 0.22, size = 146, normalized size = 1.00 \begin {gather*} \frac {5 \left (8 a b^3 B-7 A b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{9/2}}+\frac {\sqrt {a+b x} \left (264 a^4 B-279 a^3 A b-584 a^3 B (a+b x)+511 a^2 A b (a+b x)+440 a^2 B (a+b x)^2-385 a A b (a+b x)^2+105 A b (a+b x)^3-120 a B (a+b x)^3\right )}{192 a^4 b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^5*Sqrt[a + b*x]),x]

[Out]

(Sqrt[a + b*x]*(-279*a^3*A*b + 264*a^4*B + 511*a^2*A*b*(a + b*x) - 584*a^3*B*(a + b*x) - 385*a*A*b*(a + b*x)^2
+ 440*a^2*B*(a + b*x)^2 + 105*A*b*(a + b*x)^3 - 120*a*B*(a + b*x)^3))/(192*a^4*b*x^4) + (5*(-7*A*b^4 + 8*a*b^
3*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(9/2))

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fricas [A]  time = 1.39, size = 259, normalized size = 1.77 \begin {gather*} \left [-\frac {15 \, {\left (8 \, B a b^{3} - 7 \, A b^{4}\right )} \sqrt {a} x^{4} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{4} + 15 \, {\left (8 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 10 \, {\left (8 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{384 \, a^{5} x^{4}}, -\frac {15 \, {\left (8 \, B a b^{3} - 7 \, A b^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (48 \, A a^{4} + 15 \, {\left (8 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 10 \, {\left (8 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{192 \, a^{5} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(8*B*a*b^3 - 7*A*b^4)*sqrt(a)*x^4*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*A*a^4 + 15*
(8*B*a^2*b^2 - 7*A*a*b^3)*x^3 - 10*(8*B*a^3*b - 7*A*a^2*b^2)*x^2 + 8*(8*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(
a^5*x^4), -1/192*(15*(8*B*a*b^3 - 7*A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (48*A*a^4 + 15*(8*B
*a^2*b^2 - 7*A*a*b^3)*x^3 - 10*(8*B*a^3*b - 7*A*a^2*b^2)*x^2 + 8*(8*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*
x^4)]

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giac [A]  time = 1.34, size = 176, normalized size = 1.21 \begin {gather*} -\frac {\frac {15 \, {\left (8 \, B a b^{4} - 7 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {120 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{4} - 440 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{4} + 584 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{4} - 264 \, \sqrt {b x + a} B a^{4} b^{4} - 105 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{5} + 385 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{5} - 511 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{5} + 279 \, \sqrt {b x + a} A a^{3} b^{5}}{a^{4} b^{4} x^{4}}}{192 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/192*(15*(8*B*a*b^4 - 7*A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) + (120*(b*x + a)^(7/2)*B*a*b^4
- 440*(b*x + a)^(5/2)*B*a^2*b^4 + 584*(b*x + a)^(3/2)*B*a^3*b^4 - 264*sqrt(b*x + a)*B*a^4*b^4 - 105*(b*x + a)^
(7/2)*A*b^5 + 385*(b*x + a)^(5/2)*A*a*b^5 - 511*(b*x + a)^(3/2)*A*a^2*b^5 + 279*sqrt(b*x + a)*A*a^3*b^5)/(a^4*
b^4*x^4))/b

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maple [A]  time = 0.02, size = 125, normalized size = 0.86 \begin {gather*} 2 \left (-\frac {5 \left (7 A b -8 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {9}{2}}}+\frac {-\frac {\left (93 A b -88 B a \right ) \sqrt {b x +a}}{128 a}+\frac {73 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384 a^{2}}-\frac {55 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a^{3}}+\frac {5 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{4}}}{b^{4} x^{4}}\right ) b^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^5/(b*x+a)^(1/2),x)

[Out]

2*b^3*((5/128*(7*A*b-8*B*a)/a^4*(b*x+a)^(7/2)-55/384/a^3*(7*A*b-8*B*a)*(b*x+a)^(5/2)+73/384/a^2*(7*A*b-8*B*a)*
(b*x+a)^(3/2)-1/128*(93*A*b-88*B*a)/a*(b*x+a)^(1/2))/x^4/b^4-5/128*(7*A*b-8*B*a)/a^(9/2)*arctanh((b*x+a)^(1/2)
/a^(1/2)))

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maxima [A]  time = 1.97, size = 196, normalized size = 1.34 \begin {gather*} -\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (15 \, {\left (8 \, B a - 7 \, A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 55 \, {\left (8 \, B a^{2} - 7 \, A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 73 \, {\left (8 \, B a^{3} - 7 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 3 \, {\left (88 \, B a^{4} - 93 \, A a^{3} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{4} a^{4} b - 4 \, {\left (b x + a\right )}^{3} a^{5} b + 6 \, {\left (b x + a\right )}^{2} a^{6} b - 4 \, {\left (b x + a\right )} a^{7} b + a^{8} b} + \frac {15 \, {\left (8 \, B a - 7 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {9}{2}} b}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

-1/384*b^4*(2*(15*(8*B*a - 7*A*b)*(b*x + a)^(7/2) - 55*(8*B*a^2 - 7*A*a*b)*(b*x + a)^(5/2) + 73*(8*B*a^3 - 7*A
*a^2*b)*(b*x + a)^(3/2) - 3*(88*B*a^4 - 93*A*a^3*b)*sqrt(b*x + a))/((b*x + a)^4*a^4*b - 4*(b*x + a)^3*a^5*b +
6*(b*x + a)^2*a^6*b - 4*(b*x + a)*a^7*b + a^8*b) + 15*(8*B*a - 7*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x
+ a) + sqrt(a)))/(a^(9/2)*b))

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mupad [B]  time = 0.44, size = 181, normalized size = 1.24 \begin {gather*} \frac {\frac {73\,\left (7\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{3/2}}{192\,a^2}-\frac {55\,\left (7\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{5/2}}{192\,a^3}+\frac {5\,\left (7\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a^4}-\frac {\left (93\,A\,b^4-88\,B\,a\,b^3\right )\,\sqrt {a+b\,x}}{64\,a}}{{\left (a+b\,x\right )}^4-4\,a^3\,\left (a+b\,x\right )-4\,a\,{\left (a+b\,x\right )}^3+6\,a^2\,{\left (a+b\,x\right )}^2+a^4}-\frac {5\,b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (7\,A\,b-8\,B\,a\right )}{64\,a^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^5*(a + b*x)^(1/2)),x)

[Out]

((73*(7*A*b^4 - 8*B*a*b^3)*(a + b*x)^(3/2))/(192*a^2) - (55*(7*A*b^4 - 8*B*a*b^3)*(a + b*x)^(5/2))/(192*a^3) +
(5*(7*A*b^4 - 8*B*a*b^3)*(a + b*x)^(7/2))/(64*a^4) - ((93*A*b^4 - 88*B*a*b^3)*(a + b*x)^(1/2))/(64*a))/((a +
b*x)^4 - 4*a^3*(a + b*x) - 4*a*(a + b*x)^3 + 6*a^2*(a + b*x)^2 + a^4) - (5*b^3*atanh((a + b*x)^(1/2)/a^(1/2))*
(7*A*b - 8*B*a))/(64*a^(9/2))

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sympy [B]  time = 164.73, size = 303, normalized size = 2.08 \begin {gather*} - \frac {A}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A \sqrt {b}}{24 a x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {7 A b^{\frac {3}{2}}}{96 a^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {35 A b^{\frac {5}{2}}}{192 a^{3} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {35 A b^{\frac {7}{2}}}{64 a^{4} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {35 A b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {9}{2}}} - \frac {B}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {B \sqrt {b}}{12 a x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 B b^{\frac {3}{2}}}{24 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 B b^{\frac {5}{2}}}{8 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 B b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**5/(b*x+a)**(1/2),x)

[Out]

-A/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) + A*sqrt(b)/(24*a*x**(7/2)*sqrt(a/(b*x) + 1)) - 7*A*b**(3/2)/(96*a**
2*x**(5/2)*sqrt(a/(b*x) + 1)) + 35*A*b**(5/2)/(192*a**3*x**(3/2)*sqrt(a/(b*x) + 1)) + 35*A*b**(7/2)/(64*a**4*s
qrt(x)*sqrt(a/(b*x) + 1)) - 35*A*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a**(9/2)) - B/(3*sqrt(b)*x**(7/2)*s
qrt(a/(b*x) + 1)) + B*sqrt(b)/(12*a*x**(5/2)*sqrt(a/(b*x) + 1)) - 5*B*b**(3/2)/(24*a**2*x**(3/2)*sqrt(a/(b*x)
+ 1)) - 5*B*b**(5/2)/(8*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) + 5*B*b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(8*a**(7/2
))

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