3.4.39 \(\int \frac {A+B x}{x^{11/2} (a+b x)} \, dx\)

Optimal. Leaf size=136 \[ -\frac {2 b^{7/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 A}{9 a x^{9/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 51, 63, 205} \begin {gather*} \frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {2 b^{7/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 A}{9 a x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(11/2)*(a + b*x)),x]

[Out]

(-2*A)/(9*a*x^(9/2)) + (2*(A*b - a*B))/(7*a^2*x^(7/2)) - (2*b*(A*b - a*B))/(5*a^3*x^(5/2)) + (2*b^2*(A*b - a*B
))/(3*a^4*x^(3/2)) - (2*b^3*(A*b - a*B))/(a^5*Sqrt[x]) - (2*b^(7/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[
a]])/a^(11/2)

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 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^{11/2} (a+b x)} \, dx &=-\frac {2 A}{9 a x^{9/2}}+\frac {\left (2 \left (-\frac {9 A b}{2}+\frac {9 a B}{2}\right )\right ) \int \frac {1}{x^{9/2} (a+b x)} \, dx}{9 a}\\ &=-\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}+\frac {(b (A b-a B)) \int \frac {1}{x^{7/2} (a+b x)} \, dx}{a^2}\\ &=-\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}-\frac {\left (b^2 (A b-a B)\right ) \int \frac {1}{x^{5/2} (a+b x)} \, dx}{a^3}\\ &=-\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}+\frac {\left (b^3 (A b-a B)\right ) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{a^4}\\ &=-\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {\left (b^4 (A b-a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{a^5}\\ &=-\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {\left (2 b^4 (A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{a^5}\\ &=-\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {2 b^{7/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 44, normalized size = 0.32 \begin {gather*} -\frac {2 \left (\, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};-\frac {b x}{a}\right ) (9 a B x-9 A b x)+7 a A\right )}{63 a^2 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(11/2)*(a + b*x)),x]

[Out]

(-2*(7*a*A + (-9*A*b*x + 9*a*B*x)*Hypergeometric2F1[-7/2, 1, -5/2, -((b*x)/a)]))/(63*a^2*x^(9/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.14, size = 139, normalized size = 1.02 \begin {gather*} \frac {2 \left (a b^{7/2} B-A b^{9/2}\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}}-\frac {2 \left (35 a^4 A+45 a^4 B x-45 a^3 A b x-63 a^3 b B x^2+63 a^2 A b^2 x^2+105 a^2 b^2 B x^3-105 a A b^3 x^3-315 a b^3 B x^4+315 A b^4 x^4\right )}{315 a^5 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(11/2)*(a + b*x)),x]

[Out]

(-2*(35*a^4*A - 45*a^3*A*b*x + 45*a^4*B*x + 63*a^2*A*b^2*x^2 - 63*a^3*b*B*x^2 - 105*a*A*b^3*x^3 + 105*a^2*b^2*
B*x^3 + 315*A*b^4*x^4 - 315*a*b^3*B*x^4))/(315*a^5*x^(9/2)) + (2*(-(A*b^(9/2)) + a*b^(7/2)*B)*ArcTan[(Sqrt[b]*
Sqrt[x])/Sqrt[a]])/a^(11/2)

________________________________________________________________________________________

fricas [A]  time = 1.02, size = 291, normalized size = 2.14 \begin {gather*} \left [-\frac {315 \, {\left (B a b^{3} - A b^{4}\right )} x^{5} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (35 \, A a^{4} - 315 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} + 105 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 63 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 45 \, {\left (B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}}{315 \, a^{5} x^{5}}, -\frac {2 \, {\left (315 \, {\left (B a b^{3} - A b^{4}\right )} x^{5} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (35 \, A a^{4} - 315 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} + 105 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 63 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 45 \, {\left (B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}\right )}}{315 \, a^{5} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a),x, algorithm="fricas")

[Out]

[-1/315*(315*(B*a*b^3 - A*b^4)*x^5*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(35*A*a^4
- 315*(B*a*b^3 - A*b^4)*x^4 + 105*(B*a^2*b^2 - A*a*b^3)*x^3 - 63*(B*a^3*b - A*a^2*b^2)*x^2 + 45*(B*a^4 - A*a^3
*b)*x)*sqrt(x))/(a^5*x^5), -2/315*(315*(B*a*b^3 - A*b^4)*x^5*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) + (35*A
*a^4 - 315*(B*a*b^3 - A*b^4)*x^4 + 105*(B*a^2*b^2 - A*a*b^3)*x^3 - 63*(B*a^3*b - A*a^2*b^2)*x^2 + 45*(B*a^4 -
A*a^3*b)*x)*sqrt(x))/(a^5*x^5)]

________________________________________________________________________________________

giac [A]  time = 1.25, size = 128, normalized size = 0.94 \begin {gather*} \frac {2 \, {\left (B a b^{4} - A b^{5}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} + \frac {2 \, {\left (315 \, B a b^{3} x^{4} - 315 \, A b^{4} x^{4} - 105 \, B a^{2} b^{2} x^{3} + 105 \, A a b^{3} x^{3} + 63 \, B a^{3} b x^{2} - 63 \, A a^{2} b^{2} x^{2} - 45 \, B a^{4} x + 45 \, A a^{3} b x - 35 \, A a^{4}\right )}}{315 \, a^{5} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a),x, algorithm="giac")

[Out]

2*(B*a*b^4 - A*b^5)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) + 2/315*(315*B*a*b^3*x^4 - 315*A*b^4*x^4 - 105
*B*a^2*b^2*x^3 + 105*A*a*b^3*x^3 + 63*B*a^3*b*x^2 - 63*A*a^2*b^2*x^2 - 45*B*a^4*x + 45*A*a^3*b*x - 35*A*a^4)/(
a^5*x^(9/2))

________________________________________________________________________________________

maple [A]  time = 0.02, size = 150, normalized size = 1.10 \begin {gather*} -\frac {2 A \,b^{5} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{5}}+\frac {2 B \,b^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{4}}-\frac {2 A \,b^{4}}{a^{5} \sqrt {x}}+\frac {2 B \,b^{3}}{a^{4} \sqrt {x}}+\frac {2 A \,b^{3}}{3 a^{4} x^{\frac {3}{2}}}-\frac {2 B \,b^{2}}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 A \,b^{2}}{5 a^{3} x^{\frac {5}{2}}}+\frac {2 B b}{5 a^{2} x^{\frac {5}{2}}}+\frac {2 A b}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 B}{7 a \,x^{\frac {7}{2}}}-\frac {2 A}{9 a \,x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(11/2)/(b*x+a),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.04, size = 126, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (B a b^{4} - A b^{5}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} - \frac {2 \, {\left (35 \, A a^{4} - 315 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} + 105 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 63 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 45 \, {\left (B a^{4} - A a^{3} b\right )} x\right )}}{315 \, a^{5} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a),x, algorithm="maxima")

[Out]

2*(B*a*b^4 - A*b^5)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 2/315*(35*A*a^4 - 315*(B*a*b^3 - A*b^4)*x^4
+ 105*(B*a^2*b^2 - A*a*b^3)*x^3 - 63*(B*a^3*b - A*a^2*b^2)*x^2 + 45*(B*a^4 - A*a^3*b)*x)/(a^5*x^(9/2))

________________________________________________________________________________________

mupad [B]  time = 0.45, size = 109, normalized size = 0.80 \begin {gather*} -\frac {\frac {2\,A}{9\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{7\,a^2}-\frac {2\,b^2\,x^3\,\left (A\,b-B\,a\right )}{3\,a^4}+\frac {2\,b^3\,x^4\,\left (A\,b-B\,a\right )}{a^5}+\frac {2\,b\,x^2\,\left (A\,b-B\,a\right )}{5\,a^3}}{x^{9/2}}-\frac {2\,b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(11/2)*(a + b*x)),x)

[Out]

- ((2*A)/(9*a) - (2*x*(A*b - B*a))/(7*a^2) - (2*b^2*x^3*(A*b - B*a))/(3*a^4) + (2*b^3*x^4*(A*b - B*a))/a^5 + (
2*b*x^2*(A*b - B*a))/(5*a^3))/x^(9/2) - (2*b^(7/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b - B*a))/a^(11/2)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(11/2)/(b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________