3.352 \(\int \frac {A+B x}{x^{9/2} (a+b x)} \, dx\)

Optimal. Leaf size=113 \[ \frac {2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{9/2}}+\frac {2 b^2 (A b-a B)}{a^4 \sqrt {x}}-\frac {2 b (A b-a B)}{3 a^3 x^{3/2}}+\frac {2 (A b-a B)}{5 a^2 x^{5/2}}-\frac {2 A}{7 a x^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 113, 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, 205} \[ \frac {2 b^2 (A b-a B)}{a^4 \sqrt {x}}+\frac {2 b^{5/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{9/2}}-\frac {2 b (A b-a B)}{3 a^3 x^{3/2}}+\frac {2 (A b-a B)}{5 a^2 x^{5/2}}-\frac {2 A}{7 a x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(9/2)*(a + b*x)),x]

[Out]

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

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Mathematica [C]  time = 0.02, size = 44, normalized size = 0.39 \[ -\frac {2 \left (\, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\frac {b x}{a}\right ) (7 a B x-7 A b x)+5 a A\right )}{35 a^2 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(9/2)*(a + b*x)),x]

[Out]

(-2*(5*a*A + (-7*A*b*x + 7*a*B*x)*Hypergeometric2F1[-5/2, 1, -3/2, -((b*x)/a)]))/(35*a^2*x^(7/2))

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fricas [A]  time = 0.78, size = 246, normalized size = 2.18 \[ \left [-\frac {105 \, {\left (B a b^{2} - A b^{3}\right )} x^{4} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (15 \, A a^{3} + 105 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} - 35 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} + 21 \, {\left (B a^{3} - A a^{2} b\right )} x\right )} \sqrt {x}}{105 \, a^{4} x^{4}}, \frac {2 \, {\left (105 \, {\left (B a b^{2} - A b^{3}\right )} x^{4} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, A a^{3} + 105 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} - 35 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} + 21 \, {\left (B a^{3} - A a^{2} b\right )} x\right )} \sqrt {x}\right )}}{105 \, a^{4} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a),x, algorithm="fricas")

[Out]

[-1/105*(105*(B*a*b^2 - A*b^3)*x^4*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(15*A*a^3
+ 105*(B*a*b^2 - A*b^3)*x^3 - 35*(B*a^2*b - A*a*b^2)*x^2 + 21*(B*a^3 - A*a^2*b)*x)*sqrt(x))/(a^4*x^4), 2/105*(
105*(B*a*b^2 - A*b^3)*x^4*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*A*a^3 + 105*(B*a*b^2 - A*b^3)*x^3 -
35*(B*a^2*b - A*a*b^2)*x^2 + 21*(B*a^3 - A*a^2*b)*x)*sqrt(x))/(a^4*x^4)]

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giac [A]  time = 1.32, size = 104, normalized size = 0.92 \[ -\frac {2 \, {\left (B a b^{3} - A b^{4}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} - \frac {2 \, {\left (105 \, B a b^{2} x^{3} - 105 \, A b^{3} x^{3} - 35 \, B a^{2} b x^{2} + 35 \, A a b^{2} x^{2} + 21 \, B a^{3} x - 21 \, A a^{2} b x + 15 \, A a^{3}\right )}}{105 \, a^{4} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a),x, algorithm="giac")

[Out]

-2*(B*a*b^3 - A*b^4)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - 2/105*(105*B*a*b^2*x^3 - 105*A*b^3*x^3 - 35
*B*a^2*b*x^2 + 35*A*a*b^2*x^2 + 21*B*a^3*x - 21*A*a^2*b*x + 15*A*a^3)/(a^4*x^(7/2))

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maple [A]  time = 0.01, size = 126, normalized size = 1.12 \[ \frac {2 A \,b^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{4}}-\frac {2 B \,b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}+\frac {2 A \,b^{3}}{a^{4} \sqrt {x}}-\frac {2 B \,b^{2}}{a^{3} \sqrt {x}}-\frac {2 A \,b^{2}}{3 a^{3} x^{\frac {3}{2}}}+\frac {2 B b}{3 a^{2} x^{\frac {3}{2}}}+\frac {2 A b}{5 a^{2} x^{\frac {5}{2}}}-\frac {2 B}{5 a \,x^{\frac {5}{2}}}-\frac {2 A}{7 a \,x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(9/2)/(b*x+a),x)

[Out]

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

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maxima [A]  time = 1.98, size = 103, normalized size = 0.91 \[ -\frac {2 \, {\left (B a b^{3} - A b^{4}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} - \frac {2 \, {\left (15 \, A a^{3} + 105 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} - 35 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} + 21 \, {\left (B a^{3} - A a^{2} b\right )} x\right )}}{105 \, a^{4} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(b*x+a),x, algorithm="maxima")

[Out]

-2*(B*a*b^3 - A*b^4)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - 2/105*(15*A*a^3 + 105*(B*a*b^2 - A*b^3)*x^3
- 35*(B*a^2*b - A*a*b^2)*x^2 + 21*(B*a^3 - A*a^2*b)*x)/(a^4*x^(7/2))

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mupad [B]  time = 0.42, size = 90, normalized size = 0.80 \[ \frac {2\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{9/2}}-\frac {\frac {2\,A}{7\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{5\,a^2}-\frac {2\,b^2\,x^3\,\left (A\,b-B\,a\right )}{a^4}+\frac {2\,b\,x^2\,\left (A\,b-B\,a\right )}{3\,a^3}}{x^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(9/2)*(a + b*x)),x)

[Out]

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

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sympy [A]  time = 79.17, size = 326, normalized size = 2.88 \[ \begin {cases} \tilde {\infty } \left (- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{7 a x^{\frac {7}{2}}} + \frac {2 A b}{5 a^{2} x^{\frac {5}{2}}} - \frac {2 A b^{2}}{3 a^{3} x^{\frac {3}{2}}} + \frac {2 A b^{3}}{a^{4} \sqrt {x}} - \frac {i A b^{3} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {9}{2}} \sqrt {\frac {1}{b}}} + \frac {i A b^{3} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {9}{2}} \sqrt {\frac {1}{b}}} - \frac {2 B}{5 a x^{\frac {5}{2}}} + \frac {2 B b}{3 a^{2} x^{\frac {3}{2}}} - \frac {2 B b^{2}}{a^{3} \sqrt {x}} + \frac {i B b^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {i B b^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {7}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(9/2)/(b*x+a),x)

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(
7/2)))/b, Eq(a, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2)))/a, Eq(b, 0)), (-2*A/(7*a*x**(7/2)) + 2*A*b/(5*a**
2*x**(5/2)) - 2*A*b**2/(3*a**3*x**(3/2)) + 2*A*b**3/(a**4*sqrt(x)) - I*A*b**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt(
x))/(a**(9/2)*sqrt(1/b)) + I*A*b**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(9/2)*sqrt(1/b)) - 2*B/(5*a*x**(5/2
)) + 2*B*b/(3*a**2*x**(3/2)) - 2*B*b**2/(a**3*sqrt(x)) + I*B*b**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(7/2
)*sqrt(1/b)) - I*B*b**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(7/2)*sqrt(1/b)), True))

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