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

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

[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)

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Rubi [A]  time = 0.0634274, antiderivative size = 113, normalized size of antiderivative = 1., 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 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 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*}

Mathematica [C]  time = 0.0134733, 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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Maple [A]  time = 0.011, size = 126, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{7\,a}{x}^{-{\frac{7}{2}}}}+{\frac{2\,Ab}{5\,{a}^{2}}{x}^{-{\frac{5}{2}}}}-{\frac{2\,B}{5\,a}{x}^{-{\frac{5}{2}}}}-{\frac{2\,{b}^{2}A}{3\,{a}^{3}}{x}^{-{\frac{3}{2}}}}+{\frac{2\,Bb}{3\,{a}^{2}}{x}^{-{\frac{3}{2}}}}+2\,{\frac{{b}^{3}A}{{a}^{4}\sqrt{x}}}-2\,{\frac{{b}^{2}B}{{a}^{3}\sqrt{x}}}+2\,{\frac{A{b}^{4}}{{a}^{4}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) }-2\,{\frac{{b}^{3}B}{{a}^{3}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^(9/2)/(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.46819, size = 548, normalized size = 4.85 \begin{align*} \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 ] \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.18023, size = 140, normalized size = 1.24 \begin{align*} -\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}}} \end{align*}

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))