3.351 \(\int \frac{A+B x}{x^{7/2} (a+b x)} \, dx\)

Optimal. Leaf size=90 \[ -\frac{2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{2 (A b-a B)}{3 a^2 x^{3/2}}-\frac{2 b (A b-a B)}{a^3 \sqrt{x}}-\frac{2 A}{5 a x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0481859, antiderivative size = 90, normalized size of antiderivative = 1., 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, 205} \[ -\frac{2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{2 (A b-a B)}{3 a^2 x^{3/2}}-\frac{2 b (A b-a B)}{a^3 \sqrt{x}}-\frac{2 A}{5 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(a + b*x)),x]

[Out]

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

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

Mathematica [C]  time = 0.0129564, size = 44, normalized size = 0.49 \[ -\frac{2 \left (\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{b x}{a}\right ) (5 a B x-5 A b x)+3 a A\right )}{15 a^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(a + b*x)),x]

[Out]

(-2*(3*a*A + (-5*A*b*x + 5*a*B*x)*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x)/a)]))/(15*a^2*x^(5/2))

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Maple [A]  time = 0.012, size = 102, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{5\,a}{x}^{-{\frac{5}{2}}}}+{\frac{2\,Ab}{3\,{a}^{2}}{x}^{-{\frac{3}{2}}}}-{\frac{2\,B}{3\,a}{x}^{-{\frac{3}{2}}}}-2\,{\frac{{b}^{2}A}{{a}^{3}\sqrt{x}}}+2\,{\frac{Bb}{{a}^{2}\sqrt{x}}}-2\,{\frac{A{b}^{3}}{{a}^{3}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) }+2\,{\frac{{b}^{2}B}{{a}^{2}\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^(7/2)/(b*x+a),x)

[Out]

-2/5*A/a/x^(5/2)+2/3/a^2/x^(3/2)*A*b-2/3/a/x^(3/2)*B-2/a^3*b^2/x^(1/2)*A+2/a^2*b/x^(1/2)*B-2*b^3/a^3/(a*b)^(1/
2)*arctan(b*x^(1/2)/(a*b)^(1/2))*A+2*b^2/a^2/(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^(7/2)/(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.42904, size = 441, normalized size = 4.9 \begin{align*} \left [-\frac{15 \,{\left (B a b - A b^{2}\right )} x^{3} \sqrt{-\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) + 2 \,{\left (3 \, A a^{2} - 15 \,{\left (B a b - A b^{2}\right )} x^{2} + 5 \,{\left (B a^{2} - A a b\right )} x\right )} \sqrt{x}}{15 \, a^{3} x^{3}}, -\frac{2 \,{\left (15 \,{\left (B a b - A b^{2}\right )} x^{3} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) +{\left (3 \, A a^{2} - 15 \,{\left (B a b - A b^{2}\right )} x^{2} + 5 \,{\left (B a^{2} - A a b\right )} x\right )} \sqrt{x}\right )}}{15 \, a^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a),x, algorithm="fricas")

[Out]

[-1/15*(15*(B*a*b - A*b^2)*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(3*A*a^2 - 15*
(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)*sqrt(x))/(a^3*x^3), -2/15*(15*(B*a*b - A*b^2)*x^3*sqrt(b/a)*arctan(
a*sqrt(b/a)/(b*sqrt(x))) + (3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)*sqrt(x))/(a^3*x^3)]

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Sympy [A]  time = 58.7907, size = 289, normalized size = 3.21 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{2 A}{7 x^{\frac{7}{2}}} - \frac{2 B}{5 x^{\frac{5}{2}}}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{- \frac{2 A}{7 x^{\frac{7}{2}}} - \frac{2 B}{5 x^{\frac{5}{2}}}}{b} & \text{for}\: a = 0 \\\frac{- \frac{2 A}{5 x^{\frac{5}{2}}} - \frac{2 B}{3 x^{\frac{3}{2}}}}{a} & \text{for}\: b = 0 \\- \frac{2 A}{5 a x^{\frac{5}{2}}} + \frac{2 A b}{3 a^{2} x^{\frac{3}{2}}} - \frac{2 A b^{2}}{a^{3} \sqrt{x}} + \frac{i A b^{2} \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{a^{\frac{7}{2}} \sqrt{\frac{1}{b}}} - \frac{i A b^{2} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{a^{\frac{7}{2}} \sqrt{\frac{1}{b}}} - \frac{2 B}{3 a x^{\frac{3}{2}}} + \frac{2 B b}{a^{2} \sqrt{x}} - \frac{i B b \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{a^{\frac{5}{2}} \sqrt{\frac{1}{b}}} + \frac{i B b \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{a^{\frac{5}{2}} \sqrt{\frac{1}{b}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(7/2)/(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.16639, size = 108, normalized size = 1.2 \begin{align*} \frac{2 \,{\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{3}} + \frac{2 \,{\left (15 \, B a b x^{2} - 15 \, A b^{2} x^{2} - 5 \, B a^{2} x + 5 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{3} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a),x, algorithm="giac")

[Out]

2*(B*a*b^2 - A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 2/15*(15*B*a*b*x^2 - 15*A*b^2*x^2 - 5*B*a^2*
x + 5*A*a*b*x - 3*A*a^2)/(a^3*x^(5/2))