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

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

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Rubi [A]  time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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^{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*}

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Mathematica [C]  time = 0.02, size = 44, normalized size = 0.49 \begin {gather*} -\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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.10, size = 91, normalized size = 1.01 \begin {gather*} \frac {2 \left (a b^{3/2} B-A b^{5/2}\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {2 \left (3 a^2 A+5 a^2 B x-5 a A b x-15 a b B x^2+15 A b^2 x^2\right )}{15 a^3 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*a^2*A - 5*a*A*b*x + 5*a^2*B*x + 15*A*b^2*x^2 - 15*a*b*B*x^2))/(15*a^3*x^(5/2)) + (2*(-(A*b^(5/2)) + a*b
^(3/2)*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(7/2)

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fricas [A]  time = 0.89, size = 195, normalized size = 2.17 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.33, size = 80, normalized size = 0.89 \begin {gather*} \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 {gather*}

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))

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maple [A]  time = 0.01, size = 102, normalized size = 1.13 \begin {gather*} -\frac {2 A \,b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}+\frac {2 B \,b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}-\frac {2 A \,b^{2}}{a^{3} \sqrt {x}}+\frac {2 B b}{a^{2} \sqrt {x}}+\frac {2 A b}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 B}{3 a \,x^{\frac {3}{2}}}-\frac {2 A}{5 a \,x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.81, size = 80, normalized size = 0.89 \begin {gather*} \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 (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 )}}{15 \, a^{3} x^{\frac {5}{2}}} \end {gather*}

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]

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

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mupad [B]  time = 0.40, size = 71, normalized size = 0.79 \begin {gather*} -\frac {\frac {2\,A}{5\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{3\,a^2}+\frac {2\,b\,x^2\,\left (A\,b-B\,a\right )}{a^3}}{x^{5/2}}-\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(7/2)*(a + b*x)),x)

[Out]

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

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sympy [A]  time = 29.13, size = 289, normalized size = 3.21 \begin {gather*} \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 {gather*}

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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