3.2.74 \(\int \frac {A+B x}{x^{5/2} (b x+c x^2)} \, dx\)

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

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Rubi [A]  time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {781, 78, 51, 63, 205} \begin {gather*} \frac {2 c^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{7/2}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}+\frac {2 c (b B-A c)}{b^3 \sqrt {x}}-\frac {2 A}{5 b x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(5/2)*(b*x + c*x^2)),x]

[Out]

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

Rule 781

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 43, normalized size = 0.48 \begin {gather*} \frac {-10 x (b B-A c) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {c x}{b}\right )-6 A b}{15 b^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(5/2)*(b*x + c*x^2)),x]

[Out]

(-6*A*b - 10*(b*B - A*c)*x*Hypergeometric2F1[-3/2, 1, -1/2, -((c*x)/b)])/(15*b^2*x^(5/2))

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IntegrateAlgebraic [A]  time = 0.11, size = 91, normalized size = 1.01 \begin {gather*} \frac {2 \left (b B c^{3/2}-A c^{5/2}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{7/2}}-\frac {2 \left (3 A b^2-5 A b c x+15 A c^2 x^2+5 b^2 B x-15 b B c x^2\right )}{15 b^3 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(5/2)*(b*x + c*x^2)),x]

[Out]

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

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fricas [A]  time = 0.42, size = 195, normalized size = 2.17 \begin {gather*} \left [-\frac {15 \, {\left (B b c - A c^{2}\right )} x^{3} \sqrt {-\frac {c}{b}} \log \left (\frac {c x - 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (3 \, A b^{2} - 15 \, {\left (B b c - A c^{2}\right )} x^{2} + 5 \, {\left (B b^{2} - A b c\right )} x\right )} \sqrt {x}}{15 \, b^{3} x^{3}}, -\frac {2 \, {\left (15 \, {\left (B b c - A c^{2}\right )} x^{3} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) + {\left (3 \, A b^{2} - 15 \, {\left (B b c - A c^{2}\right )} x^{2} + 5 \, {\left (B b^{2} - A b c\right )} x\right )} \sqrt {x}\right )}}{15 \, b^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 80, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} + \frac {2 \, {\left (15 \, B b c x^{2} - 15 \, A c^{2} x^{2} - 5 \, B b^{2} x + 5 \, A b c x - 3 \, A b^{2}\right )}}{15 \, b^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 102, normalized size = 1.13 \begin {gather*} -\frac {2 A \,c^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{3}}+\frac {2 B \,c^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{2}}-\frac {2 A \,c^{2}}{b^{3} \sqrt {x}}+\frac {2 B c}{b^{2} \sqrt {x}}+\frac {2 A c}{3 b^{2} x^{\frac {3}{2}}}-\frac {2 B}{3 b \,x^{\frac {3}{2}}}-\frac {2 A}{5 b \,x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(5/2)/(c*x^2+b*x),x)

[Out]

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

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maxima [A]  time = 1.36, size = 80, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} - \frac {2 \, {\left (3 \, A b^{2} - 15 \, {\left (B b c - A c^{2}\right )} x^{2} + 5 \, {\left (B b^{2} - A b c\right )} x\right )}}{15 \, b^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.09, size = 71, normalized size = 0.79 \begin {gather*} -\frac {\frac {2\,A}{5\,b}-\frac {2\,x\,\left (A\,c-B\,b\right )}{3\,b^2}+\frac {2\,c\,x^2\,\left (A\,c-B\,b\right )}{b^3}}{x^{5/2}}-\frac {2\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(5/2)*(b*x + c*x^2)),x)

[Out]

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

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sympy [A]  time = 18.37, 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}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}}{c} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{b} & \text {for}\: c = 0 \\- \frac {2 A}{5 b x^{\frac {5}{2}}} + \frac {2 A c}{3 b^{2} x^{\frac {3}{2}}} - \frac {2 A c^{2}}{b^{3} \sqrt {x}} + \frac {i A c^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {7}{2}} \sqrt {\frac {1}{c}}} - \frac {i A c^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {7}{2}} \sqrt {\frac {1}{c}}} - \frac {2 B}{3 b x^{\frac {3}{2}}} + \frac {2 B c}{b^{2} \sqrt {x}} - \frac {i B c \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {5}{2}} \sqrt {\frac {1}{c}}} + \frac {i B c \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {5}{2}} \sqrt {\frac {1}{c}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(5/2)/(c*x**2+b*x),x)

[Out]

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

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