3.173 \(\int \frac {A+B x}{x^{3/2} (b x+c x^2)} \, dx\)

Optimal. Leaf size=69 \[ -\frac {2 \sqrt {c} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}}-\frac {2 (b B-A c)}{b^2 \sqrt {x}}-\frac {2 A}{3 b x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {2 (b B-A c)}{b^2 \sqrt {x}}-\frac {2 \sqrt {c} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}}-\frac {2 A}{3 b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(3/2)*(b*x + c*x^2)),x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 44, normalized size = 0.64 \[ \frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {c x}{b}\right ) (6 A c x-6 b B x)-2 A b}{3 b^2 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(3/2)*(b*x + c*x^2)),x]

[Out]

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

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fricas [A]  time = 0.88, size = 146, normalized size = 2.12 \[ \left [-\frac {3 \, {\left (B b - A c\right )} x^{2} \sqrt {-\frac {c}{b}} \log \left (\frac {c x + 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (A b + 3 \, {\left (B b - A c\right )} x\right )} \sqrt {x}}{3 \, b^{2} x^{2}}, \frac {2 \, {\left (3 \, {\left (B b - A c\right )} x^{2} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) - {\left (A b + 3 \, {\left (B b - A c\right )} x\right )} \sqrt {x}\right )}}{3 \, b^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 55, normalized size = 0.80 \[ -\frac {2 \, {\left (B b c - A c^{2}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{2}} - \frac {2 \, {\left (3 \, B b x - 3 \, A c x + A b\right )}}{3 \, b^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 78, normalized size = 1.13 \[ \frac {2 A \,c^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{2}}-\frac {2 B c \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b}+\frac {2 A c}{b^{2} \sqrt {x}}-\frac {2 B}{b \sqrt {x}}-\frac {2 A}{3 b \,x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(3/2)/(c*x^2+b*x),x)

[Out]

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

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maxima [A]  time = 1.28, size = 56, normalized size = 0.81 \[ -\frac {2 \, {\left (B b c - A c^{2}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{2}} - \frac {2 \, {\left (A b + 3 \, {\left (B b - A c\right )} x\right )}}{3 \, b^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.07, size = 54, normalized size = 0.78 \[ \frac {2\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{5/2}}-\frac {\frac {2\,A}{3\,b}-\frac {2\,x\,\left (A\,c-B\,b\right )}{b^2}}{x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(3/2)*(b*x + c*x^2)),x)

[Out]

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

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sympy [A]  time = 7.01, size = 248, normalized size = 3.59 \[ \begin {cases} \tilde {\infty } \left (- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{b} & \text {for}\: c = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{c} & \text {for}\: b = 0 \\- \frac {2 A}{3 b x^{\frac {3}{2}}} + \frac {2 A c}{b^{2} \sqrt {x}} - \frac {i A c \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {5}{2}} \sqrt {\frac {1}{c}}} + \frac {i A c \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {5}{2}} \sqrt {\frac {1}{c}}} - \frac {2 B}{b \sqrt {x}} + \frac {i B \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {3}{2}} \sqrt {\frac {1}{c}}} - \frac {i B \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {3}{2}} \sqrt {\frac {1}{c}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(3/2)/(c*x**2+b*x),x)

[Out]

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

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