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

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

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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, 80, 50, 63, 205} \begin {gather*} -\frac {2 \sqrt {x} (b B-A c)}{c^2}+\frac {2 \sqrt {b} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}}+\frac {2 B x^{3/2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

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Mathematica [A]  time = 0.03, size = 63, normalized size = 0.91 \begin {gather*} \frac {2 \sqrt {b} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}}+\frac {2 \sqrt {x} (3 A c-3 b B+B c x)}{3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 76, normalized size = 1.10 \begin {gather*} \frac {2 \left (b^{3/2} B-A \sqrt {b} c\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}}+\frac {2 \left (3 A c \sqrt {x}-3 b B \sqrt {x}+B c x^{3/2}\right )}{3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 129, normalized size = 1.87 \begin {gather*} \left [-\frac {3 \, {\left (B b - A c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x - 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (B c x - 3 \, B b + 3 \, A c\right )} \sqrt {x}}{3 \, c^{2}}, \frac {2 \, {\left (3 \, {\left (B b - A c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) + {\left (B c x - 3 \, B b + 3 \, A c\right )} \sqrt {x}\right )}}{3 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 64, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (B b^{2} - A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} + \frac {2 \, {\left (B c^{2} x^{\frac {3}{2}} - 3 \, B b c \sqrt {x} + 3 \, A c^{2} \sqrt {x}\right )}}{3 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 78, normalized size = 1.13 \begin {gather*} -\frac {2 A b \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c}+\frac {2 B \,b^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{2}}+\frac {2 B \,x^{\frac {3}{2}}}{3 c}+\frac {2 A \sqrt {x}}{c}-\frac {2 B b \sqrt {x}}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.51, size = 58, normalized size = 0.84 \begin {gather*} \frac {2 \, {\left (B b^{2} - A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} + \frac {2 \, {\left (B c x^{\frac {3}{2}} - 3 \, {\left (B b - A c\right )} \sqrt {x}\right )}}{3 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 76, normalized size = 1.10 \begin {gather*} \sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )+\frac {2\,B\,x^{3/2}}{3\,c}+\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,\sqrt {x}\,\left (A\,c-B\,b\right )}{B\,b^2-A\,b\,c}\right )\,\left (A\,c-B\,b\right )}{c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.70, size = 212, normalized size = 3.07 \begin {gather*} \begin {cases} \frac {i A \sqrt {b} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{2} \sqrt {\frac {1}{c}}} - \frac {i A \sqrt {b} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{2} \sqrt {\frac {1}{c}}} + \frac {2 A \sqrt {x}}{c} - \frac {i B b^{\frac {3}{2}} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{3} \sqrt {\frac {1}{c}}} + \frac {i B b^{\frac {3}{2}} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{3} \sqrt {\frac {1}{c}}} - \frac {2 B b \sqrt {x}}{c^{2}} + \frac {2 B x^{\frac {3}{2}}}{3 c} & \text {for}\: c \neq 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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