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

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

________________________________________________________________________________________

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, 80, 50, 63, 205} \begin {gather*} -\frac {2 b^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}}-\frac {2 x^{3/2} (b B-A c)}{3 c^2}+\frac {2 b \sqrt {x} (b B-A c)}{c^3}+\frac {2 B x^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 81, normalized size = 0.90 \begin {gather*} \frac {2 \sqrt {x} \left (-5 b c (3 A+B x)+c^2 x (5 A+3 B x)+15 b^2 B\right )}{15 c^3}-\frac {2 b^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.10, size = 88, normalized size = 0.98 \begin {gather*} \frac {2 \sqrt {x} \left (-15 A b c+5 A c^2 x+15 b^2 B-5 b B c x+3 B c^2 x^2\right )}{15 c^3}-\frac {2 \left (b^{5/2} B-A b^{3/2} c\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 180, normalized size = 2.00 \begin {gather*} \left [-\frac {15 \, {\left (B b^{2} - A b 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 (3 \, B c^{2} x^{2} + 15 \, B b^{2} - 15 \, A b c - 5 \, {\left (B b c - A c^{2}\right )} x\right )} \sqrt {x}}{15 \, c^{3}}, -\frac {2 \, {\left (15 \, {\left (B b^{2} - A b c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) - {\left (3 \, B c^{2} x^{2} + 15 \, B b^{2} - 15 \, A b c - 5 \, {\left (B b c - A c^{2}\right )} x\right )} \sqrt {x}\right )}}{15 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.17, size = 91, normalized size = 1.01 \begin {gather*} -\frac {2 \, {\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {2 \, {\left (3 \, B c^{4} x^{\frac {5}{2}} - 5 \, B b c^{3} x^{\frac {3}{2}} + 5 \, A c^{4} x^{\frac {3}{2}} + 15 \, B b^{2} c^{2} \sqrt {x} - 15 \, A b c^{3} \sqrt {x}\right )}}{15 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*b^3 - A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^3) + 2/15*(3*B*c^4*x^(5/2) - 5*B*b*c^3*x^(3/2) +
5*A*c^4*x^(3/2) + 15*B*b^2*c^2*sqrt(x) - 15*A*b*c^3*sqrt(x))/c^5

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.23, size = 82, normalized size = 0.91 \begin {gather*} -\frac {2 \, {\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {2 \, {\left (3 \, B c^{2} x^{\frac {5}{2}} - 5 \, {\left (B b c - A c^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (B b^{2} - A b c\right )} \sqrt {x}\right )}}{15 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.05, size = 101, normalized size = 1.12 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,c}-\frac {2\,B\,b}{3\,c^2}\right )+\frac {2\,B\,x^{5/2}}{5\,c}-\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,\sqrt {x}\,\left (A\,c-B\,b\right )}{B\,b^3-A\,b^2\,c}\right )\,\left (A\,c-B\,b\right )}{c^{7/2}}-\frac {b\,\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 16.35, size = 245, normalized size = 2.72 \begin {gather*} \begin {cases} - \frac {i A b^{\frac {3}{2}} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{3} \sqrt {\frac {1}{c}}} + \frac {i A b^{\frac {3}{2}} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{3} \sqrt {\frac {1}{c}}} - \frac {2 A b \sqrt {x}}{c^{2}} + \frac {2 A x^{\frac {3}{2}}}{3 c} + \frac {i B b^{\frac {5}{2}} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{4} \sqrt {\frac {1}{c}}} - \frac {i B b^{\frac {5}{2}} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{4} \sqrt {\frac {1}{c}}} + \frac {2 B b^{2} \sqrt {x}}{c^{3}} - \frac {2 B b x^{\frac {3}{2}}}{3 c^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 c} & \text {for}\: c \neq 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{b} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*A*b**(3/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(c**3*sqrt(1/c)) + I*A*b**(3/2)*log(I*sqrt(b)*sqr
t(1/c) + sqrt(x))/(c**3*sqrt(1/c)) - 2*A*b*sqrt(x)/c**2 + 2*A*x**(3/2)/(3*c) + I*B*b**(5/2)*log(-I*sqrt(b)*sqr
t(1/c) + sqrt(x))/(c**4*sqrt(1/c)) - I*B*b**(5/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(c**4*sqrt(1/c)) + 2*B*b*
*2*sqrt(x)/c**3 - 2*B*b*x**(3/2)/(3*c**2) + 2*B*x**(5/2)/(5*c), Ne(c, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/
b, True))

________________________________________________________________________________________