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

Optimal. Leaf size=88 \[ -\frac {(3 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} c^{5/2}}+\frac {\sqrt {x} (3 b B-A c)}{b c^2}-\frac {x^{3/2} (b B-A c)}{b c (b+c x)} \]

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Rubi [A]  time = 0.04, antiderivative size = 88, 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, 50, 63, 205} \begin {gather*} \frac {\sqrt {x} (3 b B-A c)}{b c^2}-\frac {(3 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} c^{5/2}}-\frac {x^{3/2} (b B-A c)}{b c (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.78 \begin {gather*} \frac {\sqrt {x} (-A c+3 b B+2 B c x)}{c^2 (b+c x)}-\frac {(3 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.10, size = 67, normalized size = 0.76 \begin {gather*} \frac {(A c-3 b B) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} c^{5/2}}+\frac {\sqrt {x} (-A c+3 b B+2 B c x)}{c^2 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 198, normalized size = 2.25 \begin {gather*} \left [\frac {{\left (3 \, B b^{2} - A b c + {\left (3 \, B b c - A c^{2}\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x - b - 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right ) + 2 \, {\left (2 \, B b c^{2} x + 3 \, B b^{2} c - A b c^{2}\right )} \sqrt {x}}{2 \, {\left (b c^{4} x + b^{2} c^{3}\right )}}, \frac {{\left (3 \, B b^{2} - A b c + {\left (3 \, B b c - A c^{2}\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (2 \, B b c^{2} x + 3 \, B b^{2} c - A b c^{2}\right )} \sqrt {x}}{b c^{4} x + b^{2} 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)^2,x, algorithm="fricas")

[Out]

[1/2*((3*B*b^2 - A*b*c + (3*B*b*c - A*c^2)*x)*sqrt(-b*c)*log((c*x - b - 2*sqrt(-b*c)*sqrt(x))/(c*x + b)) + 2*(
2*B*b*c^2*x + 3*B*b^2*c - A*b*c^2)*sqrt(x))/(b*c^4*x + b^2*c^3), ((3*B*b^2 - A*b*c + (3*B*b*c - A*c^2)*x)*sqrt
(b*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (2*B*b*c^2*x + 3*B*b^2*c - A*b*c^2)*sqrt(x))/(b*c^4*x + b^2*c^3)]

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giac [A]  time = 0.16, size = 65, normalized size = 0.74 \begin {gather*} \frac {2 \, B \sqrt {x}}{c^{2}} - \frac {{\left (3 \, B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} + \frac {B b \sqrt {x} - A c \sqrt {x}}{{\left (c x + b\right )} c^{2}} \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)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 87, normalized size = 0.99 \begin {gather*} \frac {A \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c}-\frac {3 B b \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{2}}-\frac {A \sqrt {x}}{\left (c x +b \right ) c}+\frac {B b \sqrt {x}}{\left (c x +b \right ) c^{2}}+\frac {2 B \sqrt {x}}{c^{2}} \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)^2,x)

[Out]

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

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maxima [A]  time = 1.30, size = 65, normalized size = 0.74 \begin {gather*} \frac {{\left (B b - A c\right )} \sqrt {x}}{c^{3} x + b c^{2}} + \frac {2 \, B \sqrt {x}}{c^{2}} - \frac {{\left (3 \, B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} \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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.09, size = 62, normalized size = 0.70 \begin {gather*} \frac {2\,B\,\sqrt {x}}{c^2}-\frac {\sqrt {x}\,\left (A\,c-B\,b\right )}{x\,c^3+b\,c^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-3\,B\,b\right )}{\sqrt {b}\,c^{5/2}} \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)^2,x)

[Out]

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

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sympy [A]  time = 107.25, size = 782, normalized size = 8.89 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{c^{2}} & \text {for}\: b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{b^{2}} & \text {for}\: c = 0 \\- \frac {2 i A \sqrt {b} c^{2} \sqrt {x} \sqrt {\frac {1}{c}}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} + \frac {A b c \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} - \frac {A b c \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} + \frac {A c^{2} x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} - \frac {A c^{2} x \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} + \frac {6 i B b^{\frac {3}{2}} c \sqrt {x} \sqrt {\frac {1}{c}}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} + \frac {4 i B \sqrt {b} c^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{c}}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} - \frac {3 B b^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} + \frac {3 B b^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} - \frac {3 B b c x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} + \frac {3 B b c x \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{2 i b^{\frac {3}{2}} c^{3} \sqrt {\frac {1}{c}} + 2 i \sqrt {b} c^{4} x \sqrt {\frac {1}{c}}} & \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)**2,x)

[Out]

Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(b, 0) & Eq(c, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/c**2, Eq(b, 0
)), ((2*A*x**(3/2)/3 + 2*B*x**(5/2)/5)/b**2, Eq(c, 0)), (-2*I*A*sqrt(b)*c**2*sqrt(x)*sqrt(1/c)/(2*I*b**(3/2)*c
**3*sqrt(1/c) + 2*I*sqrt(b)*c**4*x*sqrt(1/c)) + A*b*c*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*s
qrt(1/c) + 2*I*sqrt(b)*c**4*x*sqrt(1/c)) - A*b*c*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*sqrt(1/
c) + 2*I*sqrt(b)*c**4*x*sqrt(1/c)) + A*c**2*x*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*sqrt(1/c)
+ 2*I*sqrt(b)*c**4*x*sqrt(1/c)) - A*c**2*x*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*sqrt(1/c) +
2*I*sqrt(b)*c**4*x*sqrt(1/c)) + 6*I*B*b**(3/2)*c*sqrt(x)*sqrt(1/c)/(2*I*b**(3/2)*c**3*sqrt(1/c) + 2*I*sqrt(b)*
c**4*x*sqrt(1/c)) + 4*I*B*sqrt(b)*c**2*x**(3/2)*sqrt(1/c)/(2*I*b**(3/2)*c**3*sqrt(1/c) + 2*I*sqrt(b)*c**4*x*sq
rt(1/c)) - 3*B*b**2*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*sqrt(1/c) + 2*I*sqrt(b)*c**4*x*sqrt
(1/c)) + 3*B*b**2*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*sqrt(1/c) + 2*I*sqrt(b)*c**4*x*sqrt(1/
c)) - 3*B*b*c*x*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*sqrt(1/c) + 2*I*sqrt(b)*c**4*x*sqrt(1/c
)) + 3*B*b*c*x*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(3/2)*c**3*sqrt(1/c) + 2*I*sqrt(b)*c**4*x*sqrt(1/c))
, True))

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