3.2.87 \(\int \frac {x^{9/2} (A+B x)}{(b x+c x^2)^3} \, dx\)

Optimal. Leaf size=126 \[ -\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {b} c^{7/2}}+\frac {3 \sqrt {x} (5 b B-A c)}{4 b c^3}-\frac {x^{3/2} (5 b B-A c)}{4 b c^2 (b+c x)}-\frac {x^{5/2} (b B-A c)}{2 b c (b+c x)^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {781, 78, 47, 50, 63, 205} \begin {gather*} -\frac {x^{3/2} (5 b B-A c)}{4 b c^2 (b+c x)}+\frac {3 \sqrt {x} (5 b B-A c)}{4 b c^3}-\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {b} c^{7/2}}-\frac {x^{5/2} (b B-A c)}{2 b c (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(3*(5*b*B - A*c)*Sqrt[x])/(4*b*c^3) - ((b*B - A*c)*x^(5/2))/(2*b*c*(b + c*x)^2) - ((5*b*B - A*c)*x^(3/2))/(4*b
*c^2*(b + c*x)) - (3*(5*b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*Sqrt[b]*c^(7/2))

Rule 47

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

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^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac {x^{3/2} (A+B x)}{(b+c x)^3} \, dx\\ &=-\frac {(b B-A c) x^{5/2}}{2 b c (b+c x)^2}-\frac {\left (-\frac {5 b B}{2}+\frac {A c}{2}\right ) \int \frac {x^{3/2}}{(b+c x)^2} \, dx}{2 b c}\\ &=-\frac {(b B-A c) x^{5/2}}{2 b c (b+c x)^2}-\frac {(5 b B-A c) x^{3/2}}{4 b c^2 (b+c x)}+\frac {(3 (5 b B-A c)) \int \frac {\sqrt {x}}{b+c x} \, dx}{8 b c^2}\\ &=\frac {3 (5 b B-A c) \sqrt {x}}{4 b c^3}-\frac {(b B-A c) x^{5/2}}{2 b c (b+c x)^2}-\frac {(5 b B-A c) x^{3/2}}{4 b c^2 (b+c x)}-\frac {(3 (5 b B-A c)) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 c^3}\\ &=\frac {3 (5 b B-A c) \sqrt {x}}{4 b c^3}-\frac {(b B-A c) x^{5/2}}{2 b c (b+c x)^2}-\frac {(5 b B-A c) x^{3/2}}{4 b c^2 (b+c x)}-\frac {(3 (5 b B-A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 c^3}\\ &=\frac {3 (5 b B-A c) \sqrt {x}}{4 b c^3}-\frac {(b B-A c) x^{5/2}}{2 b c (b+c x)^2}-\frac {(5 b B-A c) x^{3/2}}{4 b c^2 (b+c x)}-\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {b} c^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.48 \begin {gather*} \frac {x^{5/2} \left (\frac {5 b^2 (A c-b B)}{(b+c x)^2}+(5 b B-A c) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {c x}{b}\right )\right )}{10 b^3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(x^(5/2)*((5*b^2*(-(b*B) + A*c))/(b + c*x)^2 + (5*b*B - A*c)*Hypergeometric2F1[2, 5/2, 7/2, -((c*x)/b)]))/(10*
b^3*c)

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IntegrateAlgebraic [A]  time = 0.17, size = 94, normalized size = 0.75 \begin {gather*} \frac {\sqrt {x} \left (-3 A b c-5 A c^2 x+15 b^2 B+25 b B c x+8 B c^2 x^2\right )}{4 c^3 (b+c x)^2}-\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {b} c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(Sqrt[x]*(15*b^2*B - 3*A*b*c + 25*b*B*c*x - 5*A*c^2*x + 8*B*c^2*x^2))/(4*c^3*(b + c*x)^2) - (3*(5*b*B - A*c)*A
rcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*Sqrt[b]*c^(7/2))

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fricas [A]  time = 0.43, size = 319, normalized size = 2.53 \begin {gather*} \left [\frac {3 \, {\left (5 \, B b^{3} - A b^{2} c + {\left (5 \, B b c^{2} - A c^{3}\right )} x^{2} + 2 \, {\left (5 \, B b^{2} c - A b 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 (8 \, B b c^{3} x^{2} + 15 \, B b^{3} c - 3 \, A b^{2} c^{2} + 5 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (b c^{6} x^{2} + 2 \, b^{2} c^{5} x + b^{3} c^{4}\right )}}, \frac {3 \, {\left (5 \, B b^{3} - A b^{2} c + {\left (5 \, B b c^{2} - A c^{3}\right )} x^{2} + 2 \, {\left (5 \, B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (8 \, B b c^{3} x^{2} + 15 \, B b^{3} c - 3 \, A b^{2} c^{2} + 5 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (b c^{6} x^{2} + 2 \, b^{2} c^{5} x + b^{3} c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[1/8*(3*(5*B*b^3 - A*b^2*c + (5*B*b*c^2 - A*c^3)*x^2 + 2*(5*B*b^2*c - A*b*c^2)*x)*sqrt(-b*c)*log((c*x - b - 2*
sqrt(-b*c)*sqrt(x))/(c*x + b)) + 2*(8*B*b*c^3*x^2 + 15*B*b^3*c - 3*A*b^2*c^2 + 5*(5*B*b^2*c^2 - A*b*c^3)*x)*sq
rt(x))/(b*c^6*x^2 + 2*b^2*c^5*x + b^3*c^4), 1/4*(3*(5*B*b^3 - A*b^2*c + (5*B*b*c^2 - A*c^3)*x^2 + 2*(5*B*b^2*c
- A*b*c^2)*x)*sqrt(b*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (8*B*b*c^3*x^2 + 15*B*b^3*c - 3*A*b^2*c^2 + 5*(5*B*b^
2*c^2 - A*b*c^3)*x)*sqrt(x))/(b*c^6*x^2 + 2*b^2*c^5*x + b^3*c^4)]

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giac [A]  time = 0.16, size = 87, normalized size = 0.69 \begin {gather*} \frac {2 \, B \sqrt {x}}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{3}} + \frac {9 \, B b c x^{\frac {3}{2}} - 5 \, A c^{2} x^{\frac {3}{2}} + 7 \, B b^{2} \sqrt {x} - 3 \, A b c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

2*B*sqrt(x)/c^3 - 3/4*(5*B*b - A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/4*(9*B*b*c*x^(3/2) - 5*A*c
^2*x^(3/2) + 7*B*b^2*sqrt(x) - 3*A*b*c*sqrt(x))/((c*x + b)^2*c^3)

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maple [A]  time = 0.07, size = 125, normalized size = 0.99 \begin {gather*} -\frac {5 A \,x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} c}+\frac {9 B b \,x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} c^{2}}-\frac {3 A b \sqrt {x}}{4 \left (c x +b \right )^{2} c^{2}}+\frac {7 B \,b^{2} \sqrt {x}}{4 \left (c x +b \right )^{2} c^{3}}+\frac {3 A \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, c^{2}}-\frac {15 B b \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, c^{3}}+\frac {2 B \sqrt {x}}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x)

[Out]

2*B/c^3*x^(1/2)-5/4/c/(c*x+b)^2*x^(3/2)*A+9/4/c^2/(c*x+b)^2*x^(3/2)*b*B-3/4/c^2/(c*x+b)^2*A*x^(1/2)*b+7/4/c^3/
(c*x+b)^2*B*x^(1/2)*b^2+3/4/c^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*A-15/4/c^3/(b*c)^(1/2)*arctan(1/(b
*c)^(1/2)*c*x^(1/2))*b*B

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maxima [A]  time = 1.46, size = 99, normalized size = 0.79 \begin {gather*} \frac {{\left (9 \, B b c - 5 \, A c^{2}\right )} x^{\frac {3}{2}} + {\left (7 \, B b^{2} - 3 \, A b c\right )} \sqrt {x}}{4 \, {\left (c^{5} x^{2} + 2 \, b c^{4} x + b^{2} c^{3}\right )}} + \frac {2 \, B \sqrt {x}}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

1/4*((9*B*b*c - 5*A*c^2)*x^(3/2) + (7*B*b^2 - 3*A*b*c)*sqrt(x))/(c^5*x^2 + 2*b*c^4*x + b^2*c^3) + 2*B*sqrt(x)/
c^3 - 3/4*(5*B*b - A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^3)

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mupad [B]  time = 0.11, size = 96, normalized size = 0.76 \begin {gather*} \frac {\sqrt {x}\,\left (\frac {7\,B\,b^2}{4}-\frac {3\,A\,b\,c}{4}\right )-x^{3/2}\,\left (\frac {5\,A\,c^2}{4}-\frac {9\,B\,b\,c}{4}\right )}{b^2\,c^3+2\,b\,c^4\,x+c^5\,x^2}+\frac {2\,B\,\sqrt {x}}{c^3}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-5\,B\,b\right )}{4\,\sqrt {b}\,c^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(9/2)*(A + B*x))/(b*x + c*x^2)^3,x)

[Out]

(x^(1/2)*((7*B*b^2)/4 - (3*A*b*c)/4) - x^(3/2)*((5*A*c^2)/4 - (9*B*b*c)/4))/(b^2*c^3 + c^5*x^2 + 2*b*c^4*x) +
(2*B*x^(1/2))/c^3 + (3*atan((c^(1/2)*x^(1/2))/b^(1/2))*(A*c - 5*B*b))/(4*b^(1/2)*c^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(9/2)*(B*x+A)/(c*x**2+b*x)**3,x)

[Out]

Timed out

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