3.2.86 \(\int \frac {x^{11/2} (A+B x)}{(b x+c x^2)^3} \, dx\)

Optimal. Leaf size=147 \[ \frac {5 \sqrt {b} (7 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{9/2}}-\frac {5 \sqrt {x} (7 b B-3 A c)}{4 c^4}+\frac {5 x^{3/2} (7 b B-3 A c)}{12 b c^3}-\frac {x^{5/2} (7 b B-3 A c)}{4 b c^2 (b+c x)}-\frac {x^{7/2} (b B-A c)}{2 b c (b+c x)^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, 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^{5/2} (7 b B-3 A c)}{4 b c^2 (b+c x)}+\frac {5 x^{3/2} (7 b B-3 A c)}{12 b c^3}-\frac {5 \sqrt {x} (7 b B-3 A c)}{4 c^4}+\frac {5 \sqrt {b} (7 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{9/2}}-\frac {x^{7/2} (b B-A c)}{2 b c (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(11/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(-5*(7*b*B - 3*A*c)*Sqrt[x])/(4*c^4) + (5*(7*b*B - 3*A*c)*x^(3/2))/(12*b*c^3) - ((b*B - A*c)*x^(7/2))/(2*b*c*(
b + c*x)^2) - ((7*b*B - 3*A*c)*x^(5/2))/(4*b*c^2*(b + c*x)) + (5*Sqrt[b]*(7*b*B - 3*A*c)*ArcTan[(Sqrt[c]*Sqrt[
x])/Sqrt[b]])/(4*c^(9/2))

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^{11/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac {x^{5/2} (A+B x)}{(b+c x)^3} \, dx\\ &=-\frac {(b B-A c) x^{7/2}}{2 b c (b+c x)^2}-\frac {\left (-\frac {7 b B}{2}+\frac {3 A c}{2}\right ) \int \frac {x^{5/2}}{(b+c x)^2} \, dx}{2 b c}\\ &=-\frac {(b B-A c) x^{7/2}}{2 b c (b+c x)^2}-\frac {(7 b B-3 A c) x^{5/2}}{4 b c^2 (b+c x)}+\frac {(5 (7 b B-3 A c)) \int \frac {x^{3/2}}{b+c x} \, dx}{8 b c^2}\\ &=\frac {5 (7 b B-3 A c) x^{3/2}}{12 b c^3}-\frac {(b B-A c) x^{7/2}}{2 b c (b+c x)^2}-\frac {(7 b B-3 A c) x^{5/2}}{4 b c^2 (b+c x)}-\frac {(5 (7 b B-3 A c)) \int \frac {\sqrt {x}}{b+c x} \, dx}{8 c^3}\\ &=-\frac {5 (7 b B-3 A c) \sqrt {x}}{4 c^4}+\frac {5 (7 b B-3 A c) x^{3/2}}{12 b c^3}-\frac {(b B-A c) x^{7/2}}{2 b c (b+c x)^2}-\frac {(7 b B-3 A c) x^{5/2}}{4 b c^2 (b+c x)}+\frac {(5 b (7 b B-3 A c)) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 c^4}\\ &=-\frac {5 (7 b B-3 A c) \sqrt {x}}{4 c^4}+\frac {5 (7 b B-3 A c) x^{3/2}}{12 b c^3}-\frac {(b B-A c) x^{7/2}}{2 b c (b+c x)^2}-\frac {(7 b B-3 A c) x^{5/2}}{4 b c^2 (b+c x)}+\frac {(5 b (7 b B-3 A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 c^4}\\ &=-\frac {5 (7 b B-3 A c) \sqrt {x}}{4 c^4}+\frac {5 (7 b B-3 A c) x^{3/2}}{12 b c^3}-\frac {(b B-A c) x^{7/2}}{2 b c (b+c x)^2}-\frac {(7 b B-3 A c) x^{5/2}}{4 b c^2 (b+c x)}+\frac {5 \sqrt {b} (7 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.41 \begin {gather*} \frac {x^{7/2} \left (\frac {7 b^2 (A c-b B)}{(b+c x)^2}+(7 b B-3 A c) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {c x}{b}\right )\right )}{14 b^3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(11/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(x^(7/2)*((7*b^2*(-(b*B) + A*c))/(b + c*x)^2 + (7*b*B - 3*A*c)*Hypergeometric2F1[2, 7/2, 9/2, -((c*x)/b)]))/(1
4*b^3*c)

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IntegrateAlgebraic [A]  time = 0.20, size = 122, normalized size = 0.83 \begin {gather*} \frac {5 \left (7 b^{3/2} B-3 A \sqrt {b} c\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{9/2}}+\frac {\sqrt {x} \left (45 A b^2 c+75 A b c^2 x+24 A c^3 x^2-105 b^3 B-175 b^2 B c x-56 b B c^2 x^2+8 B c^3 x^3\right )}{12 c^4 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(11/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(Sqrt[x]*(-105*b^3*B + 45*A*b^2*c - 175*b^2*B*c*x + 75*A*b*c^2*x - 56*b*B*c^2*x^2 + 24*A*c^3*x^2 + 8*B*c^3*x^3
))/(12*c^4*(b + c*x)^2) + (5*(7*b^(3/2)*B - 3*A*Sqrt[b]*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*c^(9/2))

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fricas [A]  time = 0.43, size = 349, normalized size = 2.37 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{3} - 3 \, A b^{2} c + {\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{2} + 2 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x\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 (8 \, B c^{3} x^{3} - 105 \, B b^{3} + 45 \, A b^{2} c - 8 \, {\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{2} - 25 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (c^{6} x^{2} + 2 \, b c^{5} x + b^{2} c^{4}\right )}}, \frac {15 \, {\left (7 \, B b^{3} - 3 \, A b^{2} c + {\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{2} + 2 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) + {\left (8 \, B c^{3} x^{3} - 105 \, B b^{3} + 45 \, A b^{2} c - 8 \, {\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{2} - 25 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (c^{6} x^{2} + 2 \, b c^{5} x + b^{2} c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[-1/24*(15*(7*B*b^3 - 3*A*b^2*c + (7*B*b*c^2 - 3*A*c^3)*x^2 + 2*(7*B*b^2*c - 3*A*b*c^2)*x)*sqrt(-b/c)*log((c*x
- 2*c*sqrt(x)*sqrt(-b/c) - b)/(c*x + b)) - 2*(8*B*c^3*x^3 - 105*B*b^3 + 45*A*b^2*c - 8*(7*B*b*c^2 - 3*A*c^3)*
x^2 - 25*(7*B*b^2*c - 3*A*b*c^2)*x)*sqrt(x))/(c^6*x^2 + 2*b*c^5*x + b^2*c^4), 1/12*(15*(7*B*b^3 - 3*A*b^2*c +
(7*B*b*c^2 - 3*A*c^3)*x^2 + 2*(7*B*b^2*c - 3*A*b*c^2)*x)*sqrt(b/c)*arctan(c*sqrt(x)*sqrt(b/c)/b) + (8*B*c^3*x^
3 - 105*B*b^3 + 45*A*b^2*c - 8*(7*B*b*c^2 - 3*A*c^3)*x^2 - 25*(7*B*b^2*c - 3*A*b*c^2)*x)*sqrt(x))/(c^6*x^2 + 2
*b*c^5*x + b^2*c^4)]

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giac [A]  time = 0.18, size = 119, normalized size = 0.81 \begin {gather*} \frac {5 \, {\left (7 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{4}} - \frac {13 \, B b^{2} c x^{\frac {3}{2}} - 9 \, A b c^{2} x^{\frac {3}{2}} + 11 \, B b^{3} \sqrt {x} - 7 \, A b^{2} c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} c^{4}} + \frac {2 \, {\left (B c^{6} x^{\frac {3}{2}} - 9 \, B b c^{5} \sqrt {x} + 3 \, A c^{6} \sqrt {x}\right )}}{3 \, c^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

5/4*(7*B*b^2 - 3*A*b*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) - 1/4*(13*B*b^2*c*x^(3/2) - 9*A*b*c^2*x^(3
/2) + 11*B*b^3*sqrt(x) - 7*A*b^2*c*sqrt(x))/((c*x + b)^2*c^4) + 2/3*(B*c^6*x^(3/2) - 9*B*b*c^5*sqrt(x) + 3*A*c
^6*sqrt(x))/c^9

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maple [A]  time = 0.07, size = 152, normalized size = 1.03 \begin {gather*} \frac {9 A b \,x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} c^{2}}-\frac {13 B \,b^{2} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} c^{3}}+\frac {7 A \,b^{2} \sqrt {x}}{4 \left (c x +b \right )^{2} c^{3}}-\frac {11 B \,b^{3} \sqrt {x}}{4 \left (c x +b \right )^{2} c^{4}}-\frac {15 A b \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, c^{3}}+\frac {35 B \,b^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, c^{4}}+\frac {2 B \,x^{\frac {3}{2}}}{3 c^{3}}+\frac {2 A \sqrt {x}}{c^{3}}-\frac {6 B b \sqrt {x}}{c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)*(B*x+A)/(c*x^2+b*x)^3,x)

[Out]

2/3/c^3*B*x^(3/2)+2/c^3*A*x^(1/2)-6/c^4*b*B*x^(1/2)+9/4*b/c^2/(c*x+b)^2*x^(3/2)*A-13/4*b^2/c^3/(c*x+b)^2*x^(3/
2)*B+7/4*b^2/c^3/(c*x+b)^2*A*x^(1/2)-11/4*b^3/c^4/(c*x+b)^2*B*x^(1/2)-15/4*b/c^3/(b*c)^(1/2)*arctan(1/(b*c)^(1
/2)*c*x^(1/2))*A+35/4*b^2/c^4/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*B

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maxima [A]  time = 1.19, size = 124, normalized size = 0.84 \begin {gather*} -\frac {{\left (13 \, B b^{2} c - 9 \, A b c^{2}\right )} x^{\frac {3}{2}} + {\left (11 \, B b^{3} - 7 \, A b^{2} c\right )} \sqrt {x}}{4 \, {\left (c^{6} x^{2} + 2 \, b c^{5} x + b^{2} c^{4}\right )}} + \frac {5 \, {\left (7 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{4}} + \frac {2 \, {\left (B c x^{\frac {3}{2}} - 3 \, {\left (3 \, B b - A c\right )} \sqrt {x}\right )}}{3 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

-1/4*((13*B*b^2*c - 9*A*b*c^2)*x^(3/2) + (11*B*b^3 - 7*A*b^2*c)*sqrt(x))/(c^6*x^2 + 2*b*c^5*x + b^2*c^4) + 5/4
*(7*B*b^2 - 3*A*b*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + 2/3*(B*c*x^(3/2) - 3*(3*B*b - A*c)*sqrt(x))
/c^4

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mupad [B]  time = 0.09, size = 143, normalized size = 0.97 \begin {gather*} \frac {x^{3/2}\,\left (\frac {9\,A\,b\,c^2}{4}-\frac {13\,B\,b^2\,c}{4}\right )-\sqrt {x}\,\left (\frac {11\,B\,b^3}{4}-\frac {7\,A\,b^2\,c}{4}\right )}{b^2\,c^4+2\,b\,c^5\,x+c^6\,x^2}+\sqrt {x}\,\left (\frac {2\,A}{c^3}-\frac {6\,B\,b}{c^4}\right )+\frac {2\,B\,x^{3/2}}{3\,c^3}+\frac {5\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,\sqrt {x}\,\left (3\,A\,c-7\,B\,b\right )}{7\,B\,b^2-3\,A\,b\,c}\right )\,\left (3\,A\,c-7\,B\,b\right )}{4\,c^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(11/2)*(A + B*x))/(b*x + c*x^2)^3,x)

[Out]

(x^(3/2)*((9*A*b*c^2)/4 - (13*B*b^2*c)/4) - x^(1/2)*((11*B*b^3)/4 - (7*A*b^2*c)/4))/(b^2*c^4 + c^6*x^2 + 2*b*c
^5*x) + x^(1/2)*((2*A)/c^3 - (6*B*b)/c^4) + (2*B*x^(3/2))/(3*c^3) + (5*b^(1/2)*atan((b^(1/2)*c^(1/2)*x^(1/2)*(
3*A*c - 7*B*b))/(7*B*b^2 - 3*A*b*c))*(3*A*c - 7*B*b))/(4*c^(9/2))

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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**(11/2)*(B*x+A)/(c*x**2+b*x)**3,x)

[Out]

Timed out

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