3.185 \(\int \frac {x^{13/2} (A+B x)}{(b x+c x^2)^3} \, dx\)

Optimal. Leaf size=169 \[ -\frac {7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{11/2}}+\frac {7 b \sqrt {x} (9 b B-5 A c)}{4 c^5}-\frac {7 x^{3/2} (9 b B-5 A c)}{12 c^4}+\frac {7 x^{5/2} (9 b B-5 A c)}{20 b c^3}-\frac {x^{7/2} (9 b B-5 A c)}{4 b c^2 (b+c x)}-\frac {x^{9/2} (b B-A c)}{2 b c (b+c x)^2} \]

[Out]

-7/12*(-5*A*c+9*B*b)*x^(3/2)/c^4+7/20*(-5*A*c+9*B*b)*x^(5/2)/b/c^3-1/2*(-A*c+B*b)*x^(9/2)/b/c/(c*x+b)^2-1/4*(-
5*A*c+9*B*b)*x^(7/2)/b/c^2/(c*x+b)-7/4*b^(3/2)*(-5*A*c+9*B*b)*arctan(c^(1/2)*x^(1/2)/b^(1/2))/c^(11/2)+7/4*b*(
-5*A*c+9*B*b)*x^(1/2)/c^5

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Rubi [A]  time = 0.09, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{11/2}}-\frac {x^{7/2} (9 b B-5 A c)}{4 b c^2 (b+c x)}+\frac {7 x^{5/2} (9 b B-5 A c)}{20 b c^3}-\frac {7 x^{3/2} (9 b B-5 A c)}{12 c^4}+\frac {7 b \sqrt {x} (9 b B-5 A c)}{4 c^5}-\frac {x^{9/2} (b B-A c)}{2 b c (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(13/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(7*b*(9*b*B - 5*A*c)*Sqrt[x])/(4*c^5) - (7*(9*b*B - 5*A*c)*x^(3/2))/(12*c^4) + (7*(9*b*B - 5*A*c)*x^(5/2))/(20
*b*c^3) - ((b*B - A*c)*x^(9/2))/(2*b*c*(b + c*x)^2) - ((9*b*B - 5*A*c)*x^(7/2))/(4*b*c^2*(b + c*x)) - (7*b^(3/
2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*c^(11/2))

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^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac {x^{7/2} (A+B x)}{(b+c x)^3} \, dx\\ &=-\frac {(b B-A c) x^{9/2}}{2 b c (b+c x)^2}-\frac {\left (-\frac {9 b B}{2}+\frac {5 A c}{2}\right ) \int \frac {x^{7/2}}{(b+c x)^2} \, dx}{2 b c}\\ &=-\frac {(b B-A c) x^{9/2}}{2 b c (b+c x)^2}-\frac {(9 b B-5 A c) x^{7/2}}{4 b c^2 (b+c x)}+\frac {(7 (9 b B-5 A c)) \int \frac {x^{5/2}}{b+c x} \, dx}{8 b c^2}\\ &=\frac {7 (9 b B-5 A c) x^{5/2}}{20 b c^3}-\frac {(b B-A c) x^{9/2}}{2 b c (b+c x)^2}-\frac {(9 b B-5 A c) x^{7/2}}{4 b c^2 (b+c x)}-\frac {(7 (9 b B-5 A c)) \int \frac {x^{3/2}}{b+c x} \, dx}{8 c^3}\\ &=-\frac {7 (9 b B-5 A c) x^{3/2}}{12 c^4}+\frac {7 (9 b B-5 A c) x^{5/2}}{20 b c^3}-\frac {(b B-A c) x^{9/2}}{2 b c (b+c x)^2}-\frac {(9 b B-5 A c) x^{7/2}}{4 b c^2 (b+c x)}+\frac {(7 b (9 b B-5 A c)) \int \frac {\sqrt {x}}{b+c x} \, dx}{8 c^4}\\ &=\frac {7 b (9 b B-5 A c) \sqrt {x}}{4 c^5}-\frac {7 (9 b B-5 A c) x^{3/2}}{12 c^4}+\frac {7 (9 b B-5 A c) x^{5/2}}{20 b c^3}-\frac {(b B-A c) x^{9/2}}{2 b c (b+c x)^2}-\frac {(9 b B-5 A c) x^{7/2}}{4 b c^2 (b+c x)}-\frac {\left (7 b^2 (9 b B-5 A c)\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 c^5}\\ &=\frac {7 b (9 b B-5 A c) \sqrt {x}}{4 c^5}-\frac {7 (9 b B-5 A c) x^{3/2}}{12 c^4}+\frac {7 (9 b B-5 A c) x^{5/2}}{20 b c^3}-\frac {(b B-A c) x^{9/2}}{2 b c (b+c x)^2}-\frac {(9 b B-5 A c) x^{7/2}}{4 b c^2 (b+c x)}-\frac {\left (7 b^2 (9 b B-5 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 c^5}\\ &=\frac {7 b (9 b B-5 A c) \sqrt {x}}{4 c^5}-\frac {7 (9 b B-5 A c) x^{3/2}}{12 c^4}+\frac {7 (9 b B-5 A c) x^{5/2}}{20 b c^3}-\frac {(b B-A c) x^{9/2}}{2 b c (b+c x)^2}-\frac {(9 b B-5 A c) x^{7/2}}{4 b c^2 (b+c x)}-\frac {7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.36 \[ \frac {x^{9/2} \left (\frac {9 b^2 (A c-b B)}{(b+c x)^2}+(9 b B-5 A c) \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};-\frac {c x}{b}\right )\right )}{18 b^3 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(13/2)*(A + B*x))/(b*x + c*x^2)^3,x]

[Out]

(x^(9/2)*((9*b^2*(-(b*B) + A*c))/(b + c*x)^2 + (9*b*B - 5*A*c)*Hypergeometric2F1[2, 9/2, 11/2, -((c*x)/b)]))/(
18*b^3*c)

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fricas [A]  time = 1.18, size = 408, normalized size = 2.41 \[ \left [-\frac {105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x + 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (24 \, B c^{4} x^{4} + 945 \, B b^{4} - 525 \, A b^{3} c - 8 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{3} + 56 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 175 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{120 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}, -\frac {105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) - {\left (24 \, B c^{4} x^{4} + 945 \, B b^{4} - 525 \, A b^{3} c - 8 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{3} + 56 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 175 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{60 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[-1/120*(105*(9*B*b^4 - 5*A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 2*(9*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(-b/c)*
log((c*x + 2*c*sqrt(x)*sqrt(-b/c) - b)/(c*x + b)) - 2*(24*B*c^4*x^4 + 945*B*b^4 - 525*A*b^3*c - 8*(9*B*b*c^3 -
5*A*c^4)*x^3 + 56*(9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 175*(9*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(x))/(c^7*x^2 + 2*b*c^
6*x + b^2*c^5), -1/60*(105*(9*B*b^4 - 5*A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 2*(9*B*b^3*c - 5*A*b^2*c^2)*
x)*sqrt(b/c)*arctan(c*sqrt(x)*sqrt(b/c)/b) - (24*B*c^4*x^4 + 945*B*b^4 - 525*A*b^3*c - 8*(9*B*b*c^3 - 5*A*c^4)
*x^3 + 56*(9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 175*(9*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(x))/(c^7*x^2 + 2*b*c^6*x + b^2
*c^5)]

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giac [A]  time = 0.21, size = 146, normalized size = 0.86 \[ -\frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{5}} + \frac {17 \, B b^{3} c x^{\frac {3}{2}} - 13 \, A b^{2} c^{2} x^{\frac {3}{2}} + 15 \, B b^{4} \sqrt {x} - 11 \, A b^{3} c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} c^{5}} + \frac {2 \, {\left (3 \, B c^{12} x^{\frac {5}{2}} - 15 \, B b c^{11} x^{\frac {3}{2}} + 5 \, A c^{12} x^{\frac {3}{2}} + 90 \, B b^{2} c^{10} \sqrt {x} - 45 \, A b c^{11} \sqrt {x}\right )}}{15 \, c^{15}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

-7/4*(9*B*b^3 - 5*A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^5) + 1/4*(17*B*b^3*c*x^(3/2) - 13*A*b^2*c^
2*x^(3/2) + 15*B*b^4*sqrt(x) - 11*A*b^3*c*sqrt(x))/((c*x + b)^2*c^5) + 2/15*(3*B*c^12*x^(5/2) - 15*B*b*c^11*x^
(3/2) + 5*A*c^12*x^(3/2) + 90*B*b^2*c^10*sqrt(x) - 45*A*b*c^11*sqrt(x))/c^15

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maple [A]  time = 0.07, size = 178, normalized size = 1.05 \[ -\frac {13 A \,b^{2} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} c^{3}}+\frac {17 B \,b^{3} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} c^{4}}-\frac {11 A \,b^{3} \sqrt {x}}{4 \left (c x +b \right )^{2} c^{4}}+\frac {15 B \,b^{4} \sqrt {x}}{4 \left (c x +b \right )^{2} c^{5}}+\frac {2 B \,x^{\frac {5}{2}}}{5 c^{3}}+\frac {35 A \,b^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, c^{4}}-\frac {63 B \,b^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, c^{5}}+\frac {2 A \,x^{\frac {3}{2}}}{3 c^{3}}-\frac {2 B b \,x^{\frac {3}{2}}}{c^{4}}-\frac {6 A b \sqrt {x}}{c^{4}}+\frac {12 B \,b^{2} \sqrt {x}}{c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x)

[Out]

2/5/c^3*B*x^(5/2)+2/3/c^3*A*x^(3/2)-2/c^4*B*x^(3/2)*b-6/c^4*A*b*x^(1/2)+12/c^5*b^2*B*x^(1/2)-13/4*b^2/c^3/(c*x
+b)^2*A*x^(3/2)+17/4*b^3/c^4/(c*x+b)^2*B*x^(3/2)-11/4*b^3/c^4/(c*x+b)^2*A*x^(1/2)+15/4*b^4/c^5/(c*x+b)^2*B*x^(
1/2)+35/4*b^2/c^4/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*A-63/4*b^3/c^5/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*
c*x^(1/2))*B

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maxima [A]  time = 1.13, size = 151, normalized size = 0.89 \[ \frac {{\left (17 \, B b^{3} c - 13 \, A b^{2} c^{2}\right )} x^{\frac {3}{2}} + {\left (15 \, B b^{4} - 11 \, A b^{3} c\right )} \sqrt {x}}{4 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}} - \frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{5}} + \frac {2 \, {\left (3 \, B c^{2} x^{\frac {5}{2}} - 5 \, {\left (3 \, B b c - A c^{2}\right )} x^{\frac {3}{2}} + 45 \, {\left (2 \, B b^{2} - A b c\right )} \sqrt {x}\right )}}{15 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

1/4*((17*B*b^3*c - 13*A*b^2*c^2)*x^(3/2) + (15*B*b^4 - 11*A*b^3*c)*sqrt(x))/(c^7*x^2 + 2*b*c^6*x + b^2*c^5) -
7/4*(9*B*b^3 - 5*A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^5) + 2/15*(3*B*c^2*x^(5/2) - 5*(3*B*b*c - A
*c^2)*x^(3/2) + 45*(2*B*b^2 - A*b*c)*sqrt(x))/c^5

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mupad [B]  time = 1.07, size = 183, normalized size = 1.08 \[ x^{3/2}\,\left (\frac {2\,A}{3\,c^3}-\frac {2\,B\,b}{c^4}\right )-\frac {x^{3/2}\,\left (\frac {13\,A\,b^2\,c^2}{4}-\frac {17\,B\,b^3\,c}{4}\right )-\sqrt {x}\,\left (\frac {15\,B\,b^4}{4}-\frac {11\,A\,b^3\,c}{4}\right )}{b^2\,c^5+2\,b\,c^6\,x+c^7\,x^2}-\sqrt {x}\,\left (\frac {3\,b\,\left (\frac {2\,A}{c^3}-\frac {6\,B\,b}{c^4}\right )}{c}+\frac {6\,B\,b^2}{c^5}\right )+\frac {2\,B\,x^{5/2}}{5\,c^3}-\frac {7\,b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,\sqrt {x}\,\left (5\,A\,c-9\,B\,b\right )}{9\,B\,b^3-5\,A\,b^2\,c}\right )\,\left (5\,A\,c-9\,B\,b\right )}{4\,c^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(13/2)*(A + B*x))/(b*x + c*x^2)^3,x)

[Out]

x^(3/2)*((2*A)/(3*c^3) - (2*B*b)/c^4) - (x^(3/2)*((13*A*b^2*c^2)/4 - (17*B*b^3*c)/4) - x^(1/2)*((15*B*b^4)/4 -
(11*A*b^3*c)/4))/(b^2*c^5 + c^7*x^2 + 2*b*c^6*x) - x^(1/2)*((3*b*((2*A)/c^3 - (6*B*b)/c^4))/c + (6*B*b^2)/c^5
) + (2*B*x^(5/2))/(5*c^3) - (7*b^(3/2)*atan((b^(3/2)*c^(1/2)*x^(1/2)*(5*A*c - 9*B*b))/(9*B*b^3 - 5*A*b^2*c))*(
5*A*c - 9*B*b))/(4*c^(11/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(13/2)*(B*x+A)/(c*x**2+b*x)**3,x)

[Out]

Timed out

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