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

Optimal. Leaf size=193 \[ -\frac {9 c^{5/2} (7 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{13/2}}-\frac {9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt {x}}+\frac {3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {781, 78, 51, 63, 205} \begin {gather*} -\frac {9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt {x}}-\frac {9 c^{5/2} (7 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{13/2}}+\frac {3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}+\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(9*(7*b*B - 11*A*c))/(28*b^3*c*x^(7/2)) - (9*(7*b*B - 11*A*c))/(20*b^4*x^(5/2)) + (3*c*(7*b*B - 11*A*c))/(4*b^
5*x^(3/2)) - (9*c^2*(7*b*B - 11*A*c))/(4*b^6*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(7/2)*(b + c*x)^2) - (7*b*B - 11*
A*c)/(4*b^2*c*x^(7/2)*(b + c*x)) - (9*c^(5/2)*(7*b*B - 11*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(13/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {A+B x}{x^{3/2} \left (b x+c x^2\right )^3} \, dx &=\int \frac {A+B x}{x^{9/2} (b+c x)^3} \, dx\\ &=-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {\left (\frac {7 b B}{2}-\frac {11 A c}{2}\right ) \int \frac {1}{x^{9/2} (b+c x)^2} \, dx}{2 b c}\\ &=-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac {(9 (7 b B-11 A c)) \int \frac {1}{x^{9/2} (b+c x)} \, dx}{8 b^2 c}\\ &=\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}+\frac {(9 (7 b B-11 A c)) \int \frac {1}{x^{7/2} (b+c x)} \, dx}{8 b^3}\\ &=\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac {(9 c (7 b B-11 A c)) \int \frac {1}{x^{5/2} (b+c x)} \, dx}{8 b^4}\\ &=\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac {3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}+\frac {\left (9 c^2 (7 b B-11 A c)\right ) \int \frac {1}{x^{3/2} (b+c x)} \, dx}{8 b^5}\\ &=\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac {3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac {9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt {x}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac {\left (9 c^3 (7 b B-11 A c)\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 b^6}\\ &=\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac {3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac {9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt {x}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac {\left (9 c^3 (7 b B-11 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 b^6}\\ &=\frac {9 (7 b B-11 A c)}{28 b^3 c x^{7/2}}-\frac {9 (7 b B-11 A c)}{20 b^4 x^{5/2}}+\frac {3 c (7 b B-11 A c)}{4 b^5 x^{3/2}}-\frac {9 c^2 (7 b B-11 A c)}{4 b^6 \sqrt {x}}-\frac {b B-A c}{2 b c x^{7/2} (b+c x)^2}-\frac {7 b B-11 A c}{4 b^2 c x^{7/2} (b+c x)}-\frac {9 c^{5/2} (7 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{13/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 61, normalized size = 0.32 \begin {gather*} \frac {\frac {7 b^2 (A c-b B)}{(b+c x)^2}+(7 b B-11 A c) \, _2F_1\left (-\frac {7}{2},2;-\frac {5}{2};-\frac {c x}{b}\right )}{14 b^3 c x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*b^2*(-(b*B) + A*c))/(b + c*x)^2 + (7*b*B - 11*A*c)*Hypergeometric2F1[-7/2, 2, -5/2, -((c*x)/b)])/(14*b^3*c
*x^(7/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.22, size = 173, normalized size = 0.90 \begin {gather*} \frac {-40 A b^5+88 A b^4 c x-264 A b^3 c^2 x^2+1848 A b^2 c^3 x^3+5775 A b c^4 x^4+3465 A c^5 x^5-56 b^5 B x+168 b^4 B c x^2-1176 b^3 B c^2 x^3-3675 b^2 B c^3 x^4-2205 b B c^4 x^5}{140 b^6 x^{7/2} (b+c x)^2}-\frac {9 \left (7 b B c^{5/2}-11 A c^{7/2}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-40*A*b^5 - 56*b^5*B*x + 88*A*b^4*c*x + 168*b^4*B*c*x^2 - 264*A*b^3*c^2*x^2 - 1176*b^3*B*c^2*x^3 + 1848*A*b^2
*c^3*x^3 - 3675*b^2*B*c^3*x^4 + 5775*A*b*c^4*x^4 - 2205*b*B*c^4*x^5 + 3465*A*c^5*x^5)/(140*b^6*x^(7/2)*(b + c*
x)^2) - (9*(7*b*B*c^(5/2) - 11*A*c^(7/2))*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(13/2))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 490, normalized size = 2.54 \begin {gather*} \left [-\frac {315 \, {\left ({\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{6} + 2 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{5} + {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x + 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (40 \, A b^{5} + 315 \, {\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \, {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \, {\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x\right )} \sqrt {x}}{280 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}, \frac {315 \, {\left ({\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{6} + 2 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{5} + {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) - {\left (40 \, A b^{5} + 315 \, {\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \, {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \, {\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x\right )} \sqrt {x}}{140 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/280*(315*((7*B*b*c^4 - 11*A*c^5)*x^6 + 2*(7*B*b^2*c^3 - 11*A*b*c^4)*x^5 + (7*B*b^3*c^2 - 11*A*b^2*c^3)*x^4
)*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5
+ 525*(7*B*b^2*c^3 - 11*A*b*c^4)*x^4 + 168*(7*B*b^3*c^2 - 11*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A*b^3*c^2)*x^
2 + 8*(7*B*b^5 - 11*A*b^4*c)*x)*sqrt(x))/(b^6*c^2*x^6 + 2*b^7*c*x^5 + b^8*x^4), 1/140*(315*((7*B*b*c^4 - 11*A*
c^5)*x^6 + 2*(7*B*b^2*c^3 - 11*A*b*c^4)*x^5 + (7*B*b^3*c^2 - 11*A*b^2*c^3)*x^4)*sqrt(c/b)*arctan(b*sqrt(c/b)/(
c*sqrt(x))) - (40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5 + 525*(7*B*b^2*c^3 - 11*A*b*c^4)*x^4 + 168*(7*B*b^3*c
^2 - 11*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A*b^3*c^2)*x^2 + 8*(7*B*b^5 - 11*A*b^4*c)*x)*sqrt(x))/(b^6*c^2*x^6
+ 2*b^7*c*x^5 + b^8*x^4)]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 159, normalized size = 0.82 \begin {gather*} -\frac {9 \, {\left (7 \, B b c^{3} - 11 \, A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{6}} - \frac {15 \, B b c^{4} x^{\frac {3}{2}} - 19 \, A c^{5} x^{\frac {3}{2}} + 17 \, B b^{2} c^{3} \sqrt {x} - 21 \, A b c^{4} \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{6}} - \frac {2 \, {\left (210 \, B b c^{2} x^{3} - 350 \, A c^{3} x^{3} - 35 \, B b^{2} c x^{2} + 70 \, A b c^{2} x^{2} + 7 \, B b^{3} x - 21 \, A b^{2} c x + 5 \, A b^{3}\right )}}{35 \, b^{6} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/4*(7*B*b*c^3 - 11*A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^6) - 1/4*(15*B*b*c^4*x^(3/2) - 19*A*c^5*x
^(3/2) + 17*B*b^2*c^3*sqrt(x) - 21*A*b*c^4*sqrt(x))/((c*x + b)^2*b^6) - 2/35*(210*B*b*c^2*x^3 - 350*A*c^3*x^3
- 35*B*b^2*c*x^2 + 70*A*b*c^2*x^2 + 7*B*b^3*x - 21*A*b^2*c*x + 5*A*b^3)/(b^6*x^(7/2))

________________________________________________________________________________________

maple [A]  time = 0.07, size = 202, normalized size = 1.05 \begin {gather*} \frac {19 A \,c^{5} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{6}}-\frac {15 B \,c^{4} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{5}}+\frac {21 A \,c^{4} \sqrt {x}}{4 \left (c x +b \right )^{2} b^{5}}-\frac {17 B \,c^{3} \sqrt {x}}{4 \left (c x +b \right )^{2} b^{4}}+\frac {99 A \,c^{4} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{6}}-\frac {63 B \,c^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{5}}+\frac {20 A \,c^{3}}{b^{6} \sqrt {x}}-\frac {12 B \,c^{2}}{b^{5} \sqrt {x}}-\frac {4 A \,c^{2}}{b^{5} x^{\frac {3}{2}}}+\frac {2 B c}{b^{4} x^{\frac {3}{2}}}+\frac {6 A c}{5 b^{4} x^{\frac {5}{2}}}-\frac {2 B}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 A}{7 b^{3} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

19/4/b^6*c^5/(c*x+b)^2*x^(3/2)*A-15/4/b^5*c^4/(c*x+b)^2*x^(3/2)*B+21/4/b^5*c^4/(c*x+b)^2*A*x^(1/2)-17/4/b^4*c^
3/(c*x+b)^2*B*x^(1/2)+99/4/b^6*c^4/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*A-63/4/b^5*c^3/(b*c)^(1/2)*arct
an(1/(b*c)^(1/2)*c*x^(1/2))*B-2/7/b^3*A/x^(7/2)+6/5/b^4/x^(5/2)*A*c-2/5/b^3/x^(5/2)*B-4*c^2/b^5/x^(3/2)*A+2*c/
b^4/x^(3/2)*B+20*c^3/b^6/x^(1/2)*A-12*c^2/b^5/x^(1/2)*B

________________________________________________________________________________________

maxima [A]  time = 1.37, size = 178, normalized size = 0.92 \begin {gather*} -\frac {40 \, A b^{5} + 315 \, {\left (7 \, B b c^{4} - 11 \, A c^{5}\right )} x^{5} + 525 \, {\left (7 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{4} + 168 \, {\left (7 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{3} - 24 \, {\left (7 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (7 \, B b^{5} - 11 \, A b^{4} c\right )} x}{140 \, {\left (b^{6} c^{2} x^{\frac {11}{2}} + 2 \, b^{7} c x^{\frac {9}{2}} + b^{8} x^{\frac {7}{2}}\right )}} - \frac {9 \, {\left (7 \, B b c^{3} - 11 \, A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/140*(40*A*b^5 + 315*(7*B*b*c^4 - 11*A*c^5)*x^5 + 525*(7*B*b^2*c^3 - 11*A*b*c^4)*x^4 + 168*(7*B*b^3*c^2 - 11
*A*b^2*c^3)*x^3 - 24*(7*B*b^4*c - 11*A*b^3*c^2)*x^2 + 8*(7*B*b^5 - 11*A*b^4*c)*x)/(b^6*c^2*x^(11/2) + 2*b^7*c*
x^(9/2) + b^8*x^(7/2)) - 9/4*(7*B*b*c^3 - 11*A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^6)

________________________________________________________________________________________

mupad [B]  time = 1.18, size = 154, normalized size = 0.80 \begin {gather*} \frac {\frac {2\,x\,\left (11\,A\,c-7\,B\,b\right )}{35\,b^2}-\frac {2\,A}{7\,b}+\frac {6\,c^2\,x^3\,\left (11\,A\,c-7\,B\,b\right )}{5\,b^4}+\frac {15\,c^3\,x^4\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^5}+\frac {9\,c^4\,x^5\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^6}-\frac {6\,c\,x^2\,\left (11\,A\,c-7\,B\,b\right )}{35\,b^3}}{b^2\,x^{7/2}+c^2\,x^{11/2}+2\,b\,c\,x^{9/2}}+\frac {9\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^{13/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x*(11*A*c - 7*B*b))/(35*b^2) - (2*A)/(7*b) + (6*c^2*x^3*(11*A*c - 7*B*b))/(5*b^4) + (15*c^3*x^4*(11*A*c -
7*B*b))/(4*b^5) + (9*c^4*x^5*(11*A*c - 7*B*b))/(4*b^6) - (6*c*x^2*(11*A*c - 7*B*b))/(35*b^3))/(b^2*x^(7/2) + c
^2*x^(11/2) + 2*b*c*x^(9/2)) + (9*c^(5/2)*atan((c^(1/2)*x^(1/2))/b^(1/2))*(11*A*c - 7*B*b))/(4*b^(13/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________