3.2.2 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=155 \[ c^{3/2} (2 A c+5 b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )+\frac {c^2 \sqrt {b x+c x^2} (2 A c+5 b B)}{b}-\frac {2 c \left (b x+c x^2\right )^{3/2} (2 A c+5 b B)}{3 b x^2}-\frac {2 \left (b x+c x^2\right )^{5/2} (2 A c+5 b B)}{15 b x^4}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6} \]

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {792, 662, 664, 620, 206} \begin {gather*} \frac {c^2 \sqrt {b x+c x^2} (2 A c+5 b B)}{b}+c^{3/2} (2 A c+5 b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 \left (b x+c x^2\right )^{5/2} (2 A c+5 b B)}{15 b x^4}-\frac {2 c \left (b x+c x^2\right )^{3/2} (2 A c+5 b B)}{3 b x^2}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^6,x]

[Out]

(c^2*(5*b*B + 2*A*c)*Sqrt[b*x + c*x^2])/b - (2*c*(5*b*B + 2*A*c)*(b*x + c*x^2)^(3/2))/(3*b*x^2) - (2*(5*b*B +
2*A*c)*(b*x + c*x^2)^(5/2))/(15*b*x^4) - (2*A*(b*x + c*x^2)^(7/2))/(5*b*x^6) + c^(3/2)*(5*b*B + 2*A*c)*ArcTanh
[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^6} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac {\left (2 \left (-6 (-b B+A c)+\frac {7}{2} (-b B+2 A c)\right )\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^5} \, dx}{5 b}\\ &=-\frac {2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac {(c (5 b B+2 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^3} \, dx}{3 b}\\ &=-\frac {2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac {2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac {\left (c^2 (5 b B+2 A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x} \, dx}{b}\\ &=\frac {c^2 (5 b B+2 A c) \sqrt {b x+c x^2}}{b}-\frac {2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac {2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\frac {1}{2} \left (c^2 (5 b B+2 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx\\ &=\frac {c^2 (5 b B+2 A c) \sqrt {b x+c x^2}}{b}-\frac {2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac {2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+\left (c^2 (5 b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )\\ &=\frac {c^2 (5 b B+2 A c) \sqrt {b x+c x^2}}{b}-\frac {2 c (5 b B+2 A c) \left (b x+c x^2\right )^{3/2}}{3 b x^2}-\frac {2 (5 b B+2 A c) \left (b x+c x^2\right )^{5/2}}{15 b x^4}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{5 b x^6}+c^{3/2} (5 b B+2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 87, normalized size = 0.56 \begin {gather*} -\frac {2 \sqrt {x (b+c x)} \left (b^2 x (2 A c+5 b B) \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x}{b}\right )+3 A \sqrt {\frac {c x}{b}+1} (b+c x)^3\right )}{15 b x^3 \sqrt {\frac {c x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^6,x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(3*A*(b + c*x)^3*Sqrt[1 + (c*x)/b] + b^2*(5*b*B + 2*A*c)*x*Hypergeometric2F1[-5/2, -3/2,
-1/2, -((c*x)/b)]))/(15*b*x^3*Sqrt[1 + (c*x)/b])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.54, size = 116, normalized size = 0.75 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-6 A b^2-22 A b c x-46 A c^2 x^2-10 b^2 B x-70 b B c x^2+15 B c^2 x^3\right )}{15 x^3}+\frac {1}{2} \left (-2 A c^{5/2}-5 b B c^{3/2}\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^6,x]

[Out]

(Sqrt[b*x + c*x^2]*(-6*A*b^2 - 10*b^2*B*x - 22*A*b*c*x - 70*b*B*c*x^2 - 46*A*c^2*x^2 + 15*B*c^2*x^3))/(15*x^3)
+ ((-5*b*B*c^(3/2) - 2*A*c^(5/2))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/2

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 224, normalized size = 1.45 \begin {gather*} \left [\frac {15 \, {\left (5 \, B b c + 2 \, A c^{2}\right )} \sqrt {c} x^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (15 \, B c^{2} x^{3} - 6 \, A b^{2} - 2 \, {\left (35 \, B b c + 23 \, A c^{2}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} + 11 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{30 \, x^{3}}, -\frac {15 \, {\left (5 \, B b c + 2 \, A c^{2}\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (15 \, B c^{2} x^{3} - 6 \, A b^{2} - 2 \, {\left (35 \, B b c + 23 \, A c^{2}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} + 11 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/30*(15*(5*B*b*c + 2*A*c^2)*sqrt(c)*x^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(15*B*c^2*x^3 - 6*A
*b^2 - 2*(35*B*b*c + 23*A*c^2)*x^2 - 2*(5*B*b^2 + 11*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^3, -1/15*(15*(5*B*b*c + 2*
A*c^2)*sqrt(-c)*x^3*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (15*B*c^2*x^3 - 6*A*b^2 - 2*(35*B*b*c + 23*A*c^
2)*x^2 - 2*(5*B*b^2 + 11*A*b*c)*x)*sqrt(c*x^2 + b*x))/x^3]

________________________________________________________________________________________

giac [B]  time = 0.27, size = 304, normalized size = 1.96 \begin {gather*} \sqrt {c x^{2} + b x} B c^{2} - \frac {{\left (5 \, B b c^{2} + 2 \, A c^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, \sqrt {c}} + \frac {2 \, {\left (45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{2} c^{\frac {3}{2}} + 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b c^{\frac {5}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{3} c + 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{2} c^{2} + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{4} \sqrt {c} + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{3} c^{\frac {3}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{4} c + 3 \, A b^{5} \sqrt {c}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x, algorithm="giac")

[Out]

sqrt(c*x^2 + b*x)*B*c^2 - 1/2*(5*B*b*c^2 + 2*A*c^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/s
qrt(c) + 2/15*(45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^2*c^(3/2) + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b*c
^(5/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*c + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*c^2 + 5*(sq
rt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^4*sqrt(c) + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^3*c^(3/2) + 15*(sqrt(c
)*x - sqrt(c*x^2 + b*x))*A*b^4*c + 3*A*b^5*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x))^5*sqrt(c))

________________________________________________________________________________________

maple [B]  time = 0.06, size = 460, normalized size = 2.97 \begin {gather*} A \,c^{\frac {5}{2}} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )+\frac {5 B b \,c^{\frac {3}{2}} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2}-\frac {4 \sqrt {c \,x^{2}+b x}\, A \,c^{4} x}{b^{2}}-\frac {10 \sqrt {c \,x^{2}+b x}\, B \,c^{3} x}{b}-\frac {2 \sqrt {c \,x^{2}+b x}\, A \,c^{3}}{b}+\frac {32 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{5} x}{3 b^{4}}+\frac {80 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,c^{4} x}{3 b^{3}}-5 \sqrt {c \,x^{2}+b x}\, B \,c^{2}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{4}}{3 b^{3}}+\frac {40 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,c^{3}}{3 b^{2}}+\frac {256 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,c^{5}}{15 b^{5}}+\frac {128 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,c^{4}}{3 b^{4}}-\frac {256 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A \,c^{4}}{15 b^{5} x^{2}}-\frac {128 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B \,c^{3}}{3 b^{4} x^{2}}+\frac {32 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A \,c^{3}}{5 b^{4} x^{3}}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B \,c^{2}}{b^{3} x^{3}}-\frac {16 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A \,c^{2}}{15 b^{3} x^{4}}-\frac {8 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B c}{3 b^{2} x^{4}}-\frac {4 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A c}{15 b^{2} x^{5}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B}{3 b \,x^{5}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A}{5 b \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x)

[Out]

-2/5*A*(c*x^2+b*x)^(7/2)/b/x^6-4/15*A/b^2*c/x^5*(c*x^2+b*x)^(7/2)-16/15*A/b^3*c^2/x^4*(c*x^2+b*x)^(7/2)+32/5*A
/b^4*c^3/x^3*(c*x^2+b*x)^(7/2)-256/15*A/b^5*c^4/x^2*(c*x^2+b*x)^(7/2)+256/15*A/b^5*c^5*(c*x^2+b*x)^(5/2)+32/3*
A/b^4*c^5*(c*x^2+b*x)^(3/2)*x+16/3*A/b^3*c^4*(c*x^2+b*x)^(3/2)-4*A/b^2*c^4*(c*x^2+b*x)^(1/2)*x-2*A/b*c^3*(c*x^
2+b*x)^(1/2)+A*c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))-2/3*B/b/x^5*(c*x^2+b*x)^(7/2)-8/3*B/b^2*c/x^4
*(c*x^2+b*x)^(7/2)+16*B/b^3*c^2/x^3*(c*x^2+b*x)^(7/2)-128/3*B/b^4*c^3/x^2*(c*x^2+b*x)^(7/2)+128/3*B/b^4*c^4*(c
*x^2+b*x)^(5/2)+80/3*B/b^3*c^4*(c*x^2+b*x)^(3/2)*x+40/3*B/b^2*c^3*(c*x^2+b*x)^(3/2)-10*B/b*c^3*(c*x^2+b*x)^(1/
2)*x-5*B*c^2*(c*x^2+b*x)^(1/2)+5/2*B*b*c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.98, size = 244, normalized size = 1.57 \begin {gather*} \frac {5}{2} \, B b c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + A c^{\frac {5}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {35 \, \sqrt {c x^{2} + b x} B b c}{6 \, x} - \frac {38 \, \sqrt {c x^{2} + b x} A c^{2}}{15 \, x} - \frac {5 \, \sqrt {c x^{2} + b x} B b^{2}}{6 \, x^{2}} - \frac {7 \, \sqrt {c x^{2} + b x} A b c}{30 \, x^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{6 \, x^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} A b^{2}}{10 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A c}{3 \, x^{3}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{x^{4}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{2 \, x^{4}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^6,x, algorithm="maxima")

[Out]

5/2*B*b*c^(3/2)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + A*c^(5/2)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*s
qrt(c)) - 35/6*sqrt(c*x^2 + b*x)*B*b*c/x - 38/15*sqrt(c*x^2 + b*x)*A*c^2/x - 5/6*sqrt(c*x^2 + b*x)*B*b^2/x^2 -
7/30*sqrt(c*x^2 + b*x)*A*b*c/x^2 - 5/6*(c*x^2 + b*x)^(3/2)*B*b/x^3 + 3/10*sqrt(c*x^2 + b*x)*A*b^2/x^3 - 1/3*(
c*x^2 + b*x)^(3/2)*A*c/x^3 + (c*x^2 + b*x)^(5/2)*B/x^4 - 1/2*(c*x^2 + b*x)^(3/2)*A*b/x^4 - 1/5*(c*x^2 + b*x)^(
5/2)*A/x^5

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^6,x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^6, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**6,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**6, x)

________________________________________________________________________________________