\(\int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx\) [506]

Optimal result

Integrand size = 20, antiderivative size = 152 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=\frac {5}{4} b (4 A b+3 a B) \sqrt {x} \sqrt {a+b x}+\frac {5 b (4 A b+3 a B) \sqrt {x} (a+b x)^{3/2}}{6 a}-\frac {2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {5}{4} a \sqrt {b} (4 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

[Out]

-2/3*A*(b*x+a)^(7/2)/a/x^(3/2)+5/4*a*(4*A*b+3*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))*b^(1/2)-2/3*(4*A*b+3
*B*a)*(b*x+a)^(5/2)/a/x^(1/2)+5/6*b*(4*A*b+3*B*a)*(b*x+a)^(3/2)*x^(1/2)/a+5/4*b*(4*A*b+3*B*a)*x^(1/2)*(b*x+a)^
(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {79, 49, 52, 65, 223, 212} \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=\frac {5}{4} a \sqrt {b} (3 a B+4 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 (a+b x)^{5/2} (3 a B+4 A b)}{3 a \sqrt {x}}+\frac {5 b \sqrt {x} (a+b x)^{3/2} (3 a B+4 A b)}{6 a}+\frac {5}{4} b \sqrt {x} \sqrt {a+b x} (3 a B+4 A b)-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}} \]

[In]

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

[Out]

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

Rule 49

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 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {\left (2 \left (2 A b+\frac {3 a B}{2}\right )\right ) \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx}{3 a} \\ & = -\frac {2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {(5 b (4 A b+3 a B)) \int \frac {(a+b x)^{3/2}}{\sqrt {x}} \, dx}{3 a} \\ & = \frac {5 b (4 A b+3 a B) \sqrt {x} (a+b x)^{3/2}}{6 a}-\frac {2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {1}{4} (5 b (4 A b+3 a B)) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx \\ & = \frac {5}{4} b (4 A b+3 a B) \sqrt {x} \sqrt {a+b x}+\frac {5 b (4 A b+3 a B) \sqrt {x} (a+b x)^{3/2}}{6 a}-\frac {2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {1}{8} (5 a b (4 A b+3 a B)) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx \\ & = \frac {5}{4} b (4 A b+3 a B) \sqrt {x} \sqrt {a+b x}+\frac {5 b (4 A b+3 a B) \sqrt {x} (a+b x)^{3/2}}{6 a}-\frac {2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {1}{4} (5 a b (4 A b+3 a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {5}{4} b (4 A b+3 a B) \sqrt {x} \sqrt {a+b x}+\frac {5 b (4 A b+3 a B) \sqrt {x} (a+b x)^{3/2}}{6 a}-\frac {2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {1}{4} (5 a b (4 A b+3 a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right ) \\ & = \frac {5}{4} b (4 A b+3 a B) \sqrt {x} \sqrt {a+b x}+\frac {5 b (4 A b+3 a B) \sqrt {x} (a+b x)^{3/2}}{6 a}-\frac {2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac {5}{4} a \sqrt {b} (4 A b+3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.71 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (6 b^2 x^2 (2 A+B x)-8 a^2 (A+3 B x)+a b x (-56 A+27 B x)\right )}{12 x^{3/2}}+\frac {5}{2} a \sqrt {b} (4 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right ) \]

[In]

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

[Out]

(Sqrt[a + b*x]*(6*b^2*x^2*(2*A + B*x) - 8*a^2*(A + 3*B*x) + a*b*x*(-56*A + 27*B*x)))/(12*x^(3/2)) + (5*a*Sqrt[
b]*(4*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/2

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-6 b^{2} B \,x^{3}-12 A \,b^{2} x^{2}-27 B a b \,x^{2}+56 a A b x +24 a^{2} B x +8 a^{2} A \right )}{12 x^{\frac {3}{2}}}+\frac {5 a \sqrt {b}\, \left (4 A b +3 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{8 \sqrt {x}\, \sqrt {b x +a}}\)	\(118\)
default	\(\frac {\sqrt {b x +a}\, \left (12 B \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{3}+60 A \,b^{2} \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,x^{2}+24 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}+45 B b \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} x^{2}+54 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a \,x^{2}-112 A a \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}-48 B \,a^{2} x \sqrt {x \left (b x +a \right )}\, \sqrt {b}-16 A \,a^{2} \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{24 x^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(207\)

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(b*x+a)^(1/2)*(-6*B*b^2*x^3-12*A*b^2*x^2-27*B*a*b*x^2+56*A*a*b*x+24*B*a^2*x+8*A*a^2)/x^(3/2)+5/8*a*b^(1/
2)*(4*A*b+3*B*a)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=\left [\frac {15 \, {\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt {b} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, x^{2}}, -\frac {15 \, {\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{12 \, x^{2}}\right ] \]

[In]

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

[Out]

[1/24*(15*(3*B*a^2 + 4*A*a*b)*sqrt(b)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(6*B*b^2*x^3 -
8*A*a^2 + 3*(9*B*a*b + 4*A*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2, -1/12*(15*(3*B*a^2
+ 4*A*a*b)*sqrt(-b)*x^2*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (6*B*b^2*x^3 - 8*A*a^2 + 3*(9*B*a*b + 4*A
*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2]

Sympy [A] (verification not implemented)

Time = 6.08 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.39 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=- \frac {4 A a^{\frac {3}{2}} b}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + A \sqrt {a} b^{2} \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {4 A \sqrt {a} b^{2} \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} - \frac {2 A a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} + 5 A a b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 B a^{\frac {5}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + 2 B a^{\frac {3}{2}} b \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {2 B a^{\frac {3}{2}} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 4 B a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + 2 B b^{2} \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

-4*A*a**(3/2)*b/(sqrt(x)*sqrt(1 + b*x/a)) + A*sqrt(a)*b**2*sqrt(x)*sqrt(1 + b*x/a) - 4*A*sqrt(a)*b**2*sqrt(x)/
sqrt(1 + b*x/a) - 2*A*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 2*A*a*b**(3/2)*sqrt(a/(b*x) + 1)/3 + 5*A*a*b**(3/
2)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) - 2*B*a**(5/2)/(sqrt(x)*sqrt(1 + b*x/a)) + 2*B*a**(3/2)*b*sqrt(x)*sqrt(1 + b
*x/a) - 2*B*a**(3/2)*b*sqrt(x)/sqrt(1 + b*x/a) + 4*B*a**2*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) + 2*B*b**2*Pi
ecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))
/sqrt(b*x), True))/(8*b) + a*sqrt(x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0)), (sqrt(a)*x**(3
/2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=\frac {15}{8} \, B a^{2} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + \frac {5}{2} \, A a b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {15 \, \sqrt {b x^{2} + a x} B a^{2}}{4 \, x} - \frac {35 \, \sqrt {b x^{2} + a x} A a b}{6 \, x} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{4 \, x^{2}} - \frac {5 \, \sqrt {b x^{2} + a x} A a^{2}}{6 \, x^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{2 \, x^{3}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{6 \, x^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{x^{4}} \]

[In]

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

[Out]

15/8*B*a^2*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 5/2*A*a*b^(3/2)*log(2*b*x + a + 2*sqrt(b*x^2
+ a*x)*sqrt(b)) - 15/4*sqrt(b*x^2 + a*x)*B*a^2/x - 35/6*sqrt(b*x^2 + a*x)*A*a*b/x + 5/4*(b*x^2 + a*x)^(3/2)*B
*a/x^2 - 5/6*sqrt(b*x^2 + a*x)*A*a^2/x^2 + 1/2*(b*x^2 + a*x)^(5/2)*B/x^3 - 5/6*(b*x^2 + a*x)^(3/2)*A*a/x^3 + (
b*x^2 + a*x)^(5/2)*A/x^4

Giac [A] (verification not implemented)

none

Time = 76.94 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.22 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=-\frac {{\left (\frac {15 \, {\left (3 \, B a^{2} b + 4 \, A a b^{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \frac {{\left ({\left (3 \, {\left (2 \, {\left (b x + a\right )} B b^{2} + \frac {3 \, B a^{2} b^{3} + 4 \, A a b^{4}}{a b}\right )} {\left (b x + a\right )} - \frac {20 \, {\left (3 \, B a^{3} b^{3} + 4 \, A a^{2} b^{4}\right )}}{a b}\right )} {\left (b x + a\right )} + \frac {15 \, {\left (3 \, B a^{4} b^{3} + 4 \, A a^{3} b^{4}\right )}}{a b}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}}\right )} b}{12 \, {\left | b \right |}} \]

[In]

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

[Out]

-1/12*(15*(3*B*a^2*b + 4*A*a*b^2)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/sqrt(b) - ((3*(2*
(b*x + a)*B*b^2 + (3*B*a^2*b^3 + 4*A*a*b^4)/(a*b))*(b*x + a) - 20*(3*B*a^3*b^3 + 4*A*a^2*b^4)/(a*b))*(b*x + a)
+ 15*(3*B*a^4*b^3 + 4*A*a^3*b^4)/(a*b))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(3/2))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{x^{5/2}} \,d x \]

[In]

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

[Out]

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