\(\int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx\) [507]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=\frac {b^2 (2 A b+5 a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 b (2 A b+5 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 (2 A b+5 a B) (a+b x)^{5/2}}{15 a x^{3/2}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}}+b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

[Out]

-2/15*(2*A*b+5*B*a)*(b*x+a)^(5/2)/a/x^(3/2)-2/5*A*(b*x+a)^(7/2)/a/x^(5/2)+b^(3/2)*(2*A*b+5*B*a)*arctanh(b^(1/2
)*x^(1/2)/(b*x+a)^(1/2))-2/3*b*(2*A*b+5*B*a)*(b*x+a)^(3/2)/a/x^(1/2)+b^2*(2*A*b+5*B*a)*x^(1/2)*(b*x+a)^(1/2)/a

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 150, 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} \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=b^{3/2} (5 a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )+\frac {b^2 \sqrt {x} \sqrt {a+b x} (5 a B+2 A b)}{a}-\frac {2 (a+b x)^{5/2} (5 a B+2 A b)}{15 a x^{3/2}}-\frac {2 b (a+b x)^{3/2} (5 a B+2 A b)}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{7/2}}{5 a x^{5/2}} \]

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(7/2),x]

[Out]

(b^2*(2*A*b + 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/a - (2*b*(2*A*b + 5*a*B)*(a + b*x)^(3/2))/(3*a*Sqrt[x]) - (2*(2*A*
b + 5*a*B)*(a + b*x)^(5/2))/(15*a*x^(3/2)) - (2*A*(a + b*x)^(7/2))/(5*a*x^(5/2)) + b^(3/2)*(2*A*b + 5*a*B)*Arc
Tanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]]

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=-\frac {\sqrt {a+b x} \left (b^2 x^2 (46 A-15 B x)+2 a^2 (3 A+5 B x)+2 a b x (11 A+35 B x)\right )}{15 x^{5/2}}+2 b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right ) \]

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(7/2),x]

[Out]

-1/15*(Sqrt[a + b*x]*(b^2*x^2*(46*A - 15*B*x) + 2*a^2*(3*A + 5*B*x) + 2*a*b*x*(11*A + 35*B*x)))/x^(5/2) + 2*b^
(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-15 b^{2} B \,x^{3}+46 A \,b^{2} x^{2}+70 B a b \,x^{2}+22 a A b x +10 a^{2} B x +6 a^{2} A \right )}{15 x^{\frac {5}{2}}}+\frac {b^{\frac {3}{2}} \left (2 A b +5 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{2 \sqrt {x}\, \sqrt {b x +a}}\)	\(117\)
default	\(\frac {\sqrt {b x +a}\, \left (30 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b^{3} x^{3}+75 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{2} x^{3}+30 B \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{3}-92 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}-140 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a \,x^{2}-44 A a \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}-20 B \,a^{2} x \sqrt {x \left (b x +a \right )}\, \sqrt {b}-12 A \,a^{2} \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{30 x^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(206\)

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/15*(b*x+a)^(1/2)*(-15*B*b^2*x^3+46*A*b^2*x^2+70*B*a*b*x^2+22*A*a*b*x+10*B*a^2*x+6*A*a^2)/x^(5/2)+1/2*b^(3/2
)*(2*A*b+5*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 = 1.45 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=\left [\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {b} x^{3} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{30 \, x^{3}}, -\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, x^{3}}\right ] \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (144) = 288\).

Time = 7.18 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.02 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=- \frac {2 A \sqrt {a} b^{2}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {2 A a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {22 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 x} - \frac {16 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15} + 2 A b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 A b^{3} \sqrt {x}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}} - \frac {4 B a^{\frac {3}{2}} b}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + B \sqrt {a} b^{2} \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {4 B \sqrt {a} b^{2} \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} - \frac {2 B a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} + 5 B a b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} \]

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**(7/2),x)

[Out]

-2*A*sqrt(a)*b**2/(sqrt(x)*sqrt(1 + b*x/a)) - 2*A*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 22*A*a*b**(3/2)*sq
rt(a/(b*x) + 1)/(15*x) - 16*A*b**(5/2)*sqrt(a/(b*x) + 1)/15 + 2*A*b**(5/2)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) - 2*
A*b**3*sqrt(x)/(sqrt(a)*sqrt(1 + b*x/a)) - 4*B*a**(3/2)*b/(sqrt(x)*sqrt(1 + b*x/a)) + B*sqrt(a)*b**2*sqrt(x)*s
qrt(1 + b*x/a) - 4*B*sqrt(a)*b**2*sqrt(x)/sqrt(1 + b*x/a) - 2*B*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 2*B*a*b
**(3/2)*sqrt(a/(b*x) + 1)/3 + 5*B*a*b**(3/2)*asinh(sqrt(b)*sqrt(x)/sqrt(a))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (120) = 240\).

Time = 0.20 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=\frac {5}{2} \, B a b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + A b^{\frac {5}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {35 \, \sqrt {b x^{2} + a x} B a b}{6 \, x} - \frac {38 \, \sqrt {b x^{2} + a x} A b^{2}}{15 \, x} - \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{6 \, x^{2}} - \frac {7 \, \sqrt {b x^{2} + a x} A a b}{30 \, x^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{6 \, x^{3}} + \frac {3 \, \sqrt {b x^{2} + a x} A a^{2}}{10 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A b}{3 \, x^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{2 \, x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{5 \, x^{5}} \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(7/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 75.64 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=-\frac {{\left (\frac {15 \, {\left (5 \, B a b^{2} + 2 \, A b^{3}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \frac {{\left ({\left ({\left (15 \, {\left (b x + a\right )} B b^{4} - \frac {23 \, {\left (5 \, B a^{3} b^{6} + 2 \, A a^{2} b^{7}\right )}}{a^{2} b^{2}}\right )} {\left (b x + a\right )} + \frac {35 \, {\left (5 \, B a^{4} b^{6} + 2 \, A a^{3} b^{7}\right )}}{a^{2} b^{2}}\right )} {\left (b x + a\right )} - \frac {15 \, {\left (5 \, B a^{5} b^{6} + 2 \, A a^{4} b^{7}\right )}}{a^{2} b^{2}}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}}\right )} b}{15 \, {\left | b \right |}} \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(7/2),x, algorithm="giac")

[Out]

-1/15*(15*(5*B*a*b^2 + 2*A*b^3)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/sqrt(b) - (((15*(b*
x + a)*B*b^4 - 23*(5*B*a^3*b^6 + 2*A*a^2*b^7)/(a^2*b^2))*(b*x + a) + 35*(5*B*a^4*b^6 + 2*A*a^3*b^7)/(a^2*b^2))
*(b*x + a) - 15*(5*B*a^5*b^6 + 2*A*a^4*b^7)/(a^2*b^2))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(5/2))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{7/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{x^{7/2}} \,d x \]

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^(7/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(5/2))/x^(7/2), x)