\(\int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx\) [2256]

Optimal result

Integrand size = 24, antiderivative size = 111 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx=-\frac {2 B \sqrt {d+e x}}{b^2 \sqrt {a+b x}}-\frac {2 (A b-a B) (d+e x)^{3/2}}{3 b (b d-a e) (a+b x)^{3/2}}+\frac {2 B \sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {79, 49, 65, 223, 212} \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx=-\frac {2 (d+e x)^{3/2} (A b-a B)}{3 b (a+b x)^{3/2} (b d-a e)}+\frac {2 B \sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2}}-\frac {2 B \sqrt {d+e x}}{b^2 \sqrt {a+b x}} \]

[In]

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

[Out]

(-2*B*Sqrt[d + e*x])/(b^2*Sqrt[a + b*x]) - (2*(A*b - a*B)*(d + e*x)^(3/2))/(3*b*(b*d - a*e)*(a + b*x)^(3/2)) +
(2*B*Sqrt[e]*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/b^(5/2)

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx=-\frac {2 \sqrt {d+e x} \left (A b^2 (d+e x)+B \left (-3 a^2 e+3 b^2 d x+2 a b (d-2 e x)\right )\right )}{3 b^2 (b d-a e) (a+b x)^{3/2}}+\frac {2 B \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{b^{5/2}} \]

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(A*b^2*(d + e*x) + B*(-3*a^2*e + 3*b^2*d*x + 2*a*b*(d - 2*e*x))))/(3*b^2*(b*d - a*e)*(a + b*
x)^(3/2)) + (2*B*Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])])/b^(5/2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(502\) vs. \(2(89)=178\).

Time = 1.12 (sec) , antiderivative size = 503, normalized size of antiderivative = 4.53

method	result	size
default	\(\frac {\left (3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} e^{2} x^{2}-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d e \,x^{2}+6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} b \,e^{2} x -6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} d e x +2 A \,b^{2} e x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{3} e^{2}-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} b d e -8 B a b e x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+6 B \,b^{2} d x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+2 A \,b^{2} d \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-6 B \,a^{2} e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+4 B a b d \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right ) \sqrt {e x +d}}{3 \sqrt {b e}\, \left (a e -b d \right ) \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} \left (b x +a \right )^{\frac {3}{2}}}\)	\(503\)

[In]

int((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(3*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*e^2*x^2-3*B*ln(1/2*
(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d*e*x^2+6*B*ln(1/2*(2*b*e*x+2*((b*x+a
)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*e^2*x-6*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d*e*x+2*A*b^2*e*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+3*B*ln(1/2*(2*b*e
*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^3*e^2-3*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))
^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*d*e-8*B*a*b*e*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*B*b^2*d*x
*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+2*A*b^2*d*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-6*B*a^2*e*((b*x+a)*(e*x+d))
^(1/2)*(b*e)^(1/2)+4*B*a*b*d*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))*(e*x+d)^(1/2)/(b*e)^(1/2)/(a*e-b*d)/((b*x+a)
*(e*x+d))^(1/2)/b^2/(b*x+a)^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (89) = 178\).

Time = 0.71 (sec) , antiderivative size = 525, normalized size of antiderivative = 4.73 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx=\left [\frac {3 \, {\left (B a^{2} b d - B a^{3} e + {\left (B b^{3} d - B a b^{2} e\right )} x^{2} + 2 \, {\left (B a b^{2} d - B a^{2} b e\right )} x\right )} \sqrt {\frac {e}{b}} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b^{2} e x + b^{2} d + a b e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {\frac {e}{b}} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (3 \, B a^{2} e - {\left (2 \, B a b + A b^{2}\right )} d - {\left (3 \, B b^{2} d - {\left (4 \, B a b - A b^{2}\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, {\left (a^{2} b^{3} d - a^{3} b^{2} e + {\left (b^{5} d - a b^{4} e\right )} x^{2} + 2 \, {\left (a b^{4} d - a^{2} b^{3} e\right )} x\right )}}, -\frac {3 \, {\left (B a^{2} b d - B a^{3} e + {\left (B b^{3} d - B a b^{2} e\right )} x^{2} + 2 \, {\left (B a b^{2} d - B a^{2} b e\right )} x\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {-\frac {e}{b}}}{2 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )}}\right ) - 2 \, {\left (3 \, B a^{2} e - {\left (2 \, B a b + A b^{2}\right )} d - {\left (3 \, B b^{2} d - {\left (4 \, B a b - A b^{2}\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (a^{2} b^{3} d - a^{3} b^{2} e + {\left (b^{5} d - a b^{4} e\right )} x^{2} + 2 \, {\left (a b^{4} d - a^{2} b^{3} e\right )} x\right )}}\right ] \]

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(B*a^2*b*d - B*a^3*e + (B*b^3*d - B*a*b^2*e)*x^2 + 2*(B*a*b^2*d - B*a^2*b*e)*x)*sqrt(e/b)*log(8*b^2*e^
2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b^2*e*x + b^2*d + a*b*e)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(e/b) +
8*(b^2*d*e + a*b*e^2)*x) + 4*(3*B*a^2*e - (2*B*a*b + A*b^2)*d - (3*B*b^2*d - (4*B*a*b - A*b^2)*e)*x)*sqrt(b*x
+ a)*sqrt(e*x + d))/(a^2*b^3*d - a^3*b^2*e + (b^5*d - a*b^4*e)*x^2 + 2*(a*b^4*d - a^2*b^3*e)*x), -1/3*(3*(B*a^
2*b*d - B*a^3*e + (B*b^3*d - B*a*b^2*e)*x^2 + 2*(B*a*b^2*d - B*a^2*b*e)*x)*sqrt(-e/b)*arctan(1/2*(2*b*e*x + b*
d + a*e)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(-e/b)/(b*e^2*x^2 + a*d*e + (b*d*e + a*e^2)*x)) - 2*(3*B*a^2*e - (2*B
*a*b + A*b^2)*d - (3*B*b^2*d - (4*B*a*b - A*b^2)*e)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(a^2*b^3*d - a^3*b^2*e + (
b^5*d - a*b^4*e)*x^2 + 2*(a*b^4*d - a^2*b^3*e)*x)]

Sympy [F]

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

[In]

integrate((B*x+A)*(e*x+d)**(1/2)/(b*x+a)**(5/2),x)

[Out]

Integral((A + B*x)*sqrt(d + e*x)/(a + b*x)**(5/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (89) = 178\).

Time = 0.41 (sec) , antiderivative size = 558, normalized size of antiderivative = 5.03 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{5/2}} \, dx=-\frac {\sqrt {b e} B {\left | b \right |} \log \left ({\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}{b^{4}} - \frac {4 \, {\left (3 \, \sqrt {b e} B b^{5} d^{3} {\left | b \right |} - 10 \, \sqrt {b e} B a b^{4} d^{2} e {\left | b \right |} + \sqrt {b e} A b^{5} d^{2} e {\left | b \right |} + 11 \, \sqrt {b e} B a^{2} b^{3} d e^{2} {\left | b \right |} - 2 \, \sqrt {b e} A a b^{4} d e^{2} {\left | b \right |} - 4 \, \sqrt {b e} B a^{3} b^{2} e^{3} {\left | b \right |} + \sqrt {b e} A a^{2} b^{3} e^{3} {\left | b \right |} - 6 \, \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} B b^{3} d^{2} {\left | b \right |} + 12 \, \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} B a b^{2} d e {\left | b \right |} - 6 \, \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2} B a^{2} b e^{2} {\left | b \right |} + 3 \, \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{4} B b d {\left | b \right |} - 6 \, \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{4} B a e {\left | b \right |} + 3 \, \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{4} A b e {\left | b \right |}\right )}}{3 \, {\left (b^{2} d - a b e - {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}^{3} b^{3}} \]

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-sqrt(b*e)*B*abs(b)*log((sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)/b^4 - 4/3*(3*sqrt(b
*e)*B*b^5*d^3*abs(b) - 10*sqrt(b*e)*B*a*b^4*d^2*e*abs(b) + sqrt(b*e)*A*b^5*d^2*e*abs(b) + 11*sqrt(b*e)*B*a^2*b
^3*d*e^2*abs(b) - 2*sqrt(b*e)*A*a*b^4*d*e^2*abs(b) - 4*sqrt(b*e)*B*a^3*b^2*e^3*abs(b) + sqrt(b*e)*A*a^2*b^3*e^
3*abs(b) - 6*sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*b^3*d^2*abs(b) + 12
*sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a*b^2*d*e*abs(b) - 6*sqrt(b*e)*
(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a^2*b*e^2*abs(b) + 3*sqrt(b*e)*(sqrt(b*e)*
sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*B*b*d*abs(b) - 6*sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - s
qrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*B*a*e*abs(b) + 3*sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x
+ a)*b*e - a*b*e))^4*A*b*e*abs(b))/((b^2*d - a*b*e - (sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a
*b*e))^2)^3*b^3)

Mupad [F(-1)]

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

[In]

int(((A + B*x)*(d + e*x)^(1/2))/(a + b*x)^(5/2),x)

[Out]

int(((A + B*x)*(d + e*x)^(1/2))/(a + b*x)^(5/2), x)