\(\int \frac {1}{(d+e x) (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [1611]

Optimal result

Integrand size = 28, antiderivative size = 253 \[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e}{3 (b d-a e)^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2}{2 (b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^4 (a+b x) \log (d+e x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

e^3/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)-1/4/(-a*e+b*d)/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/3*e/(-a*e+b*d)^2/(b*x+a)^2/((b
*x+a)^2)^(1/2)-1/2*e^2/(-a*e+b*d)^3/(b*x+a)/((b*x+a)^2)^(1/2)+e^4*(b*x+a)*ln(b*x+a)/(-a*e+b*d)^5/((b*x+a)^2)^(
1/2)-e^4*(b*x+a)*ln(e*x+d)/(-a*e+b*d)^5/((b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 46} \[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^4 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac {e^4 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {e^3}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {e^2}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {e}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {1}{4 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[In]

Int[1/((d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

e^3/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(4*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
) + e/(3*(b*d - a*e)^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - e^2/(2*(b*d - a*e)^3*(a + b*x)*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + (e^4*(a + b*x)*Log[a + b*x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^4*(a + b
*x)*Log[d + e*x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 660

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^5 (d+e x)} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{b^4 (b d-a e) (a+b x)^5}-\frac {e}{b^4 (b d-a e)^2 (a+b x)^4}+\frac {e^2}{b^4 (b d-a e)^3 (a+b x)^3}-\frac {e^3}{b^4 (b d-a e)^4 (a+b x)^2}+\frac {e^4}{b^4 (b d-a e)^5 (a+b x)}-\frac {e^5}{b^5 (b d-a e)^5 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {e^3}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e}{3 (b d-a e)^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2}{2 (b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^4 (a+b x) \log (d+e x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.64 \[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-\left ((b d-a e) \left (-25 a^3 e^3+a^2 b e^2 (23 d-52 e x)+a b^2 e \left (-13 d^2+20 d e x-42 e^2 x^2\right )+b^3 \left (3 d^3-4 d^2 e x+6 d e^2 x^2-12 e^3 x^3\right )\right )\right )+12 e^4 (a+b x)^4 \log (a+b x)-12 e^4 (a+b x)^4 \log (d+e x)}{12 (b d-a e)^5 (a+b x)^3 \sqrt {(a+b x)^2}} \]

[In]

Integrate[1/((d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(-((b*d - a*e)*(-25*a^3*e^3 + a^2*b*e^2*(23*d - 52*e*x) + a*b^2*e*(-13*d^2 + 20*d*e*x - 42*e^2*x^2) + b^3*(3*d
^3 - 4*d^2*e*x + 6*d*e^2*x^2 - 12*e^3*x^3))) + 12*e^4*(a + b*x)^4*Log[a + b*x] - 12*e^4*(a + b*x)^4*Log[d + e*
x])/(12*(b*d - a*e)^5*(a + b*x)^3*Sqrt[(a + b*x)^2])

Maple [A] (verified)

Time = 2.64 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.42

method	result	size
default	\(-\frac {\left (12 \ln \left (b x +a \right ) b^{4} e^{4} x^{4}-12 \ln \left (e x +d \right ) x^{4} b^{4} e^{4}+48 \ln \left (b x +a \right ) x^{3} a \,b^{3} e^{4}-48 \ln \left (e x +d \right ) a \,b^{3} e^{4} x^{3}+72 \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{4}-72 \ln \left (e x +d \right ) x^{2} a^{2} b^{2} e^{4}-12 x^{3} a \,b^{3} e^{4}+12 x^{3} b^{4} d \,e^{3}+48 \ln \left (b x +a \right ) x \,a^{3} b \,e^{4}-48 \ln \left (e x +d \right ) x \,a^{3} b \,e^{4}-42 x^{2} a^{2} b^{2} e^{4}+48 x^{2} a \,b^{3} d \,e^{3}-6 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (b x +a \right ) a^{4} e^{4}-12 \ln \left (e x +d \right ) a^{4} e^{4}-52 x \,a^{3} b \,e^{4}+72 x \,a^{2} b^{2} d \,e^{3}-24 x a \,b^{3} d^{2} e^{2}+4 x \,b^{4} d^{3} e -25 e^{4} a^{4}+48 b \,e^{3} d \,a^{3}-36 b^{2} e^{2} d^{2} a^{2}+16 a \,b^{3} d^{3} e -3 b^{4} d^{4}\right ) \left (b x +a \right )}{12 \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(359\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b^{3} e^{3} x^{3}}{e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}+\frac {\left (7 a e -b d \right ) b^{2} e^{2} x^{2}}{2 e^{4} a^{4}-8 b \,e^{3} d \,a^{3}+12 b^{2} e^{2} d^{2} a^{2}-8 a \,b^{3} d^{3} e +2 b^{4} d^{4}}+\frac {e b \left (13 a^{2} e^{2}-5 a b d e +b^{2} d^{2}\right ) x}{3 e^{4} a^{4}-12 b \,e^{3} d \,a^{3}+18 b^{2} e^{2} d^{2} a^{2}-12 a \,b^{3} d^{3} e +3 b^{4} d^{4}}+\frac {25 a^{3} e^{3}-23 a^{2} b d \,e^{2}+13 a \,b^{2} d^{2} e -3 b^{3} d^{3}}{12 e^{4} a^{4}-48 b \,e^{3} d \,a^{3}+72 b^{2} e^{2} d^{2} a^{2}-48 a \,b^{3} d^{3} e +12 b^{4} d^{4}}\right )}{\left (b x +a \right )^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{4} \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (a^{5} e^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{4} \ln \left (b x +a \right )}{\left (b x +a \right ) \left (a^{5} e^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}\)	\(508\)

[In]

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

[Out]

-1/12*(12*ln(b*x+a)*b^4*e^4*x^4-12*ln(e*x+d)*x^4*b^4*e^4+48*ln(b*x+a)*x^3*a*b^3*e^4-48*ln(e*x+d)*a*b^3*e^4*x^3
+72*ln(b*x+a)*x^2*a^2*b^2*e^4-72*ln(e*x+d)*x^2*a^2*b^2*e^4-12*x^3*a*b^3*e^4+12*x^3*b^4*d*e^3+48*ln(b*x+a)*x*a^
3*b*e^4-48*ln(e*x+d)*x*a^3*b*e^4-42*x^2*a^2*b^2*e^4+48*x^2*a*b^3*d*e^3-6*x^2*b^4*d^2*e^2+12*ln(b*x+a)*a^4*e^4-
12*ln(e*x+d)*a^4*e^4-52*x*a^3*b*e^4+72*x*a^2*b^2*d*e^3-24*x*a*b^3*d^2*e^2+4*x*b^4*d^3*e-25*e^4*a^4+48*b*e^3*d*
a^3-36*b^2*e^2*d^2*a^2+16*a*b^3*d^3*e-3*b^4*d^4)*(b*x+a)/(a*e-b*d)^5/((b*x+a)^2)^(5/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 657 vs. \(2 (181) = 362\).

Time = 0.29 (sec) , antiderivative size = 657, normalized size of antiderivative = 2.60 \[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {3 \, b^{4} d^{4} - 16 \, a b^{3} d^{3} e + 36 \, a^{2} b^{2} d^{2} e^{2} - 48 \, a^{3} b d e^{3} + 25 \, a^{4} e^{4} - 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} - 8 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \, {\left (b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} + 18 \, a^{2} b^{2} d e^{3} - 13 \, a^{3} b e^{4}\right )} x - 12 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (a^{4} b^{5} d^{5} - 5 \, a^{5} b^{4} d^{4} e + 10 \, a^{6} b^{3} d^{3} e^{2} - 10 \, a^{7} b^{2} d^{2} e^{3} + 5 \, a^{8} b d e^{4} - a^{9} e^{5} + {\left (b^{9} d^{5} - 5 \, a b^{8} d^{4} e + 10 \, a^{2} b^{7} d^{3} e^{2} - 10 \, a^{3} b^{6} d^{2} e^{3} + 5 \, a^{4} b^{5} d e^{4} - a^{5} b^{4} e^{5}\right )} x^{4} + 4 \, {\left (a b^{8} d^{5} - 5 \, a^{2} b^{7} d^{4} e + 10 \, a^{3} b^{6} d^{3} e^{2} - 10 \, a^{4} b^{5} d^{2} e^{3} + 5 \, a^{5} b^{4} d e^{4} - a^{6} b^{3} e^{5}\right )} x^{3} + 6 \, {\left (a^{2} b^{7} d^{5} - 5 \, a^{3} b^{6} d^{4} e + 10 \, a^{4} b^{5} d^{3} e^{2} - 10 \, a^{5} b^{4} d^{2} e^{3} + 5 \, a^{6} b^{3} d e^{4} - a^{7} b^{2} e^{5}\right )} x^{2} + 4 \, {\left (a^{3} b^{6} d^{5} - 5 \, a^{4} b^{5} d^{4} e + 10 \, a^{5} b^{4} d^{3} e^{2} - 10 \, a^{6} b^{3} d^{2} e^{3} + 5 \, a^{7} b^{2} d e^{4} - a^{8} b e^{5}\right )} x\right )}} \]

[In]

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

[Out]

-1/12*(3*b^4*d^4 - 16*a*b^3*d^3*e + 36*a^2*b^2*d^2*e^2 - 48*a^3*b*d*e^3 + 25*a^4*e^4 - 12*(b^4*d*e^3 - a*b^3*e
^4)*x^3 + 6*(b^4*d^2*e^2 - 8*a*b^3*d*e^3 + 7*a^2*b^2*e^4)*x^2 - 4*(b^4*d^3*e - 6*a*b^3*d^2*e^2 + 18*a^2*b^2*d*
e^3 - 13*a^3*b*e^4)*x - 12*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*log(b
*x + a) + 12*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*log(e*x + d))/(a^4*
b^5*d^5 - 5*a^5*b^4*d^4*e + 10*a^6*b^3*d^3*e^2 - 10*a^7*b^2*d^2*e^3 + 5*a^8*b*d*e^4 - a^9*e^5 + (b^9*d^5 - 5*a
*b^8*d^4*e + 10*a^2*b^7*d^3*e^2 - 10*a^3*b^6*d^2*e^3 + 5*a^4*b^5*d*e^4 - a^5*b^4*e^5)*x^4 + 4*(a*b^8*d^5 - 5*a
^2*b^7*d^4*e + 10*a^3*b^6*d^3*e^2 - 10*a^4*b^5*d^2*e^3 + 5*a^5*b^4*d*e^4 - a^6*b^3*e^5)*x^3 + 6*(a^2*b^7*d^5 -
5*a^3*b^6*d^4*e + 10*a^4*b^5*d^3*e^2 - 10*a^5*b^4*d^2*e^3 + 5*a^6*b^3*d*e^4 - a^7*b^2*e^5)*x^2 + 4*(a^3*b^6*d
^5 - 5*a^4*b^5*d^4*e + 10*a^5*b^4*d^3*e^2 - 10*a^6*b^3*d^2*e^3 + 5*a^7*b^2*d*e^4 - a^8*b*e^5)*x)

Sympy [F]

\[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral(1/((d + e*x)*((a + b*x)**2)**(5/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(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. 418 vs. \(2 (181) = 362\).

Time = 0.28 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.65 \[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {b e^{4} \log \left ({\left | b x + a \right |}\right )}{b^{6} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} b e^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {e^{5} \log \left ({\left | e x + d \right |}\right )}{b^{5} d^{5} e \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {3 \, b^{4} d^{4} - 16 \, a b^{3} d^{3} e + 36 \, a^{2} b^{2} d^{2} e^{2} - 48 \, a^{3} b d e^{3} + 25 \, a^{4} e^{4} - 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} - 8 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \, {\left (b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} + 18 \, a^{2} b^{2} d e^{3} - 13 \, a^{3} b e^{4}\right )} x}{12 \, {\left (b d - a e\right )}^{5} {\left (b x + a\right )}^{4} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

b*e^4*log(abs(b*x + a))/(b^6*d^5*sgn(b*x + a) - 5*a*b^5*d^4*e*sgn(b*x + a) + 10*a^2*b^4*d^3*e^2*sgn(b*x + a) -
10*a^3*b^3*d^2*e^3*sgn(b*x + a) + 5*a^4*b^2*d*e^4*sgn(b*x + a) - a^5*b*e^5*sgn(b*x + a)) - e^5*log(abs(e*x +
d))/(b^5*d^5*e*sgn(b*x + a) - 5*a*b^4*d^4*e^2*sgn(b*x + a) + 10*a^2*b^3*d^3*e^3*sgn(b*x + a) - 10*a^3*b^2*d^2*
e^4*sgn(b*x + a) + 5*a^4*b*d*e^5*sgn(b*x + a) - a^5*e^6*sgn(b*x + a)) - 1/12*(3*b^4*d^4 - 16*a*b^3*d^3*e + 36*
a^2*b^2*d^2*e^2 - 48*a^3*b*d*e^3 + 25*a^4*e^4 - 12*(b^4*d*e^3 - a*b^3*e^4)*x^3 + 6*(b^4*d^2*e^2 - 8*a*b^3*d*e^
3 + 7*a^2*b^2*e^4)*x^2 - 4*(b^4*d^3*e - 6*a*b^3*d^2*e^2 + 18*a^2*b^2*d*e^3 - 13*a^3*b*e^4)*x)/((b*d - a*e)^5*(
b*x + a)^4*sgn(b*x + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (d+e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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