\(\int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [1594]

Optimal result

Integrand size = 28, antiderivative size = 182 \[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a+b x}{2 (b d-a e) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{(b d-a e)^2 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (d+e x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 182, 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} \[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {b (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}+\frac {a+b x}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}+\frac {b^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

[In]

Int[1/((d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

(a + b*x)/(2*(b*d - a*e)*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (b*(a + b*x))/((b*d - a*e)^2*(d + e*x)*S
qrt[a^2 + 2*a*b*x + b^2*x^2]) + (b^2*(a + b*x)*Log[a + b*x])/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (
b^2*(a + b*x)*Log[d + e*x])/((b*d - a*e)^3*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 (a b+b^2 x\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (a b+b^2 x\right ) \int \left (\frac {b^2}{(b d-a e)^3 (a+b x)}-\frac {e}{b (b d-a e) (d+e x)^3}-\frac {e}{(b d-a e)^2 (d+e x)^2}-\frac {b e}{(b d-a e)^3 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {a+b x}{2 (b d-a e) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{(b d-a e)^2 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (d+e x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \left ((b d-a e) (3 b d-a e+2 b e x)+2 b^2 (d+e x)^2 \log (a+b x)-2 b^2 (d+e x)^2 \log (d+e x)\right )}{2 (b d-a e)^3 \sqrt {(a+b x)^2} (d+e x)^2} \]

[In]

Integrate[1/((d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

Maple [A] (verified)

Time = 2.35 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (2 \ln \left (b x +a \right ) b^{2} e^{2} x^{2}-2 \ln \left (e x +d \right ) b^{2} e^{2} x^{2}+4 \ln \left (b x +a \right ) x \,b^{2} d e -4 \ln \left (e x +d \right ) x \,b^{2} d e +2 \ln \left (b x +a \right ) b^{2} d^{2}-2 \ln \left (e x +d \right ) b^{2} d^{2}-2 x a b \,e^{2}+2 b^{2} d e x +a^{2} e^{2}-4 a b d e +3 b^{2} d^{2}\right )}{2 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{3} \left (e x +d \right )^{2}}\)	\(162\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b e x}{a^{2} e^{2}-2 a b d e +b^{2} d^{2}}-\frac {a e -3 b d}{2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\right )}{\left (b x +a \right ) \left (e x +d \right )^{2}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\)	\(219\)

[In]

int(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, b^{2} d^{2} - 4 \, a b d e + a^{2} e^{2} + 2 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (b^{3} d^{5} - 3 \, a b^{2} d^{4} e + 3 \, a^{2} b d^{3} e^{2} - a^{3} d^{2} e^{3} + {\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} x^{2} + 2 \, {\left (b^{3} d^{4} e - 3 \, a b^{2} d^{3} e^{2} + 3 \, a^{2} b d^{2} e^{3} - a^{3} d e^{4}\right )} x\right )}} \]

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\left (d + e x\right )^{3} \sqrt {\left (a + b x\right )^{2}}}\, dx \]

[In]

integrate(1/(e*x+d)**3/((b*x+a)**2)**(1/2),x)

[Out]

Integral(1/((d + e*x)**3*sqrt((a + b*x)**2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),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 [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {1}{2} \, {\left (\frac {2 \, b^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}} - \frac {2 \, b^{2} e \log \left ({\left | e x + d \right |}\right )}{b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}} + \frac {3 \, b^{2} d^{2} - 4 \, a b d e + a^{2} e^{2} + 2 \, {\left (b^{2} d e - a b e^{2}\right )} x}{{\left (b d - a e\right )}^{3} {\left (e x + d\right )}^{2}}\right )} \mathrm {sgn}\left (b x + a\right ) \]

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(2*b^3*log(abs(b*x + a))/(b^4*d^3 - 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3) - 2*b^2*e*log(abs(e*x + d
))/(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4) + (3*b^2*d^2 - 4*a*b*d*e + a^2*e^2 + 2*(b^2*d*e - a
*b*e^2)*x)/((b*d - a*e)^3*(e*x + d)^2))*sgn(b*x + a)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^3} \,d x \]

[In]

int(1/(((a + b*x)^2)^(1/2)*(d + e*x)^3),x)

[Out]

int(1/(((a + b*x)^2)^(1/2)*(d + e*x)^3), x)