3.17.2 \(\int \frac {1}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1602]

Optimal. Leaf size=217 \[ \frac {2 b e}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{2 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b e^2 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b e^2 (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 46} \begin {gather*} \frac {e^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac {3 b e^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {3 b e^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {2 b e}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*e)/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(2*(b*d - a*e)^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]) + (e^2*(a + b*x))/((b*d - a*e)^3*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*e^2*(a + b*x)*Log[a + b
*x])/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b*e^2*(a + b*x)*Log[d + e*x])/((b*d - a*e)^4*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*} \int \frac {1}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^3 (d+e x)^2} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{b (b d-a e)^2 (a+b x)^3}-\frac {2 e}{b (b d-a e)^3 (a+b x)^2}+\frac {3 e^2}{b (b d-a e)^4 (a+b x)}-\frac {e^3}{b^3 (b d-a e)^3 (d+e x)^2}-\frac {3 e^3}{b^2 (b d-a e)^4 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 b e}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{2 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b e^2 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b e^2 (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 141, normalized size = 0.65 \begin {gather*} \frac {-\left ((b d-a e) \left (-2 a^2 e^2-a b e (5 d+9 e x)+b^2 \left (d^2-3 d e x-6 e^2 x^2\right )\right )\right )+6 b e^2 (a+b x)^2 (d+e x) \log (a+b x)-6 b e^2 (a+b x)^2 (d+e x) \log (d+e x)}{2 (b d-a e)^4 (a+b x) \sqrt {(a+b x)^2} (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*d - a*e)*(-2*a^2*e^2 - a*b*e*(5*d + 9*e*x) + b^2*(d^2 - 3*d*e*x - 6*e^2*x^2))) + 6*b*e^2*(a + b*x)^2*(d
+ e*x)*Log[a + b*x] - 6*b*e^2*(a + b*x)^2*(d + e*x)*Log[d + e*x])/(2*(b*d - a*e)^4*(a + b*x)*Sqrt[(a + b*x)^2]
*(d + e*x))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(329\) vs. \(2(160)=320\).
time = 0.70, size = 330, normalized size = 1.52

method	result	size
default	\(\frac {\left (6 \ln \left (b x +a \right ) b^{3} e^{3} x^{3}-6 \ln \left (e x +d \right ) b^{3} e^{3} x^{3}+12 \ln \left (b x +a \right ) a \,b^{2} e^{3} x^{2}+6 \ln \left (b x +a \right ) b^{3} d \,e^{2} x^{2}-12 \ln \left (e x +d \right ) a \,b^{2} e^{3} x^{2}-6 \ln \left (e x +d \right ) b^{3} d \,e^{2} x^{2}+6 \ln \left (b x +a \right ) a^{2} b \,e^{3} x +12 \ln \left (b x +a \right ) a \,b^{2} d \,e^{2} x -6 \ln \left (e x +d \right ) a^{2} b \,e^{3} x -12 \ln \left (e x +d \right ) a \,b^{2} d \,e^{2} x -6 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+6 \ln \left (b x +a \right ) a^{2} b d \,e^{2}-6 \ln \left (e x +d \right ) a^{2} b d \,e^{2}-9 a^{2} b \,e^{3} x +6 a \,b^{2} d \,e^{2} x +3 b^{3} d^{2} e x -2 e^{3} a^{3}-3 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (b x +a \right )}{2 \left (e x +d \right ) \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(330\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {3 b^{2} e^{2} x^{2}}{e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}}-\frac {3 \left (3 a e +b d \right ) b e x}{2 \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}-\frac {2 a^{2} e^{2}+5 a b d e -b^{2} d^{2}}{2 \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\right )}{\left (b x +a \right )^{3} \left (e x +d \right )}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \,e^{2} \ln \left (-b x -a \right )}{\left (b x +a \right ) \left (e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \,e^{2} \ln \left (e x +d \right )}{\left (b x +a \right ) \left (e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\)	\(351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*ln(b*x+a)*b^3*e^3*x^3-6*ln(e*x+d)*b^3*e^3*x^3+12*ln(b*x+a)*a*b^2*e^3*x^2+6*ln(b*x+a)*b^3*d*e^2*x^2-12*l
n(e*x+d)*a*b^2*e^3*x^2-6*ln(e*x+d)*b^3*d*e^2*x^2+6*ln(b*x+a)*a^2*b*e^3*x+12*ln(b*x+a)*a*b^2*d*e^2*x-6*ln(e*x+d
)*a^2*b*e^3*x-12*ln(e*x+d)*a*b^2*d*e^2*x-6*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+6*ln(b*x+a)*a^2*b*d*e^2-6*ln(e*x+d)*a
^2*b*d*e^2-9*a^2*b*e^3*x+6*a*b^2*d*e^2*x+3*b^3*d^2*e*x-2*e^3*a^3-3*a^2*b*d*e^2+6*a*b^2*d^2*e-b^3*d^3)*(b*x+a)/
(e*x+d)/(a*e-b*d)^4/((b*x+a)^2)^(3/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (165) = 330\).
time = 2.81, size = 469, normalized size = 2.16 \begin {gather*} -\frac {b^{3} d^{3} + {\left (6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}\right )} e^{3} - 3 \, {\left (2 \, b^{3} d x^{2} + 2 \, a b^{2} d x - a^{2} b d\right )} e^{2} - 3 \, {\left (b^{3} d^{2} x + 2 \, a b^{2} d^{2}\right )} e - 6 \, {\left ({\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} e^{3} + {\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\right )} e^{2}\right )} \log \left (b x + a\right ) + 6 \, {\left ({\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} e^{3} + {\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\right )} e^{2}\right )} \log \left (x e + d\right )}{2 \, {\left (b^{6} d^{5} x^{2} + 2 \, a b^{5} d^{5} x + a^{2} b^{4} d^{5} + {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )} e^{5} - {\left (4 \, a^{3} b^{3} d x^{3} + 7 \, a^{4} b^{2} d x^{2} + 2 \, a^{5} b d x - a^{6} d\right )} e^{4} + 2 \, {\left (3 \, a^{2} b^{4} d^{2} x^{3} + 4 \, a^{3} b^{3} d^{2} x^{2} - a^{4} b^{2} d^{2} x - 2 \, a^{5} b d^{2}\right )} e^{3} - 2 \, {\left (2 \, a b^{5} d^{3} x^{3} + a^{2} b^{4} d^{3} x^{2} - 4 \, a^{3} b^{3} d^{3} x - 3 \, a^{4} b^{2} d^{3}\right )} e^{2} + {\left (b^{6} d^{4} x^{3} - 2 \, a b^{5} d^{4} x^{2} - 7 \, a^{2} b^{4} d^{4} x - 4 \, a^{3} b^{3} d^{4}\right )} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^3*d^3 + (6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)*e^3 - 3*(2*b^3*d*x^2 + 2*a*b^2*d*x - a^2*b*d)*e^2 - 3*(b^3*d
^2*x + 2*a*b^2*d^2)*e - 6*((b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*e^3 + (b^3*d*x^2 + 2*a*b^2*d*x + a^2*b*d)*e^2)*lo
g(b*x + a) + 6*((b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*e^3 + (b^3*d*x^2 + 2*a*b^2*d*x + a^2*b*d)*e^2)*log(x*e + d))
/(b^6*d^5*x^2 + 2*a*b^5*d^5*x + a^2*b^4*d^5 + (a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)*e^5 - (4*a^3*b^3*d*x^3 + 7*a
^4*b^2*d*x^2 + 2*a^5*b*d*x - a^6*d)*e^4 + 2*(3*a^2*b^4*d^2*x^3 + 4*a^3*b^3*d^2*x^2 - a^4*b^2*d^2*x - 2*a^5*b*d
^2)*e^3 - 2*(2*a*b^5*d^3*x^3 + a^2*b^4*d^3*x^2 - 4*a^3*b^3*d^3*x - 3*a^4*b^2*d^3)*e^2 + (b^6*d^4*x^3 - 2*a*b^5
*d^4*x^2 - 7*a^2*b^4*d^4*x - 4*a^3*b^3*d^4)*e)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x\right )^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.00, size = 313, normalized size = 1.44 \begin {gather*} \frac {3 \, b^{2} e^{2} \log \left ({\left | b x + a \right |}\right )}{b^{5} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} b e^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {3 \, b e^{3} \log \left ({\left | x e + d \right |}\right )}{b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {b^{3} d^{3} - 6 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} + 2 \, a^{3} e^{3} - 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2} - 3 \, a^{2} b e^{3}\right )} x}{2 \, {\left (b d - a e\right )}^{4} {\left (b x + a\right )}^{2} {\left (x e + d\right )} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b^2*e^2*log(abs(b*x + a))/(b^5*d^4*sgn(b*x + a) - 4*a*b^4*d^3*e*sgn(b*x + a) + 6*a^2*b^3*d^2*e^2*sgn(b*x + a
) - 4*a^3*b^2*d*e^3*sgn(b*x + a) + a^4*b*e^4*sgn(b*x + a)) - 3*b*e^3*log(abs(x*e + d))/(b^4*d^4*e*sgn(b*x + a)
- 4*a*b^3*d^3*e^2*sgn(b*x + a) + 6*a^2*b^2*d^2*e^3*sgn(b*x + a) - 4*a^3*b*d*e^4*sgn(b*x + a) + a^4*e^5*sgn(b*
x + a)) - 1/2*(b^3*d^3 - 6*a*b^2*d^2*e + 3*a^2*b*d*e^2 + 2*a^3*e^3 - 6*(b^3*d*e^2 - a*b^2*e^3)*x^2 - 3*(b^3*d^
2*e + 2*a*b^2*d*e^2 - 3*a^2*b*e^3)*x)/((b*d - a*e)^4*(b*x + a)^2*(x*e + d)*sgn(b*x + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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