3.17.3 \(\int \frac {1}{(d+e x)^3 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1603]

Optimal. Leaf size=276 \[ \frac {3 b^2 e}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2}{2 (b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x)}{2 (b d-a e)^3 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b e^2 (a+b x)}{(b d-a e)^4 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 b^2 e^2 (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 e^2 (a+b x) \log (d+e x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 276, 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 {3 b e^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^4}+\frac {e^2 (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^3}+\frac {6 b^2 e^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac {6 b^2 e^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac {b^2}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {3 b^2 e}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*e)/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b^2/(2*(b*d - a*e)^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (e^2*(a + b*x))/(2*(b*d - a*e)^3*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*e^2*(a + b*x))/
((b*d - a*e)^4*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (6*b^2*e^2*(a + b*x)*Log[a + b*x])/((b*d - a*e)^5*Sq
rt[a^2 + 2*a*b*x + b^2*x^2]) - (6*b^2*e^2*(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*} \int \frac {1}{(d+e x)^3 \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)^3} \, 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 d-a e)^3 (a+b x)^3}-\frac {3 e}{(b d-a e)^4 (a+b x)^2}+\frac {6 e^2}{(b d-a e)^5 (a+b x)}-\frac {e^3}{b^3 (b d-a e)^3 (d+e x)^3}-\frac {3 e^3}{b^2 (b d-a e)^4 (d+e x)^2}-\frac {6 e^3}{b (b d-a e)^5 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 b^2 e}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2}{2 (b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x)}{2 (b d-a e)^3 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b e^2 (a+b x)}{(b d-a e)^4 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 b^2 e^2 (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 e^2 (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*}

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Mathematica [A]
time = 0.07, size = 163, normalized size = 0.59 \begin {gather*} \frac {(a+b x) \left (-b^2 (b d-a e)^2+6 b^2 e (b d-a e) (a+b x)+\frac {e^2 (b d-a e)^2 (a+b x)^2}{(d+e x)^2}+\frac {6 b e^2 (b d-a e) (a+b x)^2}{d+e x}+12 b^2 e^2 (a+b x)^2 \log (a+b x)-12 b^2 e^2 (a+b x)^2 \log (d+e x)\right )}{2 (b d-a e)^5 \left ((a+b x)^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(-(b^2*(b*d - a*e)^2) + 6*b^2*e*(b*d - a*e)*(a + b*x) + (e^2*(b*d - a*e)^2*(a + b*x)^2)/(d + e*x)^2
+ (6*b*e^2*(b*d - a*e)*(a + b*x)^2)/(d + e*x) + 12*b^2*e^2*(a + b*x)^2*Log[a + b*x] - 12*b^2*e^2*(a + b*x)^2*
Log[d + e*x]))/(2*(b*d - a*e)^5*((a + b*x)^2)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(507\) vs. \(2(206)=412\).
time = 0.71, size = 508, normalized size = 1.84

method	result	size
default	\(-\frac {\left (8 a \,b^{3} d^{3} e +48 \ln \left (b x +a \right ) a \,b^{3} d \,e^{3} x^{2}+12 \ln \left (b x +a \right ) a^{2} b^{2} e^{4} x^{2}-12 a \,b^{3} e^{4} x^{3}+12 b^{4} d \,e^{3} x^{3}+e^{4} a^{4}+24 \ln \left (b x +a \right ) a \,b^{3} e^{4} x^{3}-b^{4} d^{4}-24 a^{2} b^{2} d \,e^{3} x +12 \ln \left (b x +a \right ) a^{2} b^{2} d^{2} e^{2}-8 a^{3} b d \,e^{3}+24 \ln \left (b x +a \right ) a \,b^{3} d^{2} e^{2} x -4 a^{3} b \,e^{4} x +4 b^{4} d^{3} e x +12 \ln \left (b x +a \right ) b^{4} e^{4} x^{4}-12 \ln \left (e x +d \right ) b^{4} e^{4} x^{4}+24 \ln \left (b x +a \right ) b^{4} d \,e^{3} x^{3}-24 \ln \left (e x +d \right ) a \,b^{3} e^{4} x^{3}-24 \ln \left (e x +d \right ) b^{4} d \,e^{3} x^{3}-12 \ln \left (e x +d \right ) a^{2} b^{2} e^{4} x^{2}-12 \ln \left (e x +d \right ) b^{4} d^{2} e^{2} x^{2}-12 \ln \left (e x +d \right ) a^{2} b^{2} d^{2} e^{2}+12 \ln \left (b x +a \right ) b^{4} d^{2} e^{2} x^{2}+24 \ln \left (b x +a \right ) a^{2} b^{2} d \,e^{3} x +24 a \,b^{3} d^{2} e^{2} x -18 a^{2} b^{2} e^{4} x^{2}+18 b^{4} d^{2} e^{2} x^{2}-48 \ln \left (e x +d \right ) a \,b^{3} d \,e^{3} x^{2}-24 \ln \left (e x +d \right ) a^{2} b^{2} d \,e^{3} x -24 \ln \left (e x +d \right ) a \,b^{3} d^{2} e^{2} x \right ) \left (b x +a \right )}{2 \left (e x +d \right )^{2} \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(508\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {6 b^{3} e^{3} x^{3}}{e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}+\frac {9 b^{2} e^{2} \left (a e +b d \right ) x^{2}}{e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}+\frac {2 \left (a^{2} e^{2}+7 a b d e +b^{2} d^{2}\right ) b e x}{e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}-\frac {e^{3} a^{3}-7 a^{2} b d \,e^{2}-7 a \,b^{2} d^{2} e +b^{3} d^{3}}{2 \left (e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\right )}{\left (b x +a \right )^{3} \left (e x +d \right )^{2}}-\frac {6 \sqrt {\left (b x +a \right )^{2}}\, b^{2} e^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) \left (a^{5} e^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}+\frac {6 \sqrt {\left (b x +a \right )^{2}}\, b^{2} e^{2} \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (a^{5} e^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}\)	\(518\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(8*a*b^3*d^3*e+48*ln(b*x+a)*a*b^3*d*e^3*x^2+12*ln(b*x+a)*a^2*b^2*e^4*x^2-12*a*b^3*e^4*x^3+12*b^4*d*e^3*x^
3+e^4*a^4+24*ln(b*x+a)*a*b^3*e^4*x^3-b^4*d^4-24*a^2*b^2*d*e^3*x+12*ln(b*x+a)*a^2*b^2*d^2*e^2-8*a^3*b*d*e^3+24*
ln(b*x+a)*a*b^3*d^2*e^2*x-4*a^3*b*e^4*x+4*b^4*d^3*e*x+12*ln(b*x+a)*b^4*e^4*x^4-12*ln(e*x+d)*b^4*e^4*x^4+24*ln(
b*x+a)*b^4*d*e^3*x^3-24*ln(e*x+d)*a*b^3*e^4*x^3-24*ln(e*x+d)*b^4*d*e^3*x^3-12*ln(e*x+d)*a^2*b^2*e^4*x^2-12*ln(
e*x+d)*b^4*d^2*e^2*x^2-12*ln(e*x+d)*a^2*b^2*d^2*e^2+12*ln(b*x+a)*b^4*d^2*e^2*x^2+24*ln(b*x+a)*a^2*b^2*d*e^3*x+
24*a*b^3*d^2*e^2*x-18*a^2*b^2*e^4*x^2+18*b^4*d^2*e^2*x^2-48*ln(e*x+d)*a*b^3*d*e^3*x^2-24*ln(e*x+d)*a^2*b^2*d*e
^3*x-24*ln(e*x+d)*a*b^3*d^2*e^2*x)*(b*x+a)/(e*x+d)^2/(a*e-b*d)^5/((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)^3/(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. 729 vs. \(2 (212) = 424\).
time = 2.73, size = 729, normalized size = 2.64 \begin {gather*} -\frac {b^{4} d^{4} + {\left (12 \, a b^{3} x^{3} + 18 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x - a^{4}\right )} e^{4} - 4 \, {\left (3 \, b^{4} d x^{3} - 6 \, a^{2} b^{2} d x - 2 \, a^{3} b d\right )} e^{3} - 6 \, {\left (3 \, b^{4} d^{2} x^{2} + 4 \, a b^{3} d^{2} x\right )} e^{2} - 4 \, {\left (b^{4} d^{3} x + 2 \, a b^{3} d^{3}\right )} e - 12 \, {\left ({\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} e^{4} + 2 \, {\left (b^{4} d x^{3} + 2 \, a b^{3} d x^{2} + a^{2} b^{2} d x\right )} e^{3} + {\left (b^{4} d^{2} x^{2} + 2 \, a b^{3} d^{2} x + a^{2} b^{2} d^{2}\right )} e^{2}\right )} \log \left (b x + a\right ) + 12 \, {\left ({\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} e^{4} + 2 \, {\left (b^{4} d x^{3} + 2 \, a b^{3} d x^{2} + a^{2} b^{2} d x\right )} e^{3} + {\left (b^{4} d^{2} x^{2} + 2 \, a b^{3} d^{2} x + a^{2} b^{2} d^{2}\right )} e^{2}\right )} \log \left (x e + d\right )}{2 \, {\left (b^{7} d^{7} x^{2} + 2 \, a b^{6} d^{7} x + a^{2} b^{5} d^{7} - {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )} e^{7} + {\left (5 \, a^{4} b^{3} d x^{4} + 8 \, a^{5} b^{2} d x^{3} + a^{6} b d x^{2} - 2 \, a^{7} d x\right )} e^{6} - {\left (10 \, a^{3} b^{4} d^{2} x^{4} + 10 \, a^{4} b^{3} d^{2} x^{3} - 9 \, a^{5} b^{2} d^{2} x^{2} - 8 \, a^{6} b d^{2} x + a^{7} d^{2}\right )} e^{5} + 5 \, {\left (2 \, a^{2} b^{5} d^{3} x^{4} - 5 \, a^{4} b^{3} d^{3} x^{2} - 2 \, a^{5} b^{2} d^{3} x + a^{6} b d^{3}\right )} e^{4} - 5 \, {\left (a b^{6} d^{4} x^{4} - 2 \, a^{2} b^{5} d^{4} x^{3} - 5 \, a^{3} b^{4} d^{4} x^{2} + 2 \, a^{5} b^{2} d^{4}\right )} e^{3} + {\left (b^{7} d^{5} x^{4} - 8 \, a b^{6} d^{5} x^{3} - 9 \, a^{2} b^{5} d^{5} x^{2} + 10 \, a^{3} b^{4} d^{5} x + 10 \, a^{4} b^{3} d^{5}\right )} e^{2} + {\left (2 \, b^{7} d^{6} x^{3} - a b^{6} d^{6} x^{2} - 8 \, a^{2} b^{5} d^{6} x - 5 \, a^{3} b^{4} d^{6}\right )} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^4*d^4 + (12*a*b^3*x^3 + 18*a^2*b^2*x^2 + 4*a^3*b*x - a^4)*e^4 - 4*(3*b^4*d*x^3 - 6*a^2*b^2*d*x - 2*a^3
*b*d)*e^3 - 6*(3*b^4*d^2*x^2 + 4*a*b^3*d^2*x)*e^2 - 4*(b^4*d^3*x + 2*a*b^3*d^3)*e - 12*((b^4*x^4 + 2*a*b^3*x^3
+ a^2*b^2*x^2)*e^4 + 2*(b^4*d*x^3 + 2*a*b^3*d*x^2 + a^2*b^2*d*x)*e^3 + (b^4*d^2*x^2 + 2*a*b^3*d^2*x + a^2*b^2
*d^2)*e^2)*log(b*x + a) + 12*((b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*e^4 + 2*(b^4*d*x^3 + 2*a*b^3*d*x^2 + a^2*b
^2*d*x)*e^3 + (b^4*d^2*x^2 + 2*a*b^3*d^2*x + a^2*b^2*d^2)*e^2)*log(x*e + d))/(b^7*d^7*x^2 + 2*a*b^6*d^7*x + a^
2*b^5*d^7 - (a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)*e^7 + (5*a^4*b^3*d*x^4 + 8*a^5*b^2*d*x^3 + a^6*b*d*x^2 - 2*a
^7*d*x)*e^6 - (10*a^3*b^4*d^2*x^4 + 10*a^4*b^3*d^2*x^3 - 9*a^5*b^2*d^2*x^2 - 8*a^6*b*d^2*x + a^7*d^2)*e^5 + 5*
(2*a^2*b^5*d^3*x^4 - 5*a^4*b^3*d^3*x^2 - 2*a^5*b^2*d^3*x + a^6*b*d^3)*e^4 - 5*(a*b^6*d^4*x^4 - 2*a^2*b^5*d^4*x
^3 - 5*a^3*b^4*d^4*x^2 + 2*a^5*b^2*d^4)*e^3 + (b^7*d^5*x^4 - 8*a*b^6*d^5*x^3 - 9*a^2*b^5*d^5*x^2 + 10*a^3*b^4*
d^5*x + 10*a^4*b^3*d^5)*e^2 + (2*b^7*d^6*x^3 - a*b^6*d^6*x^2 - 8*a^2*b^5*d^6*x - 5*a^3*b^4*d^6)*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 )^{3} \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)**3/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (212) = 424\).
time = 0.73, size = 434, normalized size = 1.57 \begin {gather*} \frac {6 \, b^{3} e^{2} \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 {6 \, b^{2} e^{3} \log \left ({\left | x e + 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 {12 \, b^{3} x^{3} e^{3} + 18 \, b^{3} d x^{2} e^{2} + 4 \, b^{3} d^{2} x e - b^{3} d^{3} + 18 \, a b^{2} x^{2} e^{3} + 28 \, a b^{2} d x e^{2} + 7 \, a b^{2} d^{2} e + 4 \, a^{2} b x e^{3} + 7 \, a^{2} b d e^{2} - a^{3} e^{3}}{2 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} {\left (b x^{2} e + b d x + a x e + a d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*b^3*e^2*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)) - 6*b^2*e^3*log(
abs(x*e + 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)) + 1/2*(12*b^3*x^3*e^3 + 18*b^3
*d*x^2*e^2 + 4*b^3*d^2*x*e - b^3*d^3 + 18*a*b^2*x^2*e^3 + 28*a*b^2*d*x*e^2 + 7*a*b^2*d^2*e + 4*a^2*b*x*e^3 + 7
*a^2*b*d*e^2 - a^3*e^3)/((b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) -
4*a^3*b*d*e^3*sgn(b*x + a) + a^4*e^4*sgn(b*x + a))*(b*x^2*e + b*d*x + a*x*e + a*d)^2)

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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 )}^3\,{\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)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)

[Out]

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

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