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

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

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Rubi [A]  time = 0.15, 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 = {646, 44} \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 44

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] && L
tQ[m + n + 2, 0])

Rule 646

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 = 0.10, size = 163, normalized size = 0.59 \begin {gather*} \frac {(a+b x) \left (-12 b^2 e^2 (a+b x)^2 \log (d+e x)+6 b^2 e (a+b x) (b d-a e)+b^2 \left (-(b d-a e)^2\right )+12 b^2 e^2 (a+b x)^2 \log (a+b x)+\frac {6 b e^2 (a+b x)^2 (b d-a e)}{d+e x}+\frac {e^2 (a+b x)^2 (b d-a e)^2}{(d+e x)^2}\right )}{2 \left ((a+b x)^2\right )^{3/2} (b d-a e)^5} \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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IntegrateAlgebraic [F]  time = 180.25, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [B]  time = 0.42, size = 760, normalized size = 2.75 \begin {gather*} -\frac {b^{4} d^{4} - 8 \, a b^{3} d^{3} e + 8 \, a^{3} b d e^{3} - a^{4} e^{4} - 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} - 18 \, {\left (b^{4} d^{2} e^{2} - a^{2} b^{2} e^{4}\right )} x^{2} - 4 \, {\left (b^{4} d^{3} e + 6 \, a b^{3} d^{2} e^{2} - 6 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x - 12 \, {\left (b^{4} e^{4} x^{4} + a^{2} b^{2} d^{2} e^{2} + 2 \, {\left (b^{4} d e^{3} + a b^{3} e^{4}\right )} x^{3} + {\left (b^{4} d^{2} e^{2} + 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} + a^{2} b^{2} d e^{3}\right )} x\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} e^{4} x^{4} + a^{2} b^{2} d^{2} e^{2} + 2 \, {\left (b^{4} d e^{3} + a b^{3} e^{4}\right )} x^{3} + {\left (b^{4} d^{2} e^{2} + 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} + a^{2} b^{2} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (a^{2} b^{5} d^{7} - 5 \, a^{3} b^{4} d^{6} e + 10 \, a^{4} b^{3} d^{5} e^{2} - 10 \, a^{5} b^{2} d^{4} e^{3} + 5 \, a^{6} b d^{3} e^{4} - a^{7} d^{2} e^{5} + {\left (b^{7} d^{5} e^{2} - 5 \, a b^{6} d^{4} e^{3} + 10 \, a^{2} b^{5} d^{3} e^{4} - 10 \, a^{3} b^{4} d^{2} e^{5} + 5 \, a^{4} b^{3} d e^{6} - a^{5} b^{2} e^{7}\right )} x^{4} + 2 \, {\left (b^{7} d^{6} e - 4 \, a b^{6} d^{5} e^{2} + 5 \, a^{2} b^{5} d^{4} e^{3} - 5 \, a^{4} b^{3} d^{2} e^{5} + 4 \, a^{5} b^{2} d e^{6} - a^{6} b e^{7}\right )} x^{3} + {\left (b^{7} d^{7} - a b^{6} d^{6} e - 9 \, a^{2} b^{5} d^{5} e^{2} + 25 \, a^{3} b^{4} d^{4} e^{3} - 25 \, a^{4} b^{3} d^{3} e^{4} + 9 \, a^{5} b^{2} d^{2} e^{5} + a^{6} b d e^{6} - a^{7} e^{7}\right )} x^{2} + 2 \, {\left (a b^{6} d^{7} - 4 \, a^{2} b^{5} d^{6} e + 5 \, a^{3} b^{4} d^{5} e^{2} - 5 \, a^{5} b^{2} d^{3} e^{4} + 4 \, a^{6} b d^{2} e^{5} - a^{7} d e^{6}\right )} x\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 - 8*a*b^3*d^3*e + 8*a^3*b*d*e^3 - a^4*e^4 - 12*(b^4*d*e^3 - a*b^3*e^4)*x^3 - 18*(b^4*d^2*e^2 - a
^2*b^2*e^4)*x^2 - 4*(b^4*d^3*e + 6*a*b^3*d^2*e^2 - 6*a^2*b^2*d*e^3 - a^3*b*e^4)*x - 12*(b^4*e^4*x^4 + a^2*b^2*
d^2*e^2 + 2*(b^4*d*e^3 + a*b^3*e^4)*x^3 + (b^4*d^2*e^2 + 4*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 2*(a*b^3*d^2*e^2 +
a^2*b^2*d*e^3)*x)*log(b*x + a) + 12*(b^4*e^4*x^4 + a^2*b^2*d^2*e^2 + 2*(b^4*d*e^3 + a*b^3*e^4)*x^3 + (b^4*d^2
*e^2 + 4*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 2*(a*b^3*d^2*e^2 + a^2*b^2*d*e^3)*x)*log(e*x + d))/(a^2*b^5*d^7 - 5*
a^3*b^4*d^6*e + 10*a^4*b^3*d^5*e^2 - 10*a^5*b^2*d^4*e^3 + 5*a^6*b*d^3*e^4 - a^7*d^2*e^5 + (b^7*d^5*e^2 - 5*a*b
^6*d^4*e^3 + 10*a^2*b^5*d^3*e^4 - 10*a^3*b^4*d^2*e^5 + 5*a^4*b^3*d*e^6 - a^5*b^2*e^7)*x^4 + 2*(b^7*d^6*e - 4*a
*b^6*d^5*e^2 + 5*a^2*b^5*d^4*e^3 - 5*a^4*b^3*d^2*e^5 + 4*a^5*b^2*d*e^6 - a^6*b*e^7)*x^3 + (b^7*d^7 - a*b^6*d^6
*e - 9*a^2*b^5*d^5*e^2 + 25*a^3*b^4*d^4*e^3 - 25*a^4*b^3*d^3*e^4 + 9*a^5*b^2*d^2*e^5 + a^6*b*d*e^6 - a^7*e^7)*
x^2 + 2*(a*b^6*d^7 - 4*a^2*b^5*d^6*e + 5*a^3*b^4*d^5*e^2 - 5*a^5*b^2*d^3*e^4 + 4*a^6*b*d^2*e^5 - a^7*d*e^6)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \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]

sage0*x

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maple [B]  time = 0.07, size = 508, normalized size = 1.84 \begin {gather*} -\frac {\left (12 b^{4} e^{4} x^{4} \ln \left (b x +a \right )-12 b^{4} e^{4} x^{4} \ln \left (e x +d \right )+24 a \,b^{3} e^{4} x^{3} \ln \left (b x +a \right )-24 a \,b^{3} e^{4} x^{3} \ln \left (e x +d \right )+24 b^{4} d \,e^{3} x^{3} \ln \left (b x +a \right )-24 b^{4} d \,e^{3} x^{3} \ln \left (e x +d \right )+12 a^{2} b^{2} e^{4} x^{2} \ln \left (b x +a \right )-12 a^{2} b^{2} e^{4} x^{2} \ln \left (e x +d \right )+48 a \,b^{3} d \,e^{3} x^{2} \ln \left (b x +a \right )-48 a \,b^{3} d \,e^{3} x^{2} \ln \left (e x +d \right )-12 a \,b^{3} e^{4} x^{3}+12 b^{4} d^{2} e^{2} x^{2} \ln \left (b x +a \right )-12 b^{4} d^{2} e^{2} x^{2} \ln \left (e x +d \right )+12 b^{4} d \,e^{3} x^{3}+24 a^{2} b^{2} d \,e^{3} x \ln \left (b x +a \right )-24 a^{2} b^{2} d \,e^{3} x \ln \left (e x +d \right )-18 a^{2} b^{2} e^{4} x^{2}+24 a \,b^{3} d^{2} e^{2} x \ln \left (b x +a \right )-24 a \,b^{3} d^{2} e^{2} x \ln \left (e x +d \right )+18 b^{4} d^{2} e^{2} x^{2}-4 a^{3} b \,e^{4} x +12 a^{2} b^{2} d^{2} e^{2} \ln \left (b x +a \right )-12 a^{2} b^{2} d^{2} e^{2} \ln \left (e x +d \right )-24 a^{2} b^{2} d \,e^{3} x +24 a \,b^{3} d^{2} e^{2} x +4 b^{4} d^{3} e x +a^{4} e^{4}-8 a^{3} b d \,e^{3}+8 a \,b^{3} d^{3} e -b^{4} d^{4}\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}}} \end {gather*}

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)

[Out]

-1/2*(8*a*b^3*d^3*e+12*a^2*b^2*e^4*x^2*ln(b*x+a)-24*ln(e*x+d)*x*a^2*b^2*d*e^3-24*ln(e*x+d)*x*a*b^3*d^2*e^2-48*
ln(e*x+d)*x^2*a*b^3*d*e^3-18*a^2*b^2*e^4*x^2+18*b^4*d^2*e^2*x^2-4*a^3*b*e^4*x+4*b^4*d^3*e*x-12*a*b^3*e^4*x^3+1
2*b^4*d*e^3*x^3+24*a*b^3*d^2*e^2*x*ln(b*x+a)+24*a^2*b^2*d*e^3*x*ln(b*x+a)+48*a*b^3*d*e^3*x^2*ln(b*x+a)-8*a^3*b
*d*e^3-b^4*d^4+12*b^4*d^2*e^2*x^2*ln(b*x+a)+12*ln(b*x+a)*x^4*b^4*e^4-12*ln(e*x+d)*x^4*b^4*e^4+a^4*e^4+12*a^2*b
^2*d^2*e^2*ln(b*x+a)-24*a^2*b^2*d*e^3*x+24*a*b^3*d^2*e^2*x+24*ln(b*x+a)*x^3*a*b^3*e^4+24*ln(b*x+a)*x^3*b^4*d*e
^3-24*ln(e*x+d)*x^3*a*b^3*e^4-24*ln(e*x+d)*x^3*b^4*d*e^3-12*ln(e*x+d)*x^2*a^2*b^2*e^4-12*ln(e*x+d)*x^2*b^4*d^2
*e^2-12*ln(e*x+d)*a^2*b^2*d^2*e^2)*(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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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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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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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