3.13.97 \(\int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=231 \[ \frac {b^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac {b (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^2}+\frac {a+b x}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}+\frac {b^3 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^3 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

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Rubi [A]  time = 0.10, antiderivative size = 231, 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 {b^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac {b (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^2}+\frac {a+b x}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}+\frac {b^3 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^3 (a+b x) \log (d+e x)}{\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)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

(a + b*x)/(3*(b*d - a*e)*(d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (b*(a + b*x))/(2*(b*d - a*e)^2*(d + e*x)
^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (b^2*(a + b*x))/((b*d - a*e)^3*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(b^3*(a + b*x)*Log[a + b*x])/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b^3*(a + b*x)*Log[d + e*x])/((b*
d - a*e)^4*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)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^4} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {b^3}{(b d-a e)^4 (a+b x)}-\frac {e}{b (b d-a e) (d+e x)^4}-\frac {e}{(b d-a e)^2 (d+e x)^3}-\frac {b e}{(b d-a e)^3 (d+e x)^2}-\frac {b^2 e}{(b d-a e)^4 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {a+b x}{3 (b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{2 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 (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 = 0.08, size = 124, normalized size = 0.54 \begin {gather*} \frac {(a+b x) \left (6 b^3 (d+e x)^3 \log (a+b x)+6 b^2 (d+e x)^2 (b d-a e)+3 b (d+e x) (b d-a e)^2+2 (b d-a e)^3-6 b^3 (d+e x)^3 \log (d+e x)\right )}{6 \sqrt {(a+b x)^2} (d+e x)^3 (b d-a e)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 180.21, 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)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

$Aborted

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fricas [B]  time = 0.43, size = 425, normalized size = 1.84 \begin {gather*} \frac {11 \, b^{3} d^{3} - 18 \, a b^{2} d^{2} e + 9 \, 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 (5 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, b^{3} d e^{2} x^{2} + 3 \, b^{3} d^{2} e x + b^{3} d^{3}\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, b^{3} d e^{2} x^{2} + 3 \, b^{3} d^{2} e x + b^{3} d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (b^{4} d^{7} - 4 \, a b^{3} d^{6} e + 6 \, a^{2} b^{2} d^{5} e^{2} - 4 \, a^{3} b d^{4} e^{3} + a^{4} d^{3} e^{4} + {\left (b^{4} d^{4} e^{3} - 4 \, a b^{3} d^{3} e^{4} + 6 \, a^{2} b^{2} d^{2} e^{5} - 4 \, a^{3} b d e^{6} + a^{4} e^{7}\right )} x^{3} + 3 \, {\left (b^{4} d^{5} e^{2} - 4 \, a b^{3} d^{4} e^{3} + 6 \, a^{2} b^{2} d^{3} e^{4} - 4 \, a^{3} b d^{2} e^{5} + a^{4} d e^{6}\right )} x^{2} + 3 \, {\left (b^{4} d^{6} e - 4 \, a b^{3} d^{5} e^{2} + 6 \, a^{2} b^{2} d^{4} e^{3} - 4 \, a^{3} b d^{3} e^{4} + a^{4} d^{2} e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*(11*b^3*d^3 - 18*a*b^2*d^2*e + 9*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*(5*b^3*d^2*e
- 6*a*b^2*d*e^2 + a^2*b*e^3)*x + 6*(b^3*e^3*x^3 + 3*b^3*d*e^2*x^2 + 3*b^3*d^2*e*x + b^3*d^3)*log(b*x + a) - 6*
(b^3*e^3*x^3 + 3*b^3*d*e^2*x^2 + 3*b^3*d^2*e*x + b^3*d^3)*log(e*x + d))/(b^4*d^7 - 4*a*b^3*d^6*e + 6*a^2*b^2*d
^5*e^2 - 4*a^3*b*d^4*e^3 + a^4*d^3*e^4 + (b^4*d^4*e^3 - 4*a*b^3*d^3*e^4 + 6*a^2*b^2*d^2*e^5 - 4*a^3*b*d*e^6 +
a^4*e^7)*x^3 + 3*(b^4*d^5*e^2 - 4*a*b^3*d^4*e^3 + 6*a^2*b^2*d^3*e^4 - 4*a^3*b*d^2*e^5 + a^4*d*e^6)*x^2 + 3*(b^
4*d^6*e - 4*a*b^3*d^5*e^2 + 6*a^2*b^2*d^4*e^3 - 4*a^3*b*d^3*e^4 + a^4*d^2*e^5)*x)

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giac [A]  time = 0.19, size = 246, normalized size = 1.06 \begin {gather*} \frac {1}{6} \, {\left (\frac {6 \, b^{4} \log \left ({\left | b x + a \right |}\right )}{b^{5} d^{4} - 4 \, a b^{4} d^{3} e + 6 \, a^{2} b^{3} d^{2} e^{2} - 4 \, a^{3} b^{2} d e^{3} + a^{4} b e^{4}} - \frac {6 \, b^{3} e \log \left ({\left | x e + d \right |}\right )}{b^{4} d^{4} e - 4 \, a b^{3} d^{3} e^{2} + 6 \, a^{2} b^{2} d^{2} e^{3} - 4 \, a^{3} b d e^{4} + a^{4} e^{5}} + \frac {11 \, b^{3} d^{3} - 18 \, a b^{2} d^{2} e + 9 \, 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 (5 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{{\left (b d - a e\right )}^{4} {\left (x e + d\right )}^{3}}\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)^4/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*(6*b^4*log(abs(b*x + a))/(b^5*d^4 - 4*a*b^4*d^3*e + 6*a^2*b^3*d^2*e^2 - 4*a^3*b^2*d*e^3 + a^4*b*e^4) - 6*b
^3*e*log(abs(x*e + d))/(b^4*d^4*e - 4*a*b^3*d^3*e^2 + 6*a^2*b^2*d^2*e^3 - 4*a^3*b*d*e^4 + a^4*e^5) + (11*b^3*d
^3 - 18*a*b^2*d^2*e + 9*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*(5*b^3*d^2*e - 6*a*b^2*d*e
^2 + a^2*b*e^3)*x)/((b*d - a*e)^4*(x*e + d)^3))*sgn(b*x + a)

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maple [A]  time = 0.06, size = 256, normalized size = 1.11 \begin {gather*} \frac {\left (b x +a \right ) \left (6 b^{3} e^{3} x^{3} \ln \left (b x +a \right )-6 b^{3} e^{3} x^{3} \ln \left (e x +d \right )+18 b^{3} d \,e^{2} x^{2} \ln \left (b x +a \right )-18 b^{3} d \,e^{2} x^{2} \ln \left (e x +d \right )-6 a \,b^{2} e^{3} x^{2}+18 b^{3} d^{2} e x \ln \left (b x +a \right )-18 b^{3} d^{2} e x \ln \left (e x +d \right )+6 b^{3} d \,e^{2} x^{2}+3 a^{2} b \,e^{3} x -18 a \,b^{2} d \,e^{2} x +6 b^{3} d^{3} \ln \left (b x +a \right )-6 b^{3} d^{3} \ln \left (e x +d \right )+15 b^{3} d^{2} e x -2 a^{3} e^{3}+9 a^{2} b d \,e^{2}-18 a \,b^{2} d^{2} e +11 b^{3} d^{3}\right )}{6 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{4} \left (e x +d \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^4/((b*x+a)^2)^(1/2),x)

[Out]

1/6*(b*x+a)*(6*ln(b*x+a)*x^3*b^3*e^3-6*b^3*e^3*x^3*ln(e*x+d)+18*ln(b*x+a)*x^2*b^3*d*e^2-18*b^3*d*e^2*x^2*ln(e*
x+d)+18*ln(b*x+a)*x*b^3*d^2*e-18*b^3*d^2*e*x*ln(e*x+d)-6*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+6*b^3*d^3*ln(b*x+a)-6*b
^3*d^3*ln(e*x+d)+3*a^2*b*e^3*x-18*a*b^2*d*e^2*x+15*b^3*d^2*e*x-2*a^3*e^3+9*a^2*b*d*e^2-18*a*b^2*d^2*e+11*b^3*d
^3)/((b*x+a)^2)^(1/2)/(a*e-b*d)^4/(e*x+d)^3

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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)^4/((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 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}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(((a + b*x)^2)^(1/2)*(d + e*x)^4),x)

[Out]

int(1/(((a + b*x)^2)^(1/2)*(d + e*x)^4), x)

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sympy [B]  time = 1.55, size = 570, normalized size = 2.47 \begin {gather*} - \frac {b^{3} \log {\left (x + \frac {- \frac {a^{5} b^{3} e^{5}}{\left (a e - b d\right )^{4}} + \frac {5 a^{4} b^{4} d e^{4}}{\left (a e - b d\right )^{4}} - \frac {10 a^{3} b^{5} d^{2} e^{3}}{\left (a e - b d\right )^{4}} + \frac {10 a^{2} b^{6} d^{3} e^{2}}{\left (a e - b d\right )^{4}} - \frac {5 a b^{7} d^{4} e}{\left (a e - b d\right )^{4}} + a b^{3} e + \frac {b^{8} d^{5}}{\left (a e - b d\right )^{4}} + b^{4} d}{2 b^{4} e} \right )}}{\left (a e - b d\right )^{4}} + \frac {b^{3} \log {\left (x + \frac {\frac {a^{5} b^{3} e^{5}}{\left (a e - b d\right )^{4}} - \frac {5 a^{4} b^{4} d e^{4}}{\left (a e - b d\right )^{4}} + \frac {10 a^{3} b^{5} d^{2} e^{3}}{\left (a e - b d\right )^{4}} - \frac {10 a^{2} b^{6} d^{3} e^{2}}{\left (a e - b d\right )^{4}} + \frac {5 a b^{7} d^{4} e}{\left (a e - b d\right )^{4}} + a b^{3} e - \frac {b^{8} d^{5}}{\left (a e - b d\right )^{4}} + b^{4} d}{2 b^{4} e} \right )}}{\left (a e - b d\right )^{4}} + \frac {- 2 a^{2} e^{2} + 7 a b d e - 11 b^{2} d^{2} - 6 b^{2} e^{2} x^{2} + x \left (3 a b e^{2} - 15 b^{2} d e\right )}{6 a^{3} d^{3} e^{3} - 18 a^{2} b d^{4} e^{2} + 18 a b^{2} d^{5} e - 6 b^{3} d^{6} + x^{3} \left (6 a^{3} e^{6} - 18 a^{2} b d e^{5} + 18 a b^{2} d^{2} e^{4} - 6 b^{3} d^{3} e^{3}\right ) + x^{2} \left (18 a^{3} d e^{5} - 54 a^{2} b d^{2} e^{4} + 54 a b^{2} d^{3} e^{3} - 18 b^{3} d^{4} e^{2}\right ) + x \left (18 a^{3} d^{2} e^{4} - 54 a^{2} b d^{3} e^{3} + 54 a b^{2} d^{4} e^{2} - 18 b^{3} d^{5} e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**4/((b*x+a)**2)**(1/2),x)

[Out]

-b**3*log(x + (-a**5*b**3*e**5/(a*e - b*d)**4 + 5*a**4*b**4*d*e**4/(a*e - b*d)**4 - 10*a**3*b**5*d**2*e**3/(a*
e - b*d)**4 + 10*a**2*b**6*d**3*e**2/(a*e - b*d)**4 - 5*a*b**7*d**4*e/(a*e - b*d)**4 + a*b**3*e + b**8*d**5/(a
*e - b*d)**4 + b**4*d)/(2*b**4*e))/(a*e - b*d)**4 + b**3*log(x + (a**5*b**3*e**5/(a*e - b*d)**4 - 5*a**4*b**4*
d*e**4/(a*e - b*d)**4 + 10*a**3*b**5*d**2*e**3/(a*e - b*d)**4 - 10*a**2*b**6*d**3*e**2/(a*e - b*d)**4 + 5*a*b*
*7*d**4*e/(a*e - b*d)**4 + a*b**3*e - b**8*d**5/(a*e - b*d)**4 + b**4*d)/(2*b**4*e))/(a*e - b*d)**4 + (-2*a**2
*e**2 + 7*a*b*d*e - 11*b**2*d**2 - 6*b**2*e**2*x**2 + x*(3*a*b*e**2 - 15*b**2*d*e))/(6*a**3*d**3*e**3 - 18*a**
2*b*d**4*e**2 + 18*a*b**2*d**5*e - 6*b**3*d**6 + x**3*(6*a**3*e**6 - 18*a**2*b*d*e**5 + 18*a*b**2*d**2*e**4 -
6*b**3*d**3*e**3) + x**2*(18*a**3*d*e**5 - 54*a**2*b*d**2*e**4 + 54*a*b**2*d**3*e**3 - 18*b**3*d**4*e**2) + x*
(18*a**3*d**2*e**4 - 54*a**2*b*d**3*e**3 + 54*a*b**2*d**4*e**2 - 18*b**3*d**5*e))

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