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

Optimal. Leaf size=92 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^2 (a+b x) (d+e x)^3}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x) (d+e x)^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 92, 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, 43} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^2 (a+b x) (d+e x)^3}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x) (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {a b+b^2 x}{(d+e x)^4} \, dx}{a b+b^2 x}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b (b d-a e)}{e (d+e x)^4}+\frac {b^2}{e (d+e x)^3}\right ) \, dx}{a b+b^2 x}\\ &=\frac {(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^3}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x) (d+e x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 0.49 \begin {gather*} -\frac {\sqrt {(a+b x)^2} (2 a e+b (d+3 e x))}{6 e^2 (a+b x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(Sqrt[(a + b*x)^2]*(2*a*e + b*(d + 3*e*x)))/(e^2*(a + b*x)*(d + e*x)^3)

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IntegrateAlgebraic [F]  time = 1.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 50, normalized size = 0.54 \begin {gather*} -\frac {3 \, b e x + b d + 2 \, a e}{6 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 45, normalized size = 0.49 \begin {gather*} -\frac {{\left (3 \, b x e \mathrm {sgn}\left (b x + a\right ) + b d \mathrm {sgn}\left (b x + a\right ) + 2 \, a e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-2\right )}}{6 \, {\left (x e + d\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 42, normalized size = 0.46 \begin {gather*} -\frac {\left (3 b e x +2 a e +b d \right ) \sqrt {\left (b x +a \right )^{2}}}{6 \left (e x +d \right )^{3} \left (b x +a \right ) e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(((b*x+a)^2)^(1/2)/(e*x+d)^4,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 [B]  time = 0.57, size = 41, normalized size = 0.45 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (2\,a\,e+b\,d+3\,b\,e\,x\right )}{6\,e^2\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.37, size = 53, normalized size = 0.58 \begin {gather*} \frac {- 2 a e - b d - 3 b e x}{6 d^{3} e^{2} + 18 d^{2} e^{3} x + 18 d e^{4} x^{2} + 6 e^{5} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*a*e - b*d - 3*b*e*x)/(6*d**3*e**2 + 18*d**2*e**3*x + 18*d*e**4*x**2 + 6*e**5*x**3)

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