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

Optimal. Leaf size=42 \[ \frac {-a-b x}{3 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 32} \begin {gather*} -\frac {a+b x}{3 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {a+b x}{(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 {a+b x}{\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 \frac {1}{(d+e x)^4} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {a+b x}{3 e (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.71 \begin {gather*} -\frac {a+b x}{3 e \sqrt {(a+b x)^2} (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 1.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 35, normalized size = 0.83 \begin {gather*} -\frac {1}{3 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 18, normalized size = 0.43 \begin {gather*} -\frac {e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right )}{3 \, {\left (x e + d\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*e^(-1)*sgn(b*x + a)/(x*e + d)^3

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maple [A]  time = 0.05, size = 27, normalized size = 0.64 \begin {gather*} -\frac {b x +a}{3 \left (e x +d \right )^{3} \sqrt {\left (b x +a \right )^{2}}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((b*x+a)/(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 [B]  time = 2.14, size = 28, normalized size = 0.67 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}}{3\,e\,\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)/(((a + b*x)^2)^(1/2)*(d + e*x)^4),x)

[Out]

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

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sympy [A]  time = 0.28, size = 37, normalized size = 0.88 \begin {gather*} - \frac {1}{3 d^{3} e + 9 d^{2} e^{2} x + 9 d e^{3} x^{2} + 3 e^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(3*d**3*e + 9*d**2*e**2*x + 9*d*e**3*x**2 + 3*e**4*x**3)

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