3.2024 \(\int \frac{a+b x}{(d+e x)^5 \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=39 \[ -\frac{a+b x}{4 e \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4} \]

[Out]

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

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Rubi [A]  time = 0.0278428, antiderivative size = 39, normalized size of antiderivative = 1., 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} \[ -\frac{a+b x}{4 e \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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]

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]

Rubi steps

\begin{align*} \int \frac{a+b x}{(d+e x)^5 \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)^5} \, 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)^5} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a+b x}{4 e (d+e x)^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0124601, size = 30, normalized size = 0.77 \[ -\frac{a+b x}{4 e \sqrt{(a+b x)^2} (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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Maple [A]  time = 0.003, size = 27, normalized size = 0.7 \begin{align*} -{\frac{bx+a}{4\,e \left ( ex+d \right ) ^{4}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(e*x+d)^5/((b*x+a)^2)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^5/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48891, size = 92, normalized size = 2.36 \begin{align*} -\frac{1}{4 \,{\left (e^{5} x^{4} + 4 \, d e^{4} x^{3} + 6 \, d^{2} e^{3} x^{2} + 4 \, d^{3} e^{2} x + d^{4} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^5/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.484596, size = 49, normalized size = 1.26 \begin{align*} - \frac{1}{4 d^{4} e + 16 d^{3} e^{2} x + 24 d^{2} e^{3} x^{2} + 16 d e^{4} x^{3} + 4 e^{5} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)**5/((b*x+a)**2)**(1/2),x)

[Out]

-1/(4*d**4*e + 16*d**3*e**2*x + 24*d**2*e**3*x**2 + 16*d*e**4*x**3 + 4*e**5*x**4)

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Giac [A]  time = 1.12267, size = 24, normalized size = 0.62 \begin{align*} -\frac{e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right )}{4 \,{\left (x e + d\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^5/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-1/4*e^(-1)*sgn(b*x + a)/(x*e + d)^4