3.2129 \(\int \frac{a+b x}{(d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{2 (a+b x)}{5 e \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0285665, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {770, 21, 32} \[ -\frac{2 (a+b x)}{5 e \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

(-2*(a + b*x))/(5*e*(d + e*x)^(5/2)*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)^{7/2} \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)^{7/2}} \, 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)^{7/2}} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 (a+b x)}{5 e (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0171365, size = 32, normalized size = 0.78 \[ -\frac{2 (a+b x)}{5 e \sqrt{(a+b x)^2} (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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Maple [A]  time = 0.003, size = 27, normalized size = 0.7 \begin{align*} -{\frac{2\,bx+2\,a}{5\,e} \left ( ex+d \right ) ^{-{\frac{5}{2}}}{\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)^(7/2)/((b*x+a)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.21639, size = 57, normalized size = 1.39 \begin{align*} -\frac{2 \, \sqrt{e x + d}}{5 \,{\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.969213, size = 89, normalized size = 2.17 \begin{align*} -\frac{2 \, \sqrt{e x + d}}{5 \,{\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)**(7/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.13856, size = 24, normalized size = 0.59 \begin{align*} -\frac{2 \, e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right )}{5 \,{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-2/5*e^(-1)*sgn(b*x + a)/(x*e + d)^(5/2)