3.2127 \(\int \frac {a+b x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 (a+b x)}{e \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x))/(e*Sqrt[d + e*x]*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)^{3/2} \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)^{3/2}} \, 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)^{3/2}} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (a+b x)}{e \sqrt {d+e x} \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 = 0.77 \[ -\frac {2 (a+b x)}{e \sqrt {(a+b x)^2} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.35, size = 20, normalized size = 0.51 \[ -\frac {2 \, \sqrt {e x + d}}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

-2*sqrt(e*x + d)/(e^2*x + d*e)

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giac [A]  time = 0.16, size = 18, normalized size = 0.46 \[ -\frac {2 \, e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right )}{\sqrt {x e + d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-2*e^(-1)*sgn(b*x + a)/sqrt(x*e + d)

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maple [A]  time = 0.05, size = 27, normalized size = 0.69 \[ -\frac {2 \left (b x +a \right )}{\sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.80, size = 20, normalized size = 0.51 \[ -\frac {2 \, \sqrt {e x + d}}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

-2*sqrt(e*x + d)/(e^2*x + d*e)

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mupad [B]  time = 2.50, size = 41, normalized size = 1.05 \[ -\frac {2\,\sqrt {{\left (a+b\,x\right )}^2}}{b\,e\,\left (x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b x}{\left (d + e x\right )^{\frac {3}{2}} \sqrt {\left (a + b x\right )^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)**(3/2)/((b*x+a)**2)**(1/2),x)

[Out]

Integral((a + b*x)/((d + e*x)**(3/2)*sqrt((a + b*x)**2)), x)

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