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

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

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Rubi [A]  time = 0.03, antiderivative size = 41, 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} \begin {gather*} -\frac {2 (a+b x)}{5 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}} \end {gather*}

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 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)^{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*}

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Mathematica [A]  time = 0.02, size = 32, normalized size = 0.78 \begin {gather*} -\frac {2 (a+b x)}{5 e \sqrt {(a+b x)^2} (d+e x)^{5/2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 36.68, size = 44, normalized size = 1.07 \begin {gather*} \frac {2 (-a e-b e x)}{5 e^2 (d+e x)^{5/2} \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 42, normalized size = 1.02 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.17, size = 18, normalized size = 0.44 \begin {gather*} -\frac {2 \, e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right )}{5 \, {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

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)

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maple [A]  time = 0.04, size = 27, normalized size = 0.66 \begin {gather*} -\frac {2 \left (b x +a \right )}{5 \left (e x +d \right )^{\frac {5}{2}} \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)^(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 = 0.83, size = 42, normalized size = 1.02 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 2.56, size = 103, normalized size = 2.51 \begin {gather*} -\frac {2\,\sqrt {{\left (a+b\,x\right )}^2}}{5\,b\,e^3\,\left (x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (a\,e^3+2\,b\,d\,e^2\right )\,\sqrt {d+e\,x}}{b\,e^3}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}\right )} \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)^(7/2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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