3.1682 \(\int \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{5 e^2 (a+b x) (d+e x)^{5/2}}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^{3/2}} \]

[Out]

(2*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)*(d + e*x)^(5/2)) - (2*b*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/(3*e^2*(a + b*x)*(d + e*x)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0352242, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {646, 43} \[ \frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{5 e^2 (a+b x) (d+e x)^{5/2}}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)*(d + e*x)^(5/2)) - (2*b*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/(3*e^2*(a + b*x)*(d + e*x)^(3/2))

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{a b+b^2 x}{(d+e x)^{7/2}} \, dx}{a b+b^2 x}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (-\frac{b (b d-a e)}{e (d+e x)^{7/2}}+\frac{b^2}{e (d+e x)^{5/2}}\right ) \, dx}{a b+b^2 x}\\ &=\frac{2 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x) (d+e x)^{5/2}}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0244522, size = 48, normalized size = 0.5 \[ -\frac{2 \sqrt{(a+b x)^2} (3 a e+2 b d+5 b e x)}{15 e^2 (a+b x) (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(2*b*d + 3*a*e + 5*b*e*x))/(15*e^2*(a + b*x)*(d + e*x)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.04, size = 43, normalized size = 0.5 \begin{align*} -{\frac{10\,bxe+6\,ae+4\,bd}{15\, \left ( bx+a \right ){e}^{2}}\sqrt{ \left ( bx+a \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/(e*x+d)^(5/2)*(5*b*e*x+3*a*e+2*b*d)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

________________________________________________________________________________________

Maxima [A]  time = 1.07257, size = 63, normalized size = 0.66 \begin{align*} -\frac{2 \,{\left (5 \, b e x + 2 \, b d + 3 \, a e\right )}}{15 \,{\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )} \sqrt{e x + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(5*b*e*x + 2*b*d + 3*a*e)/((e^4*x^2 + 2*d*e^3*x + d^2*e^2)*sqrt(e*x + d))

________________________________________________________________________________________

Fricas [A]  time = 1.53867, size = 128, normalized size = 1.33 \begin{align*} -\frac{2 \,{\left (5 \, b e x + 2 \, b d + 3 \, a e\right )} \sqrt{e x + d}}{15 \,{\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(5*b*e*x + 2*b*d + 3*a*e)*sqrt(e*x + d)/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.13031, size = 66, normalized size = 0.69 \begin{align*} -\frac{2 \,{\left (5 \,{\left (x e + d\right )} b \mathrm{sgn}\left (b x + a\right ) - 3 \, b d \mathrm{sgn}\left (b x + a\right ) + 3 \, a e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-2\right )}}{15 \,{\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)^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2/15*(5*(x*e + d)*b*sgn(b*x + a) - 3*b*d*sgn(b*x + a) + 3*a*e*sgn(b*x + a))*e^(-2)/(x*e + d)^(5/2)