3.1607 \(\int \frac{(d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=48 \[ -\frac{(d+e x)^4}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[Out]

-(d + e*x)^4/(4*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0220864, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 37} \[ -\frac{(d+e x)^4}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d + e*x)^4/(4*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^3}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^4}{4 (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [B]  time = 0.0431586, size = 106, normalized size = 2.21 \[ \frac{-a^2 b e^2 (d+4 e x)-a^3 e^3-a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )+b^3 \left (-\left (4 d^2 e x+d^3+6 d e^2 x^2+4 e^3 x^3\right )\right )}{4 b^4 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^3*e^3) - a^2*b*e^2*(d + 4*e*x) - a*b^2*e*(d^2 + 4*d*e*x + 6*e^2*x^2) - b^3*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2
+ 4*e^3*x^3))/(4*b^4*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.155, size = 119, normalized size = 2.5 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 4\,{x}^{3}{b}^{3}{e}^{3}+6\,{x}^{2}a{b}^{2}{e}^{3}+6\,{x}^{2}{b}^{3}d{e}^{2}+4\,x{a}^{2}b{e}^{3}+4\,xa{b}^{2}d{e}^{2}+4\,x{b}^{3}{d}^{2}e+{a}^{3}{e}^{3}+d{e}^{2}{a}^{2}b+a{b}^{2}{d}^{2}e+{b}^{3}{d}^{3} \right ) }{4\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

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

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Maxima [B]  time = 1.1404, size = 392, normalized size = 8.17 \begin{align*} -\frac{e^{3} x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{d^{2} e}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{2 \, a^{2} e^{3}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{4}} - \frac{3 \, a^{2} b^{2} d e^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} - \frac{a^{3} b e^{3}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, a b d e^{2}}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} + \frac{2 \, a^{2} e^{3}}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{3 \, d e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{a e^{3}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{d^{3}}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{3 \, a d^{2} e}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{4}} + \frac{a^{3} e^{3}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

-e^3*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - d^2*e/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 2/3*a^2*e^3/((b
^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) - 3/4*a^2*b^2*d*e^2/((b^2)^(9/2)*(x + a/b)^4) - 1/4*a^3*b*e^3/((b^2)^(9/2)*
(x + a/b)^4) + 2*a*b*d*e^2/((b^2)^(7/2)*(x + a/b)^3) + 2/3*a^2*e^3/((b^2)^(7/2)*(x + a/b)^3) - 3/2*d*e^2/((b^2
)^(5/2)*(x + a/b)^2) - 1/2*a*e^3/((b^2)^(5/2)*b*(x + a/b)^2) - 1/4*d^3/((b^2)^(5/2)*(x + a/b)^4) + 3/4*a*d^2*e
/((b^2)^(5/2)*b*(x + a/b)^4) + 1/2*a^3*e^3/((b^2)^(5/2)*b^3*(x + a/b)^4)

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Fricas [B]  time = 1.62099, size = 284, normalized size = 5.92 \begin{align*} -\frac{4 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + a b^{2} d^{2} e + a^{2} b d e^{2} + a^{3} e^{3} + 6 \,{\left (b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 4 \,{\left (b^{3} d^{2} e + a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{4 \,{\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/4*(4*b^3*e^3*x^3 + b^3*d^3 + a*b^2*d^2*e + a^2*b*d*e^2 + a^3*e^3 + 6*(b^3*d*e^2 + a*b^2*e^3)*x^2 + 4*(b^3*d
^2*e + a*b^2*d*e^2 + a^2*b*e^3)*x)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((d + e*x)**3/((a + b*x)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x