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

Optimal. Leaf size=41 \[ -\frac {(d+e x)^3}{3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)} \]

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {767} \begin {gather*} -\frac {(d+e x)^3}{3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 767

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Sim
p[(f*g*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)*(e*f - d*g)), x] /; FreeQ[{a, b, c, d, e, f, g,
m, p}, x] && EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && EqQ[2*c*f - b*g, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 60, normalized size = 1.46 \begin {gather*} \frac {-a^2 e^2-a b e (d+3 e x)-\left (b^2 \left (d^2+3 d e x+3 e^2 x^2\right )\right )}{3 b^3 \left ((a+b x)^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 1.13, size = 348, normalized size = 8.49 \begin {gather*} \frac {4 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (a^4 e^2-2 a^3 b d e-a^3 b e^2 x+a^2 b^2 d^2+2 a^2 b^2 d e x+a^2 b^2 e^2 x^2-a b^3 d^2 x-2 a b^3 d e x^2+b^4 d^2 x^2+3 b^4 d e x^3+3 b^4 e^2 x^4\right )+4 \left (a^5 b e^2-2 a^4 b^2 d e+a^3 b^3 d^2-a^2 b^4 e^2 x^3-a b^5 d e x^3-3 a b^5 e^2 x^4-b^6 d^2 x^3-3 b^6 d e x^4-3 b^6 e^2 x^5\right )}{3 b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2} \left (4 a^2 b^4+8 a b^5 x+4 b^6 x^2\right )+3 b^3 \sqrt {b^2} x^3 \left (-4 a^3 b^3-12 a^2 b^4 x-12 a b^5 x^2-4 b^6 x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(a^2*b^2*d^2 - 2*a^3*b*d*e + a^4*e^2 - a*b^3*d^2*x + 2*a^2*b^2*d*e*
x - a^3*b*e^2*x + b^4*d^2*x^2 - 2*a*b^3*d*e*x^2 + a^2*b^2*e^2*x^2 + 3*b^4*d*e*x^3 + 3*b^4*e^2*x^4) + 4*(a^3*b^
3*d^2 - 2*a^4*b^2*d*e + a^5*b*e^2 - b^6*d^2*x^3 - a*b^5*d*e*x^3 - a^2*b^4*e^2*x^3 - 3*b^6*d*e*x^4 - 3*a*b^5*e^
2*x^4 - 3*b^6*e^2*x^5))/(3*b^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(4*a^2*b^4 + 8*a*b^5*x + 4*b^6*x^2) + 3*b^3*S
qrt[b^2]*x^3*(-4*a^3*b^3 - 12*a^2*b^4*x - 12*a*b^5*x^2 - 4*b^6*x^3))

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fricas [B]  time = 0.44, size = 84, normalized size = 2.05 \begin {gather*} -\frac {3 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + a b d e + a^{2} e^{2} + 3 \, {\left (b^{2} d e + a b e^{2}\right )} x}{3 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 69, normalized size = 1.68 \begin {gather*} -\frac {\left (b x +a \right )^{2} \left (3 b^{2} e^{2} x^{2}+3 a b \,e^{2} x +3 b^{2} d e x +a^{2} e^{2}+a b d e +b^{2} d^{2}\right )}{3 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.73, size = 274, normalized size = 6.68 \begin {gather*} -\frac {e^{2} x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b} - \frac {2 \, a^{2} e^{2}}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{3}} - \frac {b d^{2} + 2 \, a d e}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {a e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a^{2} e^{2}}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {a d^{2}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {a^{3} e^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {2 \, b d e + a e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, {\left (2 \, b d e + a e^{2}\right )} a}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {{\left (2 \, b d e + a e^{2}\right )} a^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {{\left (b d^{2} + 2 \, a d e\right )} a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.24, size = 77, normalized size = 1.88 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^2\,e^2+a\,b\,d\,e+3\,a\,b\,e^2\,x+b^2\,d^2+3\,b^2\,d\,e\,x+3\,b^2\,e^2\,x^2\right )}{3\,b^3\,{\left (a+b\,x\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right ) \left (d + e x\right )^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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