3.18.38 \(\int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx\)

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

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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 {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 (d+e x)^3 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.03, size = 71, normalized size = 1.73 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (a^2 e^2+a b e (d+3 e x)+b^2 \left (d^2+3 d e x+3 e^2 x^2\right )\right )}{3 e^3 (a+b x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[(a + b*x)^2]*(a^2*e^2 + a*b*e*(d + 3*e*x) + b^2*(d^2 + 3*d*e*x + 3*e^2*x^2)))/(e^3*(a + b*x)*(d + e
*x)^3)

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IntegrateAlgebraic [F]  time = 1.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[IntegrateAlgebraic][((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^4, x]

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fricas [B]  time = 0.41, 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 (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [B]  time = 0.16, size = 94, normalized size = 2.29 \begin {gather*} -\frac {{\left (3 \, b^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, b^{2} d x e \mathrm {sgn}\left (b x + a\right ) + b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a b x e^{2} \mathrm {sgn}\left (b x + a\right ) + a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{3 \, {\left (x e + d\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 76, normalized size = 1.85 \begin {gather*} -\frac {\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 ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (e x +d \right )^{3} \left (b x +a \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^4,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.10, size = 75, normalized size = 1.83 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^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\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((a + b*x)^2)^(1/2)*(a + b*x))/(d + e*x)^4,x)

[Out]

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

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sympy [B]  time = 0.65, size = 88, normalized size = 2.15 \begin {gather*} \frac {- a^{2} e^{2} - a b d e - b^{2} d^{2} - 3 b^{2} e^{2} x^{2} + x \left (- 3 a b e^{2} - 3 b^{2} d e\right )}{3 d^{3} e^{3} + 9 d^{2} e^{4} x + 9 d e^{5} x^{2} + 3 e^{6} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)**2)**(1/2)/(e*x+d)**4,x)

[Out]

(-a**2*e**2 - a*b*d*e - b**2*d**2 - 3*b**2*e**2*x**2 + x*(-3*a*b*e**2 - 3*b**2*d*e))/(3*d**3*e**3 + 9*d**2*e**
4*x + 9*d*e**5*x**2 + 3*e**6*x**3)

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