3.1965 \(\int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=146 \[ -\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3}+\frac{b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^3 (a+b x) (d+e x)^4}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^3 (a+b x) (d+e x)^5} \]

[Out]

-((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)*(d + e*x)^5) + (b*(b*d - a*e)*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/(2*e^3*(a + b*x)*(d + e*x)^4) - (b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)*(d + e*x)^
3)

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Rubi [A]  time = 0.080889, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ -\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3}+\frac{b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^3 (a+b x) (d+e x)^4}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^3 (a+b x) (d+e x)^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^6,x]

[Out]

-((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)*(d + e*x)^5) + (b*(b*d - a*e)*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/(2*e^3*(a + b*x)*(d + e*x)^4) - (b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)*(d + e*x)^
3)

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]

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 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{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{(a+b x) \left (a b+b^2 x\right )}{(d+e x)^6} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{(a+b x)^2}{(d+e x)^6} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(-b d+a e)^2}{e^2 (d+e x)^6}-\frac{2 b (b d-a e)}{e^2 (d+e x)^5}+\frac{b^2}{e^2 (d+e x)^4}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{(b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}+\frac{b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x) (d+e x)^4}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0369189, size = 73, normalized size = 0.5 \[ -\frac{\sqrt{(a+b x)^2} \left (6 a^2 e^2+3 a b e (d+5 e x)+b^2 \left (d^2+5 d e x+10 e^2 x^2\right )\right )}{30 e^3 (a+b x) (d+e x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^6,x]

[Out]

-(Sqrt[(a + b*x)^2]*(6*a^2*e^2 + 3*a*b*e*(d + 5*e*x) + b^2*(d^2 + 5*d*e*x + 10*e^2*x^2)))/(30*e^3*(a + b*x)*(d
+ e*x)^5)

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Maple [A]  time = 0.007, size = 78, normalized size = 0.5 \begin{align*} -{\frac{10\,{x}^{2}{b}^{2}{e}^{2}+15\,xab{e}^{2}+5\,x{b}^{2}de+6\,{a}^{2}{e}^{2}+3\,abde+{b}^{2}{d}^{2}}{30\,{e}^{3} \left ( ex+d \right ) ^{5} \left ( bx+a \right ) }\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x)

[Out]

-1/30/e^3*(10*b^2*e^2*x^2+15*a*b*e^2*x+5*b^2*d*e*x+6*a^2*e^2+3*a*b*d*e+b^2*d^2)*((b*x+a)^2)^(1/2)/(e*x+d)^5/(b
*x+a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47635, size = 227, normalized size = 1.55 \begin{align*} -\frac{10 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 3 \, a b d e + 6 \, a^{2} e^{2} + 5 \,{\left (b^{2} d e + 3 \, a b e^{2}\right )} x}{30 \,{\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/30*(10*b^2*e^2*x^2 + b^2*d^2 + 3*a*b*d*e + 6*a^2*e^2 + 5*(b^2*d*e + 3*a*b*e^2)*x)/(e^8*x^5 + 5*d*e^7*x^4 +
10*d^2*e^6*x^3 + 10*d^3*e^5*x^2 + 5*d^4*e^4*x + d^5*e^3)

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Sympy [A]  time = 1.12258, size = 116, normalized size = 0.79 \begin{align*} - \frac{6 a^{2} e^{2} + 3 a b d e + b^{2} d^{2} + 10 b^{2} e^{2} x^{2} + x \left (15 a b e^{2} + 5 b^{2} d e\right )}{30 d^{5} e^{3} + 150 d^{4} e^{4} x + 300 d^{3} e^{5} x^{2} + 300 d^{2} e^{6} x^{3} + 150 d e^{7} x^{4} + 30 e^{8} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)**2)**(1/2)/(e*x+d)**6,x)

[Out]

-(6*a**2*e**2 + 3*a*b*d*e + b**2*d**2 + 10*b**2*e**2*x**2 + x*(15*a*b*e**2 + 5*b**2*d*e))/(30*d**5*e**3 + 150*
d**4*e**4*x + 300*d**3*e**5*x**2 + 300*d**2*e**6*x**3 + 150*d*e**7*x**4 + 30*e**8*x**5)

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Giac [A]  time = 1.15604, size = 130, normalized size = 0.89 \begin{align*} -\frac{{\left (10 \, b^{2} x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, b^{2} d x e \mathrm{sgn}\left (b x + a\right ) + b^{2} d^{2} \mathrm{sgn}\left (b x + a\right ) + 15 \, a b x e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a b d e \mathrm{sgn}\left (b x + a\right ) + 6 \, a^{2} e^{2} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{30 \,{\left (x e + d\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/30*(10*b^2*x^2*e^2*sgn(b*x + a) + 5*b^2*d*x*e*sgn(b*x + a) + b^2*d^2*sgn(b*x + a) + 15*a*b*x*e^2*sgn(b*x +
a) + 3*a*b*d*e*sgn(b*x + a) + 6*a^2*e^2*sgn(b*x + a))*e^(-3)/(x*e + d)^5