3.1565 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=98 \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3}{20 (d+e x)^4 (b d-a e)^2}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3}{5 (d+e x)^5 (b d-a e)} \]

[Out]

((a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(b*d - a*e)*(d + e*x)^5) + (b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(20*(b*d - a*e)^2*(d + e*x)^4)

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Rubi [A]  time = 0.0382047, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {646, 45, 37} \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3}{20 (d+e x)^4 (b d-a e)^2}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3}{5 (d+e x)^5 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^6,x]

[Out]

((a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(b*d - a*e)*(d + e*x)^5) + (b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(20*(b*d - a*e)^2*(d + e*x)^4)

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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^3}{(d+e x)^6} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (b d-a e) (d+e x)^5}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^3}{(d+e x)^5} \, dx}{5 b (b d-a e) \left (a b+b^2 x\right )}\\ &=\frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (b d-a e) (d+e x)^5}+\frac{b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{20 (b d-a e)^2 (d+e x)^4}\\ \end{align*}

Mathematica [A]  time = 0.0474105, size = 112, normalized size = 1.14 \[ -\frac{\sqrt{(a+b x)^2} \left (3 a^2 b e^2 (d+5 e x)+4 a^3 e^3+2 a b^2 e \left (d^2+5 d e x+10 e^2 x^2\right )+b^3 \left (5 d^2 e x+d^3+10 d e^2 x^2+10 e^3 x^3\right )\right )}{20 e^4 (a+b x) (d+e x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^6,x]

[Out]

-(Sqrt[(a + b*x)^2]*(4*a^3*e^3 + 3*a^2*b*e^2*(d + 5*e*x) + 2*a*b^2*e*(d^2 + 5*d*e*x + 10*e^2*x^2) + b^3*(d^3 +
5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3)))/(20*e^4*(a + b*x)*(d + e*x)^5)

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Maple [A]  time = 0.155, size = 131, normalized size = 1.3 \begin{align*} -{\frac{10\,{x}^{3}{b}^{3}{e}^{3}+20\,{x}^{2}a{b}^{2}{e}^{3}+10\,{x}^{2}{b}^{3}d{e}^{2}+15\,x{a}^{2}b{e}^{3}+10\,xa{b}^{2}d{e}^{2}+5\,x{b}^{3}{d}^{2}e+4\,{a}^{3}{e}^{3}+3\,d{e}^{2}{a}^{2}b+2\,a{b}^{2}{d}^{2}e+{b}^{3}{d}^{3}}{20\,{e}^{4} \left ( ex+d \right ) ^{5} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x)

[Out]

-1/20/e^4*(10*b^3*e^3*x^3+20*a*b^2*e^3*x^2+10*b^3*d*e^2*x^2+15*a^2*b*e^3*x+10*a*b^2*d*e^2*x+5*b^3*d^2*e*x+4*a^
3*e^3+3*a^2*b*d*e^2+2*a*b^2*d^2*e+b^3*d^3)*((b*x+a)^2)^(3/2)/(e*x+d)^5/(b*x+a)^3

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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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.59604, size = 328, normalized size = 3.35 \begin{align*} -\frac{10 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 2 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} + 4 \, a^{3} e^{3} + 10 \,{\left (b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + 5 \,{\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x}{20 \,{\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/20*(10*b^3*e^3*x^3 + b^3*d^3 + 2*a*b^2*d^2*e + 3*a^2*b*d*e^2 + 4*a^3*e^3 + 10*(b^3*d*e^2 + 2*a*b^2*e^3)*x^2
+ 5*(b^3*d^2*e + 2*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^7*x^3 + 10*d^3*e^6*x^2 + 5
*d^4*e^5*x + d^5*e^4)

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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**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.18876, size = 228, normalized size = 2.33 \begin{align*} -\frac{{\left (10 \, b^{3} x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 10 \, b^{3} d x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, b^{3} d^{2} x e \mathrm{sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) + 20 \, a b^{2} x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 10 \, a b^{2} d x e^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \, a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 15 \, a^{2} b x e^{3} \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) + 4 \, a^{3} e^{3} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{20 \,{\left (x e + d\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/20*(10*b^3*x^3*e^3*sgn(b*x + a) + 10*b^3*d*x^2*e^2*sgn(b*x + a) + 5*b^3*d^2*x*e*sgn(b*x + a) + b^3*d^3*sgn(
b*x + a) + 20*a*b^2*x^2*e^3*sgn(b*x + a) + 10*a*b^2*d*x*e^2*sgn(b*x + a) + 2*a*b^2*d^2*e*sgn(b*x + a) + 15*a^2
*b*x*e^3*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) + 4*a^3*e^3*sgn(b*x + a))*e^(-4)/(x*e + d)^5