\(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^6} \, dx\) [1565]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 28, antiderivative size = 98 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (b d-a e) (d+e x)^5}+\frac {b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{20 (b d-a e)^2 (d+e x)^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {660, 47, 37} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{20 (d+e x)^4 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{5 (d+e x)^5 (b d-a e)} \]

[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 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]

Rule 47

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 660

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 1.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=-\frac {\sqrt {(a+b x)^2} \left (4 a^3 e^3+3 a^2 b e^2 (d+5 e x)+2 a b^2 e \left (d^2+5 d e x+10 e^2 x^2\right )+b^3 \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )\right )}{20 e^4 (a+b x) (d+e x)^5} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.29

method result size
risch \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{3} x^{3}}{2 e}-\frac {b^{2} \left (2 a e +b d \right ) x^{2}}{2 e^{2}}-\frac {b \left (3 a^{2} e^{2}+2 a b d e +b^{2} d^{2}\right ) x}{4 e^{3}}-\frac {4 a^{3} e^{3}+3 a^{2} b d \,e^{2}+2 a \,b^{2} d^{2} e +b^{3} d^{3}}{20 e^{4}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{5}}\) \(126\)
gosper \(-\frac {\left (10 e^{3} x^{3} b^{3}+20 x^{2} a \,b^{2} e^{3}+10 x^{2} b^{3} d \,e^{2}+15 a^{2} b \,e^{3} x +10 x a \,b^{2} d \,e^{2}+5 b^{3} d^{2} e x +4 a^{3} e^{3}+3 a^{2} b d \,e^{2}+2 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 e^{4} \left (e x +d \right )^{5} \left (b x +a \right )^{3}}\) \(131\)
default \(-\frac {\left (10 e^{3} x^{3} b^{3}+20 x^{2} a \,b^{2} e^{3}+10 x^{2} b^{3} d \,e^{2}+15 a^{2} b \,e^{3} x +10 x a \,b^{2} d \,e^{2}+5 b^{3} d^{2} e x +4 a^{3} e^{3}+3 a^{2} b d \,e^{2}+2 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 e^{4} \left (e x +d \right )^{5} \left (b x +a \right )^{3}}\) \(131\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (72) = 144\).

Time = 0.33 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=-\frac {10 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 2 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} + 4 \, a^{3} e^{3} + 10 \, {\left (b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x}{20 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{6}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=\text {Exception raised: ValueError} \]

[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 >> 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 detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (72) = 144\).

Time = 0.28 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.18 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=\frac {b^{5} \mathrm {sgn}\left (b x + a\right )}{20 \, {\left (b^{2} d^{2} e^{4} - 2 \, a b d e^{5} + a^{2} e^{6}\right )}} - \frac {10 \, b^{3} e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, b^{3} d e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, a b^{2} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, b^{3} d^{2} e x \mathrm {sgn}\left (b x + a\right ) + 10 \, a b^{2} d e^{2} x \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{2} b e^{3} x \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )}{20 \, {\left (e x + d\right )}^{5} e^{4}} \]

[In]

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

[Out]

1/20*b^5*sgn(b*x + a)/(b^2*d^2*e^4 - 2*a*b*d*e^5 + a^2*e^6) - 1/20*(10*b^3*e^3*x^3*sgn(b*x + a) + 10*b^3*d*e^2
*x^2*sgn(b*x + a) + 20*a*b^2*e^3*x^2*sgn(b*x + a) + 5*b^3*d^2*e*x*sgn(b*x + a) + 10*a*b^2*d*e^2*x*sgn(b*x + a)
 + 15*a^2*b*e^3*x*sgn(b*x + a) + b^3*d^3*sgn(b*x + a) + 2*a*b^2*d^2*e*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*x + d)^5*e^4)

Mupad [B] (verification not implemented)

Time = 9.61 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx=\frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{3\,e^4}+\frac {b^3\,d}{3\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{4\,e^4}+\frac {d\,\left (\frac {b^3\,d}{4\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{4\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {a^3}{5\,e}-\frac {d\,\left (\frac {3\,a^2\,b}{5\,e}-\frac {d\,\left (\frac {3\,a\,b^2}{5\,e}-\frac {b^3\,d}{5\,e^2}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,e^4\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2} \]

[In]

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

[Out]

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