\(\int \frac {(d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [1607]

Optimal result

Integrand size = 28, antiderivative size = 48 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(d+e x)^4}{4 (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 37} \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(d+e x)^4}{4 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[In]

Int[(d + e*x)^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(106\) vs. \(2(48)=96\).

Time = 1.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.21 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-a^3 e^3-a^2 b e^2 (d+4 e x)-a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )-b^3 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )}{4 b^4 (a+b x)^3 \sqrt {(a+b x)^2}} \]

[In]

Integrate[(d + e*x)^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-(a^3*e^3) - a^2*b*e^2*(d + 4*e*x) - a*b^2*e*(d^2 + 4*d*e*x + 6*e^2*x^2) - b^3*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2
+ 4*e^3*x^3))/(4*b^4*(a + b*x)^3*Sqrt[(a + b*x)^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(112\) vs. \(2(35)=70\).

Time = 2.51 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.35

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {e^{3} x^{3}}{b}-\frac {3 e^{2} \left (a e +b d \right ) x^{2}}{2 b^{2}}-\frac {e \left (a^{2} e^{2}+a b d e +b^{2} d^{2}\right ) x}{b^{3}}-\frac {a^{3} e^{3}+a^{2} b d \,e^{2}+a \,b^{2} d^{2} e +b^{3} d^{3}}{4 b^{4}}\right )}{\left (b x +a \right )^{5}}\)	\(113\)
gosper	\(-\frac {\left (b x +a \right ) \left (4 e^{3} x^{3} b^{3}+6 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+4 a^{2} b \,e^{3} x +4 x a \,b^{2} d \,e^{2}+4 b^{3} d^{2} e x +a^{3} e^{3}+a^{2} b d \,e^{2}+a \,b^{2} d^{2} e +b^{3} d^{3}\right )}{4 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(119\)
default	\(-\frac {\left (b x +a \right ) \left (4 e^{3} x^{3} b^{3}+6 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+4 a^{2} b \,e^{3} x +4 x a \,b^{2} d \,e^{2}+4 b^{3} d^{2} e x +a^{3} e^{3}+a^{2} b d \,e^{2}+a \,b^{2} d^{2} e +b^{3} d^{3}\right )}{4 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(119\)

[In]

int((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (35) = 70\).

Time = 0.26 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.98 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + a b^{2} d^{2} e + a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 4 \, {\left (b^{3} d^{2} e + a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{4 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \]

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((d + e*x)**3/((a + b*x)**2)**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (35) = 70\).

Time = 0.26 (sec) , antiderivative size = 238, normalized size of antiderivative = 4.96 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {e^{3} x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {d^{2} e}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, a^{2} e^{3}}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} - \frac {3 \, d e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {a e^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a d e^{2}}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {2 \, a^{2} e^{3}}{3 \, b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {d^{3}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {3 \, a d^{2} e}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {3 \, a^{2} d e^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {a^{3} e^{3}}{4 \, b^{8} {\left (x + \frac {a}{b}\right )}^{4}} \]

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (35) = 70\).

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.48 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, b^{3} e^{3} x^{3} + 6 \, b^{3} d e^{2} x^{2} + 6 \, a b^{2} e^{3} x^{2} + 4 \, b^{3} d^{2} e x + 4 \, a b^{2} d e^{2} x + 4 \, a^{2} b e^{3} x + b^{3} d^{3} + a b^{2} d^{2} e + a^{2} b d e^{2} + a^{3} e^{3}}{4 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.09 (sec) , antiderivative size = 254, normalized size of antiderivative = 5.29 \[ \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\left (\frac {2\,a\,e^3-3\,b\,d\,e^2}{2\,b^4}+\frac {a\,e^3}{2\,b^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^3}-\frac {\left (\frac {d^3}{4\,b}-\frac {a\,\left (\frac {3\,d^2\,e}{4\,b}+\frac {a\,\left (\frac {a\,e^3}{4\,b^2}-\frac {3\,d\,e^2}{4\,b}\right )}{b}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^5}-\frac {\left (\frac {a^2\,e^3-3\,a\,b\,d\,e^2+3\,b^2\,d^2\,e}{3\,b^4}+\frac {a\,\left (\frac {a\,e^3}{3\,b^3}+\frac {e^2\,\left (a\,e-3\,b\,d\right )}{3\,b^3}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^4}-\frac {e^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^2} \]

[In]

int((d + e*x)^3/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

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