\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx\) [1716]

Optimal result

Integrand size = 33, antiderivative size = 104 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\frac {(A b-a B) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 (b d-a e)^2 (d+e x)^2}+\frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 (b d-a e)^2 (d+e x)^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {783, 660, 37} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (B d-A e)}{3 (d+e x)^3 (b d-a e)^2}+\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (A b-a B)}{2 (d+e x)^2 (b d-a e)^2} \]

[In]

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

[Out]

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

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]

Rule 783

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e)^2)), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], 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] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 (b d-a e)^2 (d+e x)^3}+\frac {(A b-a B) \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx}{b d-a e} \\ & = \frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 (b d-a e)^2 (d+e x)^3}+\frac {\left ((A b-a B) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {a b+b^2 x}{(d+e x)^3} \, dx}{(b d-a e) \left (a b+b^2 x\right )} \\ & = \frac {(A b-a B) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 (b d-a e)^2 (d+e x)^2}+\frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 (b d-a e)^2 (d+e x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=-\frac {\sqrt {(a+b x)^2} \left (a e (2 A e+B (d+3 e x))+b \left (A e (d+3 e x)+2 B \left (d^2+3 d e x+3 e^2 x^2\right )\right )\right )}{6 e^3 (a+b x) (d+e x)^3} \]

[In]

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

[Out]

-1/6*(Sqrt[(a + b*x)^2]*(a*e*(2*A*e + B*(d + 3*e*x)) + b*(A*e*(d + 3*e*x) + 2*B*(d^2 + 3*d*e*x + 3*e^2*x^2))))
/(e^3*(a + b*x)*(d + e*x)^3)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.61 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (6 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x +6 B b d e x +2 A a \,e^{2}+A b d e +B a d e +2 B b \,d^{2}\right )}{6 e^{3} \left (e x +d \right )^{3}}\)	\(77\)
risch	\(\frac {\left (-\frac {B b \,x^{2}}{e}-\frac {\left (A b e +B a e +2 B b d \right ) x}{2 e^{2}}-\frac {2 A a \,e^{2}+A b d e +B a d e +2 B b \,d^{2}}{6 e^{3}}\right ) \sqrt {\left (b x +a \right )^{2}}}{\left (e x +d \right )^{3} \left (b x +a \right )}\)	\(86\)
gosper	\(-\frac {\left (6 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x +6 B b d e x +2 A a \,e^{2}+A b d e +B a d e +2 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{6 \left (e x +d \right )^{3} e^{3} \left (b x +a \right )}\)	\(87\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-1/6*csgn(b*x+a)*(6*B*b*e^2*x^2+3*A*b*e^2*x+3*B*a*e^2*x+6*B*b*d*e*x+2*A*a*e^2+A*b*d*e+B*a*d*e+2*B*b*d^2)/e^3/(
e*x+d)^3

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=-\frac {6 \, B b e^{2} x^{2} + 2 \, B b d^{2} + 2 \, A a e^{2} + {\left (B a + A b\right )} d e + 3 \, {\left (2 \, B b d e + {\left (B a + A b\right )} e^{2}\right )} x}{6 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \]

[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 detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (87) = 174\).

Time = 0.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=\frac {{\left (2 \, B b^{3} d - 3 \, B a b^{2} e + A b^{3} e\right )} \mathrm {sgn}\left (b x + a\right )}{6 \, {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )}} - \frac {6 \, B b e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B b d e x \mathrm {sgn}\left (b x + a\right ) + 3 \, B a e^{2} x \mathrm {sgn}\left (b x + a\right ) + 3 \, A b e^{2} x \mathrm {sgn}\left (b x + a\right ) + 2 \, B b d^{2} \mathrm {sgn}\left (b x + a\right ) + B a d e \mathrm {sgn}\left (b x + a\right ) + A b d e \mathrm {sgn}\left (b x + a\right ) + 2 \, A a e^{2} \mathrm {sgn}\left (b x + a\right )}{6 \, {\left (e x + d\right )}^{3} e^{3}} \]

[In]

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

[Out]

1/6*(2*B*b^3*d - 3*B*a*b^2*e + A*b^3*e)*sgn(b*x + a)/(b^2*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5) - 1/6*(6*B*b*e^2*x^
2*sgn(b*x + a) + 6*B*b*d*e*x*sgn(b*x + a) + 3*B*a*e^2*x*sgn(b*x + a) + 3*A*b*e^2*x*sgn(b*x + a) + 2*B*b*d^2*sg
n(b*x + a) + B*a*d*e*sgn(b*x + a) + A*b*d*e*sgn(b*x + a) + 2*A*a*e^2*sgn(b*x + a))/((e*x + d)^3*e^3)

Mupad [B] (verification not implemented)

Time = 10.70 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^4} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (2\,A\,a\,e^2+2\,B\,b\,d^2+3\,A\,b\,e^2\,x+3\,B\,a\,e^2\,x+6\,B\,b\,e^2\,x^2+A\,b\,d\,e+B\,a\,d\,e+6\,B\,b\,d\,e\,x\right )}{6\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \]

[In]

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

[Out]

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