\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx\) [1717]

Optimal result

Integrand size = 33, antiderivative size = 158 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=-\frac {(b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4}+\frac {(2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x) (d+e x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {784, 78} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x) (d+e x)^3}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{4 e^3 (a+b x) (d+e x)^4}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x) (d+e x)^2} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 784

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]

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

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.51 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=-\frac {\sqrt {(a+b x)^2} \left (a e (3 A e+B (d+4 e x))+b \left (A e (d+4 e x)+B \left (d^2+4 d e x+6 e^2 x^2\right )\right )\right )}{12 e^3 (a+b x) (d+e x)^4} \]

[In]

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

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(a*e*(3*A*e + B*(d + 4*e*x)) + b*(A*e*(d + 4*e*x) + B*(d^2 + 4*d*e*x + 6*e^2*x^2))))/
(e^3*(a + b*x)*(d + e*x)^4)

Maple [C] (warning: unable to verify)

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

Time = 0.63 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.48

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (6 B b \,e^{2} x^{2}+4 A b \,e^{2} x +4 B a \,e^{2} x +4 B b d e x +3 A a \,e^{2}+A b d e +B a d e +B b \,d^{2}\right )}{12 e^{3} \left (e x +d \right )^{4}}\)	\(76\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B b \,x^{2}}{2 e}-\frac {\left (A b e +B a e +B b d \right ) x}{3 e^{2}}-\frac {3 A a \,e^{2}+A b d e +B a d e +B b \,d^{2}}{12 e^{3}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{4}}\)	\(84\)
gosper	\(-\frac {\left (6 B b \,e^{2} x^{2}+4 A b \,e^{2} x +4 B a \,e^{2} x +4 B b d e x +3 A a \,e^{2}+A b d e +B a d e +B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{12 e^{3} \left (e x +d \right )^{4} \left (b x +a \right )}\)	\(86\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

-1/12*csgn(b*x+a)*(6*B*b*e^2*x^2+4*A*b*e^2*x+4*B*a*e^2*x+4*B*b*d*e*x+3*A*a*e^2+A*b*d*e+B*a*d*e+B*b*d^2)/e^3/(e
*x+d)^4

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.65 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=-\frac {6 \, B b e^{2} x^{2} + B b d^{2} + 3 \, A a e^{2} + {\left (B a + A b\right )} d e + 4 \, {\left (B b d e + {\left (B a + A b\right )} e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^5,x, algorithm="fricas")

[Out]

-1/12*(6*B*b*e^2*x^2 + B*b*d^2 + 3*A*a*e^2 + (B*a + A*b)*d*e + 4*(B*b*d*e + (B*a + A*b)*e^2)*x)/(e^7*x^4 + 4*d
*e^6*x^3 + 6*d^2*e^5*x^2 + 4*d^3*e^4*x + d^4*e^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\text {Timed out} \]

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**5,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^5,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 [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.20 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=\frac {{\left (B b^{4} d - 2 \, B a b^{3} e + A b^{4} e\right )} \mathrm {sgn}\left (b x + a\right )}{12 \, {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )}} - \frac {6 \, B b e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, B b d e x \mathrm {sgn}\left (b x + a\right ) + 4 \, B a e^{2} x \mathrm {sgn}\left (b x + a\right ) + 4 \, A b e^{2} x \mathrm {sgn}\left (b x + a\right ) + B b d^{2} \mathrm {sgn}\left (b x + a\right ) + B a d e \mathrm {sgn}\left (b x + a\right ) + A b d e \mathrm {sgn}\left (b x + a\right ) + 3 \, A a e^{2} \mathrm {sgn}\left (b x + a\right )}{12 \, {\left (e x + d\right )}^{4} e^{3}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^5,x, algorithm="giac")

[Out]

1/12*(B*b^4*d - 2*B*a*b^3*e + A*b^4*e)*sgn(b*x + a)/(b^3*d^3*e^3 - 3*a*b^2*d^2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)
- 1/12*(6*B*b*e^2*x^2*sgn(b*x + a) + 4*B*b*d*e*x*sgn(b*x + a) + 4*B*a*e^2*x*sgn(b*x + a) + 4*A*b*e^2*x*sgn(b*x
+ a) + B*b*d^2*sgn(b*x + a) + B*a*d*e*sgn(b*x + a) + A*b*d*e*sgn(b*x + a) + 3*A*a*e^2*sgn(b*x + a))/((e*x + d
)^4*e^3)

Mupad [B] (verification not implemented)

Time = 10.88 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.54 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^5} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (3\,A\,a\,e^2+B\,b\,d^2+4\,A\,b\,e^2\,x+4\,B\,a\,e^2\,x+6\,B\,b\,e^2\,x^2+A\,b\,d\,e+B\,a\,d\,e+4\,B\,b\,d\,e\,x\right )}{12\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4} \]

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/(d + e*x)^5,x)

[Out]

-(((a + b*x)^2)^(1/2)*(3*A*a*e^2 + B*b*d^2 + 4*A*b*e^2*x + 4*B*a*e^2*x + 6*B*b*e^2*x^2 + A*b*d*e + B*a*d*e + 4
*B*b*d*e*x))/(12*e^3*(a + b*x)*(d + e*x)^4)