\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx\) [1718]

Optimal result

Integrand size = 33, antiderivative size = 158 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {(b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}+\frac {(2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \]

[Out]

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

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} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{4 e^3 (a+b x) (d+e x)^4}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{5 e^3 (a+b x) (d+e x)^5}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \]

[In]

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

[Out]

-1/5*((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)^5) + ((2*b*B*d - A*b*e -
a*B*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*e^3*(a + b*x)*(d + e*x)^4) - (b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*
e^3*(a + b*x)*(d + e*x)^3)

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)^6} \, 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)^6}+\frac {b (-2 b B d+A b e+a B e)}{e^2 (d+e x)^5}+\frac {b^2 B}{e^2 (d+e x)^4}\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}}{5 e^3 (a+b x) (d+e x)^5}+\frac {(2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {\sqrt {(a+b x)^2} \left (3 a e (4 A e+B (d+5 e x))+b \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )\right )}{60 e^3 (a+b x) (d+e x)^5} \]

[In]

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

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(3*a*e*(4*A*e + B*(d + 5*e*x)) + b*(3*A*e*(d + 5*e*x) + 2*B*(d^2 + 5*d*e*x + 10*e^2*x
^2))))/(e^3*(a + b*x)*(d + e*x)^5)

Maple [C] (warning: unable to verify)

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

Time = 0.86 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.50

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (20 B b \,e^{2} x^{2}+15 A b \,e^{2} x +15 B a \,e^{2} x +10 B b d e x +12 A a \,e^{2}+3 A b d e +3 B a d e +2 B b \,d^{2}\right )}{60 e^{3} \left (e x +d \right )^{5}}\)	\(79\)
gosper	\(-\frac {\left (20 B b \,e^{2} x^{2}+15 A b \,e^{2} x +15 B a \,e^{2} x +10 B b d e x +12 A a \,e^{2}+3 A b d e +3 B a d e +2 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{60 e^{3} \left (e x +d \right )^{5} \left (b x +a \right )}\)	\(89\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B b \,x^{2}}{3 e}-\frac {\left (3 A b e +3 B a e +2 B b d \right ) x}{12 e^{2}}-\frac {12 A a \,e^{2}+3 A b d e +3 B a d e +2 B b \,d^{2}}{60 e^{3}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{5}}\)	\(90\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x,method=_RETURNVERBOSE)

[Out]

-1/60*csgn(b*x+a)*(20*B*b*e^2*x^2+15*A*b*e^2*x+15*B*a*e^2*x+10*B*b*d*e*x+12*A*a*e^2+3*A*b*d*e+3*B*a*d*e+2*B*b*
d^2)/e^3/(e*x+d)^5

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {20 \, B b e^{2} x^{2} + 2 \, B b d^{2} + 12 \, A a e^{2} + 3 \, {\left (B a + A b\right )} d e + 5 \, {\left (2 \, B b d e + 3 \, {\left (B a + A b\right )} e^{2}\right )} x}{60 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/60*(20*B*b*e^2*x^2 + 2*B*b*d^2 + 12*A*a*e^2 + 3*(B*a + A*b)*d*e + 5*(2*B*b*d*e + 3*(B*a + A*b)*e^2)*x)/(e^8
*x^5 + 5*d*e^7*x^4 + 10*d^2*e^6*x^3 + 10*d^3*e^5*x^2 + 5*d^4*e^4*x + d^5*e^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\text {Timed out} \]

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**6,x)

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

none

Time = 0.28 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\frac {{\left (2 \, B b^{5} d - 5 \, B a b^{4} e + 3 \, A b^{5} e\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (b^{4} d^{4} e^{3} - 4 \, a b^{3} d^{3} e^{4} + 6 \, a^{2} b^{2} d^{2} e^{5} - 4 \, a^{3} b d e^{6} + a^{4} e^{7}\right )}} - \frac {20 \, B b e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, B b d e x \mathrm {sgn}\left (b x + a\right ) + 15 \, B a e^{2} x \mathrm {sgn}\left (b x + a\right ) + 15 \, A b e^{2} x \mathrm {sgn}\left (b x + a\right ) + 2 \, B b d^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a d e \mathrm {sgn}\left (b x + a\right ) + 3 \, A b d e \mathrm {sgn}\left (b x + a\right ) + 12 \, A a e^{2} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (e x + d\right )}^{5} e^{3}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

1/60*(2*B*b^5*d - 5*B*a*b^4*e + 3*A*b^5*e)*sgn(b*x + a)/(b^4*d^4*e^3 - 4*a*b^3*d^3*e^4 + 6*a^2*b^2*d^2*e^5 - 4
*a^3*b*d*e^6 + a^4*e^7) - 1/60*(20*B*b*e^2*x^2*sgn(b*x + a) + 10*B*b*d*e*x*sgn(b*x + a) + 15*B*a*e^2*x*sgn(b*x
+ a) + 15*A*b*e^2*x*sgn(b*x + a) + 2*B*b*d^2*sgn(b*x + a) + 3*B*a*d*e*sgn(b*x + a) + 3*A*b*d*e*sgn(b*x + a) +
12*A*a*e^2*sgn(b*x + a))/((e*x + d)^5*e^3)

Mupad [B] (verification not implemented)

Time = 11.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.56 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (12\,A\,a\,e^2+2\,B\,b\,d^2+15\,A\,b\,e^2\,x+15\,B\,a\,e^2\,x+20\,B\,b\,e^2\,x^2+3\,A\,b\,d\,e+3\,B\,a\,d\,e+10\,B\,b\,d\,e\,x\right )}{60\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5} \]

[In]

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

[Out]

-(((a + b*x)^2)^(1/2)*(12*A*a*e^2 + 2*B*b*d^2 + 15*A*b*e^2*x + 15*B*a*e^2*x + 20*B*b*e^2*x^2 + 3*A*b*d*e + 3*B
*a*d*e + 10*B*b*d*e*x))/(60*e^3*(a + b*x)*(d + e*x)^5)