\(\int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx\) [1965]

Optimal result

Integrand size = 33, antiderivative size = 146 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}+\frac {b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x) (d+e x)^4}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {784, 21, 45} \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^3 (a+b x) (d+e x)^4}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^3 (a+b x) (d+e x)^5} \]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 {(a+b x) \left (a b+b^2 x\right )}{(d+e x)^6} \, dx}{a b+b^2 x} \\ & = \frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^2}{(d+e x)^6} \, dx}{a b+b^2 x} \\ & = \frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {(-b d+a e)^2}{e^2 (d+e x)^6}-\frac {2 b (b d-a e)}{e^2 (d+e x)^5}+\frac {b^2}{e^2 (d+e x)^4}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}+\frac {b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x) (d+e x)^4}-\frac {b^2 \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 = 73, normalized size of antiderivative = 0.50 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {\sqrt {(a+b x)^2} \left (6 a^2 e^2+3 a b e (d+5 e x)+b^2 \left (d^2+5 d e x+10 e^2 x^2\right )\right )}{30 e^3 (a+b x) (d+e x)^5} \]

[In]

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

[Out]

-1/30*(Sqrt[(a + b*x)^2]*(6*a^2*e^2 + 3*a*b*e*(d + 5*e*x) + b^2*(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 = 0.73 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.47

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (10 b^{2} e^{2} x^{2}+15 a b \,e^{2} x +5 b^{2} d e x +6 e^{2} a^{2}+3 a b d e +b^{2} d^{2}\right )}{30 e^{3} \left (e x +d \right )^{5}}\)	\(68\)
gosper	\(-\frac {\left (10 b^{2} e^{2} x^{2}+15 a b \,e^{2} x +5 b^{2} d e x +6 e^{2} a^{2}+3 a b d e +b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{30 e^{3} \left (e x +d \right )^{5} \left (b x +a \right )}\)	\(78\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{2} x^{2}}{3 e}-\frac {b \left (3 a e +b d \right ) x}{6 e^{2}}-\frac {6 e^{2} a^{2}+3 a b d e +b^{2} d^{2}}{30 e^{3}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{5}}\)	\(79\)

[In]

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

[Out]

-1/30*csgn(b*x+a)*(10*b^2*e^2*x^2+15*a*b*e^2*x+5*b^2*d*e*x+6*a^2*e^2+3*a*b*d*e+b^2*d^2)/e^3/(e*x+d)^5

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {10 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 3 \, a b d e + 6 \, a^{2} e^{2} + 5 \, {\left (b^{2} d e + 3 \, a b e^{2}\right )} x}{30 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]

[In]

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

[Out]

-1/30*(10*b^2*e^2*x^2 + b^2*d^2 + 3*a*b*d*e + 6*a^2*e^2 + 5*(b^2*d*e + 3*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 = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\frac {b^{5} \mathrm {sgn}\left (b x + a\right )}{30 \, {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )}} - \frac {10 \, b^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, b^{2} d e x \mathrm {sgn}\left (b x + a\right ) + 15 \, a b e^{2} x \mathrm {sgn}\left (b x + a\right ) + b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a b d e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )}{30 \, {\left (e x + d\right )}^{5} e^{3}} \]

[In]

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

[Out]

1/30*b^5*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) - 1/30*(10*b^2*e^2*x^2*sgn(b*x
+ a) + 5*b^2*d*e*x*sgn(b*x + a) + 15*a*b*e^2*x*sgn(b*x + a) + b^2*d^2*sgn(b*x + a) + 3*a*b*d*e*sgn(b*x + a) +
6*a^2*e^2*sgn(b*x + a))/((e*x + d)^5*e^3)

Mupad [B] (verification not implemented)

Time = 10.95 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (6\,a^2\,e^2+3\,a\,b\,d\,e+15\,a\,b\,e^2\,x+b^2\,d^2+5\,b^2\,d\,e\,x+10\,b^2\,e^2\,x^2\right )}{30\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5} \]

[In]

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

[Out]

-(((a + b*x)^2)^(1/2)*(6*a^2*e^2 + b^2*d^2 + 10*b^2*e^2*x^2 + 15*a*b*e^2*x + 5*b^2*d*e*x + 3*a*b*d*e))/(30*e^3
*(a + b*x)*(d + e*x)^5)