\(\int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx\) [1966]

Optimal result

Integrand size = 33, antiderivative size = 146 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=-\frac {(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{6 e^3 (a+b x) (d+e x)^6}+\frac {2 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4} \]

[Out]

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

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} \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4}+\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{5 e^3 (a+b x) (d+e x)^5}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{6 e^3 (a+b x) (d+e x)^6} \]

[In]

Int[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^7,x]

[Out]

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

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)^7} \, 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)^7} \, 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)^7}-\frac {2 b (b d-a e)}{e^2 (d+e x)^6}+\frac {b^2}{e^2 (d+e x)^5}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{6 e^3 (a+b x) (d+e x)^6}+\frac {2 b (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 {b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.50 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=-\frac {\sqrt {(a+b x)^2} \left (10 a^2 e^2+4 a b e (d+6 e x)+b^2 \left (d^2+6 d e x+15 e^2 x^2\right )\right )}{60 e^3 (a+b x) (d+e x)^6} \]

[In]

Integrate[((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(d + e*x)^7,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(10*a^2*e^2 + 4*a*b*e*(d + 6*e*x) + b^2*(d^2 + 6*d*e*x + 15*e^2*x^2)))/(e^3*(a + b*x)
*(d + e*x)^6)

Maple [C] (warning: unable to verify)

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

Time = 0.94 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.47

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (15 b^{2} e^{2} x^{2}+24 a b \,e^{2} x +6 b^{2} d e x +10 e^{2} a^{2}+4 a b d e +b^{2} d^{2}\right )}{60 e^{3} \left (e x +d \right )^{6}}\)	\(68\)
gosper	\(-\frac {\left (15 b^{2} e^{2} x^{2}+24 a b \,e^{2} x +6 b^{2} d e x +10 e^{2} a^{2}+4 a b d e +b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{60 e^{3} \left (e x +d \right )^{6} \left (b x +a \right )}\)	\(78\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{2} x^{2}}{4 e}-\frac {b \left (4 a e +b d \right ) x}{10 e^{2}}-\frac {10 e^{2} a^{2}+4 a b d e +b^{2} d^{2}}{60 e^{3}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{6}}\)	\(79\)

[In]

int((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^7,x,method=_RETURNVERBOSE)

[Out]

-1/60*csgn(b*x+a)*(15*b^2*e^2*x^2+24*a*b*e^2*x+6*b^2*d*e*x+10*a^2*e^2+4*a*b*d*e+b^2*d^2)/e^3/(e*x+d)^6

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=-\frac {15 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 4 \, a b d e + 10 \, a^{2} e^{2} + 6 \, {\left (b^{2} d e + 4 \, a b e^{2}\right )} x}{60 \, {\left (e^{9} x^{6} + 6 \, d e^{8} x^{5} + 15 \, d^{2} e^{7} x^{4} + 20 \, d^{3} e^{6} x^{3} + 15 \, d^{4} e^{5} x^{2} + 6 \, d^{5} e^{4} x + d^{6} e^{3}\right )}} \]

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^7,x, algorithm="fricas")

[Out]

-1/60*(15*b^2*e^2*x^2 + b^2*d^2 + 4*a*b*d*e + 10*a^2*e^2 + 6*(b^2*d*e + 4*a*b*e^2)*x)/(e^9*x^6 + 6*d*e^8*x^5 +
15*d^2*e^7*x^4 + 20*d^3*e^6*x^3 + 15*d^4*e^5*x^2 + 6*d^5*e^4*x + d^6*e^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=\text {Timed out} \]

[In]

integrate((b*x+a)*((b*x+a)**2)**(1/2)/(e*x+d)**7,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^7,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 = 165, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=\frac {b^{6} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (b^{4} d^{4} e^{3} - 4 \, a b^{3} d^{3} e^{4} + 6 \, a^{2} b^{2} d^{2} e^{5} - 4 \, a^{3} b d e^{6} + a^{4} e^{7}\right )}} - \frac {15 \, b^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{2} d e x \mathrm {sgn}\left (b x + a\right ) + 24 \, a b e^{2} x \mathrm {sgn}\left (b x + a\right ) + b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, a b d e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (e x + d\right )}^{6} e^{3}} \]

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^7,x, algorithm="giac")

[Out]

1/60*b^6*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*(15
*b^2*e^2*x^2*sgn(b*x + a) + 6*b^2*d*e*x*sgn(b*x + a) + 24*a*b*e^2*x*sgn(b*x + a) + b^2*d^2*sgn(b*x + a) + 4*a*
b*d*e*sgn(b*x + a) + 10*a^2*e^2*sgn(b*x + a))/((e*x + d)^6*e^3)

Mupad [B] (verification not implemented)

Time = 10.94 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (10\,a^2\,e^2+4\,a\,b\,d\,e+24\,a\,b\,e^2\,x+b^2\,d^2+6\,b^2\,d\,e\,x+15\,b^2\,e^2\,x^2\right )}{60\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6} \]

[In]

int((((a + b*x)^2)^(1/2)*(a + b*x))/(d + e*x)^7,x)

[Out]

-(((a + b*x)^2)^(1/2)*(10*a^2*e^2 + b^2*d^2 + 15*b^2*e^2*x^2 + 24*a*b*e^2*x + 6*b^2*d*e*x + 4*a*b*d*e))/(60*e^
3*(a + b*x)*(d + e*x)^6)