\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx\) [1682]

Optimal result

Integrand size = 30, antiderivative size = 96 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=\frac {2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x) (d+e x)^{5/2}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {660, 45} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{5 e^2 (a+b x) (d+e x)^{5/2}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^{3/2}} \]

[In]

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

[Out]

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

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {a b+b^2 x}{(d+e x)^{7/2}} \, 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)}{e (d+e x)^{7/2}}+\frac {b^2}{e (d+e x)^{5/2}}\right ) \, dx}{a b+b^2 x} \\ & = \frac {2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x) (d+e x)^{5/2}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x) (d+e x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} (2 b d+3 a e+5 b e x)}{15 e^2 (a+b x) (d+e x)^{5/2}} \]

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(2*b*d + 3*a*e + 5*b*e*x))/(15*e^2*(a + b*x)*(d + e*x)^(5/2))

Maple [C] (warning: unable to verify)

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

Time = 2.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.34

method	result	size
default	\(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (5 b e x +3 a e +2 b d \right )}{15 e^{2} \left (e x +d \right )^{\frac {5}{2}}}\)	\(33\)
gosper	\(-\frac {2 \left (5 b e x +3 a e +2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{2} \left (b x +a \right )}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b e x + 2 \, b d + 3 \, a e\right )} \sqrt {e x + d}}{15 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} \]

[In]

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

[Out]

-2/15*(5*b*e*x + 2*b*d + 3*a*e)*sqrt(e*x + d)/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b e x + 2 \, b d + 3 \, a e\right )}}{15 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )} \sqrt {e x + d}} \]

[In]

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

[Out]

-2/15*(5*b*e*x + 2*b*d + 3*a*e)/((e^4*x^2 + 2*d*e^3*x + d^2*e^2)*sqrt(e*x + d))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, {\left (e x + d\right )} b \mathrm {sgn}\left (b x + a\right ) - 3 \, b d \mathrm {sgn}\left (b x + a\right ) + 3 \, a e \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.94 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,x}{3\,e^3}+\frac {\frac {2\,a\,e}{5}+\frac {4\,b\,d}{15}}{b\,e^4}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (a\,e^4+2\,b\,d\,e^3\right )\,\sqrt {d+e\,x}}{b\,e^4}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \]

[In]

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

[Out]

-(((a + b*x)^2)^(1/2)*((2*x)/(3*e^3) + ((2*a*e)/5 + (4*b*d)/15)/(b*e^4)))/(x^3*(d + e*x)^(1/2) + (a*d^2*(d + e
*x)^(1/2))/(b*e^2) + (x^2*(a*e^4 + 2*b*d*e^3)*(d + e*x)^(1/2))/(b*e^4) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(
b*e^2))