\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx\) [1680]

Optimal result

Integrand size = 30, antiderivative size = 92 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x) \sqrt {d+e x}}+\frac {2 b \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, 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} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^2 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x) \sqrt {d+e x}} \]

[In]

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

[Out]

(2*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x)*Sqrt[d + e*x]) + (2*b*Sqrt[d + e*x]*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(e^2*(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 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)^{3/2}} \, 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)^{3/2}}+\frac {b^2}{e \sqrt {d+e x}}\right ) \, dx}{a b+b^2 x} \\ & = \frac {2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x) \sqrt {d+e x}}+\frac {2 b \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} (-b d+a e-b (d+e x))}{e^2 (a+b x) \sqrt {d+e x}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 2.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.35

method	result	size
default	\(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (-b e x +a e -2 b d \right )}{e^{2} \sqrt {e x +d}}\)	\(32\)
gosper	\(-\frac {2 \left (-b e x +a e -2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{\sqrt {e x +d}\, e^{2} \left (b x +a \right )}\)	\(42\)
risch	\(\frac {2 b \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{e^{2} \left (b x +a \right )}-\frac {2 \left (a e -b d \right ) \sqrt {\left (b x +a \right )^{2}}}{e^{2} \sqrt {e x +d}\, \left (b x +a \right )}\)	\(67\)

[In]

int(((b*x+a)^2)^(1/2)/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b e x + 2 \, b d - a e\right )} \sqrt {e x + d}}{e^{3} x + d e^{2}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=\int \frac {\sqrt {\left (a + b x\right )^{2}}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(((b*x+a)**2)**(1/2)/(e*x+d)**(3/2),x)

[Out]

Integral(sqrt((a + b*x)**2)/(d + e*x)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b e x + 2 \, b d - a e\right )}}{\sqrt {e x + d} e^{2}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

2*(b*e*x + 2*b*d - a*e)/(sqrt(e*x + d)*e^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, \sqrt {e x + d} b \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {2 \, {\left (b d \mathrm {sgn}\left (b x + a\right ) - a e \mathrm {sgn}\left (b x + a\right )\right )}}{\sqrt {e x + d} e^{2}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

2*sqrt(e*x + d)*b*sgn(b*x + a)/e^2 + 2*(b*d*sgn(b*x + a) - a*e*sgn(b*x + a))/(sqrt(e*x + d)*e^2)

Mupad [B] (verification not implemented)

Time = 9.72 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,x}{e}-\frac {2\,a\,e-4\,b\,d}{b\,e^2}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]

[In]

int(((a + b*x)^2)^(1/2)/(d + e*x)^(3/2),x)

[Out]

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