\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx\) [1679]

Optimal result

Integrand size = 30, antiderivative size = 94 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=-\frac {2 (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)}+\frac {2 b (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x)} \]

[Out]

2/3*b*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^2/(b*x+a)-2*(-a*e+b*d)*(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 = 94, 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} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^2 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}{e^2 (a+b x)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} (-2 b d+3 a e+b e x)}{3 e^2 (a+b x)} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 2.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.34

method	result	size
default	\(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \sqrt {e x +d}\, \left (b e x +3 a e -2 b d \right )}{3 e^{2}}\)	\(32\)
gosper	\(\frac {2 \sqrt {e x +d}\, \left (b e x +3 a e -2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{2} \left (b x +a \right )}\)	\(42\)
risch	\(\frac {2 \sqrt {e x +d}\, \left (b e x +3 a e -2 b d \right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{2} \left (b x +a \right )}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (b e x - 2 \, b d + 3 \, a e\right )} \sqrt {e x + d}}{3 \, e^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e - {\left (b d e - 3 \, a e^{2}\right )} x\right )}}{3 \, \sqrt {e x + d} e^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.54 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, \sqrt {e x + d} a \mathrm {sgn}\left (b x + a\right ) + \frac {{\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} b \mathrm {sgn}\left (b x + a\right )}{e}\right )}}{3 \, e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.61 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,x^2}{3}-\frac {4\,b\,d^2-6\,a\,d\,e}{3\,b\,e^2}+\frac {x\,\left (6\,a\,e^2-2\,b\,d\,e\right )}{3\,b\,e^2}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]

[In]

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

[Out]

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