\(\int \sqrt {d+e x} (a^2+2 a b x+b^2 x^2) \, dx\) [1624]

Optimal result

Integrand size = 26, antiderivative size = 71 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (b d-a e)^2 (d+e x)^{3/2}}{3 e^3}-\frac {4 b (b d-a e) (d+e x)^{5/2}}{5 e^3}+\frac {2 b^2 (d+e x)^{7/2}}{7 e^3} \]

[Out]

2/3*(-a*e+b*d)^2*(e*x+d)^(3/2)/e^3-4/5*b*(-a*e+b*d)*(e*x+d)^(5/2)/e^3+2/7*b^2*(e*x+d)^(7/2)/e^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 45} \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=-\frac {4 b (d+e x)^{5/2} (b d-a e)}{5 e^3}+\frac {2 (d+e x)^{3/2} (b d-a e)^2}{3 e^3}+\frac {2 b^2 (d+e x)^{7/2}}{7 e^3} \]

[In]

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

[Out]

(2*(b*d - a*e)^2*(d + e*x)^(3/2))/(3*e^3) - (4*b*(b*d - a*e)*(d + e*x)^(5/2))/(5*e^3) + (2*b^2*(d + e*x)^(7/2)
)/(7*e^3)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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])

Rubi steps \begin{align*} \text {integral}& = \int (a+b x)^2 \sqrt {d+e x} \, dx \\ & = \int \left (\frac {(-b d+a e)^2 \sqrt {d+e x}}{e^2}-\frac {2 b (b d-a e) (d+e x)^{3/2}}{e^2}+\frac {b^2 (d+e x)^{5/2}}{e^2}\right ) \, dx \\ & = \frac {2 (b d-a e)^2 (d+e x)^{3/2}}{3 e^3}-\frac {4 b (b d-a e) (d+e x)^{5/2}}{5 e^3}+\frac {2 b^2 (d+e x)^{7/2}}{7 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (d+e x)^{3/2} \left (35 a^2 e^2+14 a b e (-2 d+3 e x)+b^2 \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3} \]

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(35*a^2*e^2 + 14*a*b*e*(-2*d + 3*e*x) + b^2*(8*d^2 - 12*d*e*x + 15*e^2*x^2)))/(105*e^3)

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {2 \left (\left (\frac {3}{7} b^{2} x^{2}+\frac {6}{5} a b x +a^{2}\right ) e^{2}-\frac {4 b \left (\frac {3 b x}{7}+a \right ) d e}{5}+\frac {8 b^{2} d^{2}}{35}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3 e^{3}}\)	\(54\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 x^{2} b^{2} e^{2}+42 x a b \,e^{2}-12 b^{2} d e x +35 a^{2} e^{2}-28 a b d e +8 b^{2} d^{2}\right )}{105 e^{3}}\)	\(63\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\)	\(70\)
default	\(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\)	\(70\)
trager	\(\frac {2 \left (15 b^{2} e^{3} x^{3}+42 a b \,e^{3} x^{2}+3 b^{2} d \,e^{2} x^{2}+35 a^{2} e^{3} x +14 a b d \,e^{2} x -4 b^{2} d^{2} e x +35 a^{2} d \,e^{2}-28 a b \,d^{2} e +8 b^{2} d^{3}\right ) \sqrt {e x +d}}{105 e^{3}}\)	\(100\)
risch	\(\frac {2 \left (15 b^{2} e^{3} x^{3}+42 a b \,e^{3} x^{2}+3 b^{2} d \,e^{2} x^{2}+35 a^{2} e^{3} x +14 a b d \,e^{2} x -4 b^{2} d^{2} e x +35 a^{2} d \,e^{2}-28 a b \,d^{2} e +8 b^{2} d^{3}\right ) \sqrt {e x +d}}{105 e^{3}}\)	\(100\)

[In]

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

[Out]

2/3*((3/7*b^2*x^2+6/5*a*b*x+a^2)*e^2-4/5*b*(3/7*b*x+a)*d*e+8/35*b^2*d^2)*(e*x+d)^(3/2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (15 \, b^{2} e^{3} x^{3} + 8 \, b^{2} d^{3} - 28 \, a b d^{2} e + 35 \, a^{2} d e^{2} + 3 \, {\left (b^{2} d e^{2} + 14 \, a b e^{3}\right )} x^{2} - {\left (4 \, b^{2} d^{2} e - 14 \, a b d e^{2} - 35 \, a^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{3}} \]

[In]

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

[Out]

2/105*(15*b^2*e^3*x^3 + 8*b^2*d^3 - 28*a*b*d^2*e + 35*a^2*d*e^2 + 3*(b^2*d*e^2 + 14*a*b*e^3)*x^2 - (4*b^2*d^2*
e - 14*a*b*d*e^2 - 35*a^2*e^3)*x)*sqrt(e*x + d)/e^3

Sympy [A] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.55 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (2 a b e - 2 b^{2} d\right )}{5 e^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a^{2} e^{2} - 2 a b d e + b^{2} d^{2}\right )}{3 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b**2*(d + e*x)**(7/2)/(7*e**2) + (d + e*x)**(5/2)*(2*a*b*e - 2*b**2*d)/(5*e**2) + (d + e*x)**(3/
2)*(a**2*e**2 - 2*a*b*d*e + b**2*d**2)/(3*e**2))/e, Ne(e, 0)), (sqrt(d)*(a**2*x + a*b*x**2 + b**2*x**3/3), Tru
e))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{2} - 42 \, {\left (b^{2} d - a b e\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{105 \, e^{3}} \]

[In]

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

[Out]

2/105*(15*(e*x + d)^(7/2)*b^2 - 42*(b^2*d - a*b*e)*(e*x + d)^(5/2) + 35*(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*(e*x +
d)^(3/2))/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (59) = 118\).

Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.82 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} a^{2} d + 35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} + \frac {70 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b d}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2} d}{e^{2}} + \frac {14 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a b}{e} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b^{2}}{e^{2}}\right )}}{105 \, e} \]

[In]

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

[Out]

2/105*(105*sqrt(e*x + d)*a^2*d + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2 + 70*((e*x + d)^(3/2) - 3*sqrt(e
*x + d)*d)*a*b*d/e + 7*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*b^2*d/e^2 + 14*(3*(e*
x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*b/e + 3*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*
d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*b^2/e^2)/e

Mupad [B] (verification not implemented)

Time = 9.56 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2\,{\left (d+e\,x\right )}^{3/2}\,\left (15\,b^2\,{\left (d+e\,x\right )}^2+35\,a^2\,e^2+35\,b^2\,d^2-42\,b^2\,d\,\left (d+e\,x\right )+42\,a\,b\,e\,\left (d+e\,x\right )-70\,a\,b\,d\,e\right )}{105\,e^3} \]

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(15*b^2*(d + e*x)^2 + 35*a^2*e^2 + 35*b^2*d^2 - 42*b^2*d*(d + e*x) + 42*a*b*e*(d + e*x) - 7
0*a*b*d*e))/(105*e^3)