\(\int (d+e x) (a^2+2 a b x+b^2 x^2)^p \, dx\) [1746]

Optimal result

Integrand size = 24, antiderivative size = 76 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(b d-a e) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (1+2 p)}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{1+p}}{2 b^2 (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {654, 623} \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(a+b x) (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+1)}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{p+1}}{2 b^2 (p+1)} \]

[In]

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

[Out]

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

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {e \left (a^2+2 a b x+b^2 x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {\left (2 b^2 d-2 a b e\right ) \int \left (a^2+2 a b x+b^2 x^2\right )^p \, dx}{2 b^2} \\ & = \frac {(b d-a e) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (1+2 p)}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{1+p}}{2 b^2 (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.71 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(a+b x) \left ((a+b x)^2\right )^p (-a e+2 b d (1+p)+b e (1+2 p) x)}{2 b^2 (1+p) (1+2 p)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86

method	result	size
gosper	\(-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} \left (-2 b e p x -2 b d p -b e x +a e -2 b d \right ) \left (b x +a \right )}{2 b^{2} \left (2 p^{2}+3 p +1\right )}\)	\(65\)
risch	\(-\frac {\left (-2 b^{2} e p \,x^{2}-2 a b e p x -2 b^{2} d p x -b^{2} e \,x^{2}-2 a b d p -2 b^{2} d x +a^{2} e -2 a b d \right ) \left (\left (b x +a \right )^{2}\right )^{p}}{2 b^{2} \left (1+p \right ) \left (1+2 p \right )}\)	\(85\)
norman	\(\frac {\left (a e p +b d p +b d \right ) x \,{\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{b \left (2 p^{2}+3 p +1\right )}+\frac {e \,x^{2} {\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{2+2 p}-\frac {a \left (-2 b d p +a e -2 b d \right ) {\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{2 b^{2} \left (2 p^{2}+3 p +1\right )}\)	\(133\)
parallelrisch	\(\frac {2 x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} e p +x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} e +2 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b e p +2 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} d p +2 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{2} d +2 \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b d p -\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{3} e +2 \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b d}{2 \left (1+p \right ) a \,b^{2} \left (1+2 p \right )}\)	\(233\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.26 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (2 \, a b d p + 2 \, a b d - a^{2} e + {\left (2 \, b^{2} e p + b^{2} e\right )} x^{2} + 2 \, {\left (b^{2} d + {\left (b^{2} d + a b e\right )} p\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (68) = 136\).

Time = 0.87 (sec) , antiderivative size = 604, normalized size of antiderivative = 7.95 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\begin {cases} \left (d x + \frac {e x^{2}}{2}\right ) \left (a^{2}\right )^{p} & \text {for}\: b = 0 \\\frac {a e \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac {a e}{a b^{2} + b^{3} x} - \frac {b d}{a b^{2} + b^{3} x} + \frac {b e x \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} & \text {for}\: p = -1 \\\begin {cases} \frac {\left (\frac {a}{b} + x\right ) \left (- \frac {a e}{b} + d\right ) \log {\left (\frac {a}{b} + x \right )}}{\sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} + \frac {e \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{b^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {2 d \sqrt {a^{2} + 2 a b x} + \frac {e \left (- a^{2} \sqrt {a^{2} + 2 a b x} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3}\right )}{a b}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {d x + \frac {e x^{2}}{2}}{\sqrt {a^{2}}} & \text {otherwise} \end {cases} & \text {for}\: p = - \frac {1}{2} \\- \frac {a^{2} e \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 a b d p \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 a b d \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 a b e p x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 b^{2} d p x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 b^{2} d x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {2 b^{2} e p x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} + \frac {b^{2} e x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 6 b^{2} p + 2 b^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((d*x + e*x**2/2)*(a**2)**p, Eq(b, 0)), (a*e*log(a/b + x)/(a*b**2 + b**3*x) + a*e/(a*b**2 + b**3*x)
- b*d/(a*b**2 + b**3*x) + b*e*x*log(a/b + x)/(a*b**2 + b**3*x), Eq(p, -1)), (Piecewise(((a/b + x)*(-a*e/b + d)
*log(a/b + x)/sqrt(b**2*(a/b + x)**2) + e*sqrt(a**2 + 2*a*b*x + b**2*x**2)/b**2, Ne(b**2, 0)), ((2*d*sqrt(a**2
+ 2*a*b*x) + e*(-a**2*sqrt(a**2 + 2*a*b*x) + (a**2 + 2*a*b*x)**(3/2)/3)/(a*b))/(2*a*b), Ne(a*b, 0)), ((d*x +
e*x**2/2)/sqrt(a**2), True)), Eq(p, -1/2)), (-a**2*e*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 6*b**2*p +
2*b**2) + 2*a*b*d*p*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 6*b**2*p + 2*b**2) + 2*a*b*d*(a**2 + 2*a*b
*x + b**2*x**2)**p/(4*b**2*p**2 + 6*b**2*p + 2*b**2) + 2*a*b*e*p*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**
2 + 6*b**2*p + 2*b**2) + 2*b**2*d*p*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 6*b**2*p + 2*b**2) + 2*b*
*2*d*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 6*b**2*p + 2*b**2) + 2*b**2*e*p*x**2*(a**2 + 2*a*b*x + b
**2*x**2)**p/(4*b**2*p**2 + 6*b**2*p + 2*b**2) + b**2*e*x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 6*
b**2*p + 2*b**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b x + a\right )}^{2 \, p} d}{b {\left (2 \, p + 1\right )}} + \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )} {\left (b x + a\right )}^{2 \, p} e}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (74) = 148\).

Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.95 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} e p x^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} d p x + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b e p x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} e x^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b d p + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} d x + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b d - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} e}{2 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.94 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.47 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\left (\frac {x\,\left (2\,b^2\,d+2\,b^2\,d\,p+2\,a\,b\,e\,p\right )}{2\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {a\,\left (2\,b\,d-a\,e+2\,b\,d\,p\right )}{2\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {e\,x^2\,\left (2\,p+1\right )}{2\,\left (2\,p^2+3\,p+1\right )}\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p \]

[In]

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

[Out]

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