3.15.40 \(\int (d+e x)^2 (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=127 \[ \frac {e (a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (p+1)}+\frac {(a+b x) (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+1)}+\frac {e^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+3)} \]

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Rubi [A]  time = 0.06, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {646, 43} \begin {gather*} \frac {e (a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (p+1)}+\frac {(a+b x) (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+1)}+\frac {e^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (2 p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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 646

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*} \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx &=\left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \left (a b+b^2 x\right )^{2 p} (d+e x)^2 \, dx\\ &=\left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \left (\frac {(b d-a e)^2 \left (a b+b^2 x\right )^{2 p}}{b^2}+\frac {2 e (b d-a e) \left (a b+b^2 x\right )^{1+2 p}}{b^3}+\frac {e^2 \left (a b+b^2 x\right )^{2+2 p}}{b^4}\right ) \, dx\\ &=\frac {(b d-a e)^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (1+2 p)}+\frac {e (b d-a e) (a+b x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (1+p)}+\frac {e^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (3+2 p)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 75, normalized size = 0.59 \begin {gather*} \frac {(a+b x) \left ((a+b x)^2\right )^p \left (\frac {e (a+b x) (b d-a e)}{p+1}+\frac {(b d-a e)^2}{2 p+1}+\frac {e^2 (a+b x)^2}{2 p+3}\right )}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 249, normalized size = 1.96 \begin {gather*} \frac {{\left (2 \, a b^{2} d^{2} p^{2} + 3 \, a b^{2} d^{2} - 3 \, a^{2} b d e + a^{3} e^{2} + {\left (2 \, b^{3} e^{2} p^{2} + 3 \, b^{3} e^{2} p + b^{3} e^{2}\right )} x^{3} + {\left (3 \, b^{3} d e + 2 \, {\left (2 \, b^{3} d e + a b^{2} e^{2}\right )} p^{2} + {\left (8 \, b^{3} d e + a b^{2} e^{2}\right )} p\right )} x^{2} + {\left (5 \, a b^{2} d^{2} - 2 \, a^{2} b d e\right )} p + {\left (3 \, b^{3} d^{2} + 2 \, {\left (b^{3} d^{2} + 2 \, a b^{2} d e\right )} p^{2} + {\left (5 \, b^{3} d^{2} + 6 \, a b^{2} d e - 2 \, a^{2} b e^{2}\right )} p\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*b^2*d^2*p^2 + 3*a*b^2*d^2 - 3*a^2*b*d*e + a^3*e^2 + (2*b^3*e^2*p^2 + 3*b^3*e^2*p + b^3*e^2)*x^3 + (3*b^3*
d*e + 2*(2*b^3*d*e + a*b^2*e^2)*p^2 + (8*b^3*d*e + a*b^2*e^2)*p)*x^2 + (5*a*b^2*d^2 - 2*a^2*b*d*e)*p + (3*b^3*
d^2 + 2*(b^3*d^2 + 2*a*b^2*d*e)*p^2 + (5*b^3*d^2 + 6*a*b^2*d*e - 2*a^2*b*e^2)*p)*x)*(b^2*x^2 + 2*a*b*x + a^2)^
p/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)

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giac [B]  time = 0.21, size = 608, normalized size = 4.79 \begin {gather*} \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} p^{2} x^{3} e^{2} + 4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d p^{2} x^{2} e + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} p^{2} x + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} p^{2} x^{2} e^{2} + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} p x^{3} e^{2} + 4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d p^{2} x e + 8 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d p x^{2} e + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} p^{2} + 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} p x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} p x^{2} e^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} x^{3} e^{2} + 6 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d p x e + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d x^{2} e + 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} p + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} x - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b p x e^{2} - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d p e + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} - 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d e + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{3} e^{2}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*p^2*x^3*e^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*p^2*x^2*e + 2*(b^2*x^2 +
2*a*b*x + a^2)^p*b^3*d^2*p^2*x + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*p^2*x^2*e^2 + 3*(b^2*x^2 + 2*a*b*x + a^2)
^p*b^3*p*x^3*e^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*p^2*x*e + 8*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*p*x^2*e
+ 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d^2*p^2 + 5*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d^2*p*x + (b^2*x^2 + 2*a*b*
x + a^2)^p*a*b^2*p*x^2*e^2 + (b^2*x^2 + 2*a*b*x + a^2)^p*b^3*x^3*e^2 + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*p
*x*e + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*x^2*e + 5*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d^2*p + 3*(b^2*x^2 + 2*
a*b*x + a^2)^p*b^3*d^2*x - 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*p*x*e^2 - 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d
*p*e + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d^2 - 3*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d*e + (b^2*x^2 + 2*a*b*x
+ a^2)^p*a^3*e^2)/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)

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maple [A]  time = 0.06, size = 175, normalized size = 1.38 \begin {gather*} \frac {\left (b x +a \right ) \left (2 b^{2} e^{2} p^{2} x^{2}+4 b^{2} d e \,p^{2} x +3 b^{2} e^{2} p \,x^{2}-2 a b \,e^{2} p x +2 b^{2} d^{2} p^{2}+8 b^{2} d e p x +b^{2} x^{2} e^{2}-2 a b d e p -a b \,e^{2} x +5 b^{2} d^{2} p +3 b^{2} d e x +a^{2} e^{2}-3 a b d e +3 b^{2} d^{2}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{\left (4 p^{3}+12 p^{2}+11 p +3\right ) b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)*(2*b^2*e^2*p^2*x^2+4*b^2*d*e*p^2*x+3*b^2*e^2*p*x^2-2*a*b*e^2*p*x+2*b^2*d^2*p^2+8*b^2*d*e*p*x+b^2*e^2*x
^2-2*a*b*d*e*p-a*b*e^2*x+5*b^2*d^2*p+3*b^2*d*e*x+a^2*e^2-3*a*b*d*e+3*b^2*d^2)*(b^2*x^2+2*a*b*x+a^2)^p/b^3/(4*p
^3+12*p^2+11*p+3)

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maxima [A]  time = 1.30, size = 157, normalized size = 1.24 \begin {gather*} \frac {{\left (b x + a\right )} {\left (b x + a\right )}^{2 \, p} d^{2}}{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} d e}{{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{3} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{2} - 2 \, a^{2} b p x + a^{3}\right )} {\left (b x + a\right )}^{2 \, p} e^{2}}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)*(b*x + a)^(2*p)*d^2/(b*(2*p + 1)) + (b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*d*e/((2*p^
2 + 3*p + 1)*b^2) + ((2*p^2 + 3*p + 1)*b^3*x^3 + (2*p^2 + p)*a*b^2*x^2 - 2*a^2*b*p*x + a^3)*(b*x + a)^(2*p)*e^
2/((4*p^3 + 12*p^2 + 11*p + 3)*b^3)

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mupad [B]  time = 0.71, size = 250, normalized size = 1.97 \begin {gather*} {\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p\,\left (\frac {e^2\,x^3\,\left (2\,p^2+3\,p+1\right )}{4\,p^3+12\,p^2+11\,p+3}+\frac {x\,\left (-2\,a^2\,b\,e^2\,p+4\,a\,b^2\,d\,e\,p^2+6\,a\,b^2\,d\,e\,p+2\,b^3\,d^2\,p^2+5\,b^3\,d^2\,p+3\,b^3\,d^2\right )}{b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {a\,\left (a^2\,e^2-2\,a\,b\,d\,e\,p-3\,a\,b\,d\,e+2\,b^2\,d^2\,p^2+5\,b^2\,d^2\,p+3\,b^2\,d^2\right )}{b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {e\,x^2\,\left (2\,p+1\right )\,\left (3\,b\,d+a\,e\,p+2\,b\,d\,p\right )}{b\,\left (4\,p^3+12\,p^2+11\,p+3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2 + b^2*x^2 + 2*a*b*x)^p*((e^2*x^3*(3*p + 2*p^2 + 1))/(11*p + 12*p^2 + 4*p^3 + 3) + (x*(3*b^3*d^2 + 5*b^3*d
^2*p + 2*b^3*d^2*p^2 - 2*a^2*b*e^2*p + 4*a*b^2*d*e*p^2 + 6*a*b^2*d*e*p))/(b^3*(11*p + 12*p^2 + 4*p^3 + 3)) + (
a*(a^2*e^2 + 3*b^2*d^2 + 5*b^2*d^2*p + 2*b^2*d^2*p^2 - 3*a*b*d*e - 2*a*b*d*e*p))/(b^3*(11*p + 12*p^2 + 4*p^3 +
3)) + (e*x^2*(2*p + 1)*(3*b*d + a*e*p + 2*b*d*p))/(b*(11*p + 12*p^2 + 4*p^3 + 3)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((d**2*x + d*e*x**2 + e**2*x**3/3)*(a**2)**p, Eq(b, 0)), (Integral((d + e*x)**2/((a + b*x)**2)**(3/2
), x), Eq(p, -3/2)), (-2*a**2*e**2*log(a/b + x)/(a*b**3 + b**4*x) - 2*a**2*e**2/(a*b**3 + b**4*x) + 2*a*b*d*e*
log(a/b + x)/(a*b**3 + b**4*x) + 2*a*b*d*e/(a*b**3 + b**4*x) - 2*a*b*e**2*x*log(a/b + x)/(a*b**3 + b**4*x) - b
**2*d**2/(a*b**3 + b**4*x) + 2*b**2*d*e*x*log(a/b + x)/(a*b**3 + b**4*x) + b**2*e**2*x**2/(a*b**3 + b**4*x), E
q(p, -1)), (Integral((d + e*x)**2/sqrt((a + b*x)**2), x), Eq(p, -1/2)), (a**3*e**2*(a**2 + 2*a*b*x + b**2*x**2
)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) - 2*a**2*b*d*e*p*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**
3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) - 3*a**2*b*d*e*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b
**3*p**2 + 11*b**3*p + 3*b**3) - 2*a**2*b*e**2*p*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2
+ 11*b**3*p + 3*b**3) + 2*a*b**2*d**2*p**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b
**3*p + 3*b**3) + 5*a*b**2*d**2*p*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*
b**3) + 3*a*b**2*d**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 4*a*
b**2*d*e*p**2*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 6*a*b**2*d
*e*p*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 2*a*b**2*e**2*p**2*
x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + a*b**2*e**2*p*x**2*(a
**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 2*b**3*d**2*p**2*x*(a**2 + 2
*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 5*b**3*d**2*p*x*(a**2 + 2*a*b*x + b
**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 3*b**3*d**2*x*(a**2 + 2*a*b*x + b**2*x**2)**p
/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 4*b**3*d*e*p**2*x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b
**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 8*b**3*d*e*p*x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3
+ 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 3*b**3*d*e*x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3
*p**2 + 11*b**3*p + 3*b**3) + 2*b**3*e**2*p**2*x**3*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**
2 + 11*b**3*p + 3*b**3) + 3*b**3*e**2*p*x**3*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*
b**3*p + 3*b**3) + b**3*e**2*x**3*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*
b**3), True))

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