3.13.74 \(\int (d+e x)^2 (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=125 \[ \frac {2 e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)}{7 b^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^2}{6 b^3}+\frac {e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \]

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Rubi [A]  time = 0.14, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {646, 43} \begin {gather*} \frac {2 e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)}{7 b^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^2}{6 b^3}+\frac {e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((b*d - a*e)^2*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3) + (2*e*(b*d - a*e)*(a + b*x)^6*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(7*b^3) + (e^2*(a + b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*b^3)

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 )^{5/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right )^5 (d+e x)^2 \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(b d-a e)^2 \left (a b+b^2 x\right )^5}{b^2}+\frac {2 e (b d-a e) \left (a b+b^2 x\right )^6}{b^3}+\frac {e^2 \left (a b+b^2 x\right )^7}{b^4}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {(b d-a e)^2 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3}+\frac {2 e (b d-a e) (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^3}+\frac {e^2 (a+b x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{8 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 187, normalized size = 1.50 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (56 a^5 \left (3 d^2+3 d e x+e^2 x^2\right )+70 a^4 b x \left (6 d^2+8 d e x+3 e^2 x^2\right )+56 a^3 b^2 x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right )+28 a^2 b^3 x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right )+8 a b^4 x^4 \left (21 d^2+35 d e x+15 e^2 x^2\right )+b^5 x^5 \left (28 d^2+48 d e x+21 e^2 x^2\right )\right )}{168 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(x*Sqrt[(a + b*x)^2]*(56*a^5*(3*d^2 + 3*d*e*x + e^2*x^2) + 70*a^4*b*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + 56*a^3*b
^2*x^2*(10*d^2 + 15*d*e*x + 6*e^2*x^2) + 28*a^2*b^3*x^3*(15*d^2 + 24*d*e*x + 10*e^2*x^2) + 8*a*b^4*x^4*(21*d^2
+ 35*d*e*x + 15*e^2*x^2) + b^5*x^5*(28*d^2 + 48*d*e*x + 21*e^2*x^2)))/(168*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.74, 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 )^{5/2} \, 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)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2), x]

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fricas [B]  time = 0.39, size = 197, normalized size = 1.58 \begin {gather*} \frac {1}{8} \, b^{5} e^{2} x^{8} + a^{5} d^{2} x + \frac {1}{7} \, {\left (2 \, b^{5} d e + 5 \, a b^{4} e^{2}\right )} x^{7} + \frac {1}{6} \, {\left (b^{5} d^{2} + 10 \, a b^{4} d e + 10 \, a^{2} b^{3} e^{2}\right )} x^{6} + {\left (a b^{4} d^{2} + 4 \, a^{2} b^{3} d e + 2 \, a^{3} b^{2} e^{2}\right )} x^{5} + \frac {5}{4} \, {\left (2 \, a^{2} b^{3} d^{2} + 4 \, a^{3} b^{2} d e + a^{4} b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (10 \, a^{3} b^{2} d^{2} + 10 \, a^{4} b d e + a^{5} e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (5 \, a^{4} b d^{2} + 2 \, a^{5} d e\right )} x^{2} \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)^(5/2),x, algorithm="fricas")

[Out]

1/8*b^5*e^2*x^8 + a^5*d^2*x + 1/7*(2*b^5*d*e + 5*a*b^4*e^2)*x^7 + 1/6*(b^5*d^2 + 10*a*b^4*d*e + 10*a^2*b^3*e^2
)*x^6 + (a*b^4*d^2 + 4*a^2*b^3*d*e + 2*a^3*b^2*e^2)*x^5 + 5/4*(2*a^2*b^3*d^2 + 4*a^3*b^2*d*e + a^4*b*e^2)*x^4
+ 1/3*(10*a^3*b^2*d^2 + 10*a^4*b*d*e + a^5*e^2)*x^3 + 1/2*(5*a^4*b*d^2 + 2*a^5*d*e)*x^2

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giac [B]  time = 0.17, size = 320, normalized size = 2.56 \begin {gather*} \frac {1}{8} \, b^{5} x^{8} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{7} \, b^{5} d x^{7} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, b^{5} d^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, a b^{4} x^{7} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, a b^{4} d x^{6} e \mathrm {sgn}\left (b x + a\right ) + a b^{4} d^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, a^{2} b^{3} x^{6} e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{2} b^{3} d x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{2} b^{3} d^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b^{2} x^{5} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{3} b^{2} d x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, a^{3} b^{2} d^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, a^{4} b x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, a^{4} b d x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{4} b d^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, a^{5} x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + a^{5} d x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{5} d^{2} x \mathrm {sgn}\left (b x + a\right ) \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)^(5/2),x, algorithm="giac")

[Out]

1/8*b^5*x^8*e^2*sgn(b*x + a) + 2/7*b^5*d*x^7*e*sgn(b*x + a) + 1/6*b^5*d^2*x^6*sgn(b*x + a) + 5/7*a*b^4*x^7*e^2
*sgn(b*x + a) + 5/3*a*b^4*d*x^6*e*sgn(b*x + a) + a*b^4*d^2*x^5*sgn(b*x + a) + 5/3*a^2*b^3*x^6*e^2*sgn(b*x + a)
+ 4*a^2*b^3*d*x^5*e*sgn(b*x + a) + 5/2*a^2*b^3*d^2*x^4*sgn(b*x + a) + 2*a^3*b^2*x^5*e^2*sgn(b*x + a) + 5*a^3*
b^2*d*x^4*e*sgn(b*x + a) + 10/3*a^3*b^2*d^2*x^3*sgn(b*x + a) + 5/4*a^4*b*x^4*e^2*sgn(b*x + a) + 10/3*a^4*b*d*x
^3*e*sgn(b*x + a) + 5/2*a^4*b*d^2*x^2*sgn(b*x + a) + 1/3*a^5*x^3*e^2*sgn(b*x + a) + a^5*d*x^2*e*sgn(b*x + a) +
a^5*d^2*x*sgn(b*x + a)

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maple [B]  time = 0.04, size = 230, normalized size = 1.84 \begin {gather*} \frac {\left (21 e^{2} b^{5} x^{7}+120 x^{6} e^{2} a \,b^{4}+48 x^{6} d e \,b^{5}+280 x^{5} e^{2} a^{2} b^{3}+280 x^{5} d e a \,b^{4}+28 x^{5} d^{2} b^{5}+336 a^{3} b^{2} e^{2} x^{4}+672 a^{2} b^{3} d e \,x^{4}+168 a \,b^{4} d^{2} x^{4}+210 x^{3} e^{2} a^{4} b +840 x^{3} d e \,a^{3} b^{2}+420 x^{3} d^{2} a^{2} b^{3}+56 x^{2} e^{2} a^{5}+560 x^{2} d e \,a^{4} b +560 x^{2} d^{2} a^{3} b^{2}+168 x d e \,a^{5}+420 x \,d^{2} a^{4} b +168 d^{2} a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x}{168 \left (b x +a \right )^{5}} \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)^(5/2),x)

[Out]

1/168*x*(21*b^5*e^2*x^7+120*a*b^4*e^2*x^6+48*b^5*d*e*x^6+280*a^2*b^3*e^2*x^5+280*a*b^4*d*e*x^5+28*b^5*d^2*x^5+
336*a^3*b^2*e^2*x^4+672*a^2*b^3*d*e*x^4+168*a*b^4*d^2*x^4+210*a^4*b*e^2*x^3+840*a^3*b^2*d*e*x^3+420*a^2*b^3*d^
2*x^3+56*a^5*e^2*x^2+560*a^4*b*d*e*x^2+560*a^3*b^2*d^2*x^2+168*a^5*d*e*x+420*a^4*b*d^2*x+168*a^5*d^2)*((b*x+a)
^2)^(5/2)/(b*x+a)^5

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maxima [B]  time = 1.14, size = 245, normalized size = 1.96 \begin {gather*} \frac {1}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} d^{2} x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a d e x}{3 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} e^{2} x}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a d^{2}}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} d e}{3 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{3} e^{2}}{6 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} e^{2} x}{8 \, b^{2}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} d e}{7 \, b^{2}} - \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a e^{2}}{56 \, 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)^(5/2),x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*d^2*x - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*d*e*x/b + 1/6*(b^2*x^2 + 2*a
*b*x + a^2)^(5/2)*a^2*e^2*x/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*d^2/b - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^
(5/2)*a^2*d*e/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3*e^2/b^3 + 1/8*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*e^2*
x/b^2 + 2/7*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*d*e/b^2 - 9/56*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*a*e^2/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \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)^(5/2),x)

[Out]

int((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right )^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}\, dx \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)**(5/2),x)

[Out]

Integral((d + e*x)**2*((a + b*x)**2)**(5/2), x)

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