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

Optimal. Leaf size=69 \[ \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (b d-a e)}{6 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {640, 609} \begin {gather*} \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (b d-a e)}{6 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 609

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 640

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

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Mathematica [A]  time = 0.04, size = 121, normalized size = 1.75 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (21 a^5 (2 d+e x)+35 a^4 b x (3 d+2 e x)+35 a^3 b^2 x^2 (4 d+3 e x)+21 a^2 b^3 x^3 (5 d+4 e x)+7 a b^4 x^4 (6 d+5 e x)+b^5 x^5 (7 d+6 e x)\right )}{42 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(21*a^5*(2*d + e*x) + 35*a^4*b*x*(3*d + 2*e*x) + 35*a^3*b^2*x^2*(4*d + 3*e*x) + 21*a^2*b^
3*x^3*(5*d + 4*e*x) + 7*a*b^4*x^4*(6*d + 5*e*x) + b^5*x^5*(7*d + 6*e*x)))/(42*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.22, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x) \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)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.40, size = 115, normalized size = 1.67 \begin {gather*} \frac {1}{7} \, b^{5} e x^{7} + a^{5} d x + \frac {1}{6} \, {\left (b^{5} d + 5 \, a b^{4} e\right )} x^{6} + {\left (a b^{4} d + 2 \, a^{2} b^{3} e\right )} x^{5} + \frac {5}{2} \, {\left (a^{2} b^{3} d + a^{3} b^{2} e\right )} x^{4} + \frac {5}{3} \, {\left (2 \, a^{3} b^{2} d + a^{4} b e\right )} x^{3} + \frac {1}{2} \, {\left (5 \, a^{4} b d + a^{5} e\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.17, size = 199, normalized size = 2.88 \begin {gather*} \frac {1}{7} \, b^{5} x^{7} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, b^{5} d x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, a b^{4} x^{6} e \mathrm {sgn}\left (b x + a\right ) + a b^{4} d x^{5} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b^{3} x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{2} b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{3} b^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, a^{3} b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, a^{4} b x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{4} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{5} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{5} d 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)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 138, normalized size = 2.00 \begin {gather*} \frac {\left (6 e \,b^{5} x^{6}+35 x^{5} e a \,b^{4}+7 x^{5} d \,b^{5}+84 a^{2} b^{3} e \,x^{4}+42 a \,b^{4} d \,x^{4}+105 x^{3} e \,a^{3} b^{2}+105 x^{3} d \,a^{2} b^{3}+70 x^{2} e \,a^{4} b +140 x^{2} d \,a^{3} b^{2}+21 x e \,a^{5}+105 x d \,a^{4} b +42 d \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x}{42 \left (b x +a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*x*(6*b^5*e*x^6+35*a*b^4*e*x^5+7*b^5*d*x^5+84*a^2*b^3*e*x^4+42*a*b^4*d*x^4+105*a^3*b^2*e*x^3+105*a^2*b^3*d
*x^3+70*a^4*b*e*x^2+140*a^3*b^2*d*x^2+21*a^5*e*x+105*a^4*b*d*x+42*a^5*d)*((b*x+a)^2)^(5/2)/(b*x+a)^5

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maxima [B]  time = 1.03, size = 125, normalized size = 1.81 \begin {gather*} \frac {1}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} d x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a e x}{6 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a d}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} e}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} e}{7 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(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*x - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*e*x/b + 1/6*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*a*d/b - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2*e/b^2 + 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*e/b^
2

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

[Out]

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

[Out]

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

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