3.7.16 \(\int (A+B x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

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

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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} (A b-a B)}{6 b^2}+\frac {B \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[(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((A*b - a*B)*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(6*b^2) + (B*(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 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx &=\frac {B \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}+\frac {\left (2 A b^2-2 a b B\right ) \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx}{2 b^2}\\ &=\frac {(A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2}+\frac {B \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 A+B x)+35 a^4 b x (3 A+2 B x)+35 a^3 b^2 x^2 (4 A+3 B x)+21 a^2 b^3 x^3 (5 A+4 B x)+7 a b^4 x^4 (6 A+5 B x)+b^5 x^5 (7 A+6 B x)\right )}{42 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*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*A + B*x) + 35*a^4*b*x*(3*A + 2*B*x) + 35*a^3*b^2*x^2*(4*A + 3*B*x) + 21*a^2*b^
3*x^3*(5*A + 4*B*x) + 7*a*b^4*x^4*(6*A + 5*B*x) + b^5*x^5*(7*A + 6*B*x)))/(42*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.16, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B 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[(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.49, size = 115, normalized size = 1.67 \begin {gather*} \frac {1}{7} \, B b^{5} x^{7} + A a^{5} x + \frac {1}{6} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{6} + {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{5} + \frac {5}{2} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{4} + \frac {5}{3} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{3} + \frac {1}{2} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.20, size = 217, normalized size = 3.14 \begin {gather*} \frac {1}{7} \, B b^{5} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, B a b^{4} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, A b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + A a b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, B a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, A a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, B a^{4} b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, A a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, A a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + A a^{5} x \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (B a^{7} - 7 \, A a^{6} b\right )} \mathrm {sgn}\left (b x + a\right )}{42 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/7*B*b^5*x^7*sgn(b*x + a) + 5/6*B*a*b^4*x^6*sgn(b*x + a) + 1/6*A*b^5*x^6*sgn(b*x + a) + 2*B*a^2*b^3*x^5*sgn(b
*x + a) + A*a*b^4*x^5*sgn(b*x + a) + 5/2*B*a^3*b^2*x^4*sgn(b*x + a) + 5/2*A*a^2*b^3*x^4*sgn(b*x + a) + 5/3*B*a
^4*b*x^3*sgn(b*x + a) + 10/3*A*a^3*b^2*x^3*sgn(b*x + a) + 1/2*B*a^5*x^2*sgn(b*x + a) + 5/2*A*a^4*b*x^2*sgn(b*x
+ a) + A*a^5*x*sgn(b*x + a) - 1/42*(B*a^7 - 7*A*a^6*b)*sgn(b*x + a)/b^2

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

[Out]

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

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maxima [B]  time = 0.52, 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}} A x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a x}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{2}}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a}{6 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{7 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

int((A + B*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 (A + B 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((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(5/2), x)

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