3.7.12 \(\int x^4 (A+B x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=306 \[ \frac {5 a^2 b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{4 (a+b x)}+\frac {b^4 x^{10} \sqrt {a^2+2 a b x+b^2 x^2} (5 a B+A b)}{10 (a+b x)}+\frac {5 a b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2} (2 a B+A b)}{9 (a+b x)}+\frac {b^5 B x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {a^5 A x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^4 x^6 \sqrt {a^2+2 a b x+b^2 x^2} (a B+5 A b)}{6 (a+b x)}+\frac {5 a^3 b x^7 \sqrt {a^2+2 a b x+b^2 x^2} (a B+2 A b)}{7 (a+b x)} \]

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Rubi [A]  time = 0.13, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 76} \begin {gather*} \frac {b^4 x^{10} \sqrt {a^2+2 a b x+b^2 x^2} (5 a B+A b)}{10 (a+b x)}+\frac {5 a b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2} (2 a B+A b)}{9 (a+b x)}+\frac {5 a^2 b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{4 (a+b x)}+\frac {5 a^3 b x^7 \sqrt {a^2+2 a b x+b^2 x^2} (a B+2 A b)}{7 (a+b x)}+\frac {a^4 x^6 \sqrt {a^2+2 a b x+b^2 x^2} (a B+5 A b)}{6 (a+b x)}+\frac {a^5 A x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {b^5 B x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(a^5*A*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (a^4*(5*A*b + a*B)*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(6*(a + b*x)) + (5*a^3*b*(2*A*b + a*B)*x^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(a + b*x)) + (5*a^2*b^2*(A*b +
a*B)*x^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x)) + (5*a*b^3*(A*b + 2*a*B)*x^9*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/(9*(a + b*x)) + (b^4*(A*b + 5*a*B)*x^10*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(10*(a + b*x)) + (b^5*B*x^11*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(11*(a + b*x))

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int x^4 (A+B x) \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 x^4 \left (a b+b^2 x\right )^5 (A+B x) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a^5 A b^5 x^4+a^4 b^5 (5 A b+a B) x^5+5 a^3 b^6 (2 A b+a B) x^6+10 a^2 b^7 (A b+a B) x^7+5 a b^8 (A b+2 a B) x^8+b^9 (A b+5 a B) x^9+b^{10} B x^{10}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {a^5 A x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^4 (5 A b+a B) x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {5 a^3 b (2 A b+a B) x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {5 a^2 b^2 (A b+a B) x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {5 a b^3 (A b+2 a B) x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {b^4 (A b+5 a B) x^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{10 (a+b x)}+\frac {b^5 B x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 125, normalized size = 0.41 \begin {gather*} \frac {x^5 \sqrt {(a+b x)^2} \left (462 a^5 (6 A+5 B x)+1650 a^4 b x (7 A+6 B x)+2475 a^3 b^2 x^2 (8 A+7 B x)+1925 a^2 b^3 x^3 (9 A+8 B x)+770 a b^4 x^4 (10 A+9 B x)+126 b^5 x^5 (11 A+10 B x)\right )}{13860 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(x^5*Sqrt[(a + b*x)^2]*(462*a^5*(6*A + 5*B*x) + 1650*a^4*b*x*(7*A + 6*B*x) + 2475*a^3*b^2*x^2*(8*A + 7*B*x) +
1925*a^2*b^3*x^3*(9*A + 8*B*x) + 770*a*b^4*x^4*(10*A + 9*B*x) + 126*b^5*x^5*(11*A + 10*B*x)))/(13860*(a + b*x)
)

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IntegrateAlgebraic [F]  time = 1.36, size = 0, normalized size = 0.00 \begin {gather*} \int x^4 (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[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2), x]

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fricas [A]  time = 0.42, size = 119, normalized size = 0.39 \begin {gather*} \frac {1}{11} \, B b^{5} x^{11} + \frac {1}{5} \, A a^{5} x^{5} + \frac {1}{10} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} + \frac {5}{9} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{9} + \frac {5}{4} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{8} + \frac {5}{7} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/11*B*b^5*x^11 + 1/5*A*a^5*x^5 + 1/10*(5*B*a*b^4 + A*b^5)*x^10 + 5/9*(2*B*a^2*b^3 + A*a*b^4)*x^9 + 5/4*(B*a^3
*b^2 + A*a^2*b^3)*x^8 + 5/7*(B*a^4*b + 2*A*a^3*b^2)*x^7 + 1/6*(B*a^5 + 5*A*a^4*b)*x^6

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giac [A]  time = 0.17, size = 222, normalized size = 0.73 \begin {gather*} \frac {1}{11} \, B b^{5} x^{11} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a b^{4} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{10} \, A b^{5} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, B a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{9} \, A a b^{4} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, B a^{3} b^{2} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, A a^{2} b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, B a^{4} b x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, A a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, B a^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, A a^{4} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A a^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (5 \, B a^{11} - 11 \, A a^{10} b\right )} \mathrm {sgn}\left (b x + a\right )}{13860 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/11*B*b^5*x^11*sgn(b*x + a) + 1/2*B*a*b^4*x^10*sgn(b*x + a) + 1/10*A*b^5*x^10*sgn(b*x + a) + 10/9*B*a^2*b^3*x
^9*sgn(b*x + a) + 5/9*A*a*b^4*x^9*sgn(b*x + a) + 5/4*B*a^3*b^2*x^8*sgn(b*x + a) + 5/4*A*a^2*b^3*x^8*sgn(b*x +
a) + 5/7*B*a^4*b*x^7*sgn(b*x + a) + 10/7*A*a^3*b^2*x^7*sgn(b*x + a) + 1/6*B*a^5*x^6*sgn(b*x + a) + 5/6*A*a^4*b
*x^6*sgn(b*x + a) + 1/5*A*a^5*x^5*sgn(b*x + a) - 1/13860*(5*B*a^11 - 11*A*a^10*b)*sgn(b*x + a)/b^6

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maple [A]  time = 0.06, size = 140, normalized size = 0.46 \begin {gather*} \frac {\left (1260 B \,b^{5} x^{6}+1386 x^{5} A \,b^{5}+6930 x^{5} B a \,b^{4}+7700 x^{4} A a \,b^{4}+15400 x^{4} B \,a^{2} b^{3}+17325 A \,a^{2} b^{3} x^{3}+17325 B \,a^{3} b^{2} x^{3}+19800 x^{2} A \,a^{3} b^{2}+9900 x^{2} B \,a^{4} b +11550 x A \,a^{4} b +2310 x B \,a^{5}+2772 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x^{5}}{13860 \left (b x +a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/13860*x^5*(1260*B*b^5*x^6+1386*A*b^5*x^5+6930*B*a*b^4*x^5+7700*A*a*b^4*x^4+15400*B*a^2*b^3*x^4+17325*A*a^2*b
^3*x^3+17325*B*a^3*b^2*x^3+19800*A*a^3*b^2*x^2+9900*B*a^4*b*x^2+11550*A*a^4*b*x+2310*B*a^5*x+2772*A*a^5)*((b*x
+a)^2)^(5/2)/(b*x+a)^5

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maxima [A]  time = 0.60, size = 361, normalized size = 1.18 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B x^{4}}{11 \, b^{2}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a x^{3}}{22 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A x^{3}}{10 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{5} x}{6 \, b^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{4} x}{6 \, b^{4}} + \frac {31 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{2} x^{2}}{198 \, b^{4}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a x^{2}}{90 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{6}}{6 \, b^{6}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{5}}{6 \, b^{5}} - \frac {65 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{3} x}{396 \, b^{5}} + \frac {29 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{2} x}{180 \, b^{4}} + \frac {461 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{4}}{2772 \, b^{6}} - \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{3}}{1260 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/11*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*x^4/b^2 - 3/22*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a*x^3/b^3 + 1/10*(b^2*
x^2 + 2*a*b*x + a^2)^(7/2)*A*x^3/b^2 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^5*x/b^5 + 1/6*(b^2*x^2 + 2*a*b*
x + a^2)^(5/2)*A*a^4*x/b^4 + 31/198*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^2*x^2/b^4 - 13/90*(b^2*x^2 + 2*a*b*x +
a^2)^(7/2)*A*a*x^2/b^3 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^6/b^6 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*
A*a^5/b^5 - 65/396*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^3*x/b^5 + 29/180*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^2*
x/b^4 + 461/2772*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^4/b^6 - 209/1260*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^3/b^
5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^4\,\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(x^4*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

int(x^4*(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 x^{4} \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(x**4*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**4*(A + B*x)*((a + b*x)**2)**(5/2), x)

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