3.2.67 \(\int \frac {(A+B x) (b x+c x^2)^3}{x^{11/2}} \, dx\)

Optimal. Leaf size=81 \[ -\frac {2 A b^3}{3 x^{3/2}}-\frac {2 b^2 (3 A c+b B)}{\sqrt {x}}+\frac {2}{3} c^2 x^{3/2} (A c+3 b B)+6 b c \sqrt {x} (A c+b B)+\frac {2}{5} B c^3 x^{5/2} \]

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Rubi [A]  time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {765} \begin {gather*} -\frac {2 b^2 (3 A c+b B)}{\sqrt {x}}-\frac {2 A b^3}{3 x^{3/2}}+\frac {2}{3} c^2 x^{3/2} (A c+3 b B)+6 b c \sqrt {x} (A c+b B)+\frac {2}{5} B c^3 x^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^(11/2),x]

[Out]

(-2*A*b^3)/(3*x^(3/2)) - (2*b^2*(b*B + 3*A*c))/Sqrt[x] + 6*b*c*(b*B + A*c)*Sqrt[x] + (2*c^2*(3*b*B + A*c)*x^(3
/2))/3 + (2*B*c^3*x^(5/2))/5

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx &=\int \left (\frac {A b^3}{x^{5/2}}+\frac {b^2 (b B+3 A c)}{x^{3/2}}+\frac {3 b c (b B+A c)}{\sqrt {x}}+c^2 (3 b B+A c) \sqrt {x}+B c^3 x^{3/2}\right ) \, dx\\ &=-\frac {2 A b^3}{3 x^{3/2}}-\frac {2 b^2 (b B+3 A c)}{\sqrt {x}}+6 b c (b B+A c) \sqrt {x}+\frac {2}{3} c^2 (3 b B+A c) x^{3/2}+\frac {2}{5} B c^3 x^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 74, normalized size = 0.91 \begin {gather*} \frac {6 B x \left (-5 b^3+15 b^2 c x+5 b c^2 x^2+c^3 x^3\right )-10 A \left (b^3+9 b^2 c x-9 b c^2 x^2-c^3 x^3\right )}{15 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^(11/2),x]

[Out]

(-10*A*(b^3 + 9*b^2*c*x - 9*b*c^2*x^2 - c^3*x^3) + 6*B*x*(-5*b^3 + 15*b^2*c*x + 5*b*c^2*x^2 + c^3*x^3))/(15*x^
(3/2))

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IntegrateAlgebraic [A]  time = 0.05, size = 79, normalized size = 0.98 \begin {gather*} \frac {2 \left (-5 A b^3-45 A b^2 c x+45 A b c^2 x^2+5 A c^3 x^3-15 b^3 B x+45 b^2 B c x^2+15 b B c^2 x^3+3 B c^3 x^4\right )}{15 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^3)/x^(11/2),x]

[Out]

(2*(-5*A*b^3 - 15*b^3*B*x - 45*A*b^2*c*x + 45*b^2*B*c*x^2 + 45*A*b*c^2*x^2 + 15*b*B*c^2*x^3 + 5*A*c^3*x^3 + 3*
B*c^3*x^4))/(15*x^(3/2))

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fricas [A]  time = 0.40, size = 73, normalized size = 0.90 \begin {gather*} \frac {2 \, {\left (3 \, B c^{3} x^{4} - 5 \, A b^{3} + 5 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 45 \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} - 15 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x\right )}}{15 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x, algorithm="fricas")

[Out]

2/15*(3*B*c^3*x^4 - 5*A*b^3 + 5*(3*B*b*c^2 + A*c^3)*x^3 + 45*(B*b^2*c + A*b*c^2)*x^2 - 15*(B*b^3 + 3*A*b^2*c)*
x)/x^(3/2)

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giac [A]  time = 0.16, size = 75, normalized size = 0.93 \begin {gather*} \frac {2}{5} \, B c^{3} x^{\frac {5}{2}} + 2 \, B b c^{2} x^{\frac {3}{2}} + \frac {2}{3} \, A c^{3} x^{\frac {3}{2}} + 6 \, B b^{2} c \sqrt {x} + 6 \, A b c^{2} \sqrt {x} - \frac {2 \, {\left (3 \, B b^{3} x + 9 \, A b^{2} c x + A b^{3}\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x, algorithm="giac")

[Out]

2/5*B*c^3*x^(5/2) + 2*B*b*c^2*x^(3/2) + 2/3*A*c^3*x^(3/2) + 6*B*b^2*c*sqrt(x) + 6*A*b*c^2*sqrt(x) - 2/3*(3*B*b
^3*x + 9*A*b^2*c*x + A*b^3)/x^(3/2)

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maple [A]  time = 0.05, size = 76, normalized size = 0.94 \begin {gather*} -\frac {2 \left (-3 B \,c^{3} x^{4}-5 A \,c^{3} x^{3}-15 B b \,c^{2} x^{3}-45 A b \,c^{2} x^{2}-45 B \,b^{2} c \,x^{2}+45 A \,b^{2} c x +15 B \,b^{3} x +5 A \,b^{3}\right )}{15 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x)

[Out]

-2/15/x^(3/2)*(-3*B*c^3*x^4-5*A*c^3*x^3-15*B*b*c^2*x^3-45*A*b*c^2*x^2-45*B*b^2*c*x^2+45*A*b^2*c*x+15*B*b^3*x+5
*A*b^3)

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maxima [A]  time = 0.57, size = 73, normalized size = 0.90 \begin {gather*} \frac {2}{5} \, B c^{3} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {3}{2}} + 6 \, {\left (B b^{2} c + A b c^{2}\right )} \sqrt {x} - \frac {2 \, {\left (A b^{3} + 3 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x, algorithm="maxima")

[Out]

2/5*B*c^3*x^(5/2) + 2/3*(3*B*b*c^2 + A*c^3)*x^(3/2) + 6*(B*b^2*c + A*b*c^2)*sqrt(x) - 2/3*(A*b^3 + 3*(B*b^3 +
3*A*b^2*c)*x)/x^(3/2)

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mupad [B]  time = 0.06, size = 70, normalized size = 0.86 \begin {gather*} x^{3/2}\,\left (\frac {2\,A\,c^3}{3}+2\,B\,b\,c^2\right )-\frac {x\,\left (2\,B\,b^3+6\,A\,c\,b^2\right )+\frac {2\,A\,b^3}{3}}{x^{3/2}}+\frac {2\,B\,c^3\,x^{5/2}}{5}+6\,b\,c\,\sqrt {x}\,\left (A\,c+B\,b\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^3*(A + B*x))/x^(11/2),x)

[Out]

x^(3/2)*((2*A*c^3)/3 + 2*B*b*c^2) - (x*(2*B*b^3 + 6*A*b^2*c) + (2*A*b^3)/3)/x^(3/2) + (2*B*c^3*x^(5/2))/5 + 6*
b*c*x^(1/2)*(A*c + B*b)

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sympy [A]  time = 10.34, size = 105, normalized size = 1.30 \begin {gather*} - \frac {2 A b^{3}}{3 x^{\frac {3}{2}}} - \frac {6 A b^{2} c}{\sqrt {x}} + 6 A b c^{2} \sqrt {x} + \frac {2 A c^{3} x^{\frac {3}{2}}}{3} - \frac {2 B b^{3}}{\sqrt {x}} + 6 B b^{2} c \sqrt {x} + 2 B b c^{2} x^{\frac {3}{2}} + \frac {2 B c^{3} x^{\frac {5}{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**(11/2),x)

[Out]

-2*A*b**3/(3*x**(3/2)) - 6*A*b**2*c/sqrt(x) + 6*A*b*c**2*sqrt(x) + 2*A*c**3*x**(3/2)/3 - 2*B*b**3/sqrt(x) + 6*
B*b**2*c*sqrt(x) + 2*B*b*c**2*x**(3/2) + 2*B*c**3*x**(5/2)/5

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