3.737 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)}{x^{9/2}} \, dx\)

Optimal. Leaf size=61 \[ -\frac {2 a^2 A}{7 x^{7/2}}-\frac {2 a (a B+2 A b)}{5 x^{5/2}}-\frac {2 b (2 a B+A b)}{3 x^{3/2}}-\frac {2 b^2 B}{\sqrt {x}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {27, 76} \[ -\frac {2 a^2 A}{7 x^{7/2}}-\frac {2 a (a B+2 A b)}{5 x^{5/2}}-\frac {2 b (2 a B+A b)}{3 x^{3/2}}-\frac {2 b^2 B}{\sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^(9/2),x]

[Out]

(-2*a^2*A)/(7*x^(7/2)) - (2*a*(2*A*b + a*B))/(5*x^(5/2)) - (2*b*(A*b + 2*a*B))/(3*x^(3/2)) - (2*b^2*B)/Sqrt[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.82 \[ -\frac {2 \left (3 a^2 (5 A+7 B x)+14 a b x (3 A+5 B x)+35 b^2 x^2 (A+3 B x)\right )}{105 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^(9/2),x]

[Out]

(-2*(35*b^2*x^2*(A + 3*B*x) + 14*a*b*x*(3*A + 5*B*x) + 3*a^2*(5*A + 7*B*x)))/(105*x^(7/2))

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fricas [A]  time = 0.85, size = 51, normalized size = 0.84 \[ -\frac {2 \, {\left (105 \, B b^{2} x^{3} + 15 \, A a^{2} + 35 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 21 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{105 \, x^{\frac {7}{2}}} \]

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)/x^(9/2),x, algorithm="fricas")

[Out]

-2/105*(105*B*b^2*x^3 + 15*A*a^2 + 35*(2*B*a*b + A*b^2)*x^2 + 21*(B*a^2 + 2*A*a*b)*x)/x^(7/2)

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giac [A]  time = 0.18, size = 51, normalized size = 0.84 \[ -\frac {2 \, {\left (105 \, B b^{2} x^{3} + 70 \, B a b x^{2} + 35 \, A b^{2} x^{2} + 21 \, B a^{2} x + 42 \, A a b x + 15 \, A a^{2}\right )}}{105 \, x^{\frac {7}{2}}} \]

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)/x^(9/2),x, algorithm="giac")

[Out]

-2/105*(105*B*b^2*x^3 + 70*B*a*b*x^2 + 35*A*b^2*x^2 + 21*B*a^2*x + 42*A*a*b*x + 15*A*a^2)/x^(7/2)

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maple [A]  time = 0.05, size = 52, normalized size = 0.85 \[ -\frac {2 \left (105 B \,b^{2} x^{3}+35 A \,b^{2} x^{2}+70 B a b \,x^{2}+42 A a b x +21 B \,a^{2} x +15 A \,a^{2}\right )}{105 x^{\frac {7}{2}}} \]

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)/x^(9/2),x)

[Out]

-2/105*(105*B*b^2*x^3+35*A*b^2*x^2+70*B*a*b*x^2+42*A*a*b*x+21*B*a^2*x+15*A*a^2)/x^(7/2)

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maxima [A]  time = 0.54, size = 51, normalized size = 0.84 \[ -\frac {2 \, {\left (105 \, B b^{2} x^{3} + 15 \, A a^{2} + 35 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 21 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{105 \, x^{\frac {7}{2}}} \]

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)/x^(9/2),x, algorithm="maxima")

[Out]

-2/105*(105*B*b^2*x^3 + 15*A*a^2 + 35*(2*B*a*b + A*b^2)*x^2 + 21*(B*a^2 + 2*A*a*b)*x)/x^(7/2)

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mupad [B]  time = 1.12, size = 51, normalized size = 0.84 \[ -\frac {x^2\,\left (\frac {2\,A\,b^2}{3}+\frac {4\,B\,a\,b}{3}\right )+\frac {2\,A\,a^2}{7}+x\,\left (\frac {2\,B\,a^2}{5}+\frac {4\,A\,b\,a}{5}\right )+2\,B\,b^2\,x^3}{x^{7/2}} \]

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))/x^(9/2),x)

[Out]

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

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sympy [A]  time = 3.07, size = 80, normalized size = 1.31 \[ - \frac {2 A a^{2}}{7 x^{\frac {7}{2}}} - \frac {4 A a b}{5 x^{\frac {5}{2}}} - \frac {2 A b^{2}}{3 x^{\frac {3}{2}}} - \frac {2 B a^{2}}{5 x^{\frac {5}{2}}} - \frac {4 B a b}{3 x^{\frac {3}{2}}} - \frac {2 B b^{2}}{\sqrt {x}} \]

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)/x**(9/2),x)

[Out]

-2*A*a**2/(7*x**(7/2)) - 4*A*a*b/(5*x**(5/2)) - 2*A*b**2/(3*x**(3/2)) - 2*B*a**2/(5*x**(5/2)) - 4*B*a*b/(3*x**
(3/2)) - 2*B*b**2/sqrt(x)

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