3.509 \(\int \frac {(a+b x)^{5/2} (A+B x)}{x^{11/2}} \, dx\)

Optimal. Leaf size=53 \[ \frac {2 (a+b x)^{7/2} (2 A b-9 a B)}{63 a^2 x^{7/2}}-\frac {2 A (a+b x)^{7/2}}{9 a x^{9/2}} \]

[Out]

-2/9*A*(b*x+a)^(7/2)/a/x^(9/2)+2/63*(2*A*b-9*B*a)*(b*x+a)^(7/2)/a^2/x^(7/2)

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Rubi [A]  time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {78, 37} \[ \frac {2 (a+b x)^{7/2} (2 A b-9 a B)}{63 a^2 x^{7/2}}-\frac {2 A (a+b x)^{7/2}}{9 a x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(11/2),x]

[Out]

(-2*A*(a + b*x)^(7/2))/(9*a*x^(9/2)) + (2*(2*A*b - 9*a*B)*(a + b*x)^(7/2))/(63*a^2*x^(7/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} (A+B x)}{x^{11/2}} \, dx &=-\frac {2 A (a+b x)^{7/2}}{9 a x^{9/2}}+\frac {\left (2 \left (-A b+\frac {9 a B}{2}\right )\right ) \int \frac {(a+b x)^{5/2}}{x^{9/2}} \, dx}{9 a}\\ &=-\frac {2 A (a+b x)^{7/2}}{9 a x^{9/2}}+\frac {2 (2 A b-9 a B) (a+b x)^{7/2}}{63 a^2 x^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.68 \[ -\frac {2 (a+b x)^{7/2} (7 a A+9 a B x-2 A b x)}{63 a^2 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(11/2),x]

[Out]

(-2*(a + b*x)^(7/2)*(7*a*A - 2*A*b*x + 9*a*B*x))/(63*a^2*x^(9/2))

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fricas [B]  time = 0.65, size = 98, normalized size = 1.85 \[ -\frac {2 \, {\left (7 \, A a^{4} + {\left (9 \, B a b^{3} - 2 \, A b^{4}\right )} x^{4} + {\left (27 \, B a^{2} b^{2} + A a b^{3}\right )} x^{3} + 3 \, {\left (9 \, B a^{3} b + 5 \, A a^{2} b^{2}\right )} x^{2} + {\left (9 \, B a^{4} + 19 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{63 \, a^{2} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(11/2),x, algorithm="fricas")

[Out]

-2/63*(7*A*a^4 + (9*B*a*b^3 - 2*A*b^4)*x^4 + (27*B*a^2*b^2 + A*a*b^3)*x^3 + 3*(9*B*a^3*b + 5*A*a^2*b^2)*x^2 +
(9*B*a^4 + 19*A*a^3*b)*x)*sqrt(b*x + a)/(a^2*x^(9/2))

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giac [A]  time = 1.18, size = 80, normalized size = 1.51 \[ -\frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}} b {\left (\frac {{\left (9 \, B a^{3} b^{8} - 2 \, A a^{2} b^{9}\right )} {\left (b x + a\right )}}{a^{4}} - \frac {9 \, {\left (B a^{4} b^{8} - A a^{3} b^{9}\right )}}{a^{4}}\right )}}{63 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {9}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(11/2),x, algorithm="giac")

[Out]

-2/63*(b*x + a)^(7/2)*b*((9*B*a^3*b^8 - 2*A*a^2*b^9)*(b*x + a)/a^4 - 9*(B*a^4*b^8 - A*a^3*b^9)/a^4)/(((b*x + a
)*b - a*b)^(9/2)*abs(b))

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maple [A]  time = 0.00, size = 31, normalized size = 0.58 \[ -\frac {2 \left (b x +a \right )^{\frac {7}{2}} \left (-2 A x b +9 B a x +7 A a \right )}{63 a^{2} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(11/2),x)

[Out]

-2/63*(b*x+a)^(7/2)*(-2*A*b*x+9*B*a*x+7*A*a)/x^(9/2)/a^2

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maxima [B]  time = 0.99, size = 258, normalized size = 4.87 \[ -\frac {2 \, \sqrt {b x^{2} + a x} B b^{3}}{7 \, a x} + \frac {4 \, \sqrt {b x^{2} + a x} A b^{4}}{63 \, a^{2} x} + \frac {\sqrt {b x^{2} + a x} B b^{2}}{7 \, x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{3}}{63 \, a x^{2}} - \frac {3 \, \sqrt {b x^{2} + a x} B a b}{28 \, x^{3}} + \frac {\sqrt {b x^{2} + a x} A b^{2}}{42 \, x^{3}} - \frac {15 \, \sqrt {b x^{2} + a x} B a^{2}}{28 \, x^{4}} - \frac {5 \, \sqrt {b x^{2} + a x} A a b}{252 \, x^{4}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{4 \, x^{5}} - \frac {5 \, \sqrt {b x^{2} + a x} A a^{2}}{36 \, x^{5}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{x^{6}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{12 \, x^{6}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{2 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(11/2),x, algorithm="maxima")

[Out]

-2/7*sqrt(b*x^2 + a*x)*B*b^3/(a*x) + 4/63*sqrt(b*x^2 + a*x)*A*b^4/(a^2*x) + 1/7*sqrt(b*x^2 + a*x)*B*b^2/x^2 -
2/63*sqrt(b*x^2 + a*x)*A*b^3/(a*x^2) - 3/28*sqrt(b*x^2 + a*x)*B*a*b/x^3 + 1/42*sqrt(b*x^2 + a*x)*A*b^2/x^3 - 1
5/28*sqrt(b*x^2 + a*x)*B*a^2/x^4 - 5/252*sqrt(b*x^2 + a*x)*A*a*b/x^4 + 5/4*(b*x^2 + a*x)^(3/2)*B*a/x^5 - 5/36*
sqrt(b*x^2 + a*x)*A*a^2/x^5 - (b*x^2 + a*x)^(5/2)*B/x^6 + 5/12*(b*x^2 + a*x)^(3/2)*A*a/x^6 - 1/2*(b*x^2 + a*x)
^(5/2)*A/x^7

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mupad [B]  time = 0.86, size = 95, normalized size = 1.79 \[ -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A\,a^2}{9}+\frac {x\,\left (18\,B\,a^4+38\,A\,b\,a^3\right )}{63\,a^2}-\frac {x^4\,\left (4\,A\,b^4-18\,B\,a\,b^3\right )}{63\,a^2}+\frac {2\,b\,x^2\,\left (5\,A\,b+9\,B\,a\right )}{21}+\frac {2\,b^2\,x^3\,\left (A\,b+27\,B\,a\right )}{63\,a}\right )}{x^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^(11/2),x)

[Out]

-((a + b*x)^(1/2)*((2*A*a^2)/9 + (x*(18*B*a^4 + 38*A*a^3*b))/(63*a^2) - (x^4*(4*A*b^4 - 18*B*a*b^3))/(63*a^2)
+ (2*b*x^2*(5*A*b + 9*B*a))/21 + (2*b^2*x^3*(A*b + 27*B*a))/(63*a)))/x^(9/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**(11/2),x)

[Out]

Timed out

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