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

Optimal. Leaf size=84 \[ -\frac {4 b (a+b x)^{5/2} (4 A b-9 a B)}{315 a^3 x^{5/2}}+\frac {2 (a+b x)^{5/2} (4 A b-9 a B)}{63 a^2 x^{7/2}}-\frac {2 A (a+b x)^{5/2}}{9 a x^{9/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {78, 45, 37} \begin {gather*} -\frac {4 b (a+b x)^{5/2} (4 A b-9 a B)}{315 a^3 x^{5/2}}+\frac {2 (a+b x)^{5/2} (4 A b-9 a B)}{63 a^2 x^{7/2}}-\frac {2 A (a+b x)^{5/2}}{9 a x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*(a + b*x)^(5/2))/(9*a*x^(9/2)) + (2*(4*A*b - 9*a*B)*(a + b*x)^(5/2))/(63*a^2*x^(7/2)) - (4*b*(4*A*b - 9*
a*B)*(a + b*x)^(5/2))/(315*a^3*x^(5/2))

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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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)^{3/2} (A+B x)}{x^{11/2}} \, dx &=-\frac {2 A (a+b x)^{5/2}}{9 a x^{9/2}}+\frac {\left (2 \left (-2 A b+\frac {9 a B}{2}\right )\right ) \int \frac {(a+b x)^{3/2}}{x^{9/2}} \, dx}{9 a}\\ &=-\frac {2 A (a+b x)^{5/2}}{9 a x^{9/2}}+\frac {2 (4 A b-9 a B) (a+b x)^{5/2}}{63 a^2 x^{7/2}}+\frac {(2 b (4 A b-9 a B)) \int \frac {(a+b x)^{3/2}}{x^{7/2}} \, dx}{63 a^2}\\ &=-\frac {2 A (a+b x)^{5/2}}{9 a x^{9/2}}+\frac {2 (4 A b-9 a B) (a+b x)^{5/2}}{63 a^2 x^{7/2}}-\frac {4 b (4 A b-9 a B) (a+b x)^{5/2}}{315 a^3 x^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.68 \begin {gather*} -\frac {2 (a+b x)^{5/2} \left (5 a^2 (7 A+9 B x)-2 a b x (10 A+9 B x)+8 A b^2 x^2\right )}{315 a^3 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x)^(5/2)*(8*A*b^2*x^2 + 5*a^2*(7*A + 9*B*x) - 2*a*b*x*(10*A + 9*B*x)))/(315*a^3*x^(9/2))

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IntegrateAlgebraic [A]  time = 0.30, size = 106, normalized size = 1.26 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-35 a^4 A-45 a^4 B x-50 a^3 A b x-72 a^3 b B x^2-3 a^2 A b^2 x^2-9 a^2 b^2 B x^3+4 a A b^3 x^3+18 a b^3 B x^4-8 A b^4 x^4\right )}{315 a^3 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*(-35*a^4*A - 50*a^3*A*b*x - 45*a^4*B*x - 3*a^2*A*b^2*x^2 - 72*a^3*b*B*x^2 + 4*a*A*b^3*x^3 - 9
*a^2*b^2*B*x^3 - 8*A*b^4*x^4 + 18*a*b^3*B*x^4))/(315*a^3*x^(9/2))

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fricas [A]  time = 1.20, size = 100, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (35 \, A a^{4} - 2 \, {\left (9 \, B a b^{3} - 4 \, A b^{4}\right )} x^{4} + {\left (9 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} + 3 \, {\left (24 \, B a^{3} b + A a^{2} b^{2}\right )} x^{2} + 5 \, {\left (9 \, B a^{4} + 10 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{315 \, a^{3} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*A*a^4 - 2*(9*B*a*b^3 - 4*A*b^4)*x^4 + (9*B*a^2*b^2 - 4*A*a*b^3)*x^3 + 3*(24*B*a^3*b + A*a^2*b^2)*x^
2 + 5*(9*B*a^4 + 10*A*a^3*b)*x)*sqrt(b*x + a)/(a^3*x^(9/2))

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giac [A]  time = 2.01, size = 110, normalized size = 1.31 \begin {gather*} \frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (9 \, B a^{2} b^{8} - 4 \, A a b^{9}\right )} {\left (b x + a\right )}}{a^{4}} - \frac {9 \, {\left (9 \, B a^{3} b^{8} - 4 \, A a^{2} b^{9}\right )}}{a^{4}}\right )} + \frac {63 \, {\left (B a^{4} b^{8} - A a^{3} b^{9}\right )}}{a^{4}}\right )} b}{315 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {9}{2}} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(b*x + a)^(5/2)*((b*x + a)*(2*(9*B*a^2*b^8 - 4*A*a*b^9)*(b*x + a)/a^4 - 9*(9*B*a^3*b^8 - 4*A*a^2*b^9)/a^
4) + 63*(B*a^4*b^8 - A*a^3*b^9)/a^4)*b/(((b*x + a)*b - a*b)^(9/2)*abs(b))

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maple [A]  time = 0.00, size = 53, normalized size = 0.63 \begin {gather*} -\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (8 A \,b^{2} x^{2}-18 B a b \,x^{2}-20 A a b x +45 B \,a^{2} x +35 A \,a^{2}\right )}{315 a^{3} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(b*x+a)^(5/2)*(8*A*b^2*x^2-18*B*a*b*x^2-20*A*a*b*x+45*B*a^2*x+35*A*a^2)/x^(9/2)/a^3

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maxima [B]  time = 0.87, size = 222, normalized size = 2.64 \begin {gather*} \frac {4 \, \sqrt {b x^{2} + a x} B b^{3}}{35 \, a^{2} x} - \frac {16 \, \sqrt {b x^{2} + a x} A b^{4}}{315 \, a^{3} x} - \frac {2 \, \sqrt {b x^{2} + a x} B b^{2}}{35 \, a x^{2}} + \frac {8 \, \sqrt {b x^{2} + a x} A b^{3}}{315 \, a^{2} x^{2}} + \frac {3 \, \sqrt {b x^{2} + a x} B b}{70 \, x^{3}} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{2}}{105 \, a x^{3}} + \frac {3 \, \sqrt {b x^{2} + a x} B a}{14 \, x^{4}} + \frac {\sqrt {b x^{2} + a x} A b}{63 \, x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{2 \, x^{5}} + \frac {\sqrt {b x^{2} + a x} A a}{9 \, x^{5}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{3 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/35*sqrt(b*x^2 + a*x)*B*b^3/(a^2*x) - 16/315*sqrt(b*x^2 + a*x)*A*b^4/(a^3*x) - 2/35*sqrt(b*x^2 + a*x)*B*b^2/(
a*x^2) + 8/315*sqrt(b*x^2 + a*x)*A*b^3/(a^2*x^2) + 3/70*sqrt(b*x^2 + a*x)*B*b/x^3 - 2/105*sqrt(b*x^2 + a*x)*A*
b^2/(a*x^3) + 3/14*sqrt(b*x^2 + a*x)*B*a/x^4 + 1/63*sqrt(b*x^2 + a*x)*A*b/x^4 - 1/2*(b*x^2 + a*x)^(3/2)*B/x^5
+ 1/9*sqrt(b*x^2 + a*x)*A*a/x^5 - 1/3*(b*x^2 + a*x)^(3/2)*A/x^6

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mupad [B]  time = 0.79, size = 96, normalized size = 1.14 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A\,a}{9}+\frac {x\,\left (90\,B\,a^4+100\,A\,b\,a^3\right )}{315\,a^3}+\frac {x^4\,\left (16\,A\,b^4-36\,B\,a\,b^3\right )}{315\,a^3}-\frac {2\,b^2\,x^3\,\left (4\,A\,b-9\,B\,a\right )}{315\,a^2}+\frac {2\,b\,x^2\,\left (A\,b+24\,B\,a\right )}{105\,a}\right )}{x^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + b*x)^(1/2)*((2*A*a)/9 + (x*(90*B*a^4 + 100*A*a^3*b))/(315*a^3) + (x^4*(16*A*b^4 - 36*B*a*b^3))/(315*a^3
) - (2*b^2*x^3*(4*A*b - 9*B*a))/(315*a^2) + (2*b*x^2*(A*b + 24*B*a))/(105*a)))/x^(9/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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