3.510 \(\int \frac {(a+b x)^{5/2} (A+B x)}{x^{13/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac {4 b (a+b x)^{7/2} (4 A b-11 a B)}{693 a^3 x^{7/2}}+\frac {2 (a+b x)^{7/2} (4 A b-11 a B)}{99 a^2 x^{9/2}}-\frac {2 A (a+b x)^{7/2}}{11 a x^{11/2}} \]

[Out]

-2/11*A*(b*x+a)^(7/2)/a/x^(11/2)+2/99*(4*A*b-11*B*a)*(b*x+a)^(7/2)/a^2/x^(9/2)-4/693*b*(4*A*b-11*B*a)*(b*x+a)^
(7/2)/a^3/x^(7/2)

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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} \[ -\frac {4 b (a+b x)^{7/2} (4 A b-11 a B)}{693 a^3 x^{7/2}}+\frac {2 (a+b x)^{7/2} (4 A b-11 a B)}{99 a^2 x^{9/2}}-\frac {2 A (a+b x)^{7/2}}{11 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(13/2),x]

[Out]

(-2*A*(a + b*x)^(7/2))/(11*a*x^(11/2)) + (2*(4*A*b - 11*a*B)*(a + b*x)^(7/2))/(99*a^2*x^(9/2)) - (4*b*(4*A*b -
11*a*B)*(a + b*x)^(7/2))/(693*a^3*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 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)^{5/2} (A+B x)}{x^{13/2}} \, dx &=-\frac {2 A (a+b x)^{7/2}}{11 a x^{11/2}}+\frac {\left (2 \left (-2 A b+\frac {11 a B}{2}\right )\right ) \int \frac {(a+b x)^{5/2}}{x^{11/2}} \, dx}{11 a}\\ &=-\frac {2 A (a+b x)^{7/2}}{11 a x^{11/2}}+\frac {2 (4 A b-11 a B) (a+b x)^{7/2}}{99 a^2 x^{9/2}}+\frac {(2 b (4 A b-11 a B)) \int \frac {(a+b x)^{5/2}}{x^{9/2}} \, dx}{99 a^2}\\ &=-\frac {2 A (a+b x)^{7/2}}{11 a x^{11/2}}+\frac {2 (4 A b-11 a B) (a+b x)^{7/2}}{99 a^2 x^{9/2}}-\frac {4 b (4 A b-11 a B) (a+b x)^{7/2}}{693 a^3 x^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.68 \[ -\frac {2 (a+b x)^{7/2} \left (7 a^2 (9 A+11 B x)-2 a b x (14 A+11 B x)+8 A b^2 x^2\right )}{693 a^3 x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(13/2),x]

[Out]

(-2*(a + b*x)^(7/2)*(8*A*b^2*x^2 + 7*a^2*(9*A + 11*B*x) - 2*a*b*x*(14*A + 11*B*x)))/(693*a^3*x^(11/2))

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fricas [A]  time = 0.63, size = 123, normalized size = 1.46 \[ -\frac {2 \, {\left (63 \, A a^{5} - 2 \, {\left (11 \, B a b^{4} - 4 \, A b^{5}\right )} x^{5} + {\left (11 \, B a^{2} b^{3} - 4 \, A a b^{4}\right )} x^{4} + 3 \, {\left (55 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + {\left (209 \, B a^{4} b + 113 \, A a^{3} b^{2}\right )} x^{2} + 7 \, {\left (11 \, B a^{5} + 23 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{693 \, a^{3} x^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(13/2),x, algorithm="fricas")

[Out]

-2/693*(63*A*a^5 - 2*(11*B*a*b^4 - 4*A*b^5)*x^5 + (11*B*a^2*b^3 - 4*A*a*b^4)*x^4 + 3*(55*B*a^3*b^2 + A*a^2*b^3
)*x^3 + (209*B*a^4*b + 113*A*a^3*b^2)*x^2 + 7*(11*B*a^5 + 23*A*a^4*b)*x)*sqrt(b*x + a)/(a^3*x^(11/2))

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giac [A]  time = 1.52, size = 112, normalized size = 1.33 \[ \frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (11 \, B a^{3} b^{10} - 4 \, A a^{2} b^{11}\right )} {\left (b x + a\right )}}{a^{5}} - \frac {11 \, {\left (11 \, B a^{4} b^{10} - 4 \, A a^{3} b^{11}\right )}}{a^{5}}\right )} + \frac {99 \, {\left (B a^{5} b^{10} - A a^{4} b^{11}\right )}}{a^{5}}\right )} b}{693 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {11}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(13/2),x, algorithm="giac")

[Out]

2/693*(b*x + a)^(7/2)*((b*x + a)*(2*(11*B*a^3*b^10 - 4*A*a^2*b^11)*(b*x + a)/a^5 - 11*(11*B*a^4*b^10 - 4*A*a^3
*b^11)/a^5) + 99*(B*a^5*b^10 - A*a^4*b^11)/a^5)*b/(((b*x + a)*b - a*b)^(11/2)*abs(b))

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maple [A]  time = 0.00, size = 53, normalized size = 0.63 \[ -\frac {2 \left (b x +a \right )^{\frac {7}{2}} \left (8 A \,b^{2} x^{2}-22 B a b \,x^{2}-28 A a b x +77 B \,a^{2} x +63 A \,a^{2}\right )}{693 a^{3} x^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(13/2),x)

[Out]

-2/693*(b*x+a)^(7/2)*(8*A*b^2*x^2-22*B*a*b*x^2-28*A*a*b*x+77*B*a^2*x+63*A*a^2)/x^(11/2)/a^3

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maxima [B]  time = 0.94, size = 304, normalized size = 3.62 \[ \frac {4 \, \sqrt {b x^{2} + a x} B b^{4}}{63 \, a^{2} x} - \frac {16 \, \sqrt {b x^{2} + a x} A b^{5}}{693 \, a^{3} x} - \frac {2 \, \sqrt {b x^{2} + a x} B b^{3}}{63 \, a x^{2}} + \frac {8 \, \sqrt {b x^{2} + a x} A b^{4}}{693 \, a^{2} x^{2}} + \frac {\sqrt {b x^{2} + a x} B b^{2}}{42 \, x^{3}} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{3}}{231 \, a x^{3}} - \frac {5 \, \sqrt {b x^{2} + a x} B a b}{252 \, x^{4}} + \frac {5 \, \sqrt {b x^{2} + a x} A b^{2}}{693 \, x^{4}} - \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{36 \, x^{5}} - \frac {5 \, \sqrt {b x^{2} + a x} A a b}{792 \, x^{5}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{12 \, x^{6}} - \frac {5 \, \sqrt {b x^{2} + a x} A a^{2}}{88 \, x^{6}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{2 \, x^{7}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{24 \, x^{7}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{3 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(13/2),x, algorithm="maxima")

[Out]

4/63*sqrt(b*x^2 + a*x)*B*b^4/(a^2*x) - 16/693*sqrt(b*x^2 + a*x)*A*b^5/(a^3*x) - 2/63*sqrt(b*x^2 + a*x)*B*b^3/(
a*x^2) + 8/693*sqrt(b*x^2 + a*x)*A*b^4/(a^2*x^2) + 1/42*sqrt(b*x^2 + a*x)*B*b^2/x^3 - 2/231*sqrt(b*x^2 + a*x)*
A*b^3/(a*x^3) - 5/252*sqrt(b*x^2 + a*x)*B*a*b/x^4 + 5/693*sqrt(b*x^2 + a*x)*A*b^2/x^4 - 5/36*sqrt(b*x^2 + a*x)
*B*a^2/x^5 - 5/792*sqrt(b*x^2 + a*x)*A*a*b/x^5 + 5/12*(b*x^2 + a*x)^(3/2)*B*a/x^6 - 5/88*sqrt(b*x^2 + a*x)*A*a
^2/x^6 - 1/2*(b*x^2 + a*x)^(5/2)*B/x^7 + 5/24*(b*x^2 + a*x)^(3/2)*A*a/x^7 - 1/3*(b*x^2 + a*x)^(5/2)*A/x^8

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mupad [B]  time = 0.83, size = 115, normalized size = 1.37 \[ -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A\,a^2}{11}+\frac {x\,\left (154\,B\,a^5+322\,A\,b\,a^4\right )}{693\,a^3}+\frac {x^5\,\left (16\,A\,b^5-44\,B\,a\,b^4\right )}{693\,a^3}+\frac {2\,b\,x^2\,\left (113\,A\,b+209\,B\,a\right )}{693}-\frac {2\,b^3\,x^4\,\left (4\,A\,b-11\,B\,a\right )}{693\,a^2}+\frac {2\,b^2\,x^3\,\left (A\,b+55\,B\,a\right )}{231\,a}\right )}{x^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^(13/2),x)

[Out]

-((a + b*x)^(1/2)*((2*A*a^2)/11 + (x*(154*B*a^5 + 322*A*a^4*b))/(693*a^3) + (x^5*(16*A*b^5 - 44*B*a*b^4))/(693
*a^3) + (2*b*x^2*(113*A*b + 209*B*a))/693 - (2*b^3*x^4*(4*A*b - 11*B*a))/(693*a^2) + (2*b^2*x^3*(A*b + 55*B*a)
)/(231*a)))/x^(11/2)

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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**(13/2),x)

[Out]

Timed out

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