\(\int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx\) [486]

Optimal result

Integrand size = 20, antiderivative size = 84 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=-\frac {2 A (a+b x)^{3/2}}{7 a x^{7/2}}+\frac {2 (4 A b-7 a B) (a+b x)^{3/2}}{35 a^2 x^{5/2}}-\frac {4 b (4 A b-7 a B) (a+b x)^{3/2}}{105 a^3 x^{3/2}} \]

[Out]

-2/7*A*(b*x+a)^(3/2)/a/x^(7/2)+2/35*(4*A*b-7*B*a)*(b*x+a)^(3/2)/a^2/x^(5/2)-4/105*b*(4*A*b-7*B*a)*(b*x+a)^(3/2
)/a^3/x^(3/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {79, 47, 37} \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=-\frac {4 b (a+b x)^{3/2} (4 A b-7 a B)}{105 a^3 x^{3/2}}+\frac {2 (a+b x)^{3/2} (4 A b-7 a B)}{35 a^2 x^{5/2}}-\frac {2 A (a+b x)^{3/2}}{7 a x^{7/2}} \]

[In]

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

[Out]

(-2*A*(a + b*x)^(3/2))/(7*a*x^(7/2)) + (2*(4*A*b - 7*a*B)*(a + b*x)^(3/2))/(35*a^2*x^(5/2)) - (4*b*(4*A*b - 7*
a*B)*(a + b*x)^(3/2))/(105*a^3*x^(3/2))

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 47

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 79

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] || L
tQ[p, n]))))

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A (a+b x)^{3/2}}{7 a x^{7/2}}+\frac {\left (2 \left (-2 A b+\frac {7 a B}{2}\right )\right ) \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx}{7 a} \\ & = -\frac {2 A (a+b x)^{3/2}}{7 a x^{7/2}}+\frac {2 (4 A b-7 a B) (a+b x)^{3/2}}{35 a^2 x^{5/2}}+\frac {(2 b (4 A b-7 a B)) \int \frac {\sqrt {a+b x}}{x^{5/2}} \, dx}{35 a^2} \\ & = -\frac {2 A (a+b x)^{3/2}}{7 a x^{7/2}}+\frac {2 (4 A b-7 a B) (a+b x)^{3/2}}{35 a^2 x^{5/2}}-\frac {4 b (4 A b-7 a B) (a+b x)^{3/2}}{105 a^3 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=-\frac {2 (a+b x)^{3/2} \left (15 a^2 A-12 a A b x+21 a^2 B x+8 A b^2 x^2-14 a b B x^2\right )}{105 a^3 x^{7/2}} \]

[In]

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

[Out]

(-2*(a + b*x)^(3/2)*(15*a^2*A - 12*a*A*b*x + 21*a^2*B*x + 8*A*b^2*x^2 - 14*a*b*B*x^2))/(105*a^3*x^(7/2))

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.63

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (8 A \,b^{2} x^{2}-14 B a b \,x^{2}-12 a A b x +21 a^{2} B x +15 a^{2} A \right )}{105 x^{\frac {7}{2}} a^{3}}\)	\(53\)
default	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (8 A \,b^{2} x^{2}-14 B a b \,x^{2}-12 a A b x +21 a^{2} B x +15 a^{2} A \right )}{105 x^{\frac {7}{2}} a^{3}}\)	\(53\)
risch	\(-\frac {2 \sqrt {b x +a}\, \left (8 A \,b^{3} x^{3}-14 B a \,b^{2} x^{3}-4 a A \,b^{2} x^{2}+7 B \,a^{2} b \,x^{2}+3 a^{2} A b x +21 a^{3} B x +15 a^{3} A \right )}{105 x^{\frac {7}{2}} a^{3}}\)	\(77\)

[In]

int((B*x+A)*(b*x+a)^(1/2)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*(b*x+a)^(3/2)*(8*A*b^2*x^2-14*B*a*b*x^2-12*A*a*b*x+21*B*a^2*x+15*A*a^2)/x^(7/2)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=-\frac {2 \, {\left (15 \, A a^{3} - 2 \, {\left (7 \, B a b^{2} - 4 \, A b^{3}\right )} x^{3} + {\left (7 \, B a^{2} b - 4 \, A a b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt {b x + a}}{105 \, a^{3} x^{\frac {7}{2}}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(9/2),x, algorithm="fricas")

[Out]

-2/105*(15*A*a^3 - 2*(7*B*a*b^2 - 4*A*b^3)*x^3 + (7*B*a^2*b - 4*A*a*b^2)*x^2 + 3*(7*B*a^3 + A*a^2*b)*x)*sqrt(b
*x + a)/(a^3*x^(7/2))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (82) = 164\).

Time = 10.80 (sec) , antiderivative size = 427, normalized size of antiderivative = 5.08 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=- \frac {30 A a^{5} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {66 A a^{4} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {34 A a^{3} b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {6 A a^{2} b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {24 A a b^{\frac {17}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {16 A b^{\frac {19}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {2 B \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {2 B b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a x} + \frac {4 B b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a^{2}} \]

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x**(9/2),x)

[Out]

-30*A*a**5*b**(9/2)*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 66*A*a*
*4*b**(11/2)*x*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 34*A*a**3*b*
*(13/2)*x**2*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 6*A*a**2*b**(1
5/2)*x**3*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 24*A*a*b**(17/2)*
x**4*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 16*A*b**(19/2)*x**5*sq
rt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 2*B*sqrt(b)*sqrt(a/(b*x) + 1)
/(5*x**2) - 2*B*b**(3/2)*sqrt(a/(b*x) + 1)/(15*a*x) + 4*B*b**(5/2)*sqrt(a/(b*x) + 1)/(15*a**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (66) = 132\).

Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.74 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=\frac {4 \, \sqrt {b x^{2} + a x} B b^{2}}{15 \, a^{2} x} - \frac {16 \, \sqrt {b x^{2} + a x} A b^{3}}{105 \, a^{3} x} - \frac {2 \, \sqrt {b x^{2} + a x} B b}{15 \, a x^{2}} + \frac {8 \, \sqrt {b x^{2} + a x} A b^{2}}{105 \, a^{2} x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} B}{5 \, x^{3}} - \frac {2 \, \sqrt {b x^{2} + a x} A b}{35 \, a x^{3}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{7 \, x^{4}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(9/2),x, algorithm="maxima")

[Out]

4/15*sqrt(b*x^2 + a*x)*B*b^2/(a^2*x) - 16/105*sqrt(b*x^2 + a*x)*A*b^3/(a^3*x) - 2/15*sqrt(b*x^2 + a*x)*B*b/(a*
x^2) + 8/105*sqrt(b*x^2 + a*x)*A*b^2/(a^2*x^2) - 2/5*sqrt(b*x^2 + a*x)*B/x^3 - 2/35*sqrt(b*x^2 + a*x)*A*b/(a*x
^3) - 2/7*sqrt(b*x^2 + a*x)*A/x^4

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (7 \, B a b^{6} - 4 \, A b^{7}\right )} {\left (b x + a\right )}}{a^{3}} - \frac {7 \, {\left (7 \, B a^{2} b^{6} - 4 \, A a b^{7}\right )}}{a^{3}}\right )} + \frac {35 \, {\left (B a^{3} b^{6} - A a^{2} b^{7}\right )}}{a^{3}}\right )} b}{105 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}} {\left | b \right |}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(9/2),x, algorithm="giac")

[Out]

2/105*(b*x + a)^(3/2)*((b*x + a)*(2*(7*B*a*b^6 - 4*A*b^7)*(b*x + a)/a^3 - 7*(7*B*a^2*b^6 - 4*A*a*b^7)/a^3) + 3
5*(B*a^3*b^6 - A*a^2*b^7)/a^3)*b/(((b*x + a)*b - a*b)^(7/2)*abs(b))

Mupad [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{9/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{7}+\frac {x\,\left (42\,B\,a^3+6\,A\,b\,a^2\right )}{105\,a^3}+\frac {x^3\,\left (16\,A\,b^3-28\,B\,a\,b^2\right )}{105\,a^3}-\frac {2\,b\,x^2\,\left (4\,A\,b-7\,B\,a\right )}{105\,a^2}\right )}{x^{7/2}} \]

[In]

int(((A + B*x)*(a + b*x)^(1/2))/x^(9/2),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/7 + (x*(42*B*a^3 + 6*A*a^2*b))/(105*a^3) + (x^3*(16*A*b^3 - 28*B*a*b^2))/(105*a^3) -
(2*b*x^2*(4*A*b - 7*B*a))/(105*a^2)))/x^(7/2)