\(\int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx\) [485]

Optimal result

Integrand size = 20, antiderivative size = 53 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=-\frac {2 A (a+b x)^{3/2}}{5 a x^{5/2}}+\frac {2 (2 A b-5 a B) (a+b x)^{3/2}}{15 a^2 x^{3/2}} \]

[Out]

-2/5*A*(b*x+a)^(3/2)/a/x^(5/2)+2/15*(2*A*b-5*B*a)*(b*x+a)^(3/2)/a^2/x^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {79, 37} \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=\frac {2 (a+b x)^{3/2} (2 A b-5 a B)}{15 a^2 x^{3/2}}-\frac {2 A (a+b x)^{3/2}}{5 a x^{5/2}} \]

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x^(7/2),x]

[Out]

(-2*A*(a + b*x)^(3/2))/(5*a*x^(5/2)) + (2*(2*A*b - 5*a*B)*(a + b*x)^(3/2))/(15*a^2*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 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}}{5 a x^{5/2}}+\frac {\left (2 \left (-A b+\frac {5 a B}{2}\right )\right ) \int \frac {\sqrt {a+b x}}{x^{5/2}} \, dx}{5 a} \\ & = -\frac {2 A (a+b x)^{3/2}}{5 a x^{5/2}}+\frac {2 (2 A b-5 a B) (a+b x)^{3/2}}{15 a^2 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=-\frac {2 (a+b x)^{3/2} (3 a A-2 A b x+5 a B x)}{15 a^2 x^{5/2}} \]

[In]

Integrate[(Sqrt[a + b*x]*(A + B*x))/x^(7/2),x]

[Out]

(-2*(a + b*x)^(3/2)*(3*a*A - 2*A*b*x + 5*a*B*x))/(15*a^2*x^(5/2))

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.58

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-2 A b x +5 B a x +3 A a \right )}{15 x^{\frac {5}{2}} a^{2}}\)	\(31\)
default	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-2 A b x +5 B a x +3 A a \right )}{15 x^{\frac {5}{2}} a^{2}}\)	\(31\)
risch	\(-\frac {2 \sqrt {b x +a}\, \left (-2 A \,b^{2} x^{2}+5 B a b \,x^{2}+a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}} a^{2}}\)	\(52\)

[In]

int((B*x+A)*(b*x+a)^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*(b*x+a)^(3/2)*(-2*A*b*x+5*B*a*x+3*A*a)/x^(5/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=-\frac {2 \, {\left (3 \, A a^{2} + {\left (5 \, B a b - 2 \, A b^{2}\right )} x^{2} + {\left (5 \, B a^{2} + A a b\right )} x\right )} \sqrt {b x + a}}{15 \, a^{2} x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

-2/15*(3*A*a^2 + (5*B*a*b - 2*A*b^2)*x^2 + (5*B*a^2 + A*a*b)*x)*sqrt(b*x + a)/(a^2*x^(5/2))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (49) = 98\).

Time = 3.61 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=- \frac {2 A \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {2 A b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a x} + \frac {4 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a^{2}} - \frac {2 B \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 B b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a} \]

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x**(7/2),x)

[Out]

-2*A*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 2*A*b**(3/2)*sqrt(a/(b*x) + 1)/(15*a*x) + 4*A*b**(5/2)*sqrt(a/(b*x)
+ 1)/(15*a**2) - 2*B*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 2*B*b**(3/2)*sqrt(a/(b*x) + 1)/(3*a)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (41) = 82\).

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.89 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=-\frac {2 \, \sqrt {b x^{2} + a x} B b}{3 \, a x} + \frac {4 \, \sqrt {b x^{2} + a x} A b^{2}}{15 \, a^{2} x} - \frac {2 \, \sqrt {b x^{2} + a x} B}{3 \, x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} A b}{15 \, a x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{5 \, x^{3}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

-2/3*sqrt(b*x^2 + a*x)*B*b/(a*x) + 4/15*sqrt(b*x^2 + a*x)*A*b^2/(a^2*x) - 2/3*sqrt(b*x^2 + a*x)*B/x^2 - 2/15*s
qrt(b*x^2 + a*x)*A*b/(a*x^2) - 2/5*sqrt(b*x^2 + a*x)*A/x^3

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=-\frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} b {\left (\frac {{\left (5 \, B a b^{4} - 2 \, A b^{5}\right )} {\left (b x + a\right )}}{a^{2}} - \frac {5 \, {\left (B a^{2} b^{4} - A a b^{5}\right )}}{a^{2}}\right )}}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}} {\left | b \right |}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

-2/15*(b*x + a)^(3/2)*b*((5*B*a*b^4 - 2*A*b^5)*(b*x + a)/a^2 - 5*(B*a^2*b^4 - A*a*b^5)/a^2)/(((b*x + a)*b - a*
b)^(5/2)*abs(b))

Mupad [B] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{7/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{5}-\frac {x^2\,\left (4\,A\,b^2-10\,B\,a\,b\right )}{15\,a^2}+\frac {x\,\left (10\,B\,a^2+2\,A\,b\,a\right )}{15\,a^2}\right )}{x^{5/2}} \]

[In]

int(((A + B*x)*(a + b*x)^(1/2))/x^(7/2),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/5 - (x^2*(4*A*b^2 - 10*B*a*b))/(15*a^2) + (x*(10*B*a^2 + 2*A*a*b))/(15*a^2)))/x^(5/2)