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

Optimal. Leaf size=53 \[ \frac {2 (a+b x)^{3/2} (2 A b-5 a B)}{15 a^2 x^{3/2}}-\frac {2 A (a+b x)^{3/2}}{5 a x^{5/2}} \]

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Rubi [A]  time = 0.02, 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} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[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 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 {\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 {\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*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.68 \begin {gather*} -\frac {2 (a+b x)^{3/2} (3 a A+5 a B x-2 A b x)}{15 a^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[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))

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IntegrateAlgebraic [A]  time = 0.16, size = 58, normalized size = 1.09 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-3 a^2 A-5 a^2 B x-a A b x-5 a b B x^2+2 A b^2 x^2\right )}{15 a^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.70, size = 51, normalized size = 0.96 \begin {gather*} -\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 1.38, size = 73, normalized size = 1.38 \begin {gather*} -\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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.00, size = 31, normalized size = 0.58 \begin {gather*} -\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-2 A x b +5 B a x +3 A a \right )}{15 a^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.84, size = 100, normalized size = 1.89 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.67, size = 54, normalized size = 1.02 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [B]  time = 61.24, size = 110, normalized size = 2.08 \begin {gather*} A \left (- \frac {2 \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {2 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a x} + \frac {4 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a^{2}}\right ) + B \left (- \frac {2 \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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