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

Optimal. Leaf size=37 \[ -\frac {2 (a B+A b)}{3 x^{3/2}}-\frac {2 a A}{5 x^{5/2}}-\frac {2 b B}{\sqrt {x}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {76} \begin {gather*} -\frac {2 (a B+A b)}{3 x^{3/2}}-\frac {2 a A}{5 x^{5/2}}-\frac {2 b B}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(A + B*x))/x^(7/2),x]

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 30, normalized size = 0.81 \begin {gather*} -\frac {2 (a (3 A+5 B x)+5 b x (A+3 B x))}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.02, size = 31, normalized size = 0.84 \begin {gather*} -\frac {2 \left (3 a A+5 a B x+5 A b x+15 b B x^2\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 27, normalized size = 0.73 \begin {gather*} -\frac {2 \, {\left (15 \, B b x^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.29, size = 27, normalized size = 0.73 \begin {gather*} -\frac {2 \, {\left (15 \, B b x^{2} + 5 \, B a x + 5 \, A b x + 3 \, A a\right )}}{15 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 28, normalized size = 0.76 \begin {gather*} -\frac {2 \left (15 B b \,x^{2}+5 A b x +5 B a x +3 A a \right )}{15 x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.83, size = 27, normalized size = 0.73 \begin {gather*} -\frac {2 \, {\left (15 \, B b x^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 28, normalized size = 0.76 \begin {gather*} -\frac {2\,B\,b\,x^2+\left (\frac {2\,A\,b}{3}+\frac {2\,B\,a}{3}\right )\,x+\frac {2\,A\,a}{5}}{x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x))/x^(7/2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.71, size = 46, normalized size = 1.24 \begin {gather*} - \frac {2 A a}{5 x^{\frac {5}{2}}} - \frac {2 A b}{3 x^{\frac {3}{2}}} - \frac {2 B a}{3 x^{\frac {3}{2}}} - \frac {2 B b}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________