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

Optimal. Leaf size=51 \[ -\frac {2 (a B+A b)}{3 x^{3/2}}-\frac {2 a A}{5 x^{5/2}}-\frac {2 (A c+b B)}{\sqrt {x}}+2 B c \sqrt {x} \]

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Rubi [A]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {765} \begin {gather*} -\frac {2 (a B+A b)}{3 x^{3/2}}-\frac {2 a A}{5 x^{5/2}}-\frac {2 (A c+b B)}{\sqrt {x}}+2 B c \sqrt {x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 42, normalized size = 0.82 \begin {gather*} -\frac {2 (a (3 A+5 B x)+5 x (A (b+3 c x)+3 B x (b-c x)))}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 45, normalized size = 0.88 \begin {gather*} \frac {2 \left (-3 a A-5 a B x-5 A b x-15 A c x^2-15 b B x^2+15 B c x^3\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 39, normalized size = 0.76 \begin {gather*} \frac {2 \, {\left (15 \, B c x^{3} - 15 \, {\left (B b + A c\right )} 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)*(c*x^2+b*x+a)/x^(7/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 42, normalized size = 0.82 \begin {gather*} 2 \, B c \sqrt {x} - \frac {2 \, {\left (15 \, B b x^{2} + 15 \, A c 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)*(c*x^2+b*x+a)/x^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 42, normalized size = 0.82 \begin {gather*} -\frac {2 \left (-15 B c \,x^{3}+15 A c \,x^{2}+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)*(c*x^2+b*x+a)/x^(7/2),x)

[Out]

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

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maxima [A]  time = 0.69, size = 40, normalized size = 0.78 \begin {gather*} 2 \, B c \sqrt {x} - \frac {2 \, {\left (15 \, {\left (B b + A c\right )} 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)*(c*x^2+b*x+a)/x^(7/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.30, size = 42, normalized size = 0.82 \begin {gather*} 2\,B\,c\,\sqrt {x}-\frac {\left (2\,A\,c+2\,B\,b\right )\,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 + c*x^2))/x^(7/2),x)

[Out]

2*B*c*x^(1/2) - ((2*A*a)/5 + x*((2*A*b)/3 + (2*B*a)/3) + x^2*(2*A*c + 2*B*b))/x^(5/2)

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sympy [A]  time = 1.43, size = 65, normalized size = 1.27 \begin {gather*} - \frac {2 A a}{5 x^{\frac {5}{2}}} - \frac {2 A b}{3 x^{\frac {3}{2}}} - \frac {2 A c}{\sqrt {x}} - \frac {2 B a}{3 x^{\frac {3}{2}}} - \frac {2 B b}{\sqrt {x}} + 2 B c \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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