3.4.34 \(\int \frac {(a+b x)^2 (A+B x)}{x^{7/2}} \, dx\) [334]

Optimal. Leaf size=59 \[ -\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a (2 A b+a B)}{3 x^{3/2}}-\frac {2 b (A b+2 a B)}{\sqrt {x}}+2 b^2 B \sqrt {x} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \begin {gather*} -\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a (a B+2 A b)}{3 x^{3/2}}-\frac {2 b (2 a B+A b)}{\sqrt {x}}+2 b^2 B \sqrt {x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

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

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Mathematica [A]
time = 0.05, size = 47, normalized size = 0.80 \begin {gather*} -\frac {2 \left (15 b^2 x^2 (A-B x)+10 a b x (A+3 B x)+a^2 (3 A+5 B x)\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*b^2*x^2*(A - B*x) + 10*a*b*x*(A + 3*B*x) + a^2*(3*A + 5*B*x)))/(15*x^(5/2))

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Maple [A]
time = 0.07, size = 48, normalized size = 0.81

method	result	size
derivativedivides	\(-\frac {2 a^{2} A}{5 x^{\frac {5}{2}}}-\frac {2 a \left (2 A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {2 b \left (A b +2 B a \right )}{\sqrt {x}}+2 b^{2} B \sqrt {x}\)	\(48\)
default	\(-\frac {2 a^{2} A}{5 x^{\frac {5}{2}}}-\frac {2 a \left (2 A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {2 b \left (A b +2 B a \right )}{\sqrt {x}}+2 b^{2} B \sqrt {x}\)	\(48\)
gosper	\(-\frac {2 \left (-15 b^{2} B \,x^{3}+15 A \,b^{2} x^{2}+30 B a b \,x^{2}+10 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}}}\)	\(52\)
trager	\(-\frac {2 \left (-15 b^{2} B \,x^{3}+15 A \,b^{2} x^{2}+30 B a b \,x^{2}+10 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}}}\)	\(52\)
risch	\(-\frac {2 \left (-15 b^{2} B \,x^{3}+15 A \,b^{2} x^{2}+30 B a b \,x^{2}+10 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 52, normalized size = 0.88 \begin {gather*} 2 \, B b^{2} \sqrt {x} - \frac {2 \, {\left (3 \, A a^{2} + 15 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} + 2 \, 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)^2*(B*x+A)/x^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.87, size = 51, normalized size = 0.86 \begin {gather*} \frac {2 \, {\left (15 \, B b^{2} x^{3} - 3 \, A a^{2} - 15 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} - 5 \, {\left (B a^{2} + 2 \, 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)^2*(B*x+A)/x^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.40, size = 75, normalized size = 1.27 \begin {gather*} - \frac {2 A a^{2}}{5 x^{\frac {5}{2}}} - \frac {4 A a b}{3 x^{\frac {3}{2}}} - \frac {2 A b^{2}}{\sqrt {x}} - \frac {2 B a^{2}}{3 x^{\frac {3}{2}}} - \frac {4 B a b}{\sqrt {x}} + 2 B b^{2} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.52, size = 52, normalized size = 0.88 \begin {gather*} 2 \, B b^{2} \sqrt {x} - \frac {2 \, {\left (30 \, B a b x^{2} + 15 \, A b^{2} x^{2} + 5 \, B a^{2} x + 10 \, A a b x + 3 \, A a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*b^2*sqrt(x) - 2/15*(30*B*a*b*x^2 + 15*A*b^2*x^2 + 5*B*a^2*x + 10*A*a*b*x + 3*A*a^2)/x^(5/2)

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Mupad [B]
time = 0.06, size = 52, normalized size = 0.88 \begin {gather*} 2\,B\,b^2\,\sqrt {x}-\frac {x^2\,\left (2\,A\,b^2+4\,B\,a\,b\right )+\frac {2\,A\,a^2}{5}+x\,\left (\frac {2\,B\,a^2}{3}+\frac {4\,A\,b\,a}{3}\right )}{x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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