3.3.63 \(\int \frac {(A+B x) (a+c x^2)^2}{x^2} \, dx\) [263]

Optimal. Leaf size=52 \[ -\frac {a^2 A}{x}+2 a A c x+a B c x^2+\frac {1}{3} A c^2 x^3+\frac {1}{4} B c^2 x^4+a^2 B \log (x) \]

[Out]

-a^2*A/x+2*a*A*c*x+a*B*c*x^2+1/3*A*c^2*x^3+1/4*B*c^2*x^4+a^2*B*ln(x)

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Rubi [A]
time = 0.03, antiderivative size = 52, 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 = {780} \begin {gather*} -\frac {a^2 A}{x}+a^2 B \log (x)+2 a A c x+a B c x^2+\frac {1}{3} A c^2 x^3+\frac {1}{4} B c^2 x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^2)/x^2,x]

[Out]

-((a^2*A)/x) + 2*a*A*c*x + a*B*c*x^2 + (A*c^2*x^3)/3 + (B*c^2*x^4)/4 + a^2*B*Log[x]

Rule 780

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^2}{x^2} \, dx &=\int \left (2 a A c+\frac {a^2 A}{x^2}+\frac {a^2 B}{x}+2 a B c x+A c^2 x^2+B c^2 x^3\right ) \, dx\\ &=-\frac {a^2 A}{x}+2 a A c x+a B c x^2+\frac {1}{3} A c^2 x^3+\frac {1}{4} B c^2 x^4+a^2 B \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 52, normalized size = 1.00 \begin {gather*} -\frac {a^2 A}{x}+2 a A c x+a B c x^2+\frac {1}{3} A c^2 x^3+\frac {1}{4} B c^2 x^4+a^2 B \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*A)/x) + 2*a*A*c*x + a*B*c*x^2 + (A*c^2*x^3)/3 + (B*c^2*x^4)/4 + a^2*B*Log[x]

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Maple [A]
time = 0.59, size = 49, normalized size = 0.94

method	result	size
default	\(-\frac {a^{2} A}{x}+2 a A c x +a B c \,x^{2}+\frac {A \,c^{2} x^{3}}{3}+\frac {B \,c^{2} x^{4}}{4}+a^{2} B \ln \left (x \right )\)	\(49\)
risch	\(-\frac {a^{2} A}{x}+2 a A c x +a B c \,x^{2}+\frac {A \,c^{2} x^{3}}{3}+\frac {B \,c^{2} x^{4}}{4}+a^{2} B \ln \left (x \right )\)	\(49\)
norman	\(\frac {a B c \,x^{3}-a^{2} A +\frac {1}{3} A \,c^{2} x^{4}+\frac {1}{4} B \,c^{2} x^{5}+2 a A c \,x^{2}}{x}+a^{2} B \ln \left (x \right )\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*A/x+2*a*A*c*x+a*B*c*x^2+1/3*A*c^2*x^3+1/4*B*c^2*x^4+a^2*B*ln(x)

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Maxima [A]
time = 0.26, size = 48, normalized size = 0.92 \begin {gather*} \frac {1}{4} \, B c^{2} x^{4} + \frac {1}{3} \, A c^{2} x^{3} + B a c x^{2} + 2 \, A a c x + B a^{2} \log \left (x\right ) - \frac {A a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*c^2*x^4 + 1/3*A*c^2*x^3 + B*a*c*x^2 + 2*A*a*c*x + B*a^2*log(x) - A*a^2/x

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Fricas [A]
time = 3.12, size = 55, normalized size = 1.06 \begin {gather*} \frac {3 \, B c^{2} x^{5} + 4 \, A c^{2} x^{4} + 12 \, B a c x^{3} + 24 \, A a c x^{2} + 12 \, B a^{2} x \log \left (x\right ) - 12 \, A a^{2}}{12 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*c^2*x^5 + 4*A*c^2*x^4 + 12*B*a*c*x^3 + 24*A*a*c*x^2 + 12*B*a^2*x*log(x) - 12*A*a^2)/x

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Sympy [A]
time = 0.05, size = 51, normalized size = 0.98 \begin {gather*} - \frac {A a^{2}}{x} + 2 A a c x + \frac {A c^{2} x^{3}}{3} + B a^{2} \log {\left (x \right )} + B a c x^{2} + \frac {B c^{2} x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a**2/x + 2*A*a*c*x + A*c**2*x**3/3 + B*a**2*log(x) + B*a*c*x**2 + B*c**2*x**4/4

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Giac [A]
time = 0.75, size = 49, normalized size = 0.94 \begin {gather*} \frac {1}{4} \, B c^{2} x^{4} + \frac {1}{3} \, A c^{2} x^{3} + B a c x^{2} + 2 \, A a c x + B a^{2} \log \left ({\left | x \right |}\right ) - \frac {A a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*c^2*x^4 + 1/3*A*c^2*x^3 + B*a*c*x^2 + 2*A*a*c*x + B*a^2*log(abs(x)) - A*a^2/x

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Mupad [B]
time = 0.03, size = 48, normalized size = 0.92 \begin {gather*} \frac {A\,c^2\,x^3}{3}-\frac {A\,a^2}{x}+\frac {B\,c^2\,x^4}{4}+B\,a^2\,\ln \left (x\right )+B\,a\,c\,x^2+2\,A\,a\,c\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^2*(A + B*x))/x^2,x)

[Out]

(A*c^2*x^3)/3 - (A*a^2)/x + (B*c^2*x^4)/4 + B*a^2*log(x) + B*a*c*x^2 + 2*A*a*c*x

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