3.2.19 \(\int \frac {7+8 x^3}{(1+x) (1+2 x)^3} \, dx\) [119]

Optimal. Leaf size=23 \[ -\frac {3}{(1+2 x)^2}+\frac {3}{1+2 x}+\log (1+x) \]

[Out]

-3/(1+2*x)^2+3/(1+2*x)+ln(1+x)

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Rubi [A]
time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1634} \begin {gather*} \frac {3}{2 x+1}-\frac {3}{(2 x+1)^2}+\log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(7 + 8*x^3)/((1 + x)*(1 + 2*x)^3),x]

[Out]

-3/(1 + 2*x)^2 + 3/(1 + 2*x) + Log[1 + x]

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {align*} \int \frac {7+8 x^3}{(1+x) (1+2 x)^3} \, dx &=\int \left (\frac {1}{1+x}+\frac {12}{(1+2 x)^3}-\frac {6}{(1+2 x)^2}\right ) \, dx\\ &=-\frac {3}{(1+2 x)^2}+\frac {3}{1+2 x}+\log (1+x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 24, normalized size = 1.04 \begin {gather*} \frac {6 x+(1+2 x)^2 \log (1+x)}{(1+2 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(7 + 8*x^3)/((1 + x)*(1 + 2*x)^3),x]

[Out]

(6*x + (1 + 2*x)^2*Log[1 + x])/(1 + 2*x)^2

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Maple [A]
time = 0.08, size = 24, normalized size = 1.04

method	result	size
norman	\(\frac {6 x}{\left (1+2 x \right )^{2}}+\ln \left (1+x \right )\)	\(16\)
risch	\(\frac {6 x}{\left (1+2 x \right )^{2}}+\ln \left (1+x \right )\)	\(16\)
default	\(-\frac {3}{\left (1+2 x \right )^{2}}+\frac {3}{1+2 x}+\ln \left (1+x \right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3+7)/(1+x)/(1+2*x)^3,x,method=_RETURNVERBOSE)

[Out]

-3/(1+2*x)^2+3/(1+2*x)+ln(1+x)

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Maxima [A]
time = 0.90, size = 20, normalized size = 0.87 \begin {gather*} \frac {6 \, x}{4 \, x^{2} + 4 \, x + 1} + \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3+7)/(1+x)/(1+2*x)^3,x, algorithm="maxima")

[Out]

6*x/(4*x^2 + 4*x + 1) + log(x + 1)

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Fricas [A]
time = 0.76, size = 32, normalized size = 1.39 \begin {gather*} \frac {{\left (4 \, x^{2} + 4 \, x + 1\right )} \log \left (x + 1\right ) + 6 \, x}{4 \, x^{2} + 4 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3+7)/(1+x)/(1+2*x)^3,x, algorithm="fricas")

[Out]

((4*x^2 + 4*x + 1)*log(x + 1) + 6*x)/(4*x^2 + 4*x + 1)

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Sympy [A]
time = 0.03, size = 17, normalized size = 0.74 \begin {gather*} \frac {6 x}{4 x^{2} + 4 x + 1} + \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**3+7)/(1+x)/(1+2*x)**3,x)

[Out]

6*x/(4*x**2 + 4*x + 1) + log(x + 1)

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Giac [A]
time = 0.43, size = 16, normalized size = 0.70 \begin {gather*} \frac {6 \, x}{{\left (2 \, x + 1\right )}^{2}} + \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^3+7)/(1+x)/(1+2*x)^3,x, algorithm="giac")

[Out]

6*x/(2*x + 1)^2 + log(abs(x + 1))

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Mupad [B]
time = 0.09, size = 15, normalized size = 0.65 \begin {gather*} \ln \left (x+1\right )+\frac {6\,x}{{\left (2\,x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3 + 7)/((2*x + 1)^3*(x + 1)),x)

[Out]

log(x + 1) + (6*x)/(2*x + 1)^2

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