3.11.49 \(\int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^5} \, dx\)

Optimal. Leaf size=55 \[ -\frac {1025}{243 (3 x+2)}+\frac {185}{162 (3 x+2)^2}-\frac {107}{729 (3 x+2)^3}+\frac {7}{972 (3 x+2)^4}-\frac {250}{243} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 55, 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 = {77} \begin {gather*} -\frac {1025}{243 (3 x+2)}+\frac {185}{162 (3 x+2)^2}-\frac {107}{729 (3 x+2)^3}+\frac {7}{972 (3 x+2)^4}-\frac {250}{243} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)*(3 + 5*x)^3)/(2 + 3*x)^5,x]

[Out]

7/(972*(2 + 3*x)^4) - 107/(729*(2 + 3*x)^3) + 185/(162*(2 + 3*x)^2) - 1025/(243*(2 + 3*x)) - (250*Log[2 + 3*x]
)/243

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^5} \, dx &=\int \left (-\frac {7}{81 (2+3 x)^5}+\frac {107}{81 (2+3 x)^4}-\frac {185}{27 (2+3 x)^3}+\frac {1025}{81 (2+3 x)^2}-\frac {250}{81 (2+3 x)}\right ) \, dx\\ &=\frac {7}{972 (2+3 x)^4}-\frac {107}{729 (2+3 x)^3}+\frac {185}{162 (2+3 x)^2}-\frac {1025}{243 (2+3 x)}-\frac {250}{243} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.75 \begin {gather*} -\frac {332100 x^3+634230 x^2+404124 x+3000 (3 x+2)^4 \log (3 x+2)+85915}{2916 (3 x+2)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)*(3 + 5*x)^3)/(2 + 3*x)^5,x]

[Out]

-1/2916*(85915 + 404124*x + 634230*x^2 + 332100*x^3 + 3000*(2 + 3*x)^4*Log[2 + 3*x])/(2 + 3*x)^4

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((1 - 2*x)*(3 + 5*x)^3)/(2 + 3*x)^5,x]

[Out]

IntegrateAlgebraic[((1 - 2*x)*(3 + 5*x)^3)/(2 + 3*x)^5, x]

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fricas [A]  time = 1.35, size = 67, normalized size = 1.22 \begin {gather*} -\frac {332100 \, x^{3} + 634230 \, x^{2} + 3000 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 404124 \, x + 85915}{2916 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^3/(2+3*x)^5,x, algorithm="fricas")

[Out]

-1/2916*(332100*x^3 + 634230*x^2 + 3000*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 404124*x + 859
15)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

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giac [A]  time = 1.18, size = 55, normalized size = 1.00 \begin {gather*} -\frac {1025}{243 \, {\left (3 \, x + 2\right )}} + \frac {185}{162 \, {\left (3 \, x + 2\right )}^{2}} - \frac {107}{729 \, {\left (3 \, x + 2\right )}^{3}} + \frac {7}{972 \, {\left (3 \, x + 2\right )}^{4}} + \frac {250}{243} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^3/(2+3*x)^5,x, algorithm="giac")

[Out]

-1025/243/(3*x + 2) + 185/162/(3*x + 2)^2 - 107/729/(3*x + 2)^3 + 7/972/(3*x + 2)^4 + 250/243*log(1/3*abs(3*x
+ 2)/(3*x + 2)^2)

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maple [A]  time = 0.01, size = 46, normalized size = 0.84 \begin {gather*} -\frac {250 \ln \left (3 x +2\right )}{243}+\frac {7}{972 \left (3 x +2\right )^{4}}-\frac {107}{729 \left (3 x +2\right )^{3}}+\frac {185}{162 \left (3 x +2\right )^{2}}-\frac {1025}{243 \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-2*x)*(5*x+3)^3/(3*x+2)^5,x)

[Out]

7/972/(3*x+2)^4-107/729/(3*x+2)^3+185/162/(3*x+2)^2-1025/243/(3*x+2)-250/243*ln(3*x+2)

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maxima [A]  time = 0.45, size = 48, normalized size = 0.87 \begin {gather*} -\frac {332100 \, x^{3} + 634230 \, x^{2} + 404124 \, x + 85915}{2916 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} - \frac {250}{243} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^3/(2+3*x)^5,x, algorithm="maxima")

[Out]

-1/2916*(332100*x^3 + 634230*x^2 + 404124*x + 85915)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) - 250/243*log(3*
x + 2)

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mupad [B]  time = 0.04, size = 31, normalized size = 0.56 \begin {gather*} -\frac {250\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {\frac {1025\,x^3}{9}+\frac {435\,x^2}{2}+\frac {33677\,x}{243}+\frac {85915}{2916}}{{\left (3\,x+2\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x - 1)*(5*x + 3)^3)/(3*x + 2)^5,x)

[Out]

- (250*log(x + 2/3))/243 - ((33677*x)/243 + (435*x^2)/2 + (1025*x^3)/9 + 85915/2916)/(3*x + 2)^4

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sympy [A]  time = 0.15, size = 46, normalized size = 0.84 \begin {gather*} - \frac {332100 x^{3} + 634230 x^{2} + 404124 x + 85915}{236196 x^{4} + 629856 x^{3} + 629856 x^{2} + 279936 x + 46656} - \frac {250 \log {\left (3 x + 2 \right )}}{243} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)**3/(2+3*x)**5,x)

[Out]

-(332100*x**3 + 634230*x**2 + 404124*x + 85915)/(236196*x**4 + 629856*x**3 + 629856*x**2 + 279936*x + 46656) -
 250*log(3*x + 2)/243

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