\(\int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx\) [1189]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=\frac {7 (3+5 x)^4}{15 (2+3 x)^5}+\frac {5 (3+5 x)^4}{12 (2+3 x)^4} \]

[Out]

7/15*(3+5*x)^4/(2+3*x)^5+5/12*(3+5*x)^4/(2+3*x)^4

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {79, 37} \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=\frac {5 (5 x+3)^4}{12 (3 x+2)^4}+\frac {7 (5 x+3)^4}{15 (3 x+2)^5} \]

[In]

Int[((1 - 2*x)*(3 + 5*x)^3)/(2 + 3*x)^6,x]

[Out]

(7*(3 + 5*x)^4)/(15*(2 + 3*x)^5) + (5*(3 + 5*x)^4)/(12*(2 + 3*x)^4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rubi steps \begin{align*} \text {integral}& = \frac {7 (3+5 x)^4}{15 (2+3 x)^5}+\frac {5}{3} \int \frac {(3+5 x)^3}{(2+3 x)^5} \, dx \\ & = \frac {7 (3+5 x)^4}{15 (2+3 x)^5}+\frac {5 (3+5 x)^4}{12 (2+3 x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=\frac {11758+153795 x+559800 x^2+803250 x^3+405000 x^4}{4860 (2+3 x)^5} \]

[In]

Integrate[((1 - 2*x)*(3 + 5*x)^3)/(2 + 3*x)^6,x]

[Out]

(11758 + 153795*x + 559800*x^2 + 803250*x^3 + 405000*x^4)/(4860*(2 + 3*x)^5)

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78

method	result	size
norman	\(\frac {\frac {250}{3} x^{4}+\frac {2975}{18} x^{3}+\frac {3110}{27} x^{2}+\frac {10253}{324} x +\frac {5879}{2430}}{\left (2+3 x \right )^{5}}\)	\(29\)
gosper	\(\frac {405000 x^{4}+803250 x^{3}+559800 x^{2}+153795 x +11758}{4860 \left (2+3 x \right )^{5}}\)	\(30\)
risch	\(\frac {\frac {250}{3} x^{4}+\frac {2975}{18} x^{3}+\frac {3110}{27} x^{2}+\frac {10253}{324} x +\frac {5879}{2430}}{\left (2+3 x \right )^{5}}\)	\(30\)
parallelrisch	\(\frac {-5879 x^{5}+7070 x^{4}+26760 x^{3}+19440 x^{2}+4320 x}{320 \left (2+3 x \right )^{5}}\)	\(34\)
default	\(\frac {250}{243 \left (2+3 x \right )}-\frac {1025}{486 \left (2+3 x \right )^{2}}-\frac {107}{972 \left (2+3 x \right )^{4}}+\frac {7}{1215 \left (2+3 x \right )^{5}}+\frac {185}{243 \left (2+3 x \right )^{3}}\)	\(47\)
meijerg	\(\frac {27 x \left (\frac {81}{16} x^{4}+\frac {135}{8} x^{3}+\frac {45}{2} x^{2}+15 x +5\right )}{320 \left (1+\frac {3 x}{2}\right )^{5}}+\frac {81 x^{2} \left (\frac {27}{8} x^{3}+\frac {45}{4} x^{2}+15 x +10\right )}{1280 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {3 x^{3} \left (\frac {9}{4} x^{2}+\frac {15}{2} x +10\right )}{128 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {65 x^{4} \left (\frac {3 x}{2}+5\right )}{256 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {25 x^{5}}{32 \left (1+\frac {3 x}{2}\right )^{5}}\)	\(110\)

[In]

int((1-2*x)*(3+5*x)^3/(2+3*x)^6,x,method=_RETURNVERBOSE)

[Out]

(250/3*x^4+2975/18*x^3+3110/27*x^2+10253/324*x+5879/2430)/(2+3*x)^5

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=\frac {405000 \, x^{4} + 803250 \, x^{3} + 559800 \, x^{2} + 153795 \, x + 11758}{4860 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

integrate((1-2*x)*(3+5*x)^3/(2+3*x)^6,x, algorithm="fricas")

[Out]

1/4860*(405000*x^4 + 803250*x^3 + 559800*x^2 + 153795*x + 11758)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240
*x + 32)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=- \frac {- 405000 x^{4} - 803250 x^{3} - 559800 x^{2} - 153795 x - 11758}{1180980 x^{5} + 3936600 x^{4} + 5248800 x^{3} + 3499200 x^{2} + 1166400 x + 155520} \]

[In]

integrate((1-2*x)*(3+5*x)**3/(2+3*x)**6,x)

[Out]

-(-405000*x**4 - 803250*x**3 - 559800*x**2 - 153795*x - 11758)/(1180980*x**5 + 3936600*x**4 + 5248800*x**3 + 3
499200*x**2 + 1166400*x + 155520)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=\frac {405000 \, x^{4} + 803250 \, x^{3} + 559800 \, x^{2} + 153795 \, x + 11758}{4860 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

integrate((1-2*x)*(3+5*x)^3/(2+3*x)^6,x, algorithm="maxima")

[Out]

1/4860*(405000*x^4 + 803250*x^3 + 559800*x^2 + 153795*x + 11758)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240
*x + 32)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=\frac {405000 \, x^{4} + 803250 \, x^{3} + 559800 \, x^{2} + 153795 \, x + 11758}{4860 \, {\left (3 \, x + 2\right )}^{5}} \]

[In]

integrate((1-2*x)*(3+5*x)^3/(2+3*x)^6,x, algorithm="giac")

[Out]

1/4860*(405000*x^4 + 803250*x^3 + 559800*x^2 + 153795*x + 11758)/(3*x + 2)^5

Mupad [B] (verification not implemented)

Time = 1.40 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^6} \, dx=\frac {250}{243\,\left (3\,x+2\right )}-\frac {1025}{486\,{\left (3\,x+2\right )}^2}+\frac {185}{243\,{\left (3\,x+2\right )}^3}-\frac {107}{972\,{\left (3\,x+2\right )}^4}+\frac {7}{1215\,{\left (3\,x+2\right )}^5} \]

[In]

int(-((2*x - 1)*(5*x + 3)^3)/(3*x + 2)^6,x)

[Out]

250/(243*(3*x + 2)) - 1025/(486*(3*x + 2)^2) + 185/(243*(3*x + 2)^3) - 107/(972*(3*x + 2)^4) + 7/(1215*(3*x +
2)^5)