3.22.93 \(\int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx\) [2193]

Optimal. Leaf size=100 \[ \frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {7559 \sqrt {1-2 x}}{146410 (3+5 x)}+\frac {10217+17296 x}{39930 \sqrt {1-2 x} (3+5 x)^2}-\frac {7559 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{73205 \sqrt {55}} \]

[Out]

7/33*(2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)^2-7559/4026275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+1/39930*(102
17+17296*x)/(3+5*x)^2/(1-2*x)^(1/2)-7559/146410*(1-2*x)^(1/2)/(3+5*x)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {100, 149, 44, 65, 212} \begin {gather*} \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)^2}+\frac {17296 x+10217}{39930 \sqrt {1-2 x} (5 x+3)^2}-\frac {7559 \sqrt {1-2 x}}{146410 (5 x+3)}-\frac {7559 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{73205 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)^2) - (7559*Sqrt[1 - 2*x])/(146410*(3 + 5*x)) + (10217 + 17296*x)
/(39930*Sqrt[1 - 2*x]*(3 + 5*x)^2) - (7559*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(73205*Sqrt[55])

Rule 44

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] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 100

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

Rule 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :>
Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(
m + n + 2)) + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m +
1) + d^2*e*g*(m + n + 2)))*x)/(b*d*(b*c - a*d)^2*(m + 1)*(n + 1)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] -
Dist[(a^2*d^2*f*h*(2 + 3*n + n^2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h
*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b
*c - a*d)^2*(m + 1)*(n + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, h
}, x] && LtQ[m, -1] && LtQ[n, -1]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {1}{33} \int \frac {(-20-9 x) (2+3 x)}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)^2}+\frac {10217+17296 x}{39930 \sqrt {1-2 x} (3+5 x)^2}+\frac {7559 \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx}{13310}\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {7559 \sqrt {1-2 x}}{146410 (3+5 x)}+\frac {10217+17296 x}{39930 \sqrt {1-2 x} (3+5 x)^2}+\frac {7559 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{146410}\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {7559 \sqrt {1-2 x}}{146410 (3+5 x)}+\frac {10217+17296 x}{39930 \sqrt {1-2 x} (3+5 x)^2}-\frac {7559 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{146410}\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {7559 \sqrt {1-2 x}}{146410 (3+5 x)}+\frac {10217+17296 x}{39930 \sqrt {1-2 x} (3+5 x)^2}-\frac {7559 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{73205 \sqrt {55}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.17, size = 63, normalized size = 0.63 \begin {gather*} \frac {-\frac {55 \left (-417036-1242261 x-639434 x^2+453540 x^3\right )}{(1-2 x)^{3/2} (3+5 x)^2}-45354 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{24157650} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*(-417036 - 1242261*x - 639434*x^2 + 453540*x^3))/((1 - 2*x)^(3/2)*(3 + 5*x)^2) - 45354*Sqrt[55]*ArcTanh[
Sqrt[5/11]*Sqrt[1 - 2*x]])/24157650

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 66, normalized size = 0.66

method	result	size
risch	\(\frac {453540 x^{3}-639434 x^{2}-1242261 x -417036}{439230 \left (3+5 x \right )^{2} \sqrt {1-2 x}\, \left (-1+2 x \right )}-\frac {7559 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{4026275}\)	\(58\)
derivativedivides	\(\frac {\frac {19 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {211 \sqrt {1-2 x}}{6655}}{\left (-6-10 x \right )^{2}}-\frac {7559 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{4026275}+\frac {343}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {294}{14641 \sqrt {1-2 x}}\)	\(66\)
default	\(\frac {\frac {19 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {211 \sqrt {1-2 x}}{6655}}{\left (-6-10 x \right )^{2}}-\frac {7559 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{4026275}+\frac {343}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {294}{14641 \sqrt {1-2 x}}\)	\(66\)
trager	\(-\frac {\left (453540 x^{3}-639434 x^{2}-1242261 x -417036\right ) \sqrt {1-2 x}}{439230 \left (10 x^{2}+x -3\right )^{2}}+\frac {7559 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{8052550}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/14641*(209/50*(1-2*x)^(3/2)-2321/250*(1-2*x)^(1/2))/(-6-10*x)^2-7559/4026275*arctanh(1/11*55^(1/2)*(1-2*x)^
(1/2))*55^(1/2)+343/3993/(1-2*x)^(3/2)+294/14641/(1-2*x)^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 92, normalized size = 0.92 \begin {gather*} \frac {7559}{8052550} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {113385 \, {\left (2 \, x - 1\right )}^{3} + 20438 \, {\left (2 \, x - 1\right )}^{2} - 3083080 \, x - 741125}{219615 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 121 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7559/8052550*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/219615*(113385*(2*
x - 1)^3 + 20438*(2*x - 1)^2 - 3083080*x - 741125)/(25*(-2*x + 1)^(7/2) - 110*(-2*x + 1)^(5/2) + 121*(-2*x + 1
)^(3/2))

________________________________________________________________________________________

Fricas [A]
time = 0.90, size = 99, normalized size = 0.99 \begin {gather*} \frac {22677 \, \sqrt {55} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (453540 \, x^{3} - 639434 \, x^{2} - 1242261 \, x - 417036\right )} \sqrt {-2 \, x + 1}}{24157650 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24157650*(22677*sqrt(55)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x
+ 3)) - 55*(453540*x^3 - 639434*x^2 - 1242261*x - 417036)*sqrt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9
)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 0.76, size = 89, normalized size = 0.89 \begin {gather*} \frac {7559}{8052550} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {49 \, {\left (36 \, x - 95\right )}}{43923 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {95 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 211 \, \sqrt {-2 \, x + 1}}{26620 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7559/8052550*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 49/43923*(
36*x - 95)/((2*x - 1)*sqrt(-2*x + 1)) + 1/26620*(95*(-2*x + 1)^(3/2) - 211*sqrt(-2*x + 1))/(5*x + 3)^2

________________________________________________________________________________________

Mupad [B]
time = 1.22, size = 71, normalized size = 0.71 \begin {gather*} \frac {\frac {5096\,x}{9075}-\frac {1858\,{\left (2\,x-1\right )}^2}{499125}-\frac {7559\,{\left (2\,x-1\right )}^3}{366025}+\frac {49}{363}}{\frac {121\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {22\,{\left (1-2\,x\right )}^{5/2}}{5}+{\left (1-2\,x\right )}^{7/2}}-\frac {7559\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{4026275} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5096*x)/9075 - (1858*(2*x - 1)^2)/499125 - (7559*(2*x - 1)^3)/366025 + 49/363)/((121*(1 - 2*x)^(3/2))/25 - (
22*(1 - 2*x)^(5/2))/5 + (1 - 2*x)^(7/2)) - (7559*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/4026275

________________________________________________________________________________________