\(\int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx\) [2030]

Optimal result

Integrand size = 24, antiderivative size = 73 \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {\sqrt {1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac {10}{189} \sqrt {1-2 x} (214+95 x)-\frac {208 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{189 \sqrt {21}} \]

[Out]

-208/3969*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+1/21*(3+5*x)^2*(1-2*x)^(1/2)/(2+3*x)-10/189*(214+95*x)*
(1-2*x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {100, 152, 65, 212} \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=-\frac {208 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{189 \sqrt {21}}+\frac {\sqrt {1-2 x} (5 x+3)^2}{21 (3 x+2)}-\frac {10}{189} \sqrt {1-2 x} (95 x+214) \]

[In]

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

[Out]

(Sqrt[1 - 2*x]*(3 + 5*x)^2)/(21*(2 + 3*x)) - (10*Sqrt[1 - 2*x]*(214 + 95*x))/189 - (208*ArcTanh[Sqrt[3/7]*Sqrt
[1 - 2*x]])/(189*Sqrt[21])

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 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(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*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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*} \text {integral}& = \frac {\sqrt {1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac {1}{21} \int \frac {(-92-190 x) (3+5 x)}{\sqrt {1-2 x} (2+3 x)} \, dx \\ & = \frac {\sqrt {1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac {10}{189} \sqrt {1-2 x} (214+95 x)+\frac {104}{189} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx \\ & = \frac {\sqrt {1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac {10}{189} \sqrt {1-2 x} (214+95 x)-\frac {104}{189} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = \frac {\sqrt {1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac {10}{189} \sqrt {1-2 x} (214+95 x)-\frac {208 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{189 \sqrt {21}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {-\frac {21 \sqrt {1-2 x} \left (4199+8050 x+2625 x^2\right )}{2+3 x}-208 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{3969} \]

[In]

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

[Out]

((-21*Sqrt[1 - 2*x]*(4199 + 8050*x + 2625*x^2))/(2 + 3*x) - 208*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/396
9

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {5250 x^{3}+13475 x^{2}+348 x -4199}{189 \left (2+3 x \right ) \sqrt {1-2 x}}-\frac {208 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3969}\)	\(51\)
pseudoelliptic	\(\frac {-208 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \sqrt {21}-21 \sqrt {1-2 x}\, \left (2625 x^{2}+8050 x +4199\right )}{7938+11907 x}\)	\(52\)
derivativedivides	\(\frac {125 \left (1-2 x \right )^{\frac {3}{2}}}{54}-\frac {725 \sqrt {1-2 x}}{54}-\frac {2 \sqrt {1-2 x}}{567 \left (-\frac {4}{3}-2 x \right )}-\frac {208 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3969}\)	\(54\)
default	\(\frac {125 \left (1-2 x \right )^{\frac {3}{2}}}{54}-\frac {725 \sqrt {1-2 x}}{54}-\frac {2 \sqrt {1-2 x}}{567 \left (-\frac {4}{3}-2 x \right )}-\frac {208 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3969}\)	\(54\)
trager	\(-\frac {\left (2625 x^{2}+8050 x +4199\right ) \sqrt {1-2 x}}{189 \left (2+3 x \right )}-\frac {104 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{3969}\)	\(72\)

[In]

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

[Out]

1/189*(5250*x^3+13475*x^2+348*x-4199)/(2+3*x)/(1-2*x)^(1/2)-208/3969*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1
/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {104 \, \sqrt {21} {\left (3 \, x + 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (2625 \, x^{2} + 8050 \, x + 4199\right )} \sqrt {-2 \, x + 1}}{3969 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

1/3969*(104*sqrt(21)*(3*x + 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(2625*x^2 + 8050*x + 41
99)*sqrt(-2*x + 1))/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 35.91 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.53 \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {125 \left (1 - 2 x\right )^{\frac {3}{2}}}{54} - \frac {725 \sqrt {1 - 2 x}}{54} + \frac {5 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{189} + \frac {4 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{27} \]

[In]

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

[Out]

125*(1 - 2*x)**(3/2)/54 - 725*sqrt(1 - 2*x)/54 + 5*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*
x) + sqrt(21)/3))/189 + 4*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/4 + log(sqrt(21)*sqrt(1 - 2*
x)/7 + 1)/4 - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (sqrt(1 - 2*x)
> -sqrt(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3)))/27

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {125}{54} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {104}{3969} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {725}{54} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{189 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

125/54*(-2*x + 1)^(3/2) + 104/3969*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1)))
- 725/54*sqrt(-2*x + 1) + 1/189*sqrt(-2*x + 1)/(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {125}{54} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {104}{3969} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {725}{54} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{189 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

125/54*(-2*x + 1)^(3/2) + 104/3969*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*
x + 1))) - 725/54*sqrt(-2*x + 1) + 1/189*sqrt(-2*x + 1)/(3*x + 2)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \frac {(3+5 x)^3}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {2\,\sqrt {1-2\,x}}{567\,\left (2\,x+\frac {4}{3}\right )}-\frac {725\,\sqrt {1-2\,x}}{54}+\frac {125\,{\left (1-2\,x\right )}^{3/2}}{54}+\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,208{}\mathrm {i}}{3969} \]

[In]

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

[Out]

(21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*208i)/3969 + (2*(1 - 2*x)^(1/2))/(567*(2*x + 4/3)) - (725*(1 -
2*x)^(1/2))/54 + (125*(1 - 2*x)^(3/2))/54