3.1997 \(\int \frac{(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^3} \, dx\)

Optimal. Leaf size=124 \[ \frac{7 (1-2 x)^{3/2}}{3 (3 x+2) (5 x+3)^2}+\frac{7103 \sqrt{1-2 x}}{30 (5 x+3)}-\frac{1133 \sqrt{1-2 x}}{30 (5 x+3)^2}+1400 \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{7209}{5} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(-1133*Sqrt[1 - 2*x])/(30*(3 + 5*x)^2) + (7*(1 - 2*x)^(3/2))/(3*(2 + 3*x)*(3 + 5*x)^2) + (7103*Sqrt[1 - 2*x])/
(30*(3 + 5*x)) + 1400*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - (7209*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1
- 2*x]])/5

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Rubi [A]  time = 0.0442867, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {98, 149, 151, 156, 63, 206} \[ \frac{7 (1-2 x)^{3/2}}{3 (3 x+2) (5 x+3)^2}+\frac{7103 \sqrt{1-2 x}}{30 (5 x+3)}-\frac{1133 \sqrt{1-2 x}}{30 (5 x+3)^2}+1400 \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{7209}{5} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1133*Sqrt[1 - 2*x])/(30*(3 + 5*x)^2) + (7*(1 - 2*x)^(3/2))/(3*(2 + 3*x)*(3 + 5*x)^2) + (7103*Sqrt[1 - 2*x])/
(30*(3 + 5*x)) + 1400*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - (7209*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1
- 2*x]])/5

Rule 98

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_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 151

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

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 206

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

Rubi steps

\begin{align*} \int \frac{(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^3} \, dx &=\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)^2}+\frac{1}{3} \int \frac{(166-101 x) \sqrt{1-2 x}}{(2+3 x) (3+5 x)^3} \, dx\\ &=-\frac{1133 \sqrt{1-2 x}}{30 (3+5 x)^2}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)^2}+\frac{1}{30} \int \frac{-9266+10601 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac{1133 \sqrt{1-2 x}}{30 (3+5 x)^2}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)^2}+\frac{7103 \sqrt{1-2 x}}{30 (3+5 x)}-\frac{1}{330} \int \frac{-382734+234399 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac{1133 \sqrt{1-2 x}}{30 (3+5 x)^2}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)^2}+\frac{7103 \sqrt{1-2 x}}{30 (3+5 x)}-4900 \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx+\frac{79299}{10} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{1133 \sqrt{1-2 x}}{30 (3+5 x)^2}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)^2}+\frac{7103 \sqrt{1-2 x}}{30 (3+5 x)}+4900 \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )-\frac{79299}{10} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{1133 \sqrt{1-2 x}}{30 (3+5 x)^2}+\frac{7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)^2}+\frac{7103 \sqrt{1-2 x}}{30 (3+5 x)}+1400 \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{7209}{5} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.114982, size = 93, normalized size = 0.75 \[ \frac{1}{50} \left (\frac{5 \sqrt{1-2 x} \left (35515 x^2+43806 x+13474\right )}{(3 x+2) (5 x+3)^2}-14418 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\right )+1400 \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

1400*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + ((5*Sqrt[1 - 2*x]*(13474 + 43806*x + 35515*x^2))/((2 + 3*x)*
(3 + 5*x)^2) - 14418*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/50

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Maple [A]  time = 0.012, size = 82, normalized size = 0.7 \begin{align*} -{\frac{98}{3}\sqrt{1-2\,x} \left ( -2\,x-{\frac{4}{3}} \right ) ^{-1}}+{\frac{1400\,\sqrt{21}}{3}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+550\,{\frac{1}{ \left ( -10\,x-6 \right ) ^{2}} \left ( -{\frac{141\, \left ( 1-2\,x \right ) ^{3/2}}{50}}+{\frac{1529\,\sqrt{1-2\,x}}{250}} \right ) }-{\frac{7209\,\sqrt{55}}{25}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-98/3*(1-2*x)^(1/2)/(-2*x-4/3)+1400/3*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+550*(-141/50*(1-2*x)^(3/2)+
1529/250*(1-2*x)^(1/2))/(-10*x-6)^2-7209/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 2.61532, size = 173, normalized size = 1.4 \begin{align*} \frac{7209}{50} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{700}{3} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) + \frac{35515 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - 158642 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + 177023 \, \sqrt{-2 \, x + 1}}{5 \,{\left (75 \,{\left (2 \, x - 1\right )}^{3} + 505 \,{\left (2 \, x - 1\right )}^{2} + 2266 \, x - 286\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7209/50*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 700/3*sqrt(21)*log(-(sqrt
(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/5*(35515*(-2*x + 1)^(5/2) - 158642*(-2*x + 1)^(3/2
) + 177023*sqrt(-2*x + 1))/(75*(2*x - 1)^3 + 505*(2*x - 1)^2 + 2266*x - 286)

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Fricas [A]  time = 1.43198, size = 428, normalized size = 3.45 \begin{align*} \frac{21627 \, \sqrt{11} \sqrt{5}{\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 35000 \, \sqrt{7} \sqrt{3}{\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (-\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 15 \,{\left (35515 \, x^{2} + 43806 \, x + 13474\right )} \sqrt{-2 \, x + 1}}{150 \,{\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/150*(21627*sqrt(11)*sqrt(5)*(75*x^3 + 140*x^2 + 87*x + 18)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(
5*x + 3)) + 35000*sqrt(7)*sqrt(3)*(75*x^3 + 140*x^2 + 87*x + 18)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x +
5)/(3*x + 2)) + 15*(35515*x^2 + 43806*x + 13474)*sqrt(-2*x + 1))/(75*x^3 + 140*x^2 + 87*x + 18)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 2.21184, size = 166, normalized size = 1.34 \begin{align*} \frac{7209}{50} \, \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{700}{3} \, \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{49 \, \sqrt{-2 \, x + 1}}{3 \, x + 2} - \frac{11 \,{\left (705 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 1529 \, \sqrt{-2 \, x + 1}\right )}}{20 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7209/50*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 700/3*sqrt(21)*
log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 49*sqrt(-2*x + 1)/(3*x + 2) - 11/
20*(705*(-2*x + 1)^(3/2) - 1529*sqrt(-2*x + 1))/(5*x + 3)^2