3.22.17 \(\int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\) [2117]

Optimal. Leaf size=100 \[ -\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}-\frac {54 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

[Out]

-54/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+7/11*(2+3*x)^3/(3+5*x)/(1-2*x)^(1/2)-36/605*(2+3*x)^2
*(1-2*x)^(1/2)/(3+5*x)+27/3025*(792+265*x)*(1-2*x)^(1/2)

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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, 154, 152, 65, 212} \begin {gather*} \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {36 \sqrt {1-2 x} (3 x+2)^2}{605 (5 x+3)}+\frac {27 \sqrt {1-2 x} (265 x+792)}{3025}-\frac {54 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^4/((1 - 2*x)^(3/2)*(3 + 5*x)^2),x]

[Out]

(-36*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(605*(3 + 5*x)) + (7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)) + (27*Sqrt[1 -
2*x]*(792 + 265*x))/3025 - (54*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*Sqrt[55])

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 154

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] && ILtQ[m, -1] && GtQ[n, 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*} \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx &=\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}-\frac {1}{11} \int \frac {(2+3 x)^2 (117+207 x)}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}-\frac {1}{605} \int \frac {(2+3 x) (4266+7155 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}+\frac {27 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{3025}\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}-\frac {27 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{3025}\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}-\frac {54 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 63, normalized size = 0.63 \begin {gather*} \frac {-\frac {55 \left (-78832-68661 x+114345 x^2+16335 x^3\right )}{\sqrt {1-2 x} (3+5 x)}-54 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{166375} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^4/((1 - 2*x)^(3/2)*(3 + 5*x)^2),x]

[Out]

((-55*(-78832 - 68661*x + 114345*x^2 + 16335*x^3))/(Sqrt[1 - 2*x]*(3 + 5*x)) - 54*Sqrt[55]*ArcTanh[Sqrt[5/11]*
Sqrt[1 - 2*x]])/166375

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Maple [A]
time = 0.12, size = 63, normalized size = 0.63

method	result	size
risch	\(-\frac {16335 x^{3}+114345 x^{2}-68661 x -78832}{3025 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {54 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}\)	\(51\)
derivativedivides	\(-\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{100}+\frac {999 \sqrt {1-2 x}}{250}+\frac {2 \sqrt {1-2 x}}{75625 \left (-\frac {6}{5}-2 x \right )}-\frac {54 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}+\frac {2401}{484 \sqrt {1-2 x}}\)	\(63\)
default	\(-\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{100}+\frac {999 \sqrt {1-2 x}}{250}+\frac {2 \sqrt {1-2 x}}{75625 \left (-\frac {6}{5}-2 x \right )}-\frac {54 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}+\frac {2401}{484 \sqrt {1-2 x}}\)	\(63\)
trager	\(\frac {\left (16335 x^{3}+114345 x^{2}-68661 x -78832\right ) \sqrt {1-2 x}}{30250 x^{2}+3025 x -9075}+\frac {27 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{166375}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^2,x,method=_RETURNVERBOSE)

[Out]

-27/100*(1-2*x)^(3/2)+999/250*(1-2*x)^(1/2)+2/75625*(1-2*x)^(1/2)/(-6/5-2*x)-54/166375*arctanh(1/11*55^(1/2)*(
1-2*x)^(1/2))*55^(1/2)+2401/484/(1-2*x)^(1/2)

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Maxima [A]
time = 0.50, size = 83, normalized size = 0.83 \begin {gather*} -\frac {27}{100} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {27}{166375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {999}{250} \, \sqrt {-2 \, x + 1} - \frac {1500633 \, x + 900371}{30250 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^2,x, algorithm="maxima")

[Out]

-27/100*(-2*x + 1)^(3/2) + 27/166375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))
) + 999/250*sqrt(-2*x + 1) - 1/30250*(1500633*x + 900371)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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Fricas [A]
time = 1.47, size = 75, normalized size = 0.75 \begin {gather*} \frac {27 \, \sqrt {55} {\left (10 \, x^{2} + x - 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (16335 \, x^{3} + 114345 \, x^{2} - 68661 \, x - 78832\right )} \sqrt {-2 \, x + 1}}{166375 \, {\left (10 \, x^{2} + x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^2,x, algorithm="fricas")

[Out]

1/166375*(27*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(16335*x^3 + 11
4345*x^2 - 68661*x - 78832)*sqrt(-2*x + 1))/(10*x^2 + x - 3)

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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)**4/(1-2*x)**(3/2)/(3+5*x)**2,x)

[Out]

Timed out

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Giac [A]
time = 1.92, size = 86, normalized size = 0.86 \begin {gather*} -\frac {27}{100} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {27}{166375} \, \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 {999}{250} \, \sqrt {-2 \, x + 1} - \frac {1500633 \, x + 900371}{30250 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^2,x, algorithm="giac")

[Out]

-27/100*(-2*x + 1)^(3/2) + 27/166375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(
-2*x + 1))) + 999/250*sqrt(-2*x + 1) - 1/30250*(1500633*x + 900371)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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Mupad [B]
time = 0.07, size = 66, normalized size = 0.66 \begin {gather*} \frac {\frac {1500633\,x}{151250}+\frac {900371}{151250}}{\frac {11\,\sqrt {1-2\,x}}{5}-{\left (1-2\,x\right )}^{3/2}}+\frac {999\,\sqrt {1-2\,x}}{250}-\frac {27\,{\left (1-2\,x\right )}^{3/2}}{100}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,54{}\mathrm {i}}{166375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 2)^4/((1 - 2*x)^(3/2)*(5*x + 3)^2),x)

[Out]

((1500633*x)/151250 + 900371/151250)/((11*(1 - 2*x)^(1/2))/5 - (1 - 2*x)^(3/2)) + (55^(1/2)*atan((55^(1/2)*(1
- 2*x)^(1/2)*1i)/11)*54i)/166375 + (999*(1 - 2*x)^(1/2))/250 - (27*(1 - 2*x)^(3/2))/100

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