3.26.40 \(\int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx\) [2540]

Optimal. Leaf size=154 \[ \frac {321709971 \sqrt {1-2 x} \sqrt {3+5 x}}{204800}+\frac {9748787 \sqrt {1-2 x} (3+5 x)^{3/2}}{51200}+\frac {33}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}+\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} (3+5 x)^{5/2} (27937+13820 x)}{6400}-\frac {3538809681 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{204800 \sqrt {10}} \]

[Out]

-3538809681/2048000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+(2+3*x)^3*(3+5*x)^(5/2)/(1-2*x)^(1/2)+9748787
/51200*(3+5*x)^(3/2)*(1-2*x)^(1/2)+33/20*(2+3*x)^2*(3+5*x)^(5/2)*(1-2*x)^(1/2)+9/6400*(3+5*x)^(5/2)*(27937+138
20*x)*(1-2*x)^(1/2)[phone]/204800*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {99, 158, 152, 52, 56, 222} \begin {gather*} -\frac {3538809681 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{204800 \sqrt {10}}+\frac {(5 x+3)^{5/2} (3 x+2)^3}{\sqrt {1-2 x}}+\frac {33}{20} \sqrt {1-2 x} (5 x+3)^{5/2} (3 x+2)^2+\frac {9748787 \sqrt {1-2 x} (5 x+3)^{3/2}}{51200}+\frac {9 \sqrt {1-2 x} (5 x+3)^{5/2} (13820 x+27937)}{6400}+\frac {321709971 \sqrt {1-2 x} \sqrt {5 x+3}}{204800} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(321709971*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/204800 + (9748787*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/51200 + (33*Sqrt[1 -
2*x]*(2 + 3*x)^2*(3 + 5*x)^(5/2))/20 + ((2 + 3*x)^3*(3 + 5*x)^(5/2))/Sqrt[1 - 2*x] + (9*Sqrt[1 - 2*x]*(3 + 5*x
)^(5/2)*(27937 + 13820*x))/6400 - (3538809681*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(204800*Sqrt[10])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (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 158

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

Rule 222

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

Rubi steps

\begin {align*} \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx &=\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}-\int \frac {(2+3 x)^2 (3+5 x)^{3/2} \left (52+\frac {165 x}{2}\right )}{\sqrt {1-2 x}} \, dx\\ &=\frac {33}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}+\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}+\frac {1}{50} \int \frac {\left (-\frac {16505}{2}-\frac {51825 x}{4}\right ) (2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx\\ &=\frac {33}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}+\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} (3+5 x)^{5/2} (27937+13820 x)}{6400}-\frac {9748787 \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx}{12800}\\ &=\frac {9748787 \sqrt {1-2 x} (3+5 x)^{3/2}}{51200}+\frac {33}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}+\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} (3+5 x)^{5/2} (27937+13820 x)}{6400}-\frac {321709971 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx}{102400}\\ &=\frac {321709971 \sqrt {1-2 x} \sqrt {3+5 x}}{204800}+\frac {9748787 \sqrt {1-2 x} (3+5 x)^{3/2}}{51200}+\frac {33}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}+\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} (3+5 x)^{5/2} (27937+13820 x)}{6400}-\frac {3538809681 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{409600}\\ &=\frac {321709971 \sqrt {1-2 x} \sqrt {3+5 x}}{204800}+\frac {9748787 \sqrt {1-2 x} (3+5 x)^{3/2}}{51200}+\frac {33}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}+\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} (3+5 x)^{5/2} (27937+13820 x)}{6400}-\frac {3538809681 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{204800 \sqrt {5}}\\ &=\frac {321709971 \sqrt {1-2 x} \sqrt {3+5 x}}{204800}+\frac {9748787 \sqrt {1-2 x} (3+5 x)^{3/2}}{51200}+\frac {33}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}+\frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} (3+5 x)^{5/2} (27937+13820 x)}{6400}-\frac {3538809681 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{204800 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 83, normalized size = 0.54 \begin {gather*} \frac {-10 \sqrt {3+5 x} \left (-538018839+381820658 x+233394520 x^2+148751040 x^3+65836800 x^4+13824000 x^5\right )[phone] \sqrt {10-20 x} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{2048000 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[3 + 5*x]*(-538018839 + 381820658*x + 233394520*x^2 + 148751040*x^3 + 65836800*x^4 + 13824000*x^5) +
3538809681*Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(2048000*Sqrt[1 - 2*x])

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Maple [A]
time = 0.08, size = 157, normalized size = 1.02

method	result	size
default	\(-\frac {\left (-276480000 x^{5} \sqrt {-10 x^{2}-x +3}-1316736000 x^{4} \sqrt {-10 x^{2}-x +3}-2975020800 x^{3} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -4667890400 x^{2} \sqrt {-10 x^{2}-x +3}-3538809681 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-7636413160 x \sqrt {-10 x^{2}-x +3}[phone] \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{4096000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4096000*(-276480000*x^5*(-10*x^2-x+3)^(1/2)-1316736000*x^4*(-10*x^2-x+3)^(1/2)-2975020800*x^3*(-10*x^2-x+3)
^(1/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)*x-4667890400*x^2*(-10*x^2-x+3)^(1/2)-3538809681*10^(1/2)*arcsi
n(20/11*x+1/11)-7636413160*x*(-10*x^2-x+3)^(1/2)[phone]*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/
(-1+2*x)/(-10*x^2-x+3)^(1/2)

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Maxima [A]
time = 0.60, size = 126, normalized size = 0.82 \begin {gather*} -\frac {675 \, x^{6}}{2 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {57915 \, x^{5}}{32 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {588291 \, x^{4}}{128 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {40330643 \, x^{3}}{5120 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {52185737 \, x^{2}}{4096 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {3538809681}{4096000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {1544632221 \, x}{204800 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1614056517}{204800 \, \sqrt {-10 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-675/2*x^6/sqrt(-10*x^2 - x + 3) - 57915/32*x^5/sqrt(-10*x^2 - x + 3) - 588291/128*x^4/sqrt(-10*x^2 - x + 3) -
40330643/5120*x^3/sqrt(-10*x^2 - x + 3) - 52185737/4096*x^2/sqrt(-10*x^2 - x + 3) + 3538809681/4096000*sqrt(1
0)*arcsin(-20/11*x - 1/11) + 1544632221/204800*x/sqrt(-10*x^2 - x + 3) + 1614056517/204800/sqrt(-10*x^2 - x +
3)

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Fricas [A]
time = 1.22, size = 96, normalized size = 0.62 \begin {gather*} \frac {3538809681 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (13824000 \, x^{5} + 65836800 \, x^{4} + 148751040 \, x^{3} + 233394520 \, x^{2} + 381820658 \, x - 538018839\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4096000 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4096000*(3538809681*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) + 20*(13824000*x^5 + 65836800*x^4 + 148751040*x^3 + 233394520*x^2 + 381820658*x - 538018839)*sqrt(5*
x + 3)*sqrt(-2*x + 1))/(2*x - 1)

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

[Out]

Timed out

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Giac [A]
time = 1.67, size = 110, normalized size = 0.71 \begin {gather*} -\frac {3538809681}{2048000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, {\left (4 \, {\left (24 \, {\left (36 \, {\left (16 \, \sqrt {5} {\left (5 \, x + 3\right )} + 141 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 42197 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 9748787 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 536183285 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 17694048405 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{25600000 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3538809681/2048000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/25600000*(2*(4*(24*(36*(16*sqrt(5)*(5*x +
3) + 141*sqrt(5))*(5*x + 3) + 42197*sqrt(5))*(5*x + 3) + 9748787*sqrt(5))*(5*x + 3) + 536183285*sqrt(5))*(5*x
+ 3) - 17694048405*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (3\,x+2\right )}^3\,{\left (5\,x+3\right )}^{5/2}}{{\left (1-2\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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