3.25.90 \(\int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac {203 (3 x+2)^2}{242 \sqrt {1-2 x} (5 x+3)^{3/2}}+\frac {\sqrt {1-2 x} (991010 x+627287)}{2196150 (5 x+3)^{3/2}}+\frac {81 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{50 \sqrt {10}} \]

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Rubi [A]  time = 0.03, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {98, 150, 145, 54, 216} \begin {gather*} \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac {203 (3 x+2)^2}{242 \sqrt {1-2 x} (5 x+3)^{3/2}}+\frac {\sqrt {1-2 x} (991010 x+627287)}{2196150 (5 x+3)^{3/2}}+\frac {81 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{50 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-203*(2 + 3*x)^2)/(242*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + (7*(2 + 3*x)^3)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))
+ (Sqrt[1 - 2*x]*(627287 + 991010*x))/(2196150*(3 + 5*x)^(3/2)) + (81*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(50*Sq
rt[10])

Rule 54

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 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 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

Rule 150

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 216

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)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx &=\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac {1}{33} \int \frac {(2+3 x)^2 \left (78+\frac {297 x}{2}\right )}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\\ &=-\frac {203 (2+3 x)^2}{242 \sqrt {1-2 x} (3+5 x)^{3/2}}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac {1}{363} \int \frac {\left (-\frac {2049}{2}-\frac {9801 x}{4}\right ) (2+3 x)}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx\\ &=-\frac {203 (2+3 x)^2}{242 \sqrt {1-2 x} (3+5 x)^{3/2}}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac {\sqrt {1-2 x} (627287+991010 x)}{2196150 (3+5 x)^{3/2}}+\frac {81}{100} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {203 (2+3 x)^2}{242 \sqrt {1-2 x} (3+5 x)^{3/2}}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac {\sqrt {1-2 x} (627287+991010 x)}{2196150 (3+5 x)^{3/2}}+\frac {81 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{50 \sqrt {5}}\\ &=-\frac {203 (2+3 x)^2}{242 \sqrt {1-2 x} (3+5 x)^{3/2}}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac {\sqrt {1-2 x} (627287+991010 x)}{2196150 (3+5 x)^{3/2}}+\frac {81 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{50 \sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 15.23, size = 65, normalized size = 0.58 \begin {gather*} \frac {49702040 x^3+51334383 x^2+7883562 x-3014813}{2196150 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac {81 \sin ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{50 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3014813 + 7883562*x + 51334383*x^2 + 49702040*x^3)/(2196150*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2)) - (81*ArcSin[Sq
rt[5/11]*Sqrt[1 - 2*x]])/(50*Sqrt[10])

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IntegrateAlgebraic [A]  time = 0.14, size = 107, normalized size = 0.95 \begin {gather*} \frac {(5 x+3)^{3/2} \left (-\frac {20 (1-2 x)^3}{(5 x+3)^3}-\frac {1656 (1-2 x)^2}{(5 x+3)^2}-\frac {694575 (1-2 x)}{5 x+3}+120050\right )}{2196150 (1-2 x)^{3/2}}-\frac {81 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{50 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3 + 5*x)^(3/2)*(120050 - (20*(1 - 2*x)^3)/(3 + 5*x)^3 - (1656*(1 - 2*x)^2)/(3 + 5*x)^2 - (694575*(1 - 2*x))/
(3 + 5*x)))/(2196150*(1 - 2*x)^(3/2)) - (81*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/(50*Sqrt[10])

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fricas [A]  time = 1.36, size = 116, normalized size = 1.03 \begin {gather*} -\frac {3557763 \, \sqrt {10} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\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 (49702040 \, x^{3} + 51334383 \, x^{2} + 7883562 \, x - 3014813\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{43923000 \, {\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)^4/(1-2*x)^(5/2)/(3+5*x)^(5/2),x, algorithm="fricas")

[Out]

-1/43923000*(3557763*sqrt(10)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x +
3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 20*(49702040*x^3 + 51334383*x^2 + 7883562*x - 3014813)*sqrt(5*x + 3)*sq
rt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)

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giac [B]  time = 1.55, size = 178, normalized size = 1.58 \begin {gather*} -\frac {1}{87846000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {3300 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} + \frac {81}{500} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {343 \, {\left (232 \, \sqrt {5} {\left (5 \, x + 3\right )} - 891 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{2196150 \, {\left (2 \, x - 1\right )}^{2}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {825 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{5490375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/87846000*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 3300*(sqrt(2)*sqrt(-10*x + 5) -
sqrt(22))/sqrt(5*x + 3)) + 81/500*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 343/2196150*(232*sqrt(5)*(5*
x + 3) - 891*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2 + 1/5490375*sqrt(10)*(5*x + 3)^(3/2)*(825*(sqr
t(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3

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maple [A]  time = 0.02, size = 165, normalized size = 1.46 \begin {gather*} \frac {\sqrt {-2 x +1}\, \left (355776300 \sqrt {10}\, x^{4} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] \sqrt {10}\, x^{3} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x^{3}-209908017 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x^{2}-21346578 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x [phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-60296260 \sqrt {-10 x^{2}-x +3}\right )}{43923000 \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}\, \left (5 x +3\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/43923000*(-2*x+1)^(1/2)*(355776300*10^(1/2)*x^4*arcsin(20/11*x+1/11)[phone]*10^(1/2)*x^3*arcsin(20/11*x+1/
11)-209908017*10^(1/2)*x^2*arcsin(20/11*x+1/11)[phone]*(-10*x^2-x+3)^(1/2)*x^3-21346578*10^(1/2)*x*arcsin(2
0/11*x+1/11)[phone]*(-10*x^2-x+3)^(1/2)*x^2+32019867*10^(1/2)*arcsin(20/11*x+1/11)[phone]*(-10*x^2-x+3)
^(1/2)*x-60296260*(-10*x^2-x+3)^(1/2))/(2*x-1)^2/(-10*x^2-x+3)^(1/2)/(5*x+3)^(3/2)

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maxima [B]  time = 1.21, size = 180, normalized size = 1.59 \begin {gather*} \frac {27}{1464100} \, x {\left (\frac {7220 \, x}{\sqrt {-10 \, x^{2} - x + 3}} + \frac {439230 \, x^{2}}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {361}{\sqrt {-10 \, x^{2} - x + 3}} + \frac {21901 \, x}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {87483}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}\right )} - \frac {81}{1000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {9747}{732050} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {1588351 \, x}{1098075 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {108 \, x^{2}}{5 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {34823}{1098075 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {86854 \, x}{9075 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {12682}{9075 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/1464100*x*(7220*x/sqrt(-10*x^2 - x + 3) + 439230*x^2/(-10*x^2 - x + 3)^(3/2) + 361/sqrt(-10*x^2 - x + 3) +
21901*x/(-10*x^2 - x + 3)^(3/2) - 87483/(-10*x^2 - x + 3)^(3/2)) - 81/1000*sqrt(10)*arcsin(-20/11*x - 1/11) +
9747/732050*sqrt(-10*x^2 - x + 3) - 1588351/1098075*x/sqrt(-10*x^2 - x + 3) + 108/5*x^2/(-10*x^2 - x + 3)^(3/2
) - 34823/1098075/sqrt(-10*x^2 - x + 3) + 86854/9075*x/(-10*x^2 - x + 3)^(3/2) - 12682/9075/(-10*x^2 - x + 3)^
(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (3\,x+2\right )}^4}{{\left (1-2\,x\right )}^{5/2}\,{\left (5\,x+3\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  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)**(5/2)/(3+5*x)**(5/2),x)

[Out]

Timed out

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