3.30.56 \(\int \frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx\) [2956]

Optimal. Leaf size=158 \[ -\frac {225}{22} \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}-\frac {34 \sqrt {2+3 x} (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}-68 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )-\frac {15}{2} \sqrt {\frac {3}{11}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right ) \]

[Out]

1/3*(2+3*x)^(3/2)*(3+5*x)^(3/2)/(1-2*x)^(3/2)-15/22*EllipticF(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(
1/2)-68/3*EllipticE(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-34/11*(3+5*x)^(3/2)*(2+3*x)^(1/2)/(1-
2*x)^(1/2)-225/22*(1-2*x)^(1/2)*(2+3*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {99, 155, 159, 164, 114, 120} \begin {gather*} -\frac {15}{2} \sqrt {\frac {3}{11}} F\left (\text {ArcSin}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )-68 \sqrt {\frac {11}{3}} E\left (\text {ArcSin}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )+\frac {(3 x+2)^{3/2} (5 x+3)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {34 \sqrt {3 x+2} (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}-\frac {225}{22} \sqrt {1-2 x} \sqrt {3 x+2} \sqrt {5 x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-225*Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*Sqrt[3 + 5*x])/22 - (34*Sqrt[2 + 3*x]*(3 + 5*x)^(3/2))/(11*Sqrt[1 - 2*x]) +
((2 + 3*x)^(3/2)*(3 + 5*x)^(3/2))/(3*(1 - 2*x)^(3/2)) - 68*Sqrt[11/3]*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]
], 35/33] - (15*Sqrt[3/11]*EllipticF[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/2

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 114

Int[Sqrt[(e_.) + (f_.)*(x_)]/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2/b)*Rt[-(b
*e - a*f)/d, 2]*EllipticE[ArcSin[Sqrt[a + b*x]/Rt[-(b*c - a*d)/d, 2]], f*((b*c - a*d)/(d*(b*e - a*f)))], x] /;
FreeQ[{a, b, c, d, e, f}, x] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !LtQ[-(b*c - a*d)/d, 0] &&
!(SimplerQ[c + d*x, a + b*x] && GtQ[-d/(b*c - a*d), 0] && GtQ[d/(d*e - c*f), 0] &&  !LtQ[(b*c - a*d)/b, 0])

Rule 120

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[2*(Rt[-b/d,
2]/(b*Sqrt[(b*e - a*f)/b]))*EllipticF[ArcSin[Sqrt[a + b*x]/(Rt[-b/d, 2]*Sqrt[(b*c - a*d)/b])], f*((b*c - a*d)
/(d*(b*e - a*f)))], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[(b*c - a*d)/b, 0] && GtQ[(b*e - a*f)/b, 0] && Po
sQ[-b/d] &&  !(SimplerQ[c + d*x, a + b*x] && GtQ[(d*e - c*f)/d, 0] && GtQ[-d/b, 0]) &&  !(SimplerQ[c + d*x, a
+ b*x] && GtQ[((-b)*e + a*f)/f, 0] && GtQ[-f/b, 0]) &&  !(SimplerQ[e + f*x, a + b*x] && GtQ[((-d)*e + c*f)/f,
0] && GtQ[((-b)*e + a*f)/f, 0] && (PosQ[-f/d] || PosQ[-f/b]))

Rule 155

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 159

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] && IntegersQ[2*m, 2
*n, 2*p]

Rule 164

Int[((g_.) + (h_.)*(x_))/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol]
:> Dist[h/f, Int[Sqrt[e + f*x]/(Sqrt[a + b*x]*Sqrt[c + d*x]), x], x] + Dist[(f*g - e*h)/f, Int[1/(Sqrt[a + b*
x]*Sqrt[c + d*x]*Sqrt[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && SimplerQ[a + b*x, e + f*x] &&
SimplerQ[c + d*x, e + f*x]

Rubi steps

\begin {align*} \int \frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx &=\frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {1}{3} \int \frac {\sqrt {2+3 x} \sqrt {3+5 x} \left (\frac {57}{2}+45 x\right )}{(1-2 x)^{3/2}} \, dx\\ &=-\frac {34 \sqrt {2+3 x} (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {1}{33} \int \frac {\left (-1974-\frac {6075 x}{2}\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx\\ &=-\frac {225}{22} \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}-\frac {34 \sqrt {2+3 x} (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}+\frac {1}{297} \int \frac {\frac {255717}{4}+100980 x}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {225}{22} \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}-\frac {34 \sqrt {2+3 x} (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}+\frac {45}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx+68 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx\\ &=-\frac {225}{22} \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}-\frac {34 \sqrt {2+3 x} (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {(2+3 x)^{3/2} (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}-68 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )-\frac {15}{2} \sqrt {\frac {3}{11}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )\\ \end {align*}

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Mathematica [A]
time = 7.73, size = 120, normalized size = 0.76 \begin {gather*} -\frac {2 \sqrt {2+3 x} \sqrt {3+5 x} \left (105-302 x+30 x^2\right )+272 \sqrt {2-4 x} (-1+2 x) E\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )|-\frac {33}{2}\right )-137 \sqrt {2-4 x} (-1+2 x) F\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )|-\frac {33}{2}\right )}{12 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(2*Sqrt[2 + 3*x]*Sqrt[3 + 5*x]*(105 - 302*x + 30*x^2) + 272*Sqrt[2 - 4*x]*(-1 + 2*x)*EllipticE[ArcSin[Sq
rt[2/11]*Sqrt[3 + 5*x]], -33/2] - 137*Sqrt[2 - 4*x]*(-1 + 2*x)*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33
/2])/(1 - 2*x)^(3/2)

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Maple [A]
time = 0.10, size = 229, normalized size = 1.45

method	result	size
default	\(-\frac {\left (270 \sqrt {2}\, \EllipticF \left (\frac {\sqrt {28+42 x}}{7}, \frac {\sqrt {70}}{2}\right ) x \sqrt {2+3 x}\, \sqrt {-3-5 x}\, \sqrt {1-2 x}-544 \sqrt {2}\, \EllipticE \left (\frac {\sqrt {28+42 x}}{7}, \frac {\sqrt {70}}{2}\right ) x \sqrt {2+3 x}\, \sqrt {-3-5 x}\, \sqrt {1-2 x}-135 \sqrt {2}\, \sqrt {2+3 x}\, \sqrt {-3-5 x}\, \sqrt {1-2 x}\, \EllipticF \left (\frac {\sqrt {28+42 x}}{7}, \frac {\sqrt {70}}{2}\right )+272 \sqrt {2}\, \sqrt {2+3 x}\, \sqrt {-3-5 x}\, \sqrt {1-2 x}\, \EllipticE \left (\frac {\sqrt {28+42 x}}{7}, \frac {\sqrt {70}}{2}\right )+900 x^{4}-7920 x^{3}-7966 x^{2}+366 x +1260\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}\, \sqrt {2+3 x}}{12 \left (-1+2 x \right )^{2} \left (15 x^{2}+19 x +6\right )}\)	\(229\)
elliptic	\(\frac {\sqrt {-\left (3+5 x \right ) \left (-1+2 x \right ) \left (2+3 x \right )}\, \sqrt {3+5 x}\, \sqrt {2+3 x}\, \left (\frac {-340 x^{2}-\frac {1292}{3} x -136}{\sqrt {\left (-\frac {1}{2}+x \right ) \left (-30 x^{2}-38 x -12\right )}}+\frac {41 \sqrt {28+42 x}\, \sqrt {-15 x -9}\, \sqrt {21-42 x}\, \EllipticF \left (\frac {\sqrt {28+42 x}}{7}, \frac {\sqrt {70}}{2}\right )}{4 \sqrt {-30 x^{3}-23 x^{2}+7 x +6}}+\frac {340 \sqrt {28+42 x}\, \sqrt {-15 x -9}\, \sqrt {21-42 x}\, \left (-\frac {\EllipticE \left (\frac {\sqrt {28+42 x}}{7}, \frac {\sqrt {70}}{2}\right )}{15}-\frac {3 \EllipticF \left (\frac {\sqrt {28+42 x}}{7}, \frac {\sqrt {70}}{2}\right )}{5}\right )}{21 \sqrt {-30 x^{3}-23 x^{2}+7 x +6}}-\frac {5 \sqrt {-30 x^{3}-23 x^{2}+7 x +6}}{4}+\frac {77 \sqrt {-30 x^{3}-23 x^{2}+7 x +6}}{48 \left (-\frac {1}{2}+x \right )^{2}}\right )}{\sqrt {1-2 x}\, \left (15 x^{2}+19 x +6\right )}\)	\(256\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(270*2^(1/2)*EllipticF(1/7*(28+42*x)^(1/2),1/2*70^(1/2))*x*(2+3*x)^(1/2)*(-3-5*x)^(1/2)*(1-2*x)^(1/2)-54
4*2^(1/2)*EllipticE(1/7*(28+42*x)^(1/2),1/2*70^(1/2))*x*(2+3*x)^(1/2)*(-3-5*x)^(1/2)*(1-2*x)^(1/2)-135*2^(1/2)
*(2+3*x)^(1/2)*(-3-5*x)^(1/2)*(1-2*x)^(1/2)*EllipticF(1/7*(28+42*x)^(1/2),1/2*70^(1/2))+272*2^(1/2)*(2+3*x)^(1
/2)*(-3-5*x)^(1/2)*(1-2*x)^(1/2)*EllipticE(1/7*(28+42*x)^(1/2),1/2*70^(1/2))+900*x^4-7920*x^3-7966*x^2+366*x+1
260)*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(2+3*x)^(1/2)/(-1+2*x)^2/(15*x^2+19*x+6)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.30, size = 45, normalized size = 0.28 \begin {gather*} -\frac {{\left (30 \, x^{2} - 302 \, x + 105\right )} \sqrt {5 \, x + 3} \sqrt {3 \, x + 2} \sqrt {-2 \, x + 1}}{6 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(30*x^2 - 302*x + 105)*sqrt(5*x + 3)*sqrt(3*x + 2)*sqrt(-2*x + 1)/(4*x^2 - 4*x + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (3 x + 2\right )^{\frac {3}{2}} \left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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