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

Optimal. Leaf size=64 \[ \frac {3 \sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}-\frac {37 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}} \]

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Rubi [A]  time = 0.01, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {96, 93, 204} \begin {gather*} \frac {3 \sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}-\frac {37 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(7*(2 + 3*x)) - (37*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 204

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}} \, dx &=\frac {3 \sqrt {1-2 x} \sqrt {3+5 x}}{7 (2+3 x)}+\frac {37}{14} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {3 \sqrt {1-2 x} \sqrt {3+5 x}}{7 (2+3 x)}+\frac {37}{7} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=\frac {3 \sqrt {1-2 x} \sqrt {3+5 x}}{7 (2+3 x)}-\frac {37 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 64, normalized size = 1.00 \begin {gather*} \frac {3 \sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}-\frac {37 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(7*(2 + 3*x)) - (37*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])

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IntegrateAlgebraic [A]  time = 0.11, size = 74, normalized size = 1.16 \begin {gather*} \frac {33 \sqrt {1-2 x}}{7 \sqrt {5 x+3} \left (\frac {1-2 x}{5 x+3}+7\right )}-\frac {37 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(33*Sqrt[1 - 2*x])/(7*Sqrt[3 + 5*x]*(7 + (1 - 2*x)/(3 + 5*x))) - (37*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*
x])])/(7*Sqrt[7])

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fricas [A]  time = 1.47, size = 71, normalized size = 1.11 \begin {gather*} -\frac {37 \, \sqrt {7} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 42 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{98 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/98*(37*sqrt(7)*(3*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 4
2*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)

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giac [B]  time = 1.33, size = 193, normalized size = 3.02 \begin {gather*} \frac {37}{980} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {66 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{7 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

37/980*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/
(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 66/7*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sq
rt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqr
t(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)

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maple [B]  time = 0.02, size = 108, normalized size = 1.69 \begin {gather*} \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (111 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+74 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+42 \sqrt {-10 x^{2}-x +3}\right )}{98 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/98*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(111*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+74*7^(1/2)
*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+42*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)

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maxima [A]  time = 1.18, size = 50, normalized size = 0.78 \begin {gather*} \frac {37}{98} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {3 \, \sqrt {-10 \, x^{2} - x + 3}}{7 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

37/98*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 3/7*sqrt(-10*x^2 - x + 3)/(3*x + 2)

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mupad [B]  time = 7.02, size = 716, normalized size = 11.19 \begin {gather*} \frac {6\,{\left (\sqrt {1-2\,x}-1\right )}^3}{7\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )}-\frac {37\,\sqrt {7}\,\mathrm {atan}\left (\frac {32856\,\sqrt {3}\,\sqrt {7}}{42875\,\left (\frac {32856\,{\left (\sqrt {1-2\,x}-1\right )}^2}{1715\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {607836\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {65712}{8575}\right )}+\frac {16428\,\sqrt {7}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )\,\left (\frac {32856\,{\left (\sqrt {1-2\,x}-1\right )}^2}{1715\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {607836\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {65712}{8575}\right )}-\frac {16428\,\sqrt {3}\,\sqrt {7}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{8575\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2\,\left (\frac {32856\,{\left (\sqrt {1-2\,x}-1\right )}^2}{1715\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {607836\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {65712}{8575}\right )}\right )}{49}-\frac {12\,\left (\sqrt {1-2\,x}-1\right )}{35\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )}+\frac {111\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{175\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*((1 - 2*x)^(1/2) - 1)^3)/(7*(3^(1/2) - (5*x + 3)^(1/2))^3*((14*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x
+ 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + (6*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/
(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + 4/25
)) - (37*7^(1/2)*atan((32856*3^(1/2)*7^(1/2))/(42875*((32856*((1 - 2*x)^(1/2) - 1)^2)/(1715*(3^(1/2) - (5*x +
3)^(1/2))^2) + (607836*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(42875*(3^(1/2) - (5*x + 3)^(1/2))) - 65712/8575)) + (16
428*7^(1/2)*((1 - 2*x)^(1/2) - 1))/(42875*(3^(1/2) - (5*x + 3)^(1/2))*((32856*((1 - 2*x)^(1/2) - 1)^2)/(1715*(
3^(1/2) - (5*x + 3)^(1/2))^2) + (607836*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(42875*(3^(1/2) - (5*x + 3)^(1/2))) - 6
5712/8575)) - (16428*3^(1/2)*7^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(8575*(3^(1/2) - (5*x + 3)^(1/2))^2*((32856*((1
- 2*x)^(1/2) - 1)^2)/(1715*(3^(1/2) - (5*x + 3)^(1/2))^2) + (607836*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(42875*(3^(
1/2) - (5*x + 3)^(1/2))) - 65712/8575))))/49 - (12*((1 - 2*x)^(1/2) - 1))/(35*(3^(1/2) - (5*x + 3)^(1/2))*((14
*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1
/2))^4 + (6*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/2)*((1 - 2*x)^(1/2)
- 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + 4/25)) + (111*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(175*(3^(1/2) - (5*x +
3)^(1/2))^2*((14*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/
2) - (5*x + 3)^(1/2))^4 + (6*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/2)*
((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + 4/25))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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