3.10.38 \(\int \frac {1}{\sqrt {x} (1+\sqrt {x}+x)^{7/2}} \, dx\) [938]

Optimal. Leaf size=76 \[ \frac {4 \left (1+2 \sqrt {x}\right )}{15 \left (1+\sqrt {x}+x\right )^{5/2}}+\frac {64 \left (1+2 \sqrt {x}\right )}{135 \left (1+\sqrt {x}+x\right )^{3/2}}+\frac {512 \left (1+2 \sqrt {x}\right )}{405 \sqrt {1+\sqrt {x}+x}} \]

[Out]

4/15*(1+2*x^(1/2))/(1+x+x^(1/2))^(5/2)+64/135*(1+2*x^(1/2))/(1+x+x^(1/2))^(3/2)+512/405*(1+2*x^(1/2))/(1+x+x^(
1/2))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1366, 628, 627} \begin {gather*} \frac {512 \left (2 \sqrt {x}+1\right )}{405 \sqrt {x+\sqrt {x}+1}}+\frac {64 \left (2 \sqrt {x}+1\right )}{135 \left (x+\sqrt {x}+1\right )^{3/2}}+\frac {4 \left (2 \sqrt {x}+1\right )}{15 \left (x+\sqrt {x}+1\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(1 + Sqrt[x] + x)^(7/2)),x]

[Out]

(4*(1 + 2*Sqrt[x]))/(15*(1 + Sqrt[x] + x)^(5/2)) + (64*(1 + 2*Sqrt[x]))/(135*(1 + Sqrt[x] + x)^(3/2)) + (512*(
1 + 2*Sqrt[x]))/(405*Sqrt[1 + Sqrt[x] + x])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {x} \left (1+\sqrt {x}+x\right )^{7/2}} \, dx &=2 \text {Subst}\left (\int \frac {1}{\left (1+x+x^2\right )^{7/2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {4 \left (1+2 \sqrt {x}\right )}{15 \left (1+\sqrt {x}+x\right )^{5/2}}+\frac {32}{15} \text {Subst}\left (\int \frac {1}{\left (1+x+x^2\right )^{5/2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {4 \left (1+2 \sqrt {x}\right )}{15 \left (1+\sqrt {x}+x\right )^{5/2}}+\frac {64 \left (1+2 \sqrt {x}\right )}{135 \left (1+\sqrt {x}+x\right )^{3/2}}+\frac {256}{135} \text {Subst}\left (\int \frac {1}{\left (1+x+x^2\right )^{3/2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {4 \left (1+2 \sqrt {x}\right )}{15 \left (1+\sqrt {x}+x\right )^{5/2}}+\frac {64 \left (1+2 \sqrt {x}\right )}{135 \left (1+\sqrt {x}+x\right )^{3/2}}+\frac {512 \left (1+2 \sqrt {x}\right )}{405 \sqrt {1+\sqrt {x}+x}}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 49, normalized size = 0.64 \begin {gather*} \frac {4 \left (1+2 \sqrt {x}\right ) \left (203+304 \sqrt {x}+432 x+256 x^{3/2}+128 x^2\right )}{405 \left (1+\sqrt {x}+x\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(1 + Sqrt[x] + x)^(7/2)),x]

[Out]

(4*(1 + 2*Sqrt[x])*(203 + 304*Sqrt[x] + 432*x + 256*x^(3/2) + 128*x^2))/(405*(1 + Sqrt[x] + x)^(5/2))

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Maple [A]
time = 0.01, size = 53, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {\frac {4}{15}+\frac {8 \sqrt {x}}{15}}{\left (1+x +\sqrt {x}\right )^{\frac {5}{2}}}+\frac {\frac {64}{135}+\frac {128 \sqrt {x}}{135}}{\left (1+x +\sqrt {x}\right )^{\frac {3}{2}}}+\frac {\frac {512}{405}+\frac {1024 \sqrt {x}}{405}}{\sqrt {1+x +\sqrt {x}}}\)	\(53\)
default	\(\frac {\frac {4}{15}+\frac {8 \sqrt {x}}{15}}{\left (1+x +\sqrt {x}\right )^{\frac {5}{2}}}+\frac {\frac {64}{135}+\frac {128 \sqrt {x}}{135}}{\left (1+x +\sqrt {x}\right )^{\frac {3}{2}}}+\frac {\frac {512}{405}+\frac {1024 \sqrt {x}}{405}}{\sqrt {1+x +\sqrt {x}}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(1/2)/(1+x+x^(1/2))^(7/2),x,method=_RETURNVERBOSE)

[Out]

4/15*(1+2*x^(1/2))/(1+x+x^(1/2))^(5/2)+64/135*(1+2*x^(1/2))/(1+x+x^(1/2))^(3/2)+512/405*(1+2*x^(1/2))/(1+x+x^(
1/2))^(1/2)

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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(1/x^(1/2)/(1+x+x^(1/2))^(7/2),x, algorithm="maxima")

[Out]

integrate(1/((x + sqrt(x) + 1)^(7/2)*sqrt(x)), x)

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Fricas [A]
time = 0.37, size = 95, normalized size = 1.25 \begin {gather*} -\frac {4 \, {\left (128 \, x^{5} + 272 \, x^{4} + 455 \, x^{3} + 232 \, x^{2} - {\left (256 \, x^{5} + 736 \, x^{4} + 1366 \, x^{3} + 1427 \, x^{2} + 839 \, x + 101\right )} \sqrt {x} - 128 \, x - 203\right )} \sqrt {x + \sqrt {x} + 1}}{405 \, {\left (x^{6} + 3 \, x^{5} + 6 \, x^{4} + 7 \, x^{3} + 6 \, x^{2} + 3 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(1+x+x^(1/2))^(7/2),x, algorithm="fricas")

[Out]

-4/405*(128*x^5 + 272*x^4 + 455*x^3 + 232*x^2 - (256*x^5 + 736*x^4 + 1366*x^3 + 1427*x^2 + 839*x + 101)*sqrt(x
) - 128*x - 203)*sqrt(x + sqrt(x) + 1)/(x^6 + 3*x^5 + 6*x^4 + 7*x^3 + 6*x^2 + 3*x + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x} \left (\sqrt {x} + x + 1\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(1/2)/(1+x+x**(1/2))**(7/2),x)

[Out]

Integral(1/(sqrt(x)*(sqrt(x) + x + 1)**(7/2)), x)

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Giac [A]
time = 3.28, size = 45, normalized size = 0.59 \begin {gather*} \frac {4 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (4 \, \sqrt {x} {\left (2 \, \sqrt {x} + 5\right )} + 35\right )} \sqrt {x} + 65\right )} \sqrt {x} + 355\right )} \sqrt {x} + 203\right )}}{405 \, {\left (x + \sqrt {x} + 1\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(1+x+x^(1/2))^(7/2),x, algorithm="giac")

[Out]

4/405*(2*(8*(2*(4*sqrt(x)*(2*sqrt(x) + 5) + 35)*sqrt(x) + 65)*sqrt(x) + 355)*sqrt(x) + 203)/(x + sqrt(x) + 1)^
(5/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {x}\,{\left (x+\sqrt {x}+1\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(x + x^(1/2) + 1)^(7/2)),x)

[Out]

int(1/(x^(1/2)*(x + x^(1/2) + 1)^(7/2)), x)

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