3.938 \(\int \frac{1}{\sqrt{x} (1+\sqrt{x}+x)^{7/2}} \, dx\)

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

[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])

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Rubi [A]  time = 0.0225249, antiderivative size = 76, normalized size of antiderivative = 1., 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 = {1352, 614, 613} \[ \frac{512 \left (2 \sqrt{x}+1\right )}{405 \sqrt{x+\sqrt{x}+1}}+\frac{64 \left (2 \sqrt{x}+1\right )}{135 \left (x+\sqrt{x}+1\right )^{3/2}}+\frac{4 \left (2 \sqrt{x}+1\right )}{15 \left (x+\sqrt{x}+1\right )^{5/2}} \]

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 1352

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]

Rule 614

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 613

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]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{x} \left (1+\sqrt{x}+x\right )^{7/2}} \, dx &=2 \operatorname{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} \operatorname{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} \operatorname{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*}

Mathematica [A]  time = 0.0159448, size = 49, normalized size = 0.64 \[ \frac{4 \left (2 \sqrt{x}+1\right ) \left (128 x^2+256 x^{3/2}+432 x+304 \sqrt{x}+203\right )}{405 \left (x+\sqrt{x}+1\right )^{5/2}} \]

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 = 0.001, size = 53, normalized size = 0.7 \begin{align*}{\frac{4}{15} \left ( 1+2\,\sqrt{x} \right ) \left ( 1+x+\sqrt{x} \right ) ^{-{\frac{5}{2}}}}+{\frac{64}{135} \left ( 1+2\,\sqrt{x} \right ) \left ( 1+x+\sqrt{x} \right ) ^{-{\frac{3}{2}}}}+{\frac{512}{405} \left ( 1+2\,\sqrt{x} \right ){\frac{1}{\sqrt{1+x+\sqrt{x}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(1/2)/(1+x+x^(1/2))^(7/2),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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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x + \sqrt{x} + 1\right )}^{\frac{7}{2}} \sqrt{x}}\,{d x} \end{align*}

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 = 1.68286, size = 267, normalized size = 3.51 \begin{align*} -\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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.10589, size = 61, normalized size = 0.8 \begin{align*} \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{align*}

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)