3.44 \(\int \frac {x^7 (2+3 x^2)}{(5+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=58 \[ -\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\left (3 x^2+2\right ) x^4}{2 \sqrt {x^4+5}}+\frac {1}{4} \left (9 x^2+8\right ) \sqrt {x^4+5} \]

[Out]

-45/4*arcsinh(1/5*x^2*5^(1/2))-1/2*x^4*(3*x^2+2)/(x^4+5)^(1/2)+1/4*(9*x^2+8)*(x^4+5)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1252, 819, 780, 215} \[ -\frac {\left (3 x^2+2\right ) x^4}{2 \sqrt {x^4+5}}+\frac {1}{4} \left (9 x^2+8\right ) \sqrt {x^4+5}-\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(2 + 3*x^2))/(5 + x^4)^(3/2),x]

[Out]

-(x^4*(2 + 3*x^2))/(2*Sqrt[5 + x^4]) + ((8 + 9*x^2)*Sqrt[5 + x^4])/4 - (45*ArcSinh[x^2/Sqrt[5]])/4

Rule 215

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rubi steps

\begin {align*} \int \frac {x^7 \left (2+3 x^2\right )}{\left (5+x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3 (2+3 x)}{\left (5+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {x^4 \left (2+3 x^2\right )}{2 \sqrt {5+x^4}}+\frac {1}{10} \operatorname {Subst}\left (\int \frac {x (20+45 x)}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^4 \left (2+3 x^2\right )}{2 \sqrt {5+x^4}}+\frac {1}{4} \left (8+9 x^2\right ) \sqrt {5+x^4}-\frac {45}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^4 \left (2+3 x^2\right )}{2 \sqrt {5+x^4}}+\frac {1}{4} \left (8+9 x^2\right ) \sqrt {5+x^4}-\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 0.88 \[ \frac {3 x^6+4 x^4+45 x^2-45 \sqrt {x^4+5} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+40}{4 \sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(2 + 3*x^2))/(5 + x^4)^(3/2),x]

[Out]

(40 + 45*x^2 + 4*x^4 + 3*x^6 - 45*Sqrt[5 + x^4]*ArcSinh[x^2/Sqrt[5]])/(4*Sqrt[5 + x^4])

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fricas [A]  time = 0.55, size = 62, normalized size = 1.07 \[ \frac {30 \, x^{4} + 45 \, {\left (x^{4} + 5\right )} \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) + {\left (3 \, x^{6} + 4 \, x^{4} + 45 \, x^{2} + 40\right )} \sqrt {x^{4} + 5} + 150}{4 \, {\left (x^{4} + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(3*x^2+2)/(x^4+5)^(3/2),x, algorithm="fricas")

[Out]

1/4*(30*x^4 + 45*(x^4 + 5)*log(-x^2 + sqrt(x^4 + 5)) + (3*x^6 + 4*x^4 + 45*x^2 + 40)*sqrt(x^4 + 5) + 150)/(x^4
+ 5)

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giac [A]  time = 0.21, size = 45, normalized size = 0.78 \[ \frac {{\left ({\left (3 \, x^{2} + 4\right )} x^{2} + 45\right )} x^{2} + 40}{4 \, \sqrt {x^{4} + 5}} + \frac {45}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(3*x^2+2)/(x^4+5)^(3/2),x, algorithm="giac")

[Out]

1/4*(((3*x^2 + 4)*x^2 + 45)*x^2 + 40)/sqrt(x^4 + 5) + 45/4*log(-x^2 + sqrt(x^4 + 5))

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maple [A]  time = 0.02, size = 50, normalized size = 0.86 \[ \frac {3 x^{6}}{4 \sqrt {x^{4}+5}}+\frac {45 x^{2}}{4 \sqrt {x^{4}+5}}-\frac {45 \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )}{4}+\frac {x^{4}+10}{\sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(3*x^2+2)/(x^4+5)^(3/2),x)

[Out]

3/4*x^6/(x^4+5)^(1/2)+45/4*x^2/(x^4+5)^(1/2)-45/4*arcsinh(1/5*5^(1/2)*x^2)+1/(x^4+5)^(1/2)*(x^4+10)

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maxima [A]  time = 1.21, size = 89, normalized size = 1.53 \[ \sqrt {x^{4} + 5} - \frac {15 \, {\left (\frac {3 \, {\left (x^{4} + 5\right )}}{x^{4}} - 2\right )}}{4 \, {\left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}}}{x^{6}}\right )}} + \frac {5}{\sqrt {x^{4} + 5}} - \frac {45}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) + \frac {45}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(3*x^2+2)/(x^4+5)^(3/2),x, algorithm="maxima")

[Out]

sqrt(x^4 + 5) - 15/4*(3*(x^4 + 5)/x^4 - 2)/(sqrt(x^4 + 5)/x^2 - (x^4 + 5)^(3/2)/x^6) + 5/sqrt(x^4 + 5) - 45/8*
log(sqrt(x^4 + 5)/x^2 + 1) + 45/8*log(sqrt(x^4 + 5)/x^2 - 1)

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mupad [B]  time = 1.11, size = 97, normalized size = 1.67 \[ \sqrt {x^4+5}\,\left (\frac {3\,x^2}{4}+1\right )-\frac {45\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{4}+\frac {\sqrt {5}\,\left (10+\sqrt {5}\,15{}\mathrm {i}\right )\,\sqrt {x^4+5}\,1{}\mathrm {i}}{20\,\left (-x^2+\sqrt {5}\,1{}\mathrm {i}\right )}-\frac {\sqrt {5}\,\left (-10+\sqrt {5}\,15{}\mathrm {i}\right )\,\sqrt {x^4+5}\,1{}\mathrm {i}}{20\,\left (x^2+\sqrt {5}\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(3*x^2 + 2))/(x^4 + 5)^(3/2),x)

[Out]

(x^4 + 5)^(1/2)*((3*x^2)/4 + 1) - (45*asinh((5^(1/2)*x^2)/5))/4 + (5^(1/2)*(5^(1/2)*15i + 10)*(x^4 + 5)^(1/2)*
1i)/(20*(5^(1/2)*1i - x^2)) - (5^(1/2)*(5^(1/2)*15i - 10)*(x^4 + 5)^(1/2)*1i)/(20*(5^(1/2)*1i + x^2))

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sympy [A]  time = 14.28, size = 66, normalized size = 1.14 \[ \frac {3 x^{6}}{4 \sqrt {x^{4} + 5}} + \frac {x^{4}}{\sqrt {x^{4} + 5}} + \frac {45 x^{2}}{4 \sqrt {x^{4} + 5}} - \frac {45 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{4} + \frac {10}{\sqrt {x^{4} + 5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(3*x**2+2)/(x**4+5)**(3/2),x)

[Out]

3*x**6/(4*sqrt(x**4 + 5)) + x**4/sqrt(x**4 + 5) + 45*x**2/(4*sqrt(x**4 + 5)) - 45*asinh(sqrt(5)*x**2/5)/4 + 10
/sqrt(x**4 + 5)

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