3.87 \(\int \cos (x) (1+\sin ^2(x))^2 \, dx\)

Optimal. Leaf size=19 \[ \frac {\sin ^5(x)}{5}+\frac {2 \sin ^3(x)}{3}+\sin (x) \]

[Out]

sin(x)+2/3*sin(x)^3+1/5*sin(x)^5

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3190, 194} \[ \frac {\sin ^5(x)}{5}+\frac {2 \sin ^3(x)}{3}+\sin (x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*(1 + Sin[x]^2)^2,x]

[Out]

Sin[x] + (2*Sin[x]^3)/3 + Sin[x]^5/5

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \cos (x) \left (1+\sin ^2(x)\right )^2 \, dx &=\operatorname {Subst}\left (\int \left (1+x^2\right )^2 \, dx,x,\sin (x)\right )\\ &=\operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\sin (x)\right )\\ &=\sin (x)+\frac {2 \sin ^3(x)}{3}+\frac {\sin ^5(x)}{5}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.00 \[ \frac {\sin ^5(x)}{5}+\frac {2 \sin ^3(x)}{3}+\sin (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*(1 + Sin[x]^2)^2,x]

[Out]

Sin[x] + (2*Sin[x]^3)/3 + Sin[x]^5/5

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fricas [A]  time = 0.42, size = 18, normalized size = 0.95 \[ \frac {1}{15} \, {\left (3 \, \cos \relax (x)^{4} - 16 \, \cos \relax (x)^{2} + 28\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(1+sin(x)^2)^2,x, algorithm="fricas")

[Out]

1/15*(3*cos(x)^4 - 16*cos(x)^2 + 28)*sin(x)

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giac [A]  time = 1.05, size = 15, normalized size = 0.79 \[ \frac {1}{5} \, \sin \relax (x)^{5} + \frac {2}{3} \, \sin \relax (x)^{3} + \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(1+sin(x)^2)^2,x, algorithm="giac")

[Out]

1/5*sin(x)^5 + 2/3*sin(x)^3 + sin(x)

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maple [A]  time = 0.02, size = 16, normalized size = 0.84 \[ \frac {\left (\sin ^{5}\relax (x )\right )}{5}+\frac {2 \left (\sin ^{3}\relax (x )\right )}{3}+\sin \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*(1+sin(x)^2)^2,x)

[Out]

sin(x)+2/3*sin(x)^3+1/5*sin(x)^5

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maxima [A]  time = 0.41, size = 15, normalized size = 0.79 \[ \frac {1}{5} \, \sin \relax (x)^{5} + \frac {2}{3} \, \sin \relax (x)^{3} + \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(1+sin(x)^2)^2,x, algorithm="maxima")

[Out]

1/5*sin(x)^5 + 2/3*sin(x)^3 + sin(x)

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mupad [B]  time = 0.04, size = 15, normalized size = 0.79 \[ \frac {{\sin \relax (x)}^5}{5}+\frac {2\,{\sin \relax (x)}^3}{3}+\sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*(sin(x)^2 + 1)^2,x)

[Out]

sin(x) + (2*sin(x)^3)/3 + sin(x)^5/5

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sympy [A]  time = 1.15, size = 17, normalized size = 0.89 \[ \frac {\sin ^{5}{\relax (x )}}{5} + \frac {2 \sin ^{3}{\relax (x )}}{3} + \sin {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(1+sin(x)**2)**2,x)

[Out]

sin(x)**5/5 + 2*sin(x)**3/3 + sin(x)

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