3.1.93 \(\int \sec ^3(x) \tan ^5(x) \, dx\) [93]

Optimal. Leaf size=25 \[ \frac {\sec ^3(x)}{3}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^7(x)}{7} \]

[Out]

1/3*sec(x)^3-2/5*sec(x)^5+1/7*sec(x)^7

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Rubi [A]
time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2686, 276} \begin {gather*} \frac {\sec ^7(x)}{7}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^3(x)}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3*Tan[x]^5,x]

[Out]

Sec[x]^3/3 - (2*Sec[x]^5)/5 + Sec[x]^7/7

Rule 276

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

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

\begin {align*} \int \sec ^3(x) \tan ^5(x) \, dx &=\text {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (x)\right )\\ &=\text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (x)\right )\\ &=\frac {\sec ^3(x)}{3}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^7(x)}{7}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 25, normalized size = 1.00 \begin {gather*} \frac {\sec ^3(x)}{3}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^7(x)}{7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3*Tan[x]^5,x]

[Out]

Sec[x]^3/3 - (2*Sec[x]^5)/5 + Sec[x]^7/7

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Mathics [A]
time = 1.87, size = 20, normalized size = 0.80 \begin {gather*} \frac {15-42 \text {Cos}\left [x\right ]^2+35 \text {Cos}\left [x\right ]^4}{105 \text {Cos}\left [x\right ]^7} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[Tan[x]^5*Sec[x]^3,x]')

[Out]

(15 - 42 Cos[x] ^ 2 + 35 Cos[x] ^ 4) / (105 Cos[x] ^ 7)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(57\) vs. \(2(19)=38\).
time = 0.04, size = 58, normalized size = 2.32

method	result	size
risch	\(\frac {\frac {8 \,{\mathrm e}^{11 i x}}{3}-\frac {32 \,{\mathrm e}^{9 i x}}{15}+\frac {304 \,{\mathrm e}^{7 i x}}{35}-\frac {32 \,{\mathrm e}^{5 i x}}{15}+\frac {8 \,{\mathrm e}^{3 i x}}{3}}{\left ({\mathrm e}^{2 i x}+1\right )^{7}}\)	\(48\)
default	\(\frac {\sin ^{6}\left (x \right )}{7 \cos \left (x \right )^{7}}+\frac {\sin ^{6}\left (x \right )}{35 \cos \left (x \right )^{5}}-\frac {\sin ^{6}\left (x \right )}{105 \cos \left (x \right )^{3}}+\frac {\sin ^{6}\left (x \right )}{35 \cos \left (x \right )}+\frac {\left (\frac {8}{3}+\sin ^{4}\left (x \right )+\frac {4 \left (\sin ^{2}\left (x \right )\right )}{3}\right ) \cos \left (x \right )}{35}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3*tan(x)^5,x,method=_RETURNVERBOSE)

[Out]

1/7*sin(x)^6/cos(x)^7+1/35*sin(x)^6/cos(x)^5-1/105*sin(x)^6/cos(x)^3+1/35*sin(x)^6/cos(x)+1/35*(8/3+sin(x)^4+4
/3*sin(x)^2)*cos(x)

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Maxima [A]
time = 0.27, size = 20, normalized size = 0.80 \begin {gather*} \frac {35 \, \cos \left (x\right )^{4} - 42 \, \cos \left (x\right )^{2} + 15}{105 \, \cos \left (x\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="maxima")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

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Fricas [A]
time = 0.34, size = 20, normalized size = 0.80 \begin {gather*} \frac {35 \, \cos \left (x\right )^{4} - 42 \, \cos \left (x\right )^{2} + 15}{105 \, \cos \left (x\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="fricas")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

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Sympy [A]
time = 0.06, size = 22, normalized size = 0.88 \begin {gather*} - \frac {- 35 \cos ^{4}{\left (x \right )} + 42 \cos ^{2}{\left (x \right )} - 15}{105 \cos ^{7}{\left (x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3*tan(x)**5,x)

[Out]

-(-35*cos(x)**4 + 42*cos(x)**2 - 15)/(105*cos(x)**7)

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Giac [A]
time = 0.00, size = 23, normalized size = 0.92 \begin {gather*} \frac {35 \cos ^{4}x-42 \cos ^{2}x+15}{105 \cos ^{7}x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3*tan(x)^5,x)

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

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Mupad [B]
time = 0.53, size = 19, normalized size = 0.76 \begin {gather*} \frac {\frac {{\cos \left (x\right )}^4}{3}-\frac {2\,{\cos \left (x\right )}^2}{5}+\frac {1}{7}}{{\cos \left (x\right )}^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^5/cos(x)^3,x)

[Out]

(cos(x)^4/3 - (2*cos(x)^2)/5 + 1/7)/cos(x)^7

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