3.3.44 \(\int (\sec ^5(x) \tan ^2(x)+\sec ^3(x) \tan ^4(x)) \, dx\) [244]

Optimal. Leaf size=12 \[ \frac {1}{3} \sec ^3(x) \tan ^3(x) \]

[Out]

1/3*sec(x)^3*tan(x)^3

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(33\) vs. \(2(12)=24\).
time = 0.05, antiderivative size = 33, normalized size of antiderivative = 2.75, number of steps used = 9, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2691, 3853, 3855} \begin {gather*} \frac {1}{6} \tan (x) \sec ^5(x)+\frac {1}{6} \tan ^3(x) \sec ^3(x)-\frac {1}{6} \tan (x) \sec ^3(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]^5*Tan[x]^2 + Sec[x]^3*Tan[x]^4,x]

[Out]

-1/6*(Sec[x]^3*Tan[x]) + (Sec[x]^5*Tan[x])/6 + (Sec[x]^3*Tan[x]^3)/6

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=\int \sec ^5(x) \tan ^2(x) \, dx+\int \sec ^3(x) \tan ^4(x) \, dx\\ &=\frac {1}{6} \sec ^5(x) \tan (x)+\frac {1}{6} \sec ^3(x) \tan ^3(x)-\frac {1}{6} \int \sec ^5(x) \, dx-\frac {1}{2} \int \sec ^3(x) \tan ^2(x) \, dx\\ &=-\frac {1}{6} \sec ^3(x) \tan (x)+\frac {1}{6} \sec ^5(x) \tan (x)+\frac {1}{6} \sec ^3(x) \tan ^3(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.02, size = 12, normalized size = 1.00 \begin {gather*} \frac {1}{3} \sec ^3(x) \tan ^3(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^5*Tan[x]^2 + Sec[x]^3*Tan[x]^4,x]

[Out]

(Sec[x]^3*Tan[x]^3)/3

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs. \(2(10)=20\).
time = 5.12, size = 68, normalized size = 5.67

method	result	size
risch	\(\frac {8 i \left ({\mathrm e}^{9 i x}-3 \,{\mathrm e}^{7 i x}+3 \,{\mathrm e}^{5 i x}-{\mathrm e}^{3 i x}\right )}{3 \left ({\mathrm e}^{2 i x}+1\right )^{6}}\)	\(40\)
default	\(\frac {\sin ^{3}\left (x \right )}{6 \cos \left (x \right )^{6}}+\frac {\sin ^{3}\left (x \right )}{8 \cos \left (x \right )^{4}}+\frac {\sin ^{3}\left (x \right )}{16 \cos \left (x \right )^{2}}+\frac {\sin ^{5}\left (x \right )}{6 \cos \left (x \right )^{6}}+\frac {\sin ^{5}\left (x \right )}{24 \cos \left (x \right )^{4}}-\frac {\sin ^{5}\left (x \right )}{48 \cos \left (x \right )^{2}}-\frac {\left (\sin ^{3}\left (x \right )\right )}{48}\)	\(68\)
parts	\(\frac {\sin ^{3}\left (x \right )}{6 \cos \left (x \right )^{6}}+\frac {\sin ^{3}\left (x \right )}{8 \cos \left (x \right )^{4}}+\frac {\sin ^{3}\left (x \right )}{16 \cos \left (x \right )^{2}}+\frac {\sin ^{5}\left (x \right )}{6 \cos \left (x \right )^{6}}+\frac {\sin ^{5}\left (x \right )}{24 \cos \left (x \right )^{4}}-\frac {\sin ^{5}\left (x \right )}{48 \cos \left (x \right )^{2}}-\frac {\left (\sin ^{3}\left (x \right )\right )}{48}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^4*sec(x)^3+tan(x)^2*sec(x)^5,x,method=_RETURNVERBOSE)

[Out]

1/6*sin(x)^3/cos(x)^6+1/8*sin(x)^3/cos(x)^4+1/16*sin(x)^3/cos(x)^2+1/6*sin(x)^5/cos(x)^6+1/24*sin(x)^5/cos(x)^
4-1/48*sin(x)^5/cos(x)^2-1/48*sin(x)^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (10) = 20\).
time = 0.43, size = 79, normalized size = 6.58 \begin {gather*} -\frac {3 \, \sin \left (x\right )^{5} + 8 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )}{48 \, {\left (\sin \left (x\right )^{6} - 3 \, \sin \left (x\right )^{4} + 3 \, \sin \left (x\right )^{2} - 1\right )}} + \frac {3 \, \sin \left (x\right )^{5} - 8 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )}{48 \, {\left (\sin \left (x\right )^{6} - 3 \, \sin \left (x\right )^{4} + 3 \, \sin \left (x\right )^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*sin(x)^5 + 8*sin(x)^3 - 3*sin(x))/(sin(x)^6 - 3*sin(x)^4 + 3*sin(x)^2 - 1) + 1/48*(3*sin(x)^5 - 8*sin
(x)^3 - 3*sin(x))/(sin(x)^6 - 3*sin(x)^4 + 3*sin(x)^2 - 1)

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Fricas [A]
time = 0.60, size = 14, normalized size = 1.17 \begin {gather*} -\frac {{\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{3 \, \cos \left (x\right )^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(cos(x)^2 - 1)*sin(x)/cos(x)^6

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (10) = 20\).
time = 0.09, size = 80, normalized size = 6.67 \begin {gather*} \frac {- 3 \sin ^{5}{\left (x \right )} - 8 \sin ^{3}{\left (x \right )} + 3 \sin {\left (x \right )}}{48 \sin ^{6}{\left (x \right )} - 144 \sin ^{4}{\left (x \right )} + 144 \sin ^{2}{\left (x \right )} - 48} + \frac {3 \sin ^{5}{\left (x \right )} - 8 \sin ^{3}{\left (x \right )} - 3 \sin {\left (x \right )}}{48 \sin ^{6}{\left (x \right )} - 144 \sin ^{4}{\left (x \right )} + 144 \sin ^{2}{\left (x \right )} - 48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**4*sec(x)**3+tan(x)**2*sec(x)**5,x)

[Out]

(-3*sin(x)**5 - 8*sin(x)**3 + 3*sin(x))/(48*sin(x)**6 - 144*sin(x)**4 + 144*sin(x)**2 - 48) + (3*sin(x)**5 - 8
*sin(x)**3 - 3*sin(x))/(48*sin(x)**6 - 144*sin(x)**4 + 144*sin(x)**2 - 48)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (10) = 20\).
time = 0.49, size = 55, normalized size = 4.58 \begin {gather*} -\frac {3 \, \sin \left (x\right )^{5} + 8 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )}{48 \, {\left (\sin \left (x\right )^{2} - 1\right )}^{3}} + \frac {3 \, \sin \left (x\right )^{5} - 8 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )}{48 \, {\left (\sin \left (x\right )^{2} - 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*sin(x)^5 + 8*sin(x)^3 - 3*sin(x))/(sin(x)^2 - 1)^3 + 1/48*(3*sin(x)^5 - 8*sin(x)^3 - 3*sin(x))/(sin(x
)^2 - 1)^3

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Mupad [B]
time = 0.33, size = 14, normalized size = 1.17 \begin {gather*} -\frac {{\sin \left (x\right )}^3}{3\,{\left ({\sin \left (x\right )}^2-1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^4/cos(x)^3 + tan(x)^2/cos(x)^5,x)

[Out]

-sin(x)^3/(3*(sin(x)^2 - 1)^3)

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Chatgpt [F] Failed to verify
time = 1.00, size = 13, normalized size = 1.08 \begin {gather*} \frac {\left (\sec ^{4}\left (x \right )\right )}{4}+\frac {\left (\sec ^{3}\left (x \right )\right )}{3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

int(tan(x)^4*sec(x)^3+tan(x)^2*sec(x)^5,x)

[Out]

1/4*sec(x)^4+1/3*sec(x)^3

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