3.322 \(\int (-\cos (x)+\sec (x))^4 \, dx\)

Optimal. Leaf size=44 \[ \frac{35 x}{8}+\frac{35 \tan ^3(x)}{24}-\frac{35 \tan (x)}{8}-\frac{1}{4} \sin ^4(x) \tan ^3(x)-\frac{7}{8} \sin ^2(x) \tan ^3(x) \]

[Out]

(35*x)/8 - (35*Tan[x])/8 + (35*Tan[x]^3)/24 - (7*Sin[x]^2*Tan[x]^3)/8 - (Sin[x]^4*Tan[x]^3)/4

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Rubi [A]  time = 0.0307354, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {288, 302, 203} \[ \frac{35 x}{8}+\frac{35 \tan ^3(x)}{24}-\frac{35 \tan (x)}{8}-\frac{1}{4} \sin ^4(x) \tan ^3(x)-\frac{7}{8} \sin ^2(x) \tan ^3(x) \]

Antiderivative was successfully verified.

[In]

Int[(-Cos[x] + Sec[x])^4,x]

[Out]

(35*x)/8 - (35*Tan[x])/8 + (35*Tan[x]^3)/24 - (7*Sin[x]^2*Tan[x]^3)/8 - (Sin[x]^4*Tan[x]^3)/4

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

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

Rubi steps

\begin{align*} \int (-\cos (x)+\sec (x))^4 \, dx &=\operatorname{Subst}\left (\int \frac{x^8}{\left (1+x^2\right )^3} \, dx,x,\tan (x)\right )\\ &=-\frac{1}{4} \sin ^4(x) \tan ^3(x)+\frac{7}{4} \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=-\frac{7}{8} \sin ^2(x) \tan ^3(x)-\frac{1}{4} \sin ^4(x) \tan ^3(x)+\frac{35}{8} \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-\frac{7}{8} \sin ^2(x) \tan ^3(x)-\frac{1}{4} \sin ^4(x) \tan ^3(x)+\frac{35}{8} \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\tan (x)\right )\\ &=-\frac{35 \tan (x)}{8}+\frac{35 \tan ^3(x)}{24}-\frac{7}{8} \sin ^2(x) \tan ^3(x)-\frac{1}{4} \sin ^4(x) \tan ^3(x)+\frac{35}{8} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac{35 x}{8}-\frac{35 \tan (x)}{8}+\frac{35 \tan ^3(x)}{24}-\frac{7}{8} \sin ^2(x) \tan ^3(x)-\frac{1}{4} \sin ^4(x) \tan ^3(x)\\ \end{align*}

Mathematica [A]  time = 0.0323595, size = 38, normalized size = 0.86 \[ \frac{35 x}{8}-\frac{3}{4} \sin (2 x)+\frac{1}{32} \sin (4 x)-\frac{10 \tan (x)}{3}+\frac{1}{3} \tan (x) \sec ^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(-Cos[x] + Sec[x])^4,x]

[Out]

(35*x)/8 - (3*Sin[2*x])/4 + Sin[4*x]/32 - (10*Tan[x])/3 + (Sec[x]^2*Tan[x])/3

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Maple [A]  time = 0.023, size = 40, normalized size = 0.9 \begin{align*} - \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( x \right ) \right ) ^{2}}{3}} \right ) \tan \left ( x \right ) -4\,\tan \left ( x \right ) +{\frac{35\,x}{8}}-2\,\cos \left ( x \right ) \sin \left ( x \right ) +{\frac{\sin \left ( x \right ) }{4} \left ( \left ( \cos \left ( x \right ) \right ) ^{3}+{\frac{3\,\cos \left ( x \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-cos(x)+sec(x))^4,x)

[Out]

-(-2/3-1/3*sec(x)^2)*tan(x)-4*tan(x)+35/8*x-2*cos(x)*sin(x)+1/4*(cos(x)^3+3/2*cos(x))*sin(x)

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Maxima [A]  time = 0.978038, size = 35, normalized size = 0.8 \begin{align*} \frac{1}{3} \, \tan \left (x\right )^{3} + \frac{35}{8} \, x + \frac{1}{32} \, \sin \left (4 \, x\right ) - \frac{3}{4} \, \sin \left (2 \, x\right ) - 3 \, \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cos(x)+sec(x))^4,x, algorithm="maxima")

[Out]

1/3*tan(x)^3 + 35/8*x + 1/32*sin(4*x) - 3/4*sin(2*x) - 3*tan(x)

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Fricas [A]  time = 2.10209, size = 116, normalized size = 2.64 \begin{align*} \frac{105 \, x \cos \left (x\right )^{3} +{\left (6 \, \cos \left (x\right )^{6} - 39 \, \cos \left (x\right )^{4} - 80 \, \cos \left (x\right )^{2} + 8\right )} \sin \left (x\right )}{24 \, \cos \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cos(x)+sec(x))^4,x, algorithm="fricas")

[Out]

1/24*(105*x*cos(x)^3 + (6*cos(x)^6 - 39*cos(x)^4 - 80*cos(x)^2 + 8)*sin(x))/cos(x)^3

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Sympy [A]  time = 32.1363, size = 44, normalized size = 1. \begin{align*} \frac{35 x}{8} - 2 \sin{\left (x \right )} \cos{\left (x \right )} - \frac{4 \sin{\left (x \right )}}{\cos{\left (x \right )}} + \frac{\sin{\left (2 x \right )}}{4} + \frac{\sin{\left (4 x \right )}}{32} + \frac{\tan ^{3}{\left (x \right )}}{3} + \tan{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cos(x)+sec(x))**4,x)

[Out]

35*x/8 - 2*sin(x)*cos(x) - 4*sin(x)/cos(x) + sin(2*x)/4 + sin(4*x)/32 + tan(x)**3/3 + tan(x)

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Giac [A]  time = 1.14325, size = 47, normalized size = 1.07 \begin{align*} \frac{1}{3} \, \tan \left (x\right )^{3} + \frac{35}{8} \, x - \frac{13 \, \tan \left (x\right )^{3} + 11 \, \tan \left (x\right )}{8 \,{\left (\tan \left (x\right )^{2} + 1\right )}^{2}} - 3 \, \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cos(x)+sec(x))^4,x, algorithm="giac")

[Out]

1/3*tan(x)^3 + 35/8*x - 1/8*(13*tan(x)^3 + 11*tan(x))/(tan(x)^2 + 1)^2 - 3*tan(x)