3.311 \(\int (a+b \sec ^2(e+f x)) \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=72 \[ \frac{(a-2 b) \sec ^4(e+f x)}{4 f}-\frac{(2 a-b) \sec ^2(e+f x)}{2 f}-\frac{a \log (\cos (e+f x))}{f}+\frac{b \sec ^6(e+f x)}{6 f} \]

[Out]

-((a*Log[Cos[e + f*x]])/f) - ((2*a - b)*Sec[e + f*x]^2)/(2*f) + ((a - 2*b)*Sec[e + f*x]^4)/(4*f) + (b*Sec[e +
f*x]^6)/(6*f)

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Rubi [A]  time = 0.0619932, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4138, 446, 76} \[ \frac{(a-2 b) \sec ^4(e+f x)}{4 f}-\frac{(2 a-b) \sec ^2(e+f x)}{2 f}-\frac{a \log (\cos (e+f x))}{f}+\frac{b \sec ^6(e+f x)}{6 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)*Tan[e + f*x]^5,x]

[Out]

-((a*Log[Cos[e + f*x]])/f) - ((2*a - b)*Sec[e + f*x]^2)/(2*f) + ((a - 2*b)*Sec[e + f*x]^4)/(4*f) + (b*Sec[e +
f*x]^6)/(6*f)

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2 \left (b+a x^2\right )}{x^7} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2 (b+a x)}{x^4} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b}{x^4}+\frac{a-2 b}{x^3}+\frac{-2 a+b}{x^2}+\frac{a}{x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{a \log (\cos (e+f x))}{f}-\frac{(2 a-b) \sec ^2(e+f x)}{2 f}+\frac{(a-2 b) \sec ^4(e+f x)}{4 f}+\frac{b \sec ^6(e+f x)}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.177252, size = 55, normalized size = 0.76 \[ \frac{b \tan ^6(e+f x)}{6 f}-\frac{a \left (-\tan ^4(e+f x)+2 \tan ^2(e+f x)+4 \log (\cos (e+f x))\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x]^2)*Tan[e + f*x]^5,x]

[Out]

(b*Tan[e + f*x]^6)/(6*f) - (a*(4*Log[Cos[e + f*x]] + 2*Tan[e + f*x]^2 - Tan[e + f*x]^4))/(4*f)

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Maple [A]  time = 0.051, size = 65, normalized size = 0.9 \begin{align*}{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{4}a}{4\,f}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}a}{2\,f}}-{\frac{a\ln \left ( \cos \left ( fx+e \right ) \right ) }{f}}+{\frac{b \left ( \sin \left ( fx+e \right ) \right ) ^{6}}{6\,f \left ( \cos \left ( fx+e \right ) \right ) ^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x)

[Out]

1/4/f*tan(f*x+e)^4*a-1/2/f*a*tan(f*x+e)^2-a*ln(cos(f*x+e))/f+1/6/f*b*sin(f*x+e)^6/cos(f*x+e)^6

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Maxima [A]  time = 0.999829, size = 128, normalized size = 1.78 \begin{align*} -\frac{6 \, a \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac{6 \,{\left (2 \, a - b\right )} \sin \left (f x + e\right )^{4} - 3 \,{\left (7 \, a - 2 \, b\right )} \sin \left (f x + e\right )^{2} + 9 \, a - 2 \, b}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

-1/12*(6*a*log(sin(f*x + e)^2 - 1) - (6*(2*a - b)*sin(f*x + e)^4 - 3*(7*a - 2*b)*sin(f*x + e)^2 + 9*a - 2*b)/(
sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(f*x + e)^2 - 1))/f

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Fricas [A]  time = 0.541261, size = 177, normalized size = 2.46 \begin{align*} -\frac{12 \, a \cos \left (f x + e\right )^{6} \log \left (-\cos \left (f x + e\right )\right ) + 6 \,{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{4} - 3 \,{\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, b}{12 \, f \cos \left (f x + e\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

-1/12*(12*a*cos(f*x + e)^6*log(-cos(f*x + e)) + 6*(2*a - b)*cos(f*x + e)^4 - 3*(a - 2*b)*cos(f*x + e)^2 - 2*b)
/(f*cos(f*x + e)^6)

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Sympy [A]  time = 8.28924, size = 116, normalized size = 1.61 \begin{align*} \begin{cases} \frac{a \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{a \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac{a \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac{b \tan ^{4}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{6 f} - \frac{b \tan ^{2}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{6 f} + \frac{b \sec ^{2}{\left (e + f x \right )}}{6 f} & \text{for}\: f \neq 0 \\x \left (a + b \sec ^{2}{\left (e \right )}\right ) \tan ^{5}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)**2)*tan(f*x+e)**5,x)

[Out]

Piecewise((a*log(tan(e + f*x)**2 + 1)/(2*f) + a*tan(e + f*x)**4/(4*f) - a*tan(e + f*x)**2/(2*f) + b*tan(e + f*
x)**4*sec(e + f*x)**2/(6*f) - b*tan(e + f*x)**2*sec(e + f*x)**2/(6*f) + b*sec(e + f*x)**2/(6*f), Ne(f, 0)), (x
*(a + b*sec(e)**2)*tan(e)**5, True))

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Giac [B]  time = 3.22155, size = 408, normalized size = 5.67 \begin{align*} \frac{6 \, a \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right ) - 6 \, a \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2\right ) + \frac{11 \, a{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{3} + 90 \, a{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{2} + 276 \, a{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 280 \, a - 128 \, b}{{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}^{3}}}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x, algorithm="giac")

[Out]

1/12*(6*a*log(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2) - 6*a*log(-(
cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2) + (11*a*((cos(f*x + e) + 1)/
(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^3 + 90*a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) +
(cos(f*x + e) - 1)/(cos(f*x + e) + 1))^2 + 276*a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(
cos(f*x + e) + 1)) + 280*a - 128*b)/((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e)
+ 1) + 2)^3)/f