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

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Rubi [A]  time = 0.06, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [A]  time = 0.49, size = 69, normalized size = 0.96 \[ -\frac {12 \, a \cos \left (f x + e\right )^{6} \log \left (-\cos \left (f x + e\right )\right ) + 6 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, b}{12 \, f \cos \left (f x + e\right )^{6}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*(a/4*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-c
os(f*x+exp(1)))*(1+cos(f*x+exp(1)))+2))-a/4*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)
))*(1+cos(f*x+exp(1)))-2))+(11*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1
))))^3*a-90*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))^2*a+276*((1-co
s(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*a+128*b-280*a)*1/24/((1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1)))-2)^3)

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maple [A]  time = 0.62, size = 65, normalized size = 0.90 \[ \frac {\left (\tan ^{4}\left (f x +e \right )\right ) a}{4 f}-\frac {a \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}-\frac {a \ln \left (\cos \left (f x +e \right )\right )}{f}+\frac {b \left (\sin ^{6}\left (f x +e \right )\right )}{6 f \cos \left (f x +e \right )^{6}} \]

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 = 0.33, size = 95, normalized size = 1.32 \[ -\frac {6 \, a \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac {6 \, {\left (2 \, a - b\right )} \sin \left (f x + e\right )^{4} - 3 \, {\left (7 \, a - 2 \, b\right )} \sin \left (f x + e\right )^{2} + 9 \, a - 2 \, b}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{12 \, f} \]

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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mupad [B]  time = 4.70, size = 52, normalized size = 0.72 \[ \frac {\frac {a\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2}-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^4}{4}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^6}{6}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.17, size = 116, normalized size = 1.61 \[ \begin {cases} \frac {a \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {a \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac {b \tan ^{4}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{6 f} - \frac {b \tan ^{2}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{6 f} + \frac {b \sec ^{2}{\left (e + f x \right )}}{6 f} & \text {for}\: f \neq 0 \\x \left (a + b \sec ^{2}{\relax (e )}\right ) \tan ^{5}{\relax (e )} & \text {otherwise} \end {cases} \]

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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