3.337 \(\int \frac{\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=69 \[ -\frac{(a+b)^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a b^2 f}+\frac{(a+2 b) \log (\cos (e+f x))}{b^2 f}+\frac{\sec ^2(e+f x)}{2 b f} \]

[Out]

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

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Rubi [A]  time = 0.099578, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4138, 446, 88} \[ -\frac{(a+b)^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a b^2 f}+\frac{(a+2 b) \log (\cos (e+f x))}{b^2 f}+\frac{\sec ^2(e+f x)}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b)*Log[Cos[e + f*x]])/(b^2*f) - ((a + b)^2*Log[b + a*Cos[e + f*x]^2])/(2*a*b^2*f) + Sec[e + f*x]^2/(2*
b*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.293114, size = 99, normalized size = 1.43 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (a b \sec ^2(e+f x)+(a+b)^2 \left (-\log \left (-a \sin ^2(e+f x)+a+b\right )\right )+2 a (a+2 b) \log (\cos (e+f x))\right )}{4 a b^2 f \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.061, size = 112, normalized size = 1.6 \begin{align*} -{\frac{a\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f{b}^{2}}}-{\frac{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{fb}}-{\frac{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,af}}+{\frac{\ln \left ( \cos \left ( fx+e \right ) \right ) a}{f{b}^{2}}}+2\,{\frac{\ln \left ( \cos \left ( fx+e \right ) \right ) }{fb}}+{\frac{1}{2\,fb \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01583, size = 109, normalized size = 1.58 \begin{align*} \frac{\frac{{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}} - \frac{1}{b \sin \left (f x + e\right )^{2} - b} - \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a b^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a + 2*b)*log(sin(f*x + e)^2 - 1)/b^2 - 1/(b*sin(f*x + e)^2 - b) - (a^2 + 2*a*b + b^2)*log(a*sin(f*x + e)
^2 - a - b)/(a*b^2))/f

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Fricas [A]  time = 0.698029, size = 205, normalized size = 2.97 \begin{align*} -\frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \,{\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} \log \left (-\cos \left (f x + e\right )\right ) - a b}{2 \, a b^{2} f \cos \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.14862, size = 539, normalized size = 7.81 \begin{align*} -\frac{\frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | -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 )} - 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}\right )} - 2 \, a + 2 \, b \right |}\right )}{a^{2} b^{2} + a b^{3}} - \frac{\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 )}{a} - \frac{{\left (a + 2 \, b\right )} \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 )}{b^{2}} + \frac{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 \, 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}\right )} + 2 \, a + 8 \, b}{b^{2}{\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 )}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(abs(-a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(
cos(f*x + e) + 1)) - b*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) - 2*a +
2*b))/(a^2*b^2 + a*b^3) - log(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1)
+ 2)/a - (a + 2*b)*log(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2)/b^2
+ (a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) + 2*b*((cos(f*x + e) + 1
)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) + 2*a + 8*b)/(b^2*((cos(f*x + e) + 1)/(cos(f*x +
e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2)))/f