\(\int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [337]

Optimal result

Integrand size = 23, antiderivative size = 69 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b) \log (\cos (e+f x))}{b^2 f}-\frac {(a+b)^2 \log \left (b+a \cos ^2(e+f x)\right )}{2 a b^2 f}+\frac {\sec ^2(e+f x)}{2 b f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4223, 457, 90} \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {(a+b)^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a b^2 f}+\frac {(a+2 b) \log (\cos (e+f x))}{b^2 f}+\frac {\sec ^2(e+f x)}{2 b f} \]

[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 90

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]))

Rule 457

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 4223

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*} \text {integral}& = -\frac {\text {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 {\text {Subst}\left (\int \frac {(1-x)^2}{x^2 (b+a x)} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {\text {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] (verified)

Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.43 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (2 a (a+2 b) \log (\cos (e+f x))-(a+b)^2 \log \left (a+b-a \sin ^2(e+f x)\right )+a b \sec ^2(e+f x)\right )}{4 a b^2 f \left (a+b \sec ^2(e+f x)\right )} \]

[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))

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\frac {\left (a +2 b \right ) \ln \left (\cos \left (f x +e \right )\right )}{b^{2}}+\frac {1}{2 b \cos \left (f x +e \right )^{2}}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b^{2} a}}{f}\)	\(67\)
default	\(\frac {\frac {\left (a +2 b \right ) \ln \left (\cos \left (f x +e \right )\right )}{b^{2}}+\frac {1}{2 b \cos \left (f x +e \right )^{2}}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b^{2} a}}{f}\)	\(67\)
risch	\(\frac {i x}{a}+\frac {2 i e}{a f}+\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )}}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{b^{2} f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}-\frac {a \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 b^{2} f}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{b f}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f}\)	\(207\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} \log \left (-\cos \left (f x + e\right )\right ) - a b}{2 \, a b^{2} f \cos \left (f x + e\right )^{2}} \]

[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)

Sympy [F]

\[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}} - \frac {1}{b \sin \left (f x + e\right )^{2} - b} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a b^{2}}}{2 \, f} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (65) = 130\).

Time = 1.52 (sec) , antiderivative size = 373, normalized size of antiderivative = 5.41 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | -a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 2 \, a + 2 \, b \right |}\right )}{a^{2} b^{2} + a b^{3}} - \frac {\log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right )}{a} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right )}{b^{2}} + \frac {a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, a + 8 \, b}{b^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}}}{2 \, f} \]

[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(abs(-(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(abs(-(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

Mupad [B] (verification not implemented)

Time = 19.46 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{b\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,a\,f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,b\,f}-\frac {a\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,b^2\,f} \]

[In]

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

[Out]

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