\(\int \tan ^5(e+f x) (a+b \tan ^2(e+f x)) \, dx\) [185]

Optimal result

Integrand size = 21, antiderivative size = 74 \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=-\frac {(a-b) \log (\cos (e+f x))}{f}-\frac {(a-b) \tan ^2(e+f x)}{2 f}+\frac {(a-b) \tan ^4(e+f x)}{4 f}+\frac {b \tan ^6(e+f x)}{6 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3712, 3554, 3556} \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {(a-b) \tan ^4(e+f x)}{4 f}-\frac {(a-b) \tan ^2(e+f x)}{2 f}-\frac {(a-b) \log (\cos (e+f x))}{f}+\frac {b \tan ^6(e+f x)}{6 f} \]

[In]

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

[Out]

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

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3712

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[A - C, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[
{a, b, e, f, A, C, m}, x] && NeQ[A*b^2 + a^2*C, 0] &&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {b \tan ^6(e+f x)}{6 f}+(a-b) \int \tan ^5(e+f x) \, dx \\ & = \frac {(a-b) \tan ^4(e+f x)}{4 f}+\frac {b \tan ^6(e+f x)}{6 f}+(-a+b) \int \tan ^3(e+f x) \, dx \\ & = -\frac {(a-b) \tan ^2(e+f x)}{2 f}+\frac {(a-b) \tan ^4(e+f x)}{4 f}+\frac {b \tan ^6(e+f x)}{6 f}+(a-b) \int \tan (e+f x) \, dx \\ & = -\frac {(a-b) \log (\cos (e+f x))}{f}-\frac {(a-b) \tan ^2(e+f x)}{2 f}+\frac {(a-b) \tan ^4(e+f x)}{4 f}+\frac {b \tan ^6(e+f x)}{6 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {12 (-a+b) \log (\cos (e+f x))-6 (a-b) \tan ^2(e+f x)+3 (a-b) \tan ^4(e+f x)+2 b \tan ^6(e+f x)}{12 f} \]

[In]

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

[Out]

(12*(-a + b)*Log[Cos[e + f*x]] - 6*(a - b)*Tan[e + f*x]^2 + 3*(a - b)*Tan[e + f*x]^4 + 2*b*Tan[e + f*x]^6)/(12
*f)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99

method	result	size
norman	\(\frac {b \tan \left (f x +e \right )^{6}}{6 f}-\frac {\left (a -b \right ) \tan \left (f x +e \right )^{2}}{2 f}+\frac {\left (a -b \right ) \tan \left (f x +e \right )^{4}}{4 f}+\frac {\left (a -b \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}\)	\(73\)
derivativedivides	\(\frac {\frac {b \tan \left (f x +e \right )^{6}}{6}+\frac {a \tan \left (f x +e \right )^{4}}{4}-\frac {b \tan \left (f x +e \right )^{4}}{4}-\frac {a \tan \left (f x +e \right )^{2}}{2}+\frac {b \tan \left (f x +e \right )^{2}}{2}+\frac {\left (a -b \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}}{f}\)	\(79\)
default	\(\frac {\frac {b \tan \left (f x +e \right )^{6}}{6}+\frac {a \tan \left (f x +e \right )^{4}}{4}-\frac {b \tan \left (f x +e \right )^{4}}{4}-\frac {a \tan \left (f x +e \right )^{2}}{2}+\frac {b \tan \left (f x +e \right )^{2}}{2}+\frac {\left (a -b \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}}{f}\)	\(79\)
parallelrisch	\(\frac {2 b \tan \left (f x +e \right )^{6}+3 a \tan \left (f x +e \right )^{4}-3 b \tan \left (f x +e \right )^{4}-6 a \tan \left (f x +e \right )^{2}+6 b \tan \left (f x +e \right )^{2}+6 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) a -6 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) b}{12 f}\)	\(90\)
parts	\(\frac {a \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {b \left (\frac {\tan \left (f x +e \right )^{6}}{6}-\frac {\tan \left (f x +e \right )^{4}}{4}+\frac {\tan \left (f x +e \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}\)	\(90\)
risch	\(i x a -i x b +\frac {2 i a e}{f}-\frac {2 i b e}{f}-\frac {2 \left (6 a \,{\mathrm e}^{10 i \left (f x +e \right )}-9 b \,{\mathrm e}^{10 i \left (f x +e \right )}+18 a \,{\mathrm e}^{8 i \left (f x +e \right )}-18 b \,{\mathrm e}^{8 i \left (f x +e \right )}+24 a \,{\mathrm e}^{6 i \left (f x +e \right )}-34 b \,{\mathrm e}^{6 i \left (f x +e \right )}+18 a \,{\mathrm e}^{4 i \left (f x +e \right )}-18 b \,{\mathrm e}^{4 i \left (f x +e \right )}+6 a \,{\mathrm e}^{2 i \left (f x +e \right )}-9 b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{6}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) b}{f}\)	\(202\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {2 \, b \tan \left (f x + e\right )^{6} + 3 \, {\left (a - b\right )} \tan \left (f x + e\right )^{4} - 6 \, {\left (a - b\right )} \tan \left (f x + e\right )^{2} - 6 \, {\left (a - b\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, f} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\begin {cases} \frac {a \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {a \tan ^{2}{\left (e + f x \right )}}{2 f} - \frac {b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {b \tan ^{6}{\left (e + f x \right )}}{6 f} - \frac {b \tan ^{4}{\left (e + f x \right )}}{4 f} + \frac {b \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right ) \tan ^{5}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.34 \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=-\frac {6 \, {\left (a - b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac {6 \, {\left (2 \, a - 3 \, b\right )} \sin \left (f x + e\right )^{4} - 3 \, {\left (7 \, a - 9 \, b\right )} \sin \left (f x + e\right )^{2} + 9 \, a - 11 \, b}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{12 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1450 vs. \(2 (68) = 136\).

Time = 3.79 (sec) , antiderivative size = 1450, normalized size of antiderivative = 19.59 \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/12*(6*a*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 +
1))*tan(f*x)^6*tan(e)^6 - 6*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f
*x)^2 + tan(e)^2 + 1))*tan(f*x)^6*tan(e)^6 + 9*a*tan(f*x)^6*tan(e)^6 - 11*b*tan(f*x)^6*tan(e)^6 - 36*a*log(4*(
tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^5*tan
(e)^5 + 36*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2
+ 1))*tan(f*x)^5*tan(e)^5 + 6*a*tan(f*x)^6*tan(e)^4 - 6*b*tan(f*x)^6*tan(e)^4 - 42*a*tan(f*x)^5*tan(e)^5 + 54*
b*tan(f*x)^5*tan(e)^5 + 6*a*tan(f*x)^4*tan(e)^6 - 6*b*tan(f*x)^4*tan(e)^6 + 90*a*log(4*(tan(f*x)^2*tan(e)^2 -
2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^4*tan(e)^4 - 90*b*log(4*(ta
n(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^4*tan(e
)^4 - 3*a*tan(f*x)^6*tan(e)^2 + 3*b*tan(f*x)^6*tan(e)^2 - 36*a*tan(f*x)^5*tan(e)^3 + 36*b*tan(f*x)^5*tan(e)^3
+ 69*a*tan(f*x)^4*tan(e)^4 - 99*b*tan(f*x)^4*tan(e)^4 - 36*a*tan(f*x)^3*tan(e)^5 + 36*b*tan(f*x)^3*tan(e)^5 -
3*a*tan(f*x)^2*tan(e)^6 + 3*b*tan(f*x)^2*tan(e)^6 - 120*a*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/
(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^3*tan(e)^3 + 120*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*
tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^3*tan(e)^3 - 2*b*tan(f*x)^6 +
6*a*tan(f*x)^5*tan(e) - 18*b*tan(f*x)^5*tan(e) + 60*a*tan(f*x)^4*tan(e)^2 - 90*b*tan(f*x)^4*tan(e)^2 - 72*a*t
an(f*x)^3*tan(e)^3 + 72*b*tan(f*x)^3*tan(e)^3 + 60*a*tan(f*x)^2*tan(e)^4 - 90*b*tan(f*x)^2*tan(e)^4 + 6*a*tan(
f*x)*tan(e)^5 - 18*b*tan(f*x)*tan(e)^5 - 2*b*tan(e)^6 + 90*a*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) +
1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^2*tan(e)^2 - 90*b*log(4*(tan(f*x)^2*tan(e)^2 -
2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^2*tan(e)^2 - 3*a*tan(f*x)^4
+ 3*b*tan(f*x)^4 - 36*a*tan(f*x)^3*tan(e) + 36*b*tan(f*x)^3*tan(e) + 69*a*tan(f*x)^2*tan(e)^2 - 99*b*tan(f*x)
^2*tan(e)^2 - 36*a*tan(f*x)*tan(e)^3 + 36*b*tan(f*x)*tan(e)^3 - 3*a*tan(e)^4 + 3*b*tan(e)^4 - 36*a*log(4*(tan(
f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)*tan(e) +
36*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*ta
n(f*x)*tan(e) + 6*a*tan(f*x)^2 - 6*b*tan(f*x)^2 - 42*a*tan(f*x)*tan(e) + 54*b*tan(f*x)*tan(e) + 6*a*tan(e)^2 -
6*b*tan(e)^2 + 6*a*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + ta
n(e)^2 + 1)) - 6*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan
(e)^2 + 1)) + 9*a - 11*b)/(f*tan(f*x)^6*tan(e)^6 - 6*f*tan(f*x)^5*tan(e)^5 + 15*f*tan(f*x)^4*tan(e)^4 - 20*f*t
an(f*x)^3*tan(e)^3 + 15*f*tan(f*x)^2*tan(e)^2 - 6*f*tan(f*x)*tan(e) + f)

Mupad [B] (verification not implemented)

Time = 11.68 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {a}{4}-\frac {b}{4}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a}{2}-\frac {b}{2}\right )+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^6}{6}+\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {a}{2}-\frac {b}{2}\right )}{f} \]

[In]

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

[Out]

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