\(\int \cot ^5(e+f x) (a+b \tan ^2(e+f x)) \, dx\) [190]

Optimal result

Integrand size = 21, antiderivative size = 53 \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {(a-b) \cot ^2(e+f x)}{2 f}-\frac {a \cot ^4(e+f x)}{4 f}+\frac {(a-b) \log (\sin (e+f x))}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3710, 12, 3554, 3556} \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {(a-b) \cot ^2(e+f x)}{2 f}+\frac {(a-b) \log (\sin (e+f x))}{f}-\frac {a \cot ^4(e+f x)}{4 f} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3710

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

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {2 (a-b) \cot ^2(e+f x)-a \cot ^4(e+f x)+4 (a-b) (\log (\cos (e+f x))+\log (\tan (e+f x)))}{4 f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {b \left (-\frac {\cot \left (f x +e \right )^{2}}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )+a \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )}{f}\)	\(58\)
default	\(\frac {b \left (-\frac {\cot \left (f x +e \right )^{2}}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )+a \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )}{f}\)	\(58\)
parallelrisch	\(\frac {\left (-2 a +2 b \right ) \ln \left (\sec \left (f x +e \right )^{2}\right )+\left (4 a -4 b \right ) \ln \left (\tan \left (f x +e \right )\right )-\cot \left (f x +e \right )^{2} \left (\cot \left (f x +e \right )^{2} a -2 a +2 b \right )}{4 f}\)	\(66\)
norman	\(\frac {-\frac {a}{4 f}+\frac {\left (a -b \right ) \tan \left (f x +e \right )^{2}}{2 f}}{\tan \left (f x +e \right )^{4}}+\frac {\left (a -b \right ) \ln \left (\tan \left (f x +e \right )\right )}{f}-\frac {\left (a -b \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}\)	\(73\)
risch	\(-i x a +i x b -\frac {2 i a e}{f}+\frac {2 i b e}{f}-\frac {2 \left (2 a \,{\mathrm e}^{6 i \left (f x +e \right )}-b \,{\mathrm e}^{6 i \left (f x +e \right )}-2 a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{f}\)	\(154\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.60 \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {2 \, {\left (a - b\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{4} + {\left (3 \, a - 2 \, b\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a - b\right )} \tan \left (f x + e\right )^{2} - a}{4 \, f \tan \left (f x + e\right )^{4}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (42) = 84\).

Time = 1.32 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.34 \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\begin {cases} \tilde {\infty } a x & \text {for}\: e = 0 \wedge f = 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right ) \cot ^{5}{\left (e \right )} & \text {for}\: f = 0 \\\tilde {\infty } a x & \text {for}\: e = - f x \\- \frac {a \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a \log {\left (\tan {\left (e + f x \right )} \right )}}{f} + \frac {a}{2 f \tan ^{2}{\left (e + f x \right )}} - \frac {a}{4 f \tan ^{4}{\left (e + f x \right )}} + \frac {b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - \frac {b \log {\left (\tan {\left (e + f x \right )} \right )}}{f} - \frac {b}{2 f \tan ^{2}{\left (e + f x \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*a*x, Eq(e, 0) & Eq(f, 0)), (x*(a + b*tan(e)**2)*cot(e)**5, Eq(f, 0)), (zoo*a*x, Eq(e, -f*x)), (
-a*log(tan(e + f*x)**2 + 1)/(2*f) + a*log(tan(e + f*x))/f + a/(2*f*tan(e + f*x)**2) - a/(4*f*tan(e + f*x)**4)
+ b*log(tan(e + f*x)**2 + 1)/(2*f) - b*log(tan(e + f*x))/f - b/(2*f*tan(e + f*x)**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98 \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {2 \, {\left (a - b\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {2 \, {\left (2 \, a - b\right )} \sin \left (f x + e\right )^{2} - a}{\sin \left (f x + e\right )^{4}}}{4 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (49) = 98\).

Time = 0.73 (sec) , antiderivative size = 245, normalized size of antiderivative = 4.62 \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {32 \, {\left (a - b\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right ) - 64 \, {\left (a - b\right )} \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a + \frac {12 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {48 \, a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {48 \, b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) - 1\right )}^{2}} - \frac {12 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{64 \, f} \]

[In]

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

[Out]

1/64*(32*(a - b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1)) - 64*(a - b)*log(abs(-(cos(f*x + e) - 1)/(c
os(f*x + e) + 1) + 1)) - (a + 12*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8*b*(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) + 48*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 48*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos
(f*x + e) + 1)^2/(cos(f*x + e) - 1)^2 - 12*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*b*(cos(f*x + e) - 1)/(c
os(f*x + e) + 1) - a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/f

Mupad [B] (verification not implemented)

Time = 11.97 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a-b\right )}{f}-\frac {\frac {a}{4}-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a}{2}-\frac {b}{2}\right )}{f\,{\mathrm {tan}\left (e+f\,x\right )}^4}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {a}{2}-\frac {b}{2}\right )}{f} \]

[In]

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

[Out]

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