3.2.89 \(\int \cot ^3(e+f x) (a+b \tan ^2(e+f x)) \, dx\) [189]

Optimal. Leaf size=34 \[ -\frac {a \cot ^2(e+f x)}{2 f}-\frac {(a-b) \log (\sin (e+f x))}{f} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3710, 12, 3556} \begin {gather*} -\frac {(a-b) \log (\sin (e+f x))}{f}-\frac {a \cot ^2(e+f x)}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2),x]

[Out]

-1/2*(a*Cot[e + f*x]^2)/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 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*} \int \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx &=-\frac {a \cot ^2(e+f x)}{2 f}-\int (a-b) \cot (e+f x) \, dx\\ &=-\frac {a \cot ^2(e+f x)}{2 f}-(a-b) \int \cot (e+f x) \, dx\\ &=-\frac {a \cot ^2(e+f x)}{2 f}-\frac {(a-b) \log (\sin (e+f x))}{f}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 56, normalized size = 1.65 \begin {gather*} \frac {b (\log (\cos (e+f x))+\log (\tan (e+f x)))}{f}-\frac {a \left (\cot ^2(e+f x)+2 \log (\cos (e+f x))+2 \log (\tan (e+f x))\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2),x]

[Out]

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

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Maple [A]
time = 0.13, size = 37, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {b \ln \left (\sin \left (f x +e \right )\right )+a \left (-\frac {\left (\cot ^{2}\left (f x +e \right )\right )}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )}{f}\)	\(37\)
default	\(\frac {b \ln \left (\sin \left (f x +e \right )\right )+a \left (-\frac {\left (\cot ^{2}\left (f x +e \right )\right )}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )}{f}\)	\(37\)
norman	\(-\frac {a}{2 f \tan \left (f x +e \right )^{2}}-\frac {\left (a -b \right ) \ln \left (\tan \left (f x +e \right )\right )}{f}+\frac {\left (a -b \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}\)	\(54\)
risch	\(i x a -i x b +\frac {2 i a e}{f}-\frac {2 i b e}{f}+\frac {2 a \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{f}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^3*(a+b*tan(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 33, normalized size = 0.97 \begin {gather*} -\frac {{\left (a - b\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {a}{\sin \left (f x + e\right )^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.41, size = 66, normalized size = 1.94 \begin {gather*} -\frac {{\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 )^{2} + a \tan \left (f x + e\right )^{2} + a}{2 \, f \tan \left (f x + e\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (27) = 54\).
time = 0.74, size = 100, normalized size = 2.94 \begin {gather*} \begin {cases} \tilde {\infty } a x & \text {for}\: \left (e = 0 \vee e = - f x\right ) \wedge \left (e = - f x \vee f = 0\right ) \\x \left (a + b \tan ^{2}{\left (e \right )}\right ) \cot ^{3}{\left (e \right )} & \text {for}\: f = 0 \\\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 {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} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3*(a+b*tan(f*x+e)**2),x)

[Out]

Piecewise((zoo*a*x, (Eq(e, 0) | Eq(e, -f*x)) & (Eq(f, 0) | Eq(e, -f*x))), (x*(a + b*tan(e)**2)*cot(e)**3, Eq(f
, 0)), (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) - b*log(tan(e + f*x
)**2 + 1)/(2*f) + b*log(tan(e + f*x))/f, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (34) = 68\).
time = 0.78, size = 165, normalized size = 4.85 \begin {gather*} -\frac {4 \, {\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 ) - 8 \, {\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 {4 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1}\right )} {\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 )}}{\cos \left (f x + e\right ) + 1}}{8 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-1/8*(4*(a - b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1)) - 8*(a - b)*log(abs(-(cos(f*x + e) - 1)/(cos
(f*x + e) + 1) + 1)) - (a + 4*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) +
1))*(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/f

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Mupad [B]
time = 11.62, size = 54, normalized size = 1.59 \begin {gather*} \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {a}{2}-\frac {b}{2}\right )}{f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a-b\right )}{f}-\frac {a\,{\mathrm {cot}\left (e+f\,x\right )}^2}{2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^3*(a + b*tan(e + f*x)^2),x)

[Out]

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

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