\(\int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx\) [432]

Optimal result

Integrand size = 21, antiderivative size = 78 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\left (a^2-b^2\right ) x+\frac {\left (a^2-b^2\right ) \cot (c+d x)}{d}-\frac {a b \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {2 a b \log (\sin (c+d x))}{d} \]

[Out]

(a^2-b^2)*x+(a^2-b^2)*cot(d*x+c)/d-a*b*cot(d*x+c)^2/d-1/3*a^2*cot(d*x+c)^3/d-2*a*b*ln(sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3623, 3610, 3612, 3556} \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right ) \cot (c+d x)}{d}+x \left (a^2-b^2\right )-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a b \cot ^2(c+d x)}{d}-\frac {2 a b \log (\sin (c+d x))}{d} \]

[In]

Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]

[Out]

(a^2 - b^2)*x + ((a^2 - b^2)*Cot[c + d*x])/d - (a*b*Cot[c + d*x]^2)/d - (a^2*Cot[c + d*x]^3)/(3*d) - (2*a*b*Lo
g[Sin[c + d*x]])/d

Rule 3556

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

Rule 3610

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

Rule 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3623

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(b*c - a*d)^2*((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*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \cot ^3(c+d x)}{3 d}+\int \cot ^3(c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx \\ & = -\frac {a b \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\int \cot ^2(c+d x) \left (-a^2+b^2-2 a b \tan (c+d x)\right ) \, dx \\ & = \frac {\left (a^2-b^2\right ) \cot (c+d x)}{d}-\frac {a b \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\int \cot (c+d x) \left (-2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx \\ & = \left (a^2-b^2\right ) x+\frac {\left (a^2-b^2\right ) \cot (c+d x)}{d}-\frac {a b \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-(2 a b) \int \cot (c+d x) \, dx \\ & = \left (a^2-b^2\right ) x+\frac {\left (a^2-b^2\right ) \cot (c+d x)}{d}-\frac {a b \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {2 a b \log (\sin (c+d x))}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.77 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.32 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {a^2 \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d}-\frac {b^2 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )}{d}-\frac {a b \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{d} \]

[In]

Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]

[Out]

-1/3*(a^2*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/d - (b^2*Cot[c + d*x]*Hypergeometr
ic2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d - (a*b*(Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/d

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+2 a b \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(75\)
default	\(\frac {a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+2 a b \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(75\)
parallelrisch	\(\frac {-\left (\cot ^{3}\left (d x +c \right )\right ) a^{2}-3 \left (\cot ^{2}\left (d x +c \right )\right ) a b +3 a^{2} d x -3 x d \,b^{2}+3 \cot \left (d x +c \right ) a^{2}-3 \cot \left (d x +c \right ) b^{2}-6 a b \ln \left (\tan \left (d x +c \right )\right )+3 a b \ln \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}\)	\(92\)
norman	\(\frac {\left (a^{2}-b^{2}\right ) x \left (\tan ^{3}\left (d x +c \right )\right )+\frac {\left (a^{2}-b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2}}{3 d}-\frac {a b \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{3}}+\frac {a b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a b \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(104\)
risch	\(2 i a b x +a^{2} x -b^{2} x +\frac {4 i a b c}{d}-\frac {2 i \left (6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 a^{2}+3 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(161\)

[In]

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c)+2*a*b*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c)))+b^2*(-cot(d*x+c)-d*x-c)
)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.37 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {3 \, a b \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} - 3 \, {\left ({\left (a^{2} - b^{2}\right )} d x - a b\right )} \tan \left (d x + c\right )^{3} + 3 \, a b \tan \left (d x + c\right ) - 3 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2}}{3 \, d \tan \left (d x + c\right )^{3}} \]

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(3*a*b*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 - 3*((a^2 - b^2)*d*x - a*b)*tan(d*x + c)^3
+ 3*a*b*tan(d*x + c) - 3*(a^2 - b^2)*tan(d*x + c)^2 + a^2)/(d*tan(d*x + c)^3)

Sympy [A] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.59 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\begin {cases} \tilde {\infty } a^{2} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } a^{2} x & \text {for}\: c = - d x \\a^{2} x + \frac {a^{2}}{d \tan {\left (c + d x \right )}} - \frac {a^{2}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {2 a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {a b}{d \tan ^{2}{\left (c + d x \right )}} - b^{2} x - \frac {b^{2}}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cot(d*x+c)**4*(a+b*tan(d*x+c))**2,x)

[Out]

Piecewise((zoo*a**2*x, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**2*cot(c)**4, Eq(d, 0)), (zoo*a**2*x, Eq(c, -d*
x)), (a**2*x + a**2/(d*tan(c + d*x)) - a**2/(3*d*tan(c + d*x)**3) + a*b*log(tan(c + d*x)**2 + 1)/d - 2*a*b*log
(tan(c + d*x))/d - a*b/(d*tan(c + d*x)**2) - b**2*x - b**2/(d*tan(c + d*x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.17 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, a b \log \left (\tan \left (d x + c\right )\right ) + 3 \, {\left (a^{2} - b^{2}\right )} {\left (d x + c\right )} - \frac {3 \, a b \tan \left (d x + c\right ) - 3 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(3*a*b*log(tan(d*x + c)^2 + 1) - 6*a*b*log(tan(d*x + c)) + 3*(a^2 - b^2)*(d*x + c) - (3*a*b*tan(d*x + c) -
3*(a^2 - b^2)*tan(d*x + c)^2 + a^2)/tan(d*x + c)^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (76) = 152\).

Time = 0.76 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.45 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 \, a b \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, {\left (a^{2} - b^{2}\right )} {\left (d x + c\right )} + \frac {88 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 - 6*a*b*tan(1/2*d*x + 1/2*c)^2 + 48*a*b*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 48*
a*b*log(abs(tan(1/2*d*x + 1/2*c))) - 15*a^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c) + 24*(a^2 - b^2
)*(d*x + c) + (88*a*b*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 -
6*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 4.91 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\frac {a^2}{3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2-b^2\right )+a\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {2\,a\,b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{2\,d} \]

[In]

int(cot(c + d*x)^4*(a + b*tan(c + d*x))^2,x)

[Out]

(log(tan(c + d*x) + 1i)*(a - b*1i)^2*1i)/(2*d) - (cot(c + d*x)^3*(a^2/3 - tan(c + d*x)^2*(a^2 - b^2) + a*b*tan
(c + d*x)))/d + (log(tan(c + d*x) - 1i)*(a*1i - b)^2*1i)/(2*d) - (2*a*b*log(tan(c + d*x)))/d