3.5.66 \(\int \frac {\cot ^4(c+d x)}{a+b \tan (c+d x)} \, dx\) [466]

Optimal. Leaf size=133 \[ \frac {a x}{a^2+b^2}+\frac {\left (a^2-b^2\right ) \cot (c+d x)}{a^3 d}+\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {b \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac {b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 \left (a^2+b^2\right ) d} \]

[Out]

a*x/(a^2+b^2)+(a^2-b^2)*cot(d*x+c)/a^3/d+1/2*b*cot(d*x+c)^2/a^2/d-1/3*cot(d*x+c)^3/a/d+b*(a^2-b^2)*ln(sin(d*x+
c))/a^4/d+b^5*ln(a*cos(d*x+c)+b*sin(d*x+c))/a^4/(a^2+b^2)/d

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Rubi [A]
time = 0.34, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3650, 3730, 3731, 3732, 3611, 3556} \begin {gather*} \frac {a x}{a^2+b^2}+\frac {b \cot ^2(c+d x)}{2 a^2 d}+\frac {b \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac {b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 d \left (a^2+b^2\right )}+\frac {\left (a^2-b^2\right ) \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/(a^2 + b^2) + ((a^2 - b^2)*Cot[c + d*x])/(a^3*d) + (b*Cot[c + d*x]^2)/(2*a^2*d) - Cot[c + d*x]^3/(3*a*d)
+ (b*(a^2 - b^2)*Log[Sin[c + d*x]])/(a^4*d) + (b^5*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a^4*(a^2 + b^2)*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 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[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] && EqQ[a*c + b*d, 0]

Rule 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3731

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*t
an[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 + a^2*C)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x]
)^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[
e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2)) - a*C*(b*c*(m + 1)
+ a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 2)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^
2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3732

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

Rubi steps

\begin {align*} \int \frac {\cot ^4(c+d x)}{a+b \tan (c+d x)} \, dx &=-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\int \frac {\cot ^3(c+d x) \left (3 b+3 a \tan (c+d x)+3 b \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{3 a}\\ &=\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {\int \frac {\cot ^2(c+d x) \left (-6 \left (a^2-b^2\right )+6 b^2 \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{6 a^2}\\ &=\frac {\left (a^2-b^2\right ) \cot (c+d x)}{a^3 d}+\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\int \frac {\cot (c+d x) \left (-6 b \left (a^2-b^2\right )-6 a^3 \tan (c+d x)-6 b \left (a^2-b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{6 a^3}\\ &=\frac {a x}{a^2+b^2}+\frac {\left (a^2-b^2\right ) \cot (c+d x)}{a^3 d}+\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {\left (b \left (a^2-b^2\right )\right ) \int \cot (c+d x) \, dx}{a^4}+\frac {b^5 \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^4 \left (a^2+b^2\right )}\\ &=\frac {a x}{a^2+b^2}+\frac {\left (a^2-b^2\right ) \cot (c+d x)}{a^3 d}+\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {b \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac {b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 \left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.79, size = 131, normalized size = 0.98 \begin {gather*} -\frac {-\frac {6 \left (a^2-b^2\right ) \cot (c+d x)}{a^3}-\frac {3 b \cot ^2(c+d x)}{a^2}+\frac {2 \cot ^3(c+d x)}{a}+\frac {3 \log (i-\cot (c+d x))}{i a+b}+\frac {3 i \log (i+\cot (c+d x))}{a+i b}-\frac {6 b^5 \log (b+a \cot (c+d x))}{a^4 \left (a^2+b^2\right )}}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((-6*(a^2 - b^2)*Cot[c + d*x])/a^3 - (3*b*Cot[c + d*x]^2)/a^2 + (2*Cot[c + d*x]^3)/a + (3*Log[I - Cot[c +
d*x]])/(I*a + b) + ((3*I)*Log[I + Cot[c + d*x]])/(a + I*b) - (6*b^5*Log[b + a*Cot[c + d*x]])/(a^4*(a^2 + b^2)
))/d

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Maple [A]
time = 0.29, size = 137, normalized size = 1.03

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {-a^{2}+b^{2}}{a^{3} \tan \left (d x +c \right )}+\frac {\left (a^{2}-b^{2}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}+\frac {b}{2 a^{2} \tan \left (d x +c \right )^{2}}+\frac {b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{4}}+\frac {-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(137\)
default	\(\frac {-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {-a^{2}+b^{2}}{a^{3} \tan \left (d x +c \right )}+\frac {\left (a^{2}-b^{2}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}+\frac {b}{2 a^{2} \tan \left (d x +c \right )^{2}}+\frac {b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{4}}+\frac {-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(137\)
norman	\(\frac {\frac {a x \left (\tan ^{3}\left (d x +c \right )\right )}{a^{2}+b^{2}}+\frac {\left (a^{2}-b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{a^{3} d}-\frac {1}{3 d a}+\frac {b \tan \left (d x +c \right )}{2 a^{2} d}}{\tan \left (d x +c \right )^{3}}+\frac {b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) a^{4} d}+\frac {\left (a^{2}-b^{2}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}\)	\(159\)
risch	\(-\frac {x}{i b -a}-\frac {2 i b x}{a^{2}}-\frac {2 i b c}{a^{2} d}+\frac {2 i b^{3} x}{a^{4}}+\frac {2 i b^{3} c}{a^{4} d}-\frac {2 i b^{5} x}{\left (a^{2}+b^{2}\right ) a^{4}}-\frac {2 i b^{5} c}{\left (a^{2}+b^{2}\right ) a^{4} d}-\frac {2 i \left (-3 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 a^{2}+3 b^{2}\right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{2}+b^{2}\right ) a^{4} d}\)	\(306\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/3/a/tan(d*x+c)^3-(-a^2+b^2)/a^3/tan(d*x+c)+(a^2-b^2)/a^4*b*ln(tan(d*x+c))+1/2*b/a^2/tan(d*x+c)^2+b^5/(
a^2+b^2)/a^4*ln(a+b*tan(d*x+c))+1/(a^2+b^2)*(-1/2*b*ln(1+tan(d*x+c)^2)+a*arctan(tan(d*x+c))))

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Maxima [A]
time = 1.02, size = 145, normalized size = 1.09 \begin {gather*} \frac {\frac {6 \, b^{5} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + a^{4} b^{2}} + \frac {6 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac {3 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {6 \, {\left (a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}} + \frac {3 \, a b \tan \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 2 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*b^5*log(b*tan(d*x + c) + a)/(a^6 + a^4*b^2) + 6*(d*x + c)*a/(a^2 + b^2) - 3*b*log(tan(d*x + c)^2 + 1)/(
a^2 + b^2) + 6*(a^2*b - b^3)*log(tan(d*x + c))/a^4 + (3*a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan(d*x + c)^2 - 2*a^
2)/(a^3*tan(d*x + c)^3))/d

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Fricas [A]
time = 1.83, size = 207, normalized size = 1.56 \begin {gather*} \frac {3 \, b^{5} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} - 2 \, a^{5} - 2 \, a^{3} b^{2} + 3 \, {\left (a^{4} b - b^{5}\right )} \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 (2 \, a^{5} d x + a^{4} b + a^{2} b^{3}\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (a^{5} - a b^{4}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{4} b + a^{2} b^{3}\right )} \tan \left (d x + c\right )}{6 \, {\left (a^{6} + a^{4} b^{2}\right )} d \tan \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*b^5*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 - 2*a^5 -
2*a^3*b^2 + 3*(a^4*b - b^5)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 + 3*(2*a^5*d*x + a^4*b + a
^2*b^3)*tan(d*x + c)^3 + 6*(a^5 - a*b^4)*tan(d*x + c)^2 + 3*(a^4*b + a^2*b^3)*tan(d*x + c))/((a^6 + a^4*b^2)*d
*tan(d*x + c)^3)

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Sympy [C] Result contains complex when optimal does not.
time = 3.24, size = 1533, normalized size = 11.53 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(c, 0) & Eq(d, 0)), ((x - cot(c + d*x)**3/(3*d) + cot(c + d*x)/d)/a,
Eq(b, 0)), ((-log(tan(c + d*x)**2 + 1)/(2*d) + log(tan(c + d*x))/d + 1/(2*d*tan(c + d*x)**2) - 1/(4*d*tan(c +
d*x)**4))/b, Eq(a, 0)), (15*I*d*x*tan(c + d*x)**4/(6*b*d*tan(c + d*x)**4 - 6*I*b*d*tan(c + d*x)**3) + 15*d*x*
tan(c + d*x)**3/(6*b*d*tan(c + d*x)**4 - 6*I*b*d*tan(c + d*x)**3) + 6*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**4
/(6*b*d*tan(c + d*x)**4 - 6*I*b*d*tan(c + d*x)**3) - 6*I*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**3/(6*b*d*tan(c
+ d*x)**4 - 6*I*b*d*tan(c + d*x)**3) - 12*log(tan(c + d*x))*tan(c + d*x)**4/(6*b*d*tan(c + d*x)**4 - 6*I*b*d*
tan(c + d*x)**3) + 12*I*log(tan(c + d*x))*tan(c + d*x)**3/(6*b*d*tan(c + d*x)**4 - 6*I*b*d*tan(c + d*x)**3) +
15*I*tan(c + d*x)**3/(6*b*d*tan(c + d*x)**4 - 6*I*b*d*tan(c + d*x)**3) + 9*tan(c + d*x)**2/(6*b*d*tan(c + d*x)
**4 - 6*I*b*d*tan(c + d*x)**3) + I*tan(c + d*x)/(6*b*d*tan(c + d*x)**4 - 6*I*b*d*tan(c + d*x)**3) - 2/(6*b*d*t
an(c + d*x)**4 - 6*I*b*d*tan(c + d*x)**3), Eq(a, -I*b)), (-15*I*d*x*tan(c + d*x)**4/(6*b*d*tan(c + d*x)**4 + 6
*I*b*d*tan(c + d*x)**3) + 15*d*x*tan(c + d*x)**3/(6*b*d*tan(c + d*x)**4 + 6*I*b*d*tan(c + d*x)**3) + 6*log(tan
(c + d*x)**2 + 1)*tan(c + d*x)**4/(6*b*d*tan(c + d*x)**4 + 6*I*b*d*tan(c + d*x)**3) + 6*I*log(tan(c + d*x)**2
+ 1)*tan(c + d*x)**3/(6*b*d*tan(c + d*x)**4 + 6*I*b*d*tan(c + d*x)**3) - 12*log(tan(c + d*x))*tan(c + d*x)**4/
(6*b*d*tan(c + d*x)**4 + 6*I*b*d*tan(c + d*x)**3) - 12*I*log(tan(c + d*x))*tan(c + d*x)**3/(6*b*d*tan(c + d*x)
**4 + 6*I*b*d*tan(c + d*x)**3) - 15*I*tan(c + d*x)**3/(6*b*d*tan(c + d*x)**4 + 6*I*b*d*tan(c + d*x)**3) + 9*ta
n(c + d*x)**2/(6*b*d*tan(c + d*x)**4 + 6*I*b*d*tan(c + d*x)**3) - I*tan(c + d*x)/(6*b*d*tan(c + d*x)**4 + 6*I*
b*d*tan(c + d*x)**3) - 2/(6*b*d*tan(c + d*x)**4 + 6*I*b*d*tan(c + d*x)**3), Eq(a, I*b)), (zoo*x/a, Eq(c, -d*x)
), (x*cot(c)**4/(a + b*tan(c)), Eq(d, 0)), (6*a**5*d*x*tan(c + d*x)**3/(6*a**6*d*tan(c + d*x)**3 + 6*a**4*b**2
*d*tan(c + d*x)**3) + 6*a**5*tan(c + d*x)**2/(6*a**6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) - 2*a*
*5/(6*a**6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) - 3*a**4*b*log(tan(c + d*x)**2 + 1)*tan(c + d*x)
**3/(6*a**6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) + 6*a**4*b*log(tan(c + d*x))*tan(c + d*x)**3/(6
*a**6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) + 3*a**4*b*tan(c + d*x)/(6*a**6*d*tan(c + d*x)**3 + 6
*a**4*b**2*d*tan(c + d*x)**3) - 2*a**3*b**2/(6*a**6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) + 3*a**
2*b**3*tan(c + d*x)/(6*a**6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) - 6*a*b**4*tan(c + d*x)**2/(6*a
**6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) + 6*b**5*log(a/b + tan(c + d*x))*tan(c + d*x)**3/(6*a**
6*d*tan(c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3) - 6*b**5*log(tan(c + d*x))*tan(c + d*x)**3/(6*a**6*d*tan(
c + d*x)**3 + 6*a**4*b**2*d*tan(c + d*x)**3), True))

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Giac [A]
time = 0.79, size = 187, normalized size = 1.41 \begin {gather*} \frac {\frac {6 \, b^{6} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + a^{4} b^{3}} + \frac {6 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac {3 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {6 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {11 \, a^{2} b \tan \left (d x + c\right )^{3} - 11 \, b^{3} \tan \left (d x + c\right )^{3} - 6 \, a^{3} \tan \left (d x + c\right )^{2} + 6 \, a b^{2} \tan \left (d x + c\right )^{2} - 3 \, a^{2} b \tan \left (d x + c\right ) + 2 \, a^{3}}{a^{4} \tan \left (d x + c\right )^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*b^6*log(abs(b*tan(d*x + c) + a))/(a^6*b + a^4*b^3) + 6*(d*x + c)*a/(a^2 + b^2) - 3*b*log(tan(d*x + c)^2
+ 1)/(a^2 + b^2) + 6*(a^2*b - b^3)*log(abs(tan(d*x + c)))/a^4 - (11*a^2*b*tan(d*x + c)^3 - 11*b^3*tan(d*x + c
)^3 - 6*a^3*tan(d*x + c)^2 + 6*a*b^2*tan(d*x + c)^2 - 3*a^2*b*tan(d*x + c) + 2*a^3)/(a^4*tan(d*x + c)^3))/d

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Mupad [B]
time = 4.26, size = 153, normalized size = 1.15 \begin {gather*} \frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2-b^2\right )}{a^3}-\frac {1}{3\,a}+\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{2\,a^2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}+\frac {b^5\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a^4\,d\,\left (a^2+b^2\right )}+\frac {b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cot(c + d*x)^3*((tan(c + d*x)^2*(a^2 - b^2))/a^3 - 1/(3*a) + (b*tan(c + d*x))/(2*a^2)))/d - log(tan(c + d*x)
+ 1i)/(2*d*(a*1i + b)) - (log(tan(c + d*x) - 1i)*1i)/(2*d*(a + b*1i)) + (b^5*log(a + b*tan(c + d*x)))/(a^4*d*(
a^2 + b^2)) + (b*log(tan(c + d*x))*(a^2 - b^2))/(a^4*d)

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