3.274 \(\int \frac{\cot ^4(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=169 \[ \frac{\left (a^2 A+a b B-A b^2\right ) \cot (c+d x)}{a^3 d}+\frac{\left (a^2-b^2\right ) (A b-a B) \log (\sin (c+d x))}{a^4 d}+\frac{b^4 (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 d \left (a^2+b^2\right )}+\frac{x (a A+b B)}{a^2+b^2}+\frac{(A b-a B) \cot ^2(c+d x)}{2 a^2 d}-\frac{A \cot ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.832577, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3609, 3649, 3651, 3530, 3475} \[ \frac{\left (a^2 A+a b B-A b^2\right ) \cot (c+d x)}{a^3 d}+\frac{\left (a^2-b^2\right ) (A b-a B) \log (\sin (c+d x))}{a^4 d}+\frac{b^4 (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 d \left (a^2+b^2\right )}+\frac{x (a A+b B)}{a^2+b^2}+\frac{(A b-a B) \cot ^2(c+d x)}{2 a^2 d}-\frac{A \cot ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3609

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

Rule 3649

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*T
an[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 3651

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]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), 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 3475

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

Rubi steps

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

Mathematica [C]  time = 2.48529, size = 194, normalized size = 1.15 \[ \frac{\frac{6 \left (a^2 A+a b B-A b^2\right ) \cot (c+d x)}{a^3}+\frac{6 b^4 (A b-a B) \log (a+b \tan (c+d x))}{a^4 \left (a^2+b^2\right )}+\frac{3 (A b-a B) \cot ^2(c+d x)}{a^2}+\frac{6 (a-b) (a+b) (A b-a B) \log (\tan (c+d x))}{a^4}+\frac{3 (B-i A) \log (-\tan (c+d x)+i)}{a+i b}+\frac{3 (B+i A) \log (\tan (c+d x)+i)}{a-i b}-\frac{2 A \cot ^3(c+d x)}{a}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.11, size = 337, normalized size = 2. \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Ab}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) aB}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{A}{3\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{Ab}{2\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{B}{2\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{A}{ad\tan \left ( dx+c \right ) }}-{\frac{A{b}^{2}}{{a}^{3}d\tan \left ( dx+c \right ) }}+{\frac{Bb}{{a}^{2}d\tan \left ( dx+c \right ) }}+{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) Ab}{{a}^{2}d}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) A{b}^{3}}{{a}^{4}d}}-{\frac{B\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{{a}^{3}d}}+{\frac{{b}^{5}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{{a}^{4}d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{{b}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{{a}^{3}d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.48449, size = 270, normalized size = 1.6 \begin{align*} \frac{\frac{6 \,{\left (A a + B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{6 \,{\left (B a b^{4} - A b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + a^{4} b^{2}} + \frac{3 \,{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{6 \,{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}} - \frac{2 \, A a^{2} - 6 \,{\left (A a^{2} + B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (B a^{2} - A a b\right )} \tan \left (d x + c\right )}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.0828, size = 644, normalized size = 3.81 \begin{align*} -\frac{2 \, A a^{5} + 2 \, A a^{3} b^{2} + 3 \,{\left (B a^{5} - A a^{4} b - B a b^{4} + A 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 (B a b^{4} - A b^{5}\right )} \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} + 3 \,{\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \,{\left (A a^{5} + B a^{4} b\right )} d x\right )} \tan \left (d x + c\right )^{3} - 6 \,{\left (A a^{5} + B a^{4} b + B a^{2} b^{3} - A a b^{4}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A 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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*A*a^5 + 2*A*a^3*b^2 + 3*(B*a^5 - A*a^4*b - B*a*b^4 + A*b^5)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*t
an(d*x + c)^3 + 3*(B*a*b^4 - A*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 + 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*(A*a^5 + B*a^4*b)*d*x)*tan(d*x + c)^3 - 6*(A*a
^5 + B*a^4*b + B*a^2*b^3 - A*a*b^4)*tan(d*x + c)^2 + 3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.27118, size = 385, normalized size = 2.28 \begin{align*} \frac{\frac{6 \,{\left (A a + B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{3 \,{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{6 \,{\left (B a b^{5} - A b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + a^{4} b^{3}} - \frac{6 \,{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} + \frac{11 \, B a^{3} \tan \left (d x + c\right )^{3} - 11 \, A a^{2} b \tan \left (d x + c\right )^{3} - 11 \, B a b^{2} \tan \left (d x + c\right )^{3} + 11 \, A b^{3} \tan \left (d x + c\right )^{3} + 6 \, A a^{3} \tan \left (d x + c\right )^{2} + 6 \, B a^{2} b \tan \left (d x + c\right )^{2} - 6 \, A a b^{2} \tan \left (d x + c\right )^{2} - 3 \, B a^{3} \tan \left (d x + c\right ) + 3 \, A a^{2} b \tan \left (d x + c\right ) - 2 \, A a^{3}}{a^{4} \tan \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(A*a + B*b)*(d*x + c)/(a^2 + b^2) + 3*(B*a - A*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) - 6*(B*a*b^5 - A*
b^6)*log(abs(b*tan(d*x + c) + a))/(a^6*b + a^4*b^3) - 6*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(abs(tan(d*x +
c)))/a^4 + (11*B*a^3*tan(d*x + c)^3 - 11*A*a^2*b*tan(d*x + c)^3 - 11*B*a*b^2*tan(d*x + c)^3 + 11*A*b^3*tan(d*x
+ c)^3 + 6*A*a^3*tan(d*x + c)^2 + 6*B*a^2*b*tan(d*x + c)^2 - 6*A*a*b^2*tan(d*x + c)^2 - 3*B*a^3*tan(d*x + c)
+ 3*A*a^2*b*tan(d*x + c) - 2*A*a^3)/(a^4*tan(d*x + c)^3))/d