3.281 \(\int \frac{\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=250 \[ \frac{b \left (2 a^2 A b+a^3 (-B)-2 a b^2 B+3 A b^3\right )}{a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (a^2 A+2 a b B-3 A b^2\right ) \log (\sin (c+d x))}{a^4 d}-\frac{b^3 \left (5 a^2 A b-4 a^3 B-2 a b^2 B+3 A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 d \left (a^2+b^2\right )^2}+\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2}+\frac{(3 A b-2 a B) \cot (c+d x)}{2 a^2 d (a+b \tan (c+d x))}-\frac{A \cot ^2(c+d x)}{2 a d (a+b \tan (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.859716, antiderivative size = 250, 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} \[ \frac{b \left (2 a^2 A b+a^3 (-B)-2 a b^2 B+3 A b^3\right )}{a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (a^2 A+2 a b B-3 A b^2\right ) \log (\sin (c+d x))}{a^4 d}-\frac{b^3 \left (5 a^2 A b-4 a^3 B-2 a b^2 B+3 A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^4 d \left (a^2+b^2\right )^2}+\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2}+\frac{(3 A b-2 a B) \cot (c+d x)}{2 a^2 d (a+b \tan (c+d x))}-\frac{A \cot ^2(c+d x)}{2 a d (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

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

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

Mathematica [C]  time = 4.33774, size = 220, normalized size = 0.88 \[ \frac{\frac{2 b^3 (A b-a B)}{a^3 \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{2 b^3 \left (-5 a^2 A b+4 a^3 B+2 a b^2 B-3 A b^3\right ) \log (a+b \tan (c+d x))}{a^4 \left (a^2+b^2\right )^2}-\frac{2 \left (a^2 A+2 a b B-3 A b^2\right ) \log (\tan (c+d x))}{a^4}-\frac{2 (a B-2 A b) \cot (c+d x)}{a^3}-\frac{A \cot ^2(c+d x)}{a^2}+\frac{(A+i B) \log (-\tan (c+d x)+i)}{(a+i b)^2}+\frac{(A-i B) \log (\tan (c+d x)+i)}{(a-i b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.15, size = 457, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{2}A}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) A{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) ab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{A}{2\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{Ab}{{a}^{3}d\tan \left ( dx+c \right ) }}-{\frac{B}{{a}^{2}d\tan \left ( dx+c \right ) }}-{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}}+3\,{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{{a}^{4}d}}-2\,{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) Bb}{{a}^{3}d}}-5\,{\frac{{b}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{{a}^{2}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-3\,{\frac{{b}^{6}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{{a}^{4}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+4\,{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{ad \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{{b}^{5}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{{a}^{3}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{{b}^{4}A}{{a}^{3}d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{B{b}^{3}}{{a}^{2}d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.57438, size = 439, normalized size = 1.76 \begin{align*} -\frac{\frac{2 \,{\left (B a^{2} - 2 \, A a b - B b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (4 \, B a^{3} b^{3} - 5 \, A a^{2} b^{4} + 2 \, B a b^{5} - 3 \, A b^{6}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}} - \frac{{\left (A a^{2} + 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{A a^{4} + A a^{2} b^{2} + 2 \,{\left (B a^{3} b - 2 \, A a^{2} b^{2} + 2 \, B a b^{3} - 3 \, A b^{4}\right )} \tan \left (d x + c\right )^{2} +{\left (2 \, B a^{4} - 3 \, A a^{3} b + 2 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} \tan \left (d x + c\right )}{{\left (a^{5} b + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{3} +{\left (a^{6} + a^{4} b^{2}\right )} \tan \left (d x + c\right )^{2}} + \frac{2 \,{\left (A a^{2} + 2 \, B a b - 3 \, A b^{2}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(2*(B*a^2 - 2*A*a*b - B*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - 2*(4*B*a^3*b^3 - 5*A*a^2*b^4 + 2*B*a*b^5
- 3*A*b^6)*log(b*tan(d*x + c) + a)/(a^8 + 2*a^6*b^2 + a^4*b^4) - (A*a^2 + 2*B*a*b - A*b^2)*log(tan(d*x + c)^2
+ 1)/(a^4 + 2*a^2*b^2 + b^4) + (A*a^4 + A*a^2*b^2 + 2*(B*a^3*b - 2*A*a^2*b^2 + 2*B*a*b^3 - 3*A*b^4)*tan(d*x +
c)^2 + (2*B*a^4 - 3*A*a^3*b + 2*B*a^2*b^2 - 3*A*a*b^3)*tan(d*x + c))/((a^5*b + a^3*b^3)*tan(d*x + c)^3 + (a^6
+ a^4*b^2)*tan(d*x + c)^2) + 2*(A*a^2 + 2*B*a*b - 3*A*b^2)*log(tan(d*x + c))/a^4)/d

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Fricas [B]  time = 2.51781, size = 1283, normalized size = 5.13 \begin{align*} -\frac{A a^{7} + 2 \, A a^{5} b^{2} + A a^{3} b^{4} +{\left (A a^{6} b + 2 \, A a^{4} b^{3} - 2 \, B a^{3} b^{4} + 3 \, A a^{2} b^{5} + 2 \,{\left (B a^{6} b - 2 \, A a^{5} b^{2} - B a^{4} b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{3} +{\left (A a^{7} + 2 \, B a^{6} b - 2 \, A a^{5} b^{2} + 4 \, B a^{4} b^{3} - 7 \, A a^{3} b^{4} + 4 \, B a^{2} b^{5} - 6 \, A a b^{6} + 2 \,{\left (B a^{7} - 2 \, A a^{6} b - B a^{5} b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{2} +{\left ({\left (A a^{6} b + 2 \, B a^{5} b^{2} - A a^{4} b^{3} + 4 \, B a^{3} b^{4} - 5 \, A a^{2} b^{5} + 2 \, B a b^{6} - 3 \, A b^{7}\right )} \tan \left (d x + c\right )^{3} +{\left (A a^{7} + 2 \, B a^{6} b - A a^{5} b^{2} + 4 \, B a^{4} b^{3} - 5 \, A a^{3} b^{4} + 2 \, B a^{2} b^{5} - 3 \, A a b^{6}\right )} \tan \left (d x + c\right )^{2}\right )} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) -{\left ({\left (4 \, B a^{3} b^{4} - 5 \, A a^{2} b^{5} + 2 \, B a b^{6} - 3 \, A b^{7}\right )} \tan \left (d x + c\right )^{3} +{\left (4 \, B a^{4} b^{3} - 5 \, A a^{3} b^{4} + 2 \, B a^{2} b^{5} - 3 \, A a b^{6}\right )} \tan \left (d x + c\right )^{2}\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 ) +{\left (2 \, B a^{7} - 3 \, A a^{6} b + 4 \, B a^{5} b^{2} - 6 \, A a^{4} b^{3} + 2 \, B a^{3} b^{4} - 3 \, A a^{2} b^{5}\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{8} b + 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d \tan \left (d x + c\right )^{3} +{\left (a^{9} + 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d \tan \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(A*a^7 + 2*A*a^5*b^2 + A*a^3*b^4 + (A*a^6*b + 2*A*a^4*b^3 - 2*B*a^3*b^4 + 3*A*a^2*b^5 + 2*(B*a^6*b - 2*A*
a^5*b^2 - B*a^4*b^3)*d*x)*tan(d*x + c)^3 + (A*a^7 + 2*B*a^6*b - 2*A*a^5*b^2 + 4*B*a^4*b^3 - 7*A*a^3*b^4 + 4*B*
a^2*b^5 - 6*A*a*b^6 + 2*(B*a^7 - 2*A*a^6*b - B*a^5*b^2)*d*x)*tan(d*x + c)^2 + ((A*a^6*b + 2*B*a^5*b^2 - A*a^4*
b^3 + 4*B*a^3*b^4 - 5*A*a^2*b^5 + 2*B*a*b^6 - 3*A*b^7)*tan(d*x + c)^3 + (A*a^7 + 2*B*a^6*b - A*a^5*b^2 + 4*B*a
^4*b^3 - 5*A*a^3*b^4 + 2*B*a^2*b^5 - 3*A*a*b^6)*tan(d*x + c)^2)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) - ((4
*B*a^3*b^4 - 5*A*a^2*b^5 + 2*B*a*b^6 - 3*A*b^7)*tan(d*x + c)^3 + (4*B*a^4*b^3 - 5*A*a^3*b^4 + 2*B*a^2*b^5 - 3*
A*a*b^6)*tan(d*x + c)^2)*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) + (2*B*a^7
- 3*A*a^6*b + 4*B*a^5*b^2 - 6*A*a^4*b^3 + 2*B*a^3*b^4 - 3*A*a^2*b^5)*tan(d*x + c))/((a^8*b + 2*a^6*b^3 + a^4*b
^5)*d*tan(d*x + c)^3 + (a^9 + 2*a^7*b^2 + a^5*b^4)*d*tan(d*x + c)^2)

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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)**3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.33389, size = 543, normalized size = 2.17 \begin{align*} -\frac{\frac{2 \,{\left (B a^{2} - 2 \, A a b - B b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (A a^{2} + 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (4 \, B a^{3} b^{4} - 5 \, A a^{2} b^{5} + 2 \, B a b^{6} - 3 \, A b^{7}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 2 \, a^{6} b^{3} + a^{4} b^{5}} + \frac{2 \,{\left (4 \, B a^{3} b^{4} \tan \left (d x + c\right ) - 5 \, A a^{2} b^{5} \tan \left (d x + c\right ) + 2 \, B a b^{6} \tan \left (d x + c\right ) - 3 \, A b^{7} \tan \left (d x + c\right ) + 5 \, B a^{4} b^{3} - 6 \, A a^{3} b^{4} + 3 \, B a^{2} b^{5} - 4 \, A a b^{6}\right )}}{{\left (a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}} + \frac{2 \,{\left (A a^{2} + 2 \, B a b - 3 \, A b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac{3 \, A a^{2} \tan \left (d x + c\right )^{2} + 6 \, B a b \tan \left (d x + c\right )^{2} - 9 \, A b^{2} \tan \left (d x + c\right )^{2} - 2 \, B a^{2} \tan \left (d x + c\right ) + 4 \, A a b \tan \left (d x + c\right ) - A a^{2}}{a^{4} \tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(2*(B*a^2 - 2*A*a*b - B*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - (A*a^2 + 2*B*a*b - A*b^2)*log(tan(d*x +
c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) - 2*(4*B*a^3*b^4 - 5*A*a^2*b^5 + 2*B*a*b^6 - 3*A*b^7)*log(abs(b*tan(d*x + c)
+ a))/(a^8*b + 2*a^6*b^3 + a^4*b^5) + 2*(4*B*a^3*b^4*tan(d*x + c) - 5*A*a^2*b^5*tan(d*x + c) + 2*B*a*b^6*tan(
d*x + c) - 3*A*b^7*tan(d*x + c) + 5*B*a^4*b^3 - 6*A*a^3*b^4 + 3*B*a^2*b^5 - 4*A*a*b^6)/((a^8 + 2*a^6*b^2 + a^4
*b^4)*(b*tan(d*x + c) + a)) + 2*(A*a^2 + 2*B*a*b - 3*A*b^2)*log(abs(tan(d*x + c)))/a^4 - (3*A*a^2*tan(d*x + c)
^2 + 6*B*a*b*tan(d*x + c)^2 - 9*A*b^2*tan(d*x + c)^2 - 2*B*a^2*tan(d*x + c) + 4*A*a*b*tan(d*x + c) - A*a^2)/(a
^4*tan(d*x + c)^2))/d