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

Optimal. Leaf size=175 \[ -\frac{b B-a C}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{a^2 (-C)+2 a b B+b^2 C}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b B+a^3 (-C)+3 a b^2 C-b^3 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (3 a^2 b C+a^3 B-3 a b^2 B-b^3 C\right )}{\left (a^2+b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.315501, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3632, 3529, 3531, 3530} \[ -\frac{b B-a C}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{a^2 (-C)+2 a b B+b^2 C}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{\left (3 a^2 b B+a^3 (-C)+3 a b^2 C-b^3 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (3 a^2 b C+a^3 B-3 a b^2 B-b^3 C\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3632

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

Rule 3529

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 3531

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 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]

Rubi steps

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

Mathematica [C]  time = 3.80715, size = 243, normalized size = 1.39 \[ -\frac{(b B-a C) \left (\frac{b \left (\frac{\left (a^2+b^2\right ) \left (5 a^2+4 a b \tan (c+d x)+b^2\right )}{(a+b \tan (c+d x))^2}+\left (2 b^2-6 a^2\right ) \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^3}+\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^3}-\frac{\log (\tan (c+d x)+i)}{(b+i a)^3}\right )+C \left (\frac{2 b \left (\frac{a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^2}+\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac{i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.154, size = 483, normalized size = 2.8 \begin{align*} -{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) B{a}^{2}b}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) B{b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) C{a}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Ca{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{C\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{C\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{Bb}{2\,d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Ca}{2\,d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-2\,{\frac{Bab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+{\frac{C{a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{b}^{2}C}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+3\,{\frac{b{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) C}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) Ca{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*B*a^2*b+1/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*B*b^3+1/2/d/(a^2+b^2)^3*ln(
1+tan(d*x+c)^2)*C*a^3-3/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*C*a*b^2+1/d/(a^2+b^2)^3*B*arctan(tan(d*x+c))*a^3-3/
d/(a^2+b^2)^3*B*arctan(tan(d*x+c))*a*b^2+3/d/(a^2+b^2)^3*C*arctan(tan(d*x+c))*a^2*b-1/d/(a^2+b^2)^3*C*arctan(t
an(d*x+c))*b^3-1/2/d/(a^2+b^2)/(a+b*tan(d*x+c))^2*B*b+1/2/d/(a^2+b^2)/(a+b*tan(d*x+c))^2*C*a-2/d*a/(a^2+b^2)^2
*b/(a+b*tan(d*x+c))*B+1/d*a^2/(a^2+b^2)^2/(a+b*tan(d*x+c))*C-1/d/(a^2+b^2)^2/(a+b*tan(d*x+c))*b^2*C+3/d*b*a^2/
(a^2+b^2)^3*ln(a+b*tan(d*x+c))*B-1/d/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*B*b^3-1/d*a^3/(a^2+b^2)^3*ln(a+b*tan(d*x+c
))*C+3/d/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*C*a*b^2

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Maxima [A]  time = 1.62061, size = 433, normalized size = 2.47 \begin{align*} \frac{\frac{2 \,{\left (B a^{3} + 3 \, C a^{2} b - 3 \, B a b^{2} - C b^{3}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (C a^{3} - 3 \, B a^{2} b - 3 \, C a b^{2} + B b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (C a^{3} - 3 \, B a^{2} b - 3 \, C a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{3 \, C a^{3} - 5 \, B a^{2} b - C a b^{2} - B b^{3} + 2 \,{\left (C a^{2} b - 2 \, B a b^{2} - C b^{3}\right )} \tan \left (d x + c\right )}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} +{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.19085, size = 1038, normalized size = 5.93 \begin{align*} \frac{5 \, C a^{3} b^{2} - 7 \, B a^{2} b^{3} - C a b^{4} - B b^{5} + 2 \,{\left (B a^{5} + 3 \, C a^{4} b - 3 \, B a^{3} b^{2} - C a^{2} b^{3}\right )} d x -{\left (3 \, C a^{3} b^{2} - 5 \, B a^{2} b^{3} - 3 \, C a b^{4} + B b^{5} - 2 \,{\left (B a^{3} b^{2} + 3 \, C a^{2} b^{3} - 3 \, B a b^{4} - C b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} -{\left (C a^{5} - 3 \, B a^{4} b - 3 \, C a^{3} b^{2} + B a^{2} b^{3} +{\left (C a^{3} b^{2} - 3 \, B a^{2} b^{3} - 3 \, C a b^{4} + B b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (C a^{4} b - 3 \, B a^{3} b^{2} - 3 \, C a^{2} b^{3} + B a b^{4}\right )} \tan \left (d x + c\right )\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 ) - 2 \,{\left (2 \, C a^{4} b - 3 \, B a^{3} b^{2} - 3 \, C a^{2} b^{3} + 3 \, B a b^{4} + C b^{5} - 2 \,{\left (B a^{4} b + 3 \, C a^{3} b^{2} - 3 \, B a^{2} b^{3} - C a b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) +{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.50252, size = 552, normalized size = 3.15 \begin{align*} \frac{\frac{2 \,{\left (B a^{3} + 3 \, C a^{2} b - 3 \, B a b^{2} - C b^{3}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (C a^{3} - 3 \, B a^{2} b - 3 \, C a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (C a^{3} b - 3 \, B a^{2} b^{2} - 3 \, C a b^{3} + B b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} + \frac{3 \, C a^{3} b^{2} \tan \left (d x + c\right )^{2} - 9 \, B a^{2} b^{3} \tan \left (d x + c\right )^{2} - 9 \, C a b^{4} \tan \left (d x + c\right )^{2} + 3 \, B b^{5} \tan \left (d x + c\right )^{2} + 8 \, C a^{4} b \tan \left (d x + c\right ) - 22 \, B a^{3} b^{2} \tan \left (d x + c\right ) - 18 \, C a^{2} b^{3} \tan \left (d x + c\right ) + 2 \, B a b^{4} \tan \left (d x + c\right ) - 2 \, C b^{5} \tan \left (d x + c\right ) + 6 \, C a^{5} - 14 \, B a^{4} b - 7 \, C a^{3} b^{2} - 3 \, B a^{2} b^{3} - C a b^{4} - B b^{5}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(B*a^3 + 3*C*a^2*b - 3*B*a*b^2 - C*b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (C*a^3 - 3*B*a^
2*b - 3*C*a*b^2 + B*b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(C*a^3*b - 3*B*a^2*b^
2 - 3*C*a*b^3 + B*b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) + (3*C*a^3*b^2*tan(d
*x + c)^2 - 9*B*a^2*b^3*tan(d*x + c)^2 - 9*C*a*b^4*tan(d*x + c)^2 + 3*B*b^5*tan(d*x + c)^2 + 8*C*a^4*b*tan(d*x
+ c) - 22*B*a^3*b^2*tan(d*x + c) - 18*C*a^2*b^3*tan(d*x + c) + 2*B*a*b^4*tan(d*x + c) - 2*C*b^5*tan(d*x + c)
+ 6*C*a^5 - 14*B*a^4*b - 7*C*a^3*b^2 - 3*B*a^2*b^3 - C*a*b^4 - B*b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(b*
tan(d*x + c) + a)^2))/d