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

Optimal. Leaf size=189 \[ -\frac{a^2 (A b-a B)}{2 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac{x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.371693, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3604, 3628, 3531, 3530} \[ -\frac{a^2 (A b-a B)}{2 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac{x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3604

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

Rule 3628

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

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

Mathematica [C]  time = 4.60584, size = 288, normalized size = 1.52 \[ \frac{(A b-a B) \left (\frac{b \left (\frac{\left (a^2+b^2\right ) \left (5 a^2+4 a b \tan (c+d x)+b^2\right )}{(a+b \tan (c+d x))^2}+\left (2 b^2-6 a^2\right ) \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^3}+\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^3}-\frac{\log (\tan (c+d x)+i)}{(b+i a)^3}\right )+B \left (\frac{2 b \left (\frac{a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^2}+\frac{i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac{i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )-\frac{a B+A b}{b (a+b \tan (c+d x))^2}-\frac{2 B \tan (c+d x)}{(a+b \tan (c+d x))^2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((A*b + a*B)/(b*(a + b*Tan[c + d*x])^2)) - (2*B*Tan[c + d*x])/(a + b*Tan[c + d*x])^2 + B*((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) + (A*b - a*B)*((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 = 0.051, size = 495, normalized size = 2.6 \begin{align*}{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) A{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 ) A{b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) B{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 ) Ba{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{{a}^{2}A}{2\,bd \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B{a}^{3}}{2\,{b}^{2}d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-3\,{\frac{b{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) Ba{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+2\,{\frac{abA}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{B{a}^{4}}{{b}^{2}d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-3\,{\frac{{a}^{2}B}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.56641, size = 450, normalized size = 2.38 \begin{align*} -\frac{\frac{2 \,{\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B 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 (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A 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 (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A 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{B a^{5} + A a^{4} b + 5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} + 2 \,{\left (B a^{4} b + 3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} \tan \left (d x + c\right )}{a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6} +{\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.88033, size = 1038, normalized size = 5.49 \begin{align*} \frac{B a^{5} - 3 \, A a^{4} b - 5 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3} - 2 \,{\left (A a^{5} + 3 \, B a^{4} b - 3 \, A a^{3} b^{2} - B a^{2} b^{3}\right )} d x +{\left (B a^{5} + A a^{4} b + 7 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} - 2 \,{\left (A a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4} - B b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} +{\left (B a^{5} - 3 \, A a^{4} b - 3 \, B a^{3} b^{2} + A a^{2} b^{3} +{\left (B a^{3} b^{2} - 3 \, A a^{2} b^{3} - 3 \, B a b^{4} + A b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (B a^{4} b - 3 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + A 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 (A a^{5} + 3 \, B a^{4} b - 3 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + 2 \, A a b^{4} - 2 \,{\left (A a^{4} b + 3 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} - B 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(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(B*a^5 - 3*A*a^4*b - 5*B*a^3*b^2 + 3*A*a^2*b^3 - 2*(A*a^5 + 3*B*a^4*b - 3*A*a^3*b^2 - B*a^2*b^3)*d*x + (B*
a^5 + A*a^4*b + 7*B*a^3*b^2 - 5*A*a^2*b^3 - 2*(A*a^3*b^2 + 3*B*a^2*b^3 - 3*A*a*b^4 - B*b^5)*d*x)*tan(d*x + c)^
2 + (B*a^5 - 3*A*a^4*b - 3*B*a^3*b^2 + A*a^2*b^3 + (B*a^3*b^2 - 3*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*tan(d*x + c)^
2 + 2*(B*a^4*b - 3*A*a^3*b^2 - 3*B*a^2*b^3 + A*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*(A*a^5 + 3*B*a^4*b - 3*A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*b^4 - 2*(A*a^4*b +
3*B*a^3*b^2 - 3*A*a^2*b^3 - B*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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.42936, size = 554, normalized size = 2.93 \begin{align*} -\frac{\frac{2 \,{\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A 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 (B a^{3} b - 3 \, A a^{2} b^{2} - 3 \, B a b^{3} + A 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 \, B a^{3} b^{4} \tan \left (d x + c\right )^{2} - 9 \, A a^{2} b^{5} \tan \left (d x + c\right )^{2} - 9 \, B a b^{6} \tan \left (d x + c\right )^{2} + 3 \, A b^{7} \tan \left (d x + c\right )^{2} + 2 \, B a^{6} b \tan \left (d x + c\right ) + 14 \, B a^{4} b^{3} \tan \left (d x + c\right ) - 22 \, A a^{3} b^{4} \tan \left (d x + c\right ) - 12 \, B a^{2} b^{5} \tan \left (d x + c\right ) + 2 \, A a b^{6} \tan \left (d x + c\right ) + B a^{7} + A a^{6} b + 9 \, B a^{5} b^{2} - 11 \, A a^{4} b^{3} - 4 \, B a^{3} b^{4}}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\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(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*(A*a^3 + 3*B*a^2*b - 3*A*a*b^2 - B*b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (B*a^3 - 3*A*a
^2*b - 3*B*a*b^2 + A*b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(B*a^3*b - 3*A*a^2*b
^2 - 3*B*a*b^3 + A*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*B*a^3*b^4*tan(
d*x + c)^2 - 9*A*a^2*b^5*tan(d*x + c)^2 - 9*B*a*b^6*tan(d*x + c)^2 + 3*A*b^7*tan(d*x + c)^2 + 2*B*a^6*b*tan(d*
x + c) + 14*B*a^4*b^3*tan(d*x + c) - 22*A*a^3*b^4*tan(d*x + c) - 12*B*a^2*b^5*tan(d*x + c) + 2*A*a*b^6*tan(d*x
+ c) + B*a^7 + A*a^6*b + 9*B*a^5*b^2 - 11*A*a^4*b^3 - 4*B*a^3*b^4)/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*(
b*tan(d*x + c) + a)^2))/d