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

Optimal. Leaf size=208 \[ \frac{a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (-2 a^2 B+a A b-b^2 B\right ) \tan (c+d x)}{b^2 d \left (a^2+b^2\right )}+\frac{a^2 \left (a^2 A b-2 a^3 B-4 a b^2 B+3 A b^3\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^2}+\frac{\left (a^2 A+2 a b B-A b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.454827, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3605, 3647, 3626, 3617, 31, 3475} \[ \frac{a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\left (-2 a^2 B+a A b-b^2 B\right ) \tan (c+d x)}{b^2 d \left (a^2+b^2\right )}+\frac{a^2 \left (a^2 A b-2 a^3 B-4 a b^2 B+3 A b^3\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^2}+\frac{\left (a^2 A+2 a b B-A b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[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 - A*b^2 + 2*a*b*B)*Log[Cos[c + d*x]])/((a^2 + b^2)^2*
d) + (a^2*(a^2*A*b + 3*A*b^3 - 2*a^3*B - 4*a*b^2*B)*Log[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^2)^2*d) - ((a*A*b -
2*a^2*B - b^2*B)*Tan[c + d*x])/(b^2*(a^2 + b^2)*d) + (a*(A*b - a*B)*Tan[c + d*x]^2)/(b*(a^2 + b^2)*d*(a + b*T
an[c + d*x]))

Rule 3605

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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e
+ f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3647

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] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3626

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

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=\frac{a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{\tan (c+d x) \left (-2 a (A b-a B)+b (A b-a B) \tan (c+d x)-\left (a A b-2 a^2 B-b^2 B\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{\left (a A b-2 a^2 B-b^2 B\right ) \tan (c+d x)}{b^2 \left (a^2+b^2\right ) d}+\frac{a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{a \left (a A b-2 a^2 B-b^2 B\right )-b^2 (a A+b B) \tan (c+d x)+\left (a^2+b^2\right ) (A b-2 a B) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \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{\left (a A b-2 a^2 B-b^2 B\right ) \tan (c+d x)}{b^2 \left (a^2+b^2\right ) d}+\frac{a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\left (a^2 A-A b^2+2 a b B\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^2}+\frac{\left (a^2 \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right )\right ) \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \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-A b^2+2 a b B\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac{\left (a A b-2 a^2 B-b^2 B\right ) \tan (c+d x)}{b^2 \left (a^2+b^2\right ) d}+\frac{a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (a^2 \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 \left (a^2+b^2\right )^2 d}\\ &=-\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-A b^2+2 a b B\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac{a^2 \left (a^2 A b+3 A b^3-2 a^3 B-4 a b^2 B\right ) \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right )^2 d}-\frac{\left (a A b-2 a^2 B-b^2 B\right ) \tan (c+d x)}{b^2 \left (a^2+b^2\right ) d}+\frac{a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 3.7792, size = 444, normalized size = 2.13 \[ \frac{2 i a^2 \left (-a^2 A b+2 a^3 B+4 a b^2 B-3 A b^3\right ) \tan ^{-1}(\tan (c+d x)) (a+b \tan (c+d x))+a \left (2 (a+i b)^2 (c+d x) \left (i a^2 b (A+4 i B)-2 i a^3 B+2 a b^2 (A+i B)+b^3 B\right )+2 \left (a^2+b^2\right )^2 (2 a B-A b) \log (\cos (c+d x))+a^2 \left (a^2 A b-2 a^3 B-4 a b^2 B+3 A b^3\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )\right )+b \tan (c+d x) \left (2 \left (a^2 b^3 (B (c+d x)+i A (3 c+3 d x+i))+i a^4 A b (c+d x+i)+a^3 b^2 B (-4 i c-4 i d x+3)-2 i a^5 B (c+d x+i)+a b^4 (B-2 A (c+d x))-b^5 B (c+d x)\right )+2 \left (a^2+b^2\right )^2 (2 a B-A b) \log (\cos (c+d x))+a^2 \left (a^2 A b-2 a^3 B-4 a b^2 B+3 A b^3\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )\right )+2 b^2 B \left (a^2+b^2\right )^2 \tan ^2(c+d x)}{2 b^3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.044, size = 364, normalized size = 1.8 \begin{align*}{\frac{B\tan \left ( dx+c \right ) }{{b}^{2}d}}-{\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}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{{b}^{2}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+3\,{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-2\,{\frac{{a}^{5}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{{b}^{3}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-4\,{\frac{{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{bd \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{A{a}^{3}}{{b}^{2}d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{B{a}^{4}}{{b}^{3}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(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.49544, size = 297, normalized size = 1.43 \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 (2 \, B a^{5} - A a^{4} b + 4 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}} - \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 (B a^{4} - A a^{3} b\right )}}{a^{3} b^{3} + a b^{5} +{\left (a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )} + \frac{2 \, B \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [B]  time = 2.37061, size = 936, normalized size = 4.5 \begin{align*} -\frac{2 \, B a^{4} b^{2} - 2 \, A a^{3} b^{3} - 2 \,{\left (B a^{3} b^{3} - 2 \, A a^{2} b^{4} - B a b^{5}\right )} d x - 2 \,{\left (B a^{4} b^{2} + 2 \, B a^{2} b^{4} + B b^{6}\right )} \tan \left (d x + c\right )^{2} +{\left (2 \, B a^{6} - A a^{5} b + 4 \, B a^{4} b^{2} - 3 \, A a^{3} b^{3} +{\left (2 \, B a^{5} b - A a^{4} b^{2} + 4 \, B a^{3} b^{3} - 3 \, A a^{2} 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 ) -{\left (2 \, B a^{6} - A a^{5} b + 4 \, B a^{4} b^{2} - 2 \, A a^{3} b^{3} + 2 \, B a^{2} b^{4} - A a b^{5} +{\left (2 \, B a^{5} b - A a^{4} b^{2} + 4 \, B a^{3} b^{3} - 2 \, A a^{2} b^{4} + 2 \, B a b^{5} - A b^{6}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (2 \, B a^{5} b - A a^{4} b^{2} + 2 \, B a^{3} b^{3} + B a b^{5} +{\left (B a^{2} b^{4} - 2 \, A a b^{5} - B b^{6}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right ) +{\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [A]  time = 1.83639, size = 392, normalized size = 1.88 \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 (2 \, B a^{5} - A a^{4} b + 4 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}} + \frac{2 \, B \tan \left (d x + c\right )}{b^{2}} + \frac{2 \,{\left (2 \, B a^{5} b \tan \left (d x + c\right ) - A a^{4} b^{2} \tan \left (d x + c\right ) + 4 \, B a^{3} b^{3} \tan \left (d x + c\right ) - 3 \, A a^{2} b^{4} \tan \left (d x + c\right ) + B a^{6} + 3 \, B a^{4} b^{2} - 2 \, A a^{3} b^{3}\right )}}{{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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