3.55 \(\int \frac{(c+d \tan (e+f x)) (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=265 \[ -\frac{(b c-a d) \left (A b^2-a (b B-a C)\right )}{b^2 f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{\left (-a^2 b^2 (d (A-3 C)+B c)+a^4 C d+2 a b^3 (A c-B d-c C)+b^4 (A d+B c)\right ) \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )^2}+\frac{\log (\cos (e+f x)) \left (a^2 (-(d (A-C)+B c))+2 a b (A c-B d-c C)+b^2 (d (A-C)+B c)\right )}{f \left (a^2+b^2\right )^2}+\frac{x \left (a^2 (A c-B d-c C)+2 a b (d (A-C)+B c)-b^2 (A c-B d-c C)\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.473547, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3635, 3626, 3617, 31, 3475} \[ -\frac{(b c-a d) \left (A b^2-a (b B-a C)\right )}{b^2 f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{\left (-a^2 b^2 (d (A-3 C)+B c)+a^4 C d+2 a b^3 (A c-B d-c C)+b^4 (A d+B c)\right ) \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )^2}+\frac{\log (\cos (e+f x)) \left (a^2 (-(d (A-C)+B c))+2 a b (A c-B d-c C)+b^2 (d (A-C)+B c)\right )}{f \left (a^2+b^2\right )^2}+\frac{x \left (a^2 (A c-B d-c C)+2 a b (d (A-C)+B c)-b^2 (A c-B d-c C)\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3635

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

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

Mathematica [C]  time = 6.48375, size = 589, normalized size = 2.22 \[ \frac{-2 i a \tan ^{-1}(\tan (e+f x)) (a+b \tan (e+f x)) \left (-a^2 b^2 (d (A-3 C)+B c)+a^4 C d+2 a b^3 (A c-B d-c C)+b^4 (A d+B c)\right )+a^2 \left (2 (a+i b)^2 (e+f x) \left (i a^2 C d+2 a b C d+A b^2 (c-i d)+b^2 (-i B c-B d-c C)\right )+\left (-a^2 b^2 (d (A-3 C)+B c)+a^4 C d+2 a b^3 (A c-B d-c C)+b^4 (A d+B c)\right ) \log \left ((a \cos (e+f x)+b \sin (e+f x))^2\right )-2 C d \left (a^2+b^2\right )^2 \log (\cos (e+f x))\right )+b \tan (e+f x) \left (2 (a+i b) \left (-i a^2 b^2 (i A c (e+f x)+A d (e+f x-i)+B c (e+f x-i)-i B d (e+f x+i)-i c C (e+f x+i)-2 C d (e+f x))+a^3 b (c C+d (B+C (e+f x+i)))+i a^4 C d (e+f x+i)+a b^3 (A c (i e+i f x+1)+A d (e+f x+i)+B c (e+f x+i)-i B d (e+f x)-i c C (e+f x))-i A b^4 c\right )+a \left (-a^2 b^2 (d (A-3 C)+B c)+a^4 C d+2 a b^3 (A c-B d-c C)+b^4 (A d+B c)\right ) \log \left ((a \cos (e+f x)+b \sin (e+f x))^2\right )-2 a C d \left (a^2+b^2\right )^2 \log (\cos (e+f x))\right )}{2 a b^2 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(2*(a + I*b)^2*(A*b^2*(c - I*d) + I*a^2*C*d + 2*a*b*C*d + b^2*((-I)*B*c - c*C - B*d))*(e + f*x) - 2*(a^2
+ b^2)^2*C*d*Log[Cos[e + f*x]] + (a^4*C*d + b^4*(B*c + A*d) + 2*a*b^3*(A*c - c*C - B*d) - a^2*b^2*(B*c + (A -
3*C)*d))*Log[(a*Cos[e + f*x] + b*Sin[e + f*x])^2]) + b*(2*(a + I*b)*((-I)*A*b^4*c + I*a^4*C*d*(I + e + f*x) +
a*b^3*(A*c*(1 + I*e + I*f*x) - I*c*C*(e + f*x) - I*B*d*(e + f*x) + B*c*(I + e + f*x) + A*d*(I + e + f*x)) - I*
a^2*b^2*(I*A*c*(e + f*x) - 2*C*d*(e + f*x) + B*c*(-I + e + f*x) + A*d*(-I + e + f*x) - I*c*C*(I + e + f*x) - I
*B*d*(I + e + f*x)) + a^3*b*(c*C + d*(B + C*(I + e + f*x)))) - 2*a*(a^2 + b^2)^2*C*d*Log[Cos[e + f*x]] + a*(a^
4*C*d + b^4*(B*c + A*d) + 2*a*b^3*(A*c - c*C - B*d) - a^2*b^2*(B*c + (A - 3*C)*d))*Log[(a*Cos[e + f*x] + b*Sin
[e + f*x])^2])*Tan[e + f*x] - (2*I)*a*(a^4*C*d + b^4*(B*c + A*d) + 2*a*b^3*(A*c - c*C - B*d) - a^2*b^2*(B*c +
(A - 3*C)*d))*ArcTan[Tan[e + f*x]]*(a + b*Tan[e + f*x]))/(2*a*b^2*(a^2 + b^2)^2*f*(a + b*Tan[e + f*x]))

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Maple [B]  time = 0.056, size = 948, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.48858, size = 456, normalized size = 1.72 \begin{align*} \frac{\frac{2 \,{\left ({\left ({\left (A - C\right )} a^{2} + 2 \, B a b -{\left (A - C\right )} b^{2}\right )} c -{\left (B a^{2} - 2 \,{\left (A - C\right )} a b - B b^{2}\right )} d\right )}{\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left ({\left (B a^{2} b^{2} - 2 \,{\left (A - C\right )} a b^{3} - B b^{4}\right )} c -{\left (C a^{4} -{\left (A - 3 \, C\right )} a^{2} b^{2} - 2 \, B a b^{3} + A b^{4}\right )} d\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac{{\left ({\left (B a^{2} - 2 \,{\left (A - C\right )} a b - B b^{2}\right )} c +{\left ({\left (A - C\right )} a^{2} + 2 \, B a b -{\left (A - C\right )} b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left ({\left (C a^{2} b - B a b^{2} + A b^{3}\right )} c -{\left (C a^{3} - B a^{2} b + A a b^{2}\right )} d\right )}}{a^{3} b^{2} + a b^{4} +{\left (a^{2} b^{3} + b^{5}\right )} \tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.32446, size = 1160, normalized size = 4.38 \begin{align*} \frac{2 \,{\left ({\left ({\left (A - C\right )} a^{3} b^{2} + 2 \, B a^{2} b^{3} -{\left (A - C\right )} a b^{4}\right )} c -{\left (B a^{3} b^{2} - 2 \,{\left (A - C\right )} a^{2} b^{3} - B a b^{4}\right )} d\right )} f x - 2 \,{\left (C a^{2} b^{3} - B a b^{4} + A b^{5}\right )} c + 2 \,{\left (C a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} d -{\left ({\left (B a^{3} b^{2} - 2 \,{\left (A - C\right )} a^{2} b^{3} - B a b^{4}\right )} c -{\left (C a^{5} -{\left (A - 3 \, C\right )} a^{3} b^{2} - 2 \, B a^{2} b^{3} + A a b^{4}\right )} d +{\left ({\left (B a^{2} b^{3} - 2 \,{\left (A - C\right )} a b^{4} - B b^{5}\right )} c -{\left (C a^{4} b -{\left (A - 3 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + A b^{5}\right )} d\right )} \tan \left (f x + e\right )\right )} \log \left (\frac{b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) -{\left ({\left (C a^{4} b + 2 \, C a^{2} b^{3} + C b^{5}\right )} d \tan \left (f x + e\right ) +{\left (C a^{5} + 2 \, C a^{3} b^{2} + C a b^{4}\right )} d\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \,{\left ({\left ({\left ({\left (A - C\right )} a^{2} b^{3} + 2 \, B a b^{4} -{\left (A - C\right )} b^{5}\right )} c -{\left (B a^{2} b^{3} - 2 \,{\left (A - C\right )} a b^{4} - B b^{5}\right )} d\right )} f x +{\left (C a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} c -{\left (C a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} d\right )} \tan \left (f x + e\right )}{2 \,{\left ({\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} f \tan \left (f x + e\right ) +{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((c+d*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(a+b*tan(f*x+e))**2,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.39674, size = 717, normalized size = 2.71 \begin{align*} \frac{\frac{2 \,{\left (A a^{2} c - C a^{2} c + 2 \, B a b c - A b^{2} c + C b^{2} c - B a^{2} d + 2 \, A a b d - 2 \, C a b d + B b^{2} d\right )}{\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (B a^{2} c - 2 \, A a b c + 2 \, C a b c - B b^{2} c + A a^{2} d - C a^{2} d + 2 \, B a b d - A b^{2} d + C b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \,{\left (B a^{2} b^{2} c - 2 \, A a b^{3} c + 2 \, C a b^{3} c - B b^{4} c - C a^{4} d + A a^{2} b^{2} d - 3 \, C a^{2} b^{2} d + 2 \, B a b^{3} d - A b^{4} d\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (B a^{2} b^{2} c \tan \left (f x + e\right ) - 2 \, A a b^{3} c \tan \left (f x + e\right ) + 2 \, C a b^{3} c \tan \left (f x + e\right ) - B b^{4} c \tan \left (f x + e\right ) - C a^{4} d \tan \left (f x + e\right ) + A a^{2} b^{2} d \tan \left (f x + e\right ) - 3 \, C a^{2} b^{2} d \tan \left (f x + e\right ) + 2 \, B a b^{3} d \tan \left (f x + e\right ) - A b^{4} d \tan \left (f x + e\right ) - C a^{4} c + 2 \, B a^{3} b c - 3 \, A a^{2} b^{2} c + C a^{2} b^{2} c - A b^{4} c - B a^{4} d + 2 \, A a^{3} b d - 2 \, C a^{3} b d + B a^{2} b^{2} d\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )}{\left (b \tan \left (f x + e\right ) + a\right )}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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