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

Optimal. Leaf size=156 \[ \frac{(b c-a d) \left (A b^2-a (b B-a C)\right ) \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )}+\frac{\log (\cos (e+f x)) (-a A d-a B c+a C d+A b c-b B d-b c C)}{f \left (a^2+b^2\right )}+\frac{x (a (A c-B d-c C)+b (d (A-C)+B c))}{a^2+b^2}+\frac{C d \tan (e+f x)}{b f} \]

[Out]

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

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Rubi [A]  time = 0.349431, antiderivative size = 155, normalized size of antiderivative = 0.99, 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 = {3637, 3626, 3617, 31, 3475} \[ \frac{(b c-a d) \left (A b^2-a (b B-a C)\right ) \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )}+\frac{\log (\cos (e+f x)) (-a A d-a B c+a C d+A b c-b B d-b c C)}{f \left (a^2+b^2\right )}+\frac{x (a (A c-B d-c C)+b d (A-C)+b B c)}{a^2+b^2}+\frac{C d \tan (e+f x)}{b f} \]

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]),x]

[Out]

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

Rule 3637

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*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, 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)} \, dx &=\frac{C d \tan (e+f x)}{b f}-\frac{\int \frac{-A b c+a C d-b (B c+(A-C) d) \tan (e+f x)-(b c C+b B d-a C d) \tan ^2(e+f x)}{a+b \tan (e+f x)} \, dx}{b}\\ &=\frac{(b B c+b (A-C) d+a (A c-c C-B d)) x}{a^2+b^2}+\frac{C d \tan (e+f x)}{b f}+\frac{\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)\right ) \int \frac{1+\tan ^2(e+f x)}{a+b \tan (e+f x)} \, dx}{b \left (a^2+b^2\right )}-\frac{(A b c-a B c-b c C-a A d-b B d+a C d) \int \tan (e+f x) \, dx}{a^2+b^2}\\ &=\frac{(b B c+b (A-C) d+a (A c-c C-B d)) x}{a^2+b^2}+\frac{(A b c-a B c-b c C-a A d-b B d+a C d) \log (\cos (e+f x))}{\left (a^2+b^2\right ) f}+\frac{C d \tan (e+f x)}{b f}+\frac{\left (\left (A b^2-a (b B-a C)\right ) (b c-a 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 ) f}\\ &=\frac{(b B c+b (A-C) d+a (A c-c C-B d)) x}{a^2+b^2}+\frac{(A b c-a B c-b c C-a A d-b B d+a C d) \log (\cos (e+f x))}{\left (a^2+b^2\right ) f}+\frac{\left (A b^2-a (b B-a C)\right ) (b c-a d) \log (a+b \tan (e+f x))}{b^2 \left (a^2+b^2\right ) f}+\frac{C d \tan (e+f x)}{b f}\\ \end{align*}

Mathematica [C]  time = 1.11109, size = 148, normalized size = 0.95 \[ \frac{\frac{2 (b c-a d) \left (a (a C-b B)+A b^2\right ) \log (a+b \tan (e+f x))}{b^2 \left (a^2+b^2\right )}+\frac{(d-i c) (A+i B-C) \log (-\tan (e+f x)+i)}{a+i b}+\frac{(d+i c) (A-i B-C) \log (\tan (e+f x)+i)}{a-i b}+\frac{2 C d \tan (e+f x)}{b}}{2 f} \]

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]),x]

[Out]

(((A + I*B - C)*((-I)*c + d)*Log[I - Tan[e + f*x]])/(a + I*b) + ((A - I*B - C)*(I*c + d)*Log[I + Tan[e + f*x]]
)/(a - I*b) + (2*(A*b^2 + a*(-(b*B) + a*C))*(b*c - a*d)*Log[a + b*Tan[e + f*x]])/(b^2*(a^2 + b^2)) + (2*C*d*Ta
n[e + f*x])/b)/(2*f)

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Maple [B]  time = 0.041, size = 506, normalized size = 3.2 \begin{align*}{\frac{Cd\tan \left ( fx+e \right ) }{bf}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) Aad}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) Abc}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) Bac}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) Bbd}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) aCd}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) Cbc}{2\,f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{A\arctan \left ( \tan \left ( fx+e \right ) \right ) ac}{f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{A\arctan \left ( \tan \left ( fx+e \right ) \right ) bd}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B\arctan \left ( \tan \left ( fx+e \right ) \right ) ad}{f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{B\arctan \left ( \tan \left ( fx+e \right ) \right ) bc}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{C\arctan \left ( \tan \left ( fx+e \right ) \right ) ac}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{C\arctan \left ( \tan \left ( fx+e \right ) \right ) bd}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) Aad}{f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{b\ln \left ( a+b\tan \left ( fx+e \right ) \right ) Ac}{f \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) B{a}^{2}d}{bf \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) Bac}{f \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ){a}^{3}Cd}{f{b}^{2} \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( a+b\tan \left ( fx+e \right ) \right ) C{a}^{2}c}{bf \left ({a}^{2}+{b}^{2} \right ) }} \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)),x)

[Out]

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

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Maxima [A]  time = 1.46592, size = 247, normalized size = 1.58 \begin{align*} \frac{\frac{2 \, C d \tan \left (f x + e\right )}{b} + \frac{2 \,{\left ({\left ({\left (A - C\right )} a + B b\right )} c -{\left (B a -{\left (A - C\right )} b\right )} d\right )}{\left (f x + e\right )}}{a^{2} + b^{2}} + \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 )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{2} b^{2} + b^{4}} + \frac{{\left ({\left (B a -{\left (A - C\right )} b\right )} c +{\left ({\left (A - C\right )} a + B b\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}}}{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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.96562, size = 483, normalized size = 3.1 \begin{align*} \frac{2 \,{\left ({\left ({\left (A - C\right )} a b^{2} + B b^{3}\right )} c -{\left (B a b^{2} -{\left (A - C\right )} b^{3}\right )} d\right )} f x + 2 \,{\left (C a^{2} b + C b^{3}\right )} d \tan \left (f x + e\right ) +{\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 )} \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^{2} b + C b^{3}\right )} c -{\left (C a^{3} - B a^{2} b + C a b^{2} - B b^{3}\right )} d\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left (a^{2} b^{2} + b^{4}\right )} 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 23.0993, size = 2387, normalized size = 15.3 \begin{align*} \text{result too large to display} \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)),x)

[Out]

Piecewise((zoo*x*(c + d*tan(e))*(A + B*tan(e) + C*tan(e)**2)/tan(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((A*c*x
+ A*d*log(tan(e + f*x)**2 + 1)/(2*f) + B*c*log(tan(e + f*x)**2 + 1)/(2*f) - B*d*x + B*d*tan(e + f*x)/f - C*c*x
+ C*c*tan(e + f*x)/f - C*d*log(tan(e + f*x)**2 + 1)/(2*f) + C*d*tan(e + f*x)**2/(2*f))/a, Eq(b, 0)), (-I*A*c*
f*x*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) - A*c*f*x/(-2*b*f*tan(e + f*x) + 2*I*b*f) - I*A*c/(-2*b*f*tan
(e + f*x) + 2*I*b*f) - A*d*f*x*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) + I*A*d*f*x/(-2*b*f*tan(e + f*x) +
2*I*b*f) + A*d/(-2*b*f*tan(e + f*x) + 2*I*b*f) - B*c*f*x*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) + I*B*c
*f*x/(-2*b*f*tan(e + f*x) + 2*I*b*f) + B*c/(-2*b*f*tan(e + f*x) + 2*I*b*f) - I*B*d*f*x*tan(e + f*x)/(-2*b*f*ta
n(e + f*x) + 2*I*b*f) - B*d*f*x/(-2*b*f*tan(e + f*x) + 2*I*b*f) - B*d*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(-
2*b*f*tan(e + f*x) + 2*I*b*f) + I*B*d*log(tan(e + f*x)**2 + 1)/(-2*b*f*tan(e + f*x) + 2*I*b*f) + I*B*d/(-2*b*f
*tan(e + f*x) + 2*I*b*f) - I*C*c*f*x*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) - C*c*f*x/(-2*b*f*tan(e + f*
x) + 2*I*b*f) - C*c*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) + I*C*c*log(tan(e +
f*x)**2 + 1)/(-2*b*f*tan(e + f*x) + 2*I*b*f) + I*C*c/(-2*b*f*tan(e + f*x) + 2*I*b*f) + 3*C*d*f*x*tan(e + f*x)/
(-2*b*f*tan(e + f*x) + 2*I*b*f) - 3*I*C*d*f*x/(-2*b*f*tan(e + f*x) + 2*I*b*f) - I*C*d*log(tan(e + f*x)**2 + 1)
*tan(e + f*x)/(-2*b*f*tan(e + f*x) + 2*I*b*f) - C*d*log(tan(e + f*x)**2 + 1)/(-2*b*f*tan(e + f*x) + 2*I*b*f) -
2*C*d*tan(e + f*x)**2/(-2*b*f*tan(e + f*x) + 2*I*b*f) - 3*C*d/(-2*b*f*tan(e + f*x) + 2*I*b*f), Eq(a, -I*b)),
(-I*A*c*f*x*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + A*c*f*x/(2*b*f*tan(e + f*x) + 2*I*b*f) - I*A*c/(2*b*
f*tan(e + f*x) + 2*I*b*f) + A*d*f*x*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + I*A*d*f*x/(2*b*f*tan(e + f*x
) + 2*I*b*f) - A*d/(2*b*f*tan(e + f*x) + 2*I*b*f) + B*c*f*x*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + I*B*
c*f*x/(2*b*f*tan(e + f*x) + 2*I*b*f) - B*c/(2*b*f*tan(e + f*x) + 2*I*b*f) - I*B*d*f*x*tan(e + f*x)/(2*b*f*tan(
e + f*x) + 2*I*b*f) + B*d*f*x/(2*b*f*tan(e + f*x) + 2*I*b*f) + B*d*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*b*
f*tan(e + f*x) + 2*I*b*f) + I*B*d*log(tan(e + f*x)**2 + 1)/(2*b*f*tan(e + f*x) + 2*I*b*f) + I*B*d/(2*b*f*tan(e
+ f*x) + 2*I*b*f) - I*C*c*f*x*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + C*c*f*x/(2*b*f*tan(e + f*x) + 2*I
*b*f) + C*c*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*b*f*tan(e + f*x) + 2*I*b*f) + I*C*c*log(tan(e + f*x)**2 +
1)/(2*b*f*tan(e + f*x) + 2*I*b*f) + I*C*c/(2*b*f*tan(e + f*x) + 2*I*b*f) - 3*C*d*f*x*tan(e + f*x)/(2*b*f*tan(
e + f*x) + 2*I*b*f) - 3*I*C*d*f*x/(2*b*f*tan(e + f*x) + 2*I*b*f) - I*C*d*log(tan(e + f*x)**2 + 1)*tan(e + f*x)
/(2*b*f*tan(e + f*x) + 2*I*b*f) + C*d*log(tan(e + f*x)**2 + 1)/(2*b*f*tan(e + f*x) + 2*I*b*f) + 2*C*d*tan(e +
f*x)**2/(2*b*f*tan(e + f*x) + 2*I*b*f) + 3*C*d/(2*b*f*tan(e + f*x) + 2*I*b*f), Eq(a, I*b)), (x*(c + d*tan(e))*
(A + B*tan(e) + C*tan(e)**2)/(a + b*tan(e)), Eq(f, 0)), (2*A*a*b**2*c*f*x/(2*a**2*b**2*f + 2*b**4*f) - 2*A*a*b
**2*d*log(a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4*f) + A*a*b**2*d*log(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f
+ 2*b**4*f) + 2*A*b**3*c*log(a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4*f) - A*b**3*c*log(tan(e + f*x)**2 + 1
)/(2*a**2*b**2*f + 2*b**4*f) + 2*A*b**3*d*f*x/(2*a**2*b**2*f + 2*b**4*f) + 2*B*a**2*b*d*log(a/b + tan(e + f*x)
)/(2*a**2*b**2*f + 2*b**4*f) - 2*B*a*b**2*c*log(a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4*f) + B*a*b**2*c*lo
g(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f + 2*b**4*f) - 2*B*a*b**2*d*f*x/(2*a**2*b**2*f + 2*b**4*f) + 2*B*b**3*c*f
*x/(2*a**2*b**2*f + 2*b**4*f) + B*b**3*d*log(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f + 2*b**4*f) - 2*C*a**3*d*log(
a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4*f) + 2*C*a**2*b*c*log(a/b + tan(e + f*x))/(2*a**2*b**2*f + 2*b**4*
f) + 2*C*a**2*b*d*tan(e + f*x)/(2*a**2*b**2*f + 2*b**4*f) - 2*C*a*b**2*c*f*x/(2*a**2*b**2*f + 2*b**4*f) - C*a*
b**2*d*log(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f + 2*b**4*f) + C*b**3*c*log(tan(e + f*x)**2 + 1)/(2*a**2*b**2*f
+ 2*b**4*f) - 2*C*b**3*d*f*x/(2*a**2*b**2*f + 2*b**4*f) + 2*C*b**3*d*tan(e + f*x)/(2*a**2*b**2*f + 2*b**4*f),
True))

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Giac [A]  time = 1.40056, size = 251, normalized size = 1.61 \begin{align*} \frac{\frac{2 \, C d \tan \left (f x + e\right )}{b} + \frac{2 \,{\left (A a c - C a c + B b c - B a d + A b d - C b d\right )}{\left (f x + e\right )}}{a^{2} + b^{2}} + \frac{{\left (B a c - A b c + C b c + A a d - C a d + B b d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (C a^{2} b c - B a b^{2} c + A b^{3} c - C a^{3} d + B a^{2} b d - A a b^{2} d\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{2} b^{2} + b^{4}}}{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)),x, algorithm="giac")

[Out]

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