3.87 \(\int \frac{A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=209 \[ -\frac{A d^2-B c d+c^2 C}{2 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}-\frac{2 c d (A-C)-B \left (c^2-d^2\right )}{f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}+\frac{\left (d (A-C) \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )^3}-\frac{x \left (-A \left (c^3-3 c d^2\right )-3 B c^2 d+B d^3+c^3 C-3 c C d^2\right )}{\left (c^2+d^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.375888, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {3628, 3529, 3531, 3530} \[ -\frac{A d^2-B c d+c^2 C}{2 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}-\frac{2 c d (A-C)-B \left (c^2-d^2\right )}{f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}+\frac{\left (d (A-C) \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )^3}-\frac{x \left (-A \left (c^3-3 c d^2\right )-3 B c^2 d+B d^3+c^3 C-3 c C d^2\right )}{\left (c^2+d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/(c + d*Tan[e + f*x])^3,x]

[Out]

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

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

Mathematica [C]  time = 4.63189, size = 261, normalized size = 1.25 \[ -\frac{-(d (C-A)+B c) \left (\frac{d \left (\frac{\left (c^2+d^2\right ) \left (5 c^2+4 c d \tan (e+f x)+d^2\right )}{(c+d \tan (e+f x))^2}+\left (2 d^2-6 c^2\right ) \log (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^3}+\frac{i \log (-\tan (e+f x)+i)}{(c+i d)^3}-\frac{\log (\tan (e+f x)+i)}{(d+i c)^3}\right )+B \left (\frac{2 d \left (\frac{c^2+d^2}{c+d \tan (e+f x)}-2 c \log (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^2}+\frac{i \log (-\tan (e+f x)+i)}{(c+i d)^2}-\frac{i \log (\tan (e+f x)+i)}{(c-i d)^2}\right )+\frac{C}{(c+d \tan (e+f x))^2}}{2 d f} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/(c + d*Tan[e + f*x])^3,x]

[Out]

-(C/(c + d*Tan[e + f*x])^2 + B*((I*Log[I - Tan[e + f*x]])/(c + I*d)^2 - (I*Log[I + Tan[e + f*x]])/(c - I*d)^2
+ (2*d*(-2*c*Log[c + d*Tan[e + f*x]] + (c^2 + d^2)/(c + d*Tan[e + f*x])))/(c^2 + d^2)^2) - (B*c + (-A + C)*d)*
((I*Log[I - Tan[e + f*x]])/(c + I*d)^3 - Log[I + Tan[e + f*x]]/(I*c + d)^3 + (d*((-6*c^2 + 2*d^2)*Log[c + d*Ta
n[e + f*x]] + ((c^2 + d^2)*(5*c^2 + d^2 + 4*c*d*Tan[e + f*x]))/(c + d*Tan[e + f*x])^2))/(c^2 + d^2)^3))/(2*d*f
)

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Maple [B]  time = 0.055, size = 713, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 1.51475, size = 495, normalized size = 2.37 \begin{align*} \frac{\frac{2 \,{\left ({\left (A - C\right )} c^{3} + 3 \, B c^{2} d - 3 \,{\left (A - C\right )} c d^{2} - B d^{3}\right )}{\left (f x + e\right )}}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} - \frac{2 \,{\left (B c^{3} - 3 \,{\left (A - C\right )} c^{2} d - 3 \, B c d^{2} +{\left (A - C\right )} d^{3}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{{\left (B c^{3} - 3 \,{\left (A - C\right )} c^{2} d - 3 \, B c d^{2} +{\left (A - C\right )} d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} - \frac{C c^{4} - 3 \, B c^{3} d +{\left (5 \, A - 3 \, C\right )} c^{2} d^{2} + B c d^{3} + A d^{4} - 2 \,{\left (B c^{2} d^{2} - 2 \,{\left (A - C\right )} c d^{3} - B d^{4}\right )} \tan \left (f x + e\right )}{c^{6} d + 2 \, c^{4} d^{3} + c^{2} d^{5} +{\left (c^{4} d^{3} + 2 \, c^{2} d^{5} + d^{7}\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left (c^{5} d^{2} + 2 \, c^{3} d^{4} + c d^{6}\right )} \tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.35637, size = 1215, normalized size = 5.81 \begin{align*} -\frac{3 \, C c^{4} d - 5 \, B c^{3} d^{2} +{\left (7 \, A - 3 \, C\right )} c^{2} d^{3} + B c d^{4} + A d^{5} - 2 \,{\left ({\left (A - C\right )} c^{5} + 3 \, B c^{4} d - 3 \,{\left (A - C\right )} c^{3} d^{2} - B c^{2} d^{3}\right )} f x -{\left (C c^{4} d - 3 \, B c^{3} d^{2} + 5 \,{\left (A - C\right )} c^{2} d^{3} + 3 \, B c d^{4} - A d^{5} + 2 \,{\left ({\left (A - C\right )} c^{3} d^{2} + 3 \, B c^{2} d^{3} - 3 \,{\left (A - C\right )} c d^{4} - B d^{5}\right )} f x\right )} \tan \left (f x + e\right )^{2} +{\left (B c^{5} - 3 \,{\left (A - C\right )} c^{4} d - 3 \, B c^{3} d^{2} +{\left (A - C\right )} c^{2} d^{3} +{\left (B c^{3} d^{2} - 3 \,{\left (A - C\right )} c^{2} d^{3} - 3 \, B c d^{4} +{\left (A - C\right )} d^{5}\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left (B c^{4} d - 3 \,{\left (A - C\right )} c^{3} d^{2} - 3 \, B c^{2} d^{3} +{\left (A - C\right )} c d^{4}\right )} \tan \left (f x + e\right )\right )} \log \left (\frac{d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \,{\left (C c^{5} - 2 \, B c^{4} d + 3 \,{\left (A - C\right )} c^{3} d^{2} + 3 \, B c^{2} d^{3} -{\left (3 \, A - 2 \, C\right )} c d^{4} - B d^{5} + 2 \,{\left ({\left (A - C\right )} c^{4} d + 3 \, B c^{3} d^{2} - 3 \,{\left (A - C\right )} c^{2} d^{3} - B c d^{4}\right )} f x\right )} \tan \left (f x + e\right )}{2 \,{\left ({\left (c^{6} d^{2} + 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} + d^{8}\right )} f \tan \left (f x + e\right )^{2} + 2 \,{\left (c^{7} d + 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} + c d^{7}\right )} f \tan \left (f x + e\right ) +{\left (c^{8} + 3 \, c^{6} d^{2} + 3 \, c^{4} d^{4} + c^{2} d^{6}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*C*c^4*d - 5*B*c^3*d^2 + (7*A - 3*C)*c^2*d^3 + B*c*d^4 + A*d^5 - 2*((A - C)*c^5 + 3*B*c^4*d - 3*(A - C)
*c^3*d^2 - B*c^2*d^3)*f*x - (C*c^4*d - 3*B*c^3*d^2 + 5*(A - C)*c^2*d^3 + 3*B*c*d^4 - A*d^5 + 2*((A - C)*c^3*d^
2 + 3*B*c^2*d^3 - 3*(A - C)*c*d^4 - B*d^5)*f*x)*tan(f*x + e)^2 + (B*c^5 - 3*(A - C)*c^4*d - 3*B*c^3*d^2 + (A -
C)*c^2*d^3 + (B*c^3*d^2 - 3*(A - C)*c^2*d^3 - 3*B*c*d^4 + (A - C)*d^5)*tan(f*x + e)^2 + 2*(B*c^4*d - 3*(A - C
)*c^3*d^2 - 3*B*c^2*d^3 + (A - C)*c*d^4)*tan(f*x + e))*log((d^2*tan(f*x + e)^2 + 2*c*d*tan(f*x + e) + c^2)/(ta
n(f*x + e)^2 + 1)) - 2*(C*c^5 - 2*B*c^4*d + 3*(A - C)*c^3*d^2 + 3*B*c^2*d^3 - (3*A - 2*C)*c*d^4 - B*d^5 + 2*((
A - C)*c^4*d + 3*B*c^3*d^2 - 3*(A - C)*c^2*d^3 - B*c*d^4)*f*x)*tan(f*x + e))/((c^6*d^2 + 3*c^4*d^4 + 3*c^2*d^6
+ d^8)*f*tan(f*x + e)^2 + 2*(c^7*d + 3*c^5*d^3 + 3*c^3*d^5 + c*d^7)*f*tan(f*x + e) + (c^8 + 3*c^6*d^2 + 3*c^4
*d^4 + c^2*d^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((A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f*x+e))**3,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.96573, size = 740, normalized size = 3.54 \begin{align*} \frac{\frac{2 \,{\left (A c^{3} - C c^{3} + 3 \, B c^{2} d - 3 \, A c d^{2} + 3 \, C c d^{2} - B d^{3}\right )}{\left (f x + e\right )}}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{{\left (B c^{3} - 3 \, A c^{2} d + 3 \, C c^{2} d - 3 \, B c d^{2} + A d^{3} - C d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} - \frac{2 \,{\left (B c^{3} d - 3 \, A c^{2} d^{2} + 3 \, C c^{2} d^{2} - 3 \, B c d^{3} + A d^{4} - C d^{4}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{6} d + 3 \, c^{4} d^{3} + 3 \, c^{2} d^{5} + d^{7}} + \frac{3 \, B c^{3} d^{3} \tan \left (f x + e\right )^{2} - 9 \, A c^{2} d^{4} \tan \left (f x + e\right )^{2} + 9 \, C c^{2} d^{4} \tan \left (f x + e\right )^{2} - 9 \, B c d^{5} \tan \left (f x + e\right )^{2} + 3 \, A d^{6} \tan \left (f x + e\right )^{2} - 3 \, C d^{6} \tan \left (f x + e\right )^{2} + 8 \, B c^{4} d^{2} \tan \left (f x + e\right ) - 22 \, A c^{3} d^{3} \tan \left (f x + e\right ) + 22 \, C c^{3} d^{3} \tan \left (f x + e\right ) - 18 \, B c^{2} d^{4} \tan \left (f x + e\right ) + 2 \, A c d^{5} \tan \left (f x + e\right ) - 2 \, C c d^{5} \tan \left (f x + e\right ) - 2 \, B d^{6} \tan \left (f x + e\right ) - C c^{6} + 6 \, B c^{5} d - 14 \, A c^{4} d^{2} + 11 \, C c^{4} d^{2} - 7 \, B c^{3} d^{3} - 3 \, A c^{2} d^{4} - B c d^{5} - A d^{6}}{{\left (c^{6} d + 3 \, c^{4} d^{3} + 3 \, c^{2} d^{5} + d^{7}\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^3,x, algorithm="giac")

[Out]

1/2*(2*(A*c^3 - C*c^3 + 3*B*c^2*d - 3*A*c*d^2 + 3*C*c*d^2 - B*d^3)*(f*x + e)/(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^
6) + (B*c^3 - 3*A*c^2*d + 3*C*c^2*d - 3*B*c*d^2 + A*d^3 - C*d^3)*log(tan(f*x + e)^2 + 1)/(c^6 + 3*c^4*d^2 + 3*
c^2*d^4 + d^6) - 2*(B*c^3*d - 3*A*c^2*d^2 + 3*C*c^2*d^2 - 3*B*c*d^3 + A*d^4 - C*d^4)*log(abs(d*tan(f*x + e) +
c))/(c^6*d + 3*c^4*d^3 + 3*c^2*d^5 + d^7) + (3*B*c^3*d^3*tan(f*x + e)^2 - 9*A*c^2*d^4*tan(f*x + e)^2 + 9*C*c^2
*d^4*tan(f*x + e)^2 - 9*B*c*d^5*tan(f*x + e)^2 + 3*A*d^6*tan(f*x + e)^2 - 3*C*d^6*tan(f*x + e)^2 + 8*B*c^4*d^2
*tan(f*x + e) - 22*A*c^3*d^3*tan(f*x + e) + 22*C*c^3*d^3*tan(f*x + e) - 18*B*c^2*d^4*tan(f*x + e) + 2*A*c*d^5*
tan(f*x + e) - 2*C*c*d^5*tan(f*x + e) - 2*B*d^6*tan(f*x + e) - C*c^6 + 6*B*c^5*d - 14*A*c^4*d^2 + 11*C*c^4*d^2
- 7*B*c^3*d^3 - 3*A*c^2*d^4 - B*c*d^5 - A*d^6)/((c^6*d + 3*c^4*d^3 + 3*c^2*d^5 + d^7)*(d*tan(f*x + e) + c)^2)
)/f