3.1207 \(\int \frac{(c+d \tan (e+f x))^3}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=230 \[ \frac{\left (a^2 \left (-\left (3 c^2 d-d^3\right )\right )+2 a b c \left (c^2-3 d^2\right )+b^2 d \left (3 c^2-d^2\right )\right ) \log (\cos (e+f x))}{f \left (a^2+b^2\right )^2}-\frac{x \left (a^2 \left (-\left (c^3-3 c d^2\right )\right )-2 a b d \left (3 c^2-d^2\right )+b^2 c \left (c^2-3 d^2\right )\right )}{\left (a^2+b^2\right )^2}-\frac{(b c-a d)^2 (c+d \tan (e+f x))}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{\left (a^2 d+2 a b c+3 b^2 d\right ) (b c-a d)^2 \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.341654, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3565, 3626, 3617, 31, 3475} \[ \frac{\left (a^2 \left (-\left (3 c^2 d-d^3\right )\right )+2 a b c \left (c^2-3 d^2\right )+b^2 d \left (3 c^2-d^2\right )\right ) \log (\cos (e+f x))}{f \left (a^2+b^2\right )^2}-\frac{x \left (a^2 \left (-\left (c^3-3 c d^2\right )\right )-2 a b d \left (3 c^2-d^2\right )+b^2 c \left (c^2-3 d^2\right )\right )}{\left (a^2+b^2\right )^2}-\frac{(b c-a d)^2 (c+d \tan (e+f x))}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{\left (a^2 d+2 a b c+3 b^2 d\right ) (b c-a d)^2 \log (a+b \tan (e+f x))}{b^2 f \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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

Mathematica [C]  time = 4.63305, size = 535, normalized size = 2.33 \[ \frac{\cos (e+f x) (c+d \tan (e+f x))^3 (a \cos (e+f x)+b \sin (e+f x)) \left (a^2 \cos (e+f x) \left (2 (a+i b)^2 (e+f x) \left (i a^2 d^3+2 a b d^3+b^2 c \left (c^2-3 i c d-3 d^2\right )\right )+(b c-a d)^2 \left (a^2 d+2 a b c+3 b^2 d\right ) \log \left ((a \cos (e+f x)+b \sin (e+f x))^2\right )-2 d^3 \left (a^2+b^2\right )^2 \log (\cos (e+f x))\right )+b \sin (e+f x) \left (2 (a+i b) \left (a^2 b^2 \left (-3 i c^2 d (e+f x-i)+c^3 (e+f x)-3 c d^2 (e+f x+i)+2 i d^3 (e+f x)\right )+a^3 b d^2 (3 c+d (e+f x+i))+i a^4 d^3 (e+f x+i)+a b^3 c \left (c^2 (i e+i f x+1)+3 c d (e+f x+i)-3 i d^2 (e+f x)\right )-i b^4 c^3\right )+a (b c-a d)^2 \left (a^2 d+2 a b c+3 b^2 d\right ) \log \left ((a \cos (e+f x)+b \sin (e+f x))^2\right )-2 a d^3 \left (a^2+b^2\right )^2 \log (\cos (e+f x))\right )-2 i a (b c-a d)^2 \left (a^2 d+2 a b c+3 b^2 d\right ) \tan ^{-1}(\tan (e+f x)) (a \cos (e+f x)+b \sin (e+f x))\right )}{2 a b^2 f (a-i b)^2 (a+i b)^2 (a+b \tan (e+f x))^2 (c \cos (e+f x)+d \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.038, size = 671, normalized size = 2.9 \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))^3/(a+b*tan(f*x+e))^2,x)

[Out]

3/2/f/(a^2+b^2)^2*ln(1+tan(f*x+e)^2)*a^2*c^2*d-1/2/f/(a^2+b^2)^2*ln(1+tan(f*x+e)^2)*a^2*d^3-1/f/(a^2+b^2)^2*ln
(1+tan(f*x+e)^2)*a*b*c^3+3/f/(a^2+b^2)^2*ln(1+tan(f*x+e)^2)*a*b*c*d^2-3/2/f/(a^2+b^2)^2*ln(1+tan(f*x+e)^2)*b^2
*c^2*d+1/2/f/(a^2+b^2)^2*ln(1+tan(f*x+e)^2)*b^2*d^3+1/f/(a^2+b^2)^2*arctan(tan(f*x+e))*a^2*c^3-3/f/(a^2+b^2)^2
*arctan(tan(f*x+e))*a^2*c*d^2+6/f/(a^2+b^2)^2*arctan(tan(f*x+e))*a*b*c^2*d-2/f/(a^2+b^2)^2*arctan(tan(f*x+e))*
a*b*d^3-1/f/(a^2+b^2)^2*arctan(tan(f*x+e))*b^2*c^3+3/f/(a^2+b^2)^2*arctan(tan(f*x+e))*b^2*c*d^2+1/f/b^2/(a^2+b
^2)/(a+b*tan(f*x+e))*a^3*d^3-3/f/b/(a^2+b^2)/(a+b*tan(f*x+e))*a^2*c*d^2+3/f/(a^2+b^2)/(a+b*tan(f*x+e))*a*c^2*d
-1/f*b/(a^2+b^2)/(a+b*tan(f*x+e))*c^3+1/f/(a^2+b^2)^2/b^2*ln(a+b*tan(f*x+e))*a^4*d^3-3/f/(a^2+b^2)^2*ln(a+b*ta
n(f*x+e))*a^2*c^2*d+3/f/(a^2+b^2)^2*ln(a+b*tan(f*x+e))*a^2*d^3+2/f/(a^2+b^2)^2*b*ln(a+b*tan(f*x+e))*a*c^3-6/f/
(a^2+b^2)^2*b*ln(a+b*tan(f*x+e))*a*c*d^2+3/f/(a^2+b^2)^2*b^2*ln(a+b*tan(f*x+e))*c^2*d

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Maxima [A]  time = 1.57135, size = 416, normalized size = 1.81 \begin{align*} \frac{\frac{2 \,{\left (6 \, a b c^{2} d - 2 \, a b d^{3} +{\left (a^{2} - b^{2}\right )} c^{3} - 3 \,{\left (a^{2} - b^{2}\right )} c d^{2}\right )}{\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (2 \, a b^{3} c^{3} - 6 \, a b^{3} c d^{2} - 3 \,{\left (a^{2} b^{2} - b^{4}\right )} c^{2} d +{\left (a^{4} + 3 \, a^{2} b^{2}\right )} d^{3}\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 (2 \, a b c^{3} - 6 \, a b c d^{2} - 3 \,{\left (a^{2} - b^{2}\right )} c^{2} d +{\left (a^{2} - b^{2}\right )} d^{3}\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^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\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))^3/(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

1/2*(2*(6*a*b*c^2*d - 2*a*b*d^3 + (a^2 - b^2)*c^3 - 3*(a^2 - b^2)*c*d^2)*(f*x + e)/(a^4 + 2*a^2*b^2 + b^4) + 2
*(2*a*b^3*c^3 - 6*a*b^3*c*d^2 - 3*(a^2*b^2 - b^4)*c^2*d + (a^4 + 3*a^2*b^2)*d^3)*log(b*tan(f*x + e) + a)/(a^4*
b^2 + 2*a^2*b^4 + b^6) - (2*a*b*c^3 - 6*a*b*c*d^2 - 3*(a^2 - b^2)*c^2*d + (a^2 - b^2)*d^3)*log(tan(f*x + e)^2
+ 1)/(a^4 + 2*a^2*b^2 + b^4) - 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/(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 = 2.10335, size = 1037, normalized size = 4.51 \begin{align*} -\frac{2 \, b^{5} c^{3} - 6 \, a b^{4} c^{2} d + 6 \, a^{2} b^{3} c d^{2} - 2 \, a^{3} b^{2} d^{3} - 2 \,{\left (6 \, a^{2} b^{3} c^{2} d - 2 \, a^{2} b^{3} d^{3} +{\left (a^{3} b^{2} - a b^{4}\right )} c^{3} - 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} c d^{2}\right )} f x -{\left (2 \, a^{2} b^{3} c^{3} - 6 \, a^{2} b^{3} c d^{2} - 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} c^{2} d +{\left (a^{5} + 3 \, a^{3} b^{2}\right )} d^{3} +{\left (2 \, a b^{4} c^{3} - 6 \, a b^{4} c d^{2} - 3 \,{\left (a^{2} b^{3} - b^{5}\right )} c^{2} d +{\left (a^{4} b + 3 \, a^{2} b^{3}\right )} d^{3}\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 (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d^{3} \tan \left (f x + e\right ) +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d^{3}\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \,{\left (a b^{4} c^{3} - 3 \, a^{2} b^{3} c^{2} d + 3 \, a^{3} b^{2} c d^{2} - a^{4} b d^{3} +{\left (6 \, a b^{4} c^{2} d - 2 \, a b^{4} d^{3} +{\left (a^{2} b^{3} - b^{5}\right )} c^{3} - 3 \,{\left (a^{2} b^{3} - b^{5}\right )} c d^{2}\right )} f x\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))^3/(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.92834, size = 605, normalized size = 2.63 \begin{align*} \frac{\frac{2 \,{\left (a^{2} c^{3} - b^{2} c^{3} + 6 \, a b c^{2} d - 3 \, a^{2} c d^{2} + 3 \, b^{2} c d^{2} - 2 \, a b d^{3}\right )}{\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (2 \, a b c^{3} - 3 \, a^{2} c^{2} d + 3 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3} - b^{2} d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (2 \, a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, b^{4} c^{2} d - 6 \, a b^{3} c d^{2} + a^{4} d^{3} + 3 \, a^{2} b^{2} d^{3}\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 (2 \, a b^{3} c^{3} \tan \left (f x + e\right ) - 3 \, a^{2} b^{2} c^{2} d \tan \left (f x + e\right ) + 3 \, b^{4} c^{2} d \tan \left (f x + e\right ) - 6 \, a b^{3} c d^{2} \tan \left (f x + e\right ) + a^{4} d^{3} \tan \left (f x + e\right ) + 3 \, a^{2} b^{2} d^{3} \tan \left (f x + e\right ) + 3 \, a^{2} b^{2} c^{3} + b^{4} c^{3} - 6 \, a^{3} b c^{2} d + 3 \, a^{4} c d^{2} - 3 \, a^{2} b^{2} c d^{2} + 2 \, a^{3} b d^{3}\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))^3/(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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