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

Optimal. Leaf size=443 \[ \frac{(c+d \tan (e+f x))^3 \left (8 a^2 C d^2-10 a b d (c C-4 B d)+b^2 \left (20 d^2 (A-C)-5 B c d+2 c^2 C\right )\right )}{60 d^3 f}+\frac{\log (\cos (e+f x)) \left (a^2 \left (-\left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )+2 a b \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )+b^2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )}{f}-x \left (a^2 \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )+2 a b \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )-b^2 \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )\right )+\frac{\left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^2}{2 f}+\frac{d \tan (e+f x) \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac{b \tan (e+f x) (-2 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^3}{20 d^2 f}+\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f} \]

[Out]

-((a^2*(c^2*C + 2*B*c*d - C*d^2 - A*(c^2 - d^2)) - b^2*(c^2*C + 2*B*c*d - C*d^2 - A*(c^2 - d^2)) + 2*a*b*(2*c*
(A - C)*d + B*(c^2 - d^2)))*x) + ((2*a*b*(c^2*C + 2*B*c*d - C*d^2 - A*(c^2 - d^2)) - a^2*(2*c*(A - C)*d + B*(c
^2 - d^2)) + b^2*(2*c*(A - C)*d + B*(c^2 - d^2)))*Log[Cos[e + f*x]])/f + (d*(2*a*b*(A*c - c*C - B*d) + a^2*(B*
c + (A - C)*d) - b^2*(B*c + (A - C)*d))*Tan[e + f*x])/f + ((a^2*B - b^2*B + 2*a*b*(A - C))*(c + d*Tan[e + f*x]
)^2)/(2*f) + ((8*a^2*C*d^2 - 10*a*b*d*(c*C - 4*B*d) + b^2*(2*c^2*C - 5*B*c*d + 20*(A - C)*d^2))*(c + d*Tan[e +
f*x])^3)/(60*d^3*f) - (b*(2*b*c*C - 5*b*B*d - 2*a*C*d)*Tan[e + f*x]*(c + d*Tan[e + f*x])^3)/(20*d^2*f) + (C*(
a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^3)/(5*d*f)

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Rubi [A]  time = 1.27812, antiderivative size = 443, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3647, 3637, 3630, 3528, 3525, 3475} \[ \frac{(c+d \tan (e+f x))^3 \left (8 a^2 C d^2-10 a b d (c C-4 B d)+b^2 \left (20 d^2 (A-C)-5 B c d+2 c^2 C\right )\right )}{60 d^3 f}+\frac{\log (\cos (e+f x)) \left (a^2 \left (-\left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )+2 a b \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )+b^2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right )}{f}-x \left (a^2 \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )+2 a b \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )-b^2 \left (-A \left (c^2-d^2\right )+2 B c d+c^2 C-C d^2\right )\right )+\frac{\left (a^2 B+2 a b (A-C)-b^2 B\right ) (c+d \tan (e+f x))^2}{2 f}+\frac{d \tan (e+f x) \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac{b \tan (e+f x) (-2 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^3}{20 d^2 f}+\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(c^2*C + 2*B*c*d - C*d^2 - A*(c^2 - d^2)) - b^2*(c^2*C + 2*B*c*d - C*d^2 - A*(c^2 - d^2)) + 2*a*b*(2*c*
(A - C)*d + B*(c^2 - d^2)))*x) + ((2*a*b*(c^2*C + 2*B*c*d - C*d^2 - A*(c^2 - d^2)) - a^2*(2*c*(A - C)*d + B*(c
^2 - d^2)) + b^2*(2*c*(A - C)*d + B*(c^2 - d^2)))*Log[Cos[e + f*x]])/f + (d*(2*a*b*(A*c - c*C - B*d) + a^2*(B*
c + (A - C)*d) - b^2*(B*c + (A - C)*d))*Tan[e + f*x])/f + ((a^2*B - b^2*B + 2*a*b*(A - C))*(c + d*Tan[e + f*x]
)^2)/(2*f) + ((8*a^2*C*d^2 - 10*a*b*d*(c*C - 4*B*d) + b^2*(2*c^2*C - 5*B*c*d + 20*(A - C)*d^2))*(c + d*Tan[e +
f*x])^3)/(60*d^3*f) - (b*(2*b*c*C - 5*b*B*d - 2*a*C*d)*Tan[e + f*x]*(c + d*Tan[e + f*x])^3)/(20*d^2*f) + (C*(
a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^3)/(5*d*f)

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 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 3630

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

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 (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f}+\frac{\int (a+b \tan (e+f x)) (c+d \tan (e+f x))^2 \left (-2 b c C+a (5 A-3 C) d+5 (A b+a B-b C) d \tan (e+f x)-(2 b c C-5 b B d-2 a C d) \tan ^2(e+f x)\right ) \, dx}{5 d}\\ &=-\frac{b (2 b c C-5 b B d-2 a C d) \tan (e+f x) (c+d \tan (e+f x))^3}{20 d^2 f}+\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f}-\frac{\int (c+d \tan (e+f x))^2 \left (10 a b c C d-4 a^2 (5 A-3 C) d^2-b^2 c (2 c C-5 B d)-20 \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)-\left (8 a^2 C d^2-10 a b d (c C-4 B d)+b^2 \left (2 c^2 C-5 B c d+20 (A-C) d^2\right )\right ) \tan ^2(e+f x)\right ) \, dx}{20 d^2}\\ &=\frac{\left (8 a^2 C d^2-10 a b d (c C-4 B d)+b^2 \left (2 c^2 C-5 B c d+20 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^3}{60 d^3 f}-\frac{b (2 b c C-5 b B d-2 a C d) \tan (e+f x) (c+d \tan (e+f x))^3}{20 d^2 f}+\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f}-\frac{\int (c+d \tan (e+f x))^2 \left (20 \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2-20 \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2 \tan (e+f x)\right ) \, dx}{20 d^2}\\ &=\frac{\left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^2}{2 f}+\frac{\left (8 a^2 C d^2-10 a b d (c C-4 B d)+b^2 \left (2 c^2 C-5 B c d+20 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^3}{60 d^3 f}-\frac{b (2 b c C-5 b B d-2 a C d) \tan (e+f x) (c+d \tan (e+f x))^3}{20 d^2 f}+\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f}-\frac{\int (c+d \tan (e+f x)) \left (-20 d^2 \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 )-20 d^2 \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 ) \tan (e+f x)\right ) \, dx}{20 d^2}\\ &=-\left (a^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+2 a b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) x+\frac{d \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 ) \tan (e+f x)}{f}+\frac{\left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^2}{2 f}+\frac{\left (8 a^2 C d^2-10 a b d (c C-4 B d)+b^2 \left (2 c^2 C-5 B c d+20 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^3}{60 d^3 f}-\frac{b (2 b c C-5 b B d-2 a C d) \tan (e+f x) (c+d \tan (e+f x))^3}{20 d^2 f}+\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f}-\left (2 a b \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-a^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \int \tan (e+f x) \, dx\\ &=-\left (a^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-b^2 \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )+2 a b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) x+\frac{\left (2 a b \left (c^2 C+2 B c d-C d^2-A \left (c^2-d^2\right )\right )-a^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right ) \log (\cos (e+f x))}{f}+\frac{d \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 ) \tan (e+f x)}{f}+\frac{\left (a^2 B-b^2 B+2 a b (A-C)\right ) (c+d \tan (e+f x))^2}{2 f}+\frac{\left (8 a^2 C d^2-10 a b d (c C-4 B d)+b^2 \left (2 c^2 C-5 B c d+20 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^3}{60 d^3 f}-\frac{b (2 b c C-5 b B d-2 a C d) \tan (e+f x) (c+d \tan (e+f x))^3}{20 d^2 f}+\frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f}\\ \end{align*}

Mathematica [C]  time = 6.5013, size = 383, normalized size = 0.86 \[ \frac{C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^3}{5 d f}+\frac{\frac{b \tan (e+f x) (2 a C d+5 b B d-2 b c C) (c+d \tan (e+f x))^3}{4 d f}-\frac{\frac{(c+d \tan (e+f x))^3 \left (-8 a^2 C d^2+10 a b d (c C-4 B d)+b^2 \left (-\left (20 d^2 (A-C)-5 B c d+2 c^2 C\right )\right )\right )}{3 d f}-\frac{10 \left (d \left (a^2 B+2 a b (A-C)-b^2 B\right ) \left (6 c d^2 \tan (e+f x)+(-d+i c)^3 \log (-\tan (e+f x)+i)-(d+i c)^3 \log (\tan (e+f x)+i)+d^3 \tan ^2(e+f x)\right )+d \left (a^2 (B c-d (A-C))+2 a b (A c+B d-c C)-b^2 (B c-d (A-C))\right ) \left (-i (c-i d)^2 \log (\tan (e+f x)+i)+i (c+i d)^2 \log (-\tan (e+f x)+i)-2 d^2 \tan (e+f x)\right )\right )}{f}}{4 d}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^3)/(5*d*f) + ((b*(-2*b*c*C + 5*b*B*d + 2*a*C*d)*Tan[e + f*x]*(c
+ d*Tan[e + f*x])^3)/(4*d*f) - (((-8*a^2*C*d^2 + 10*a*b*d*(c*C - 4*B*d) - b^2*(2*c^2*C - 5*B*c*d + 20*(A - C)
*d^2))*(c + d*Tan[e + f*x])^3)/(3*d*f) - (10*(d*(2*a*b*(A*c - c*C + B*d) + a^2*(B*c - (A - C)*d) - b^2*(B*c -
(A - C)*d))*(I*(c + I*d)^2*Log[I - Tan[e + f*x]] - I*(c - I*d)^2*Log[I + Tan[e + f*x]] - 2*d^2*Tan[e + f*x]) +
(a^2*B - b^2*B + 2*a*b*(A - C))*d*((I*c - d)^3*Log[I - Tan[e + f*x]] - (I*c + d)^3*Log[I + Tan[e + f*x]] + 6*
c*d^2*Tan[e + f*x] + d^3*Tan[e + f*x]^2)))/f)/(4*d))/(5*d)

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Maple [B]  time = 0.017, size = 1165, normalized size = 2.6 \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))^2*(c+d*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

[Out]

4/f*C*arctan(tan(f*x+e))*a*b*c*d-2/f*ln(1+tan(f*x+e)^2)*B*a*b*c*d-4/f*C*a*b*c*d*tan(f*x+e)+4/3/f*C*tan(f*x+e)^
3*a*b*c*d+2/f*B*tan(f*x+e)^2*a*b*c*d+4/f*A*a*b*c*d*tan(f*x+e)-4/f*A*arctan(tan(f*x+e))*a*b*c*d+1/3/f*A*tan(f*x
+e)^3*b^2*d^2-1/2/f*B*tan(f*x+e)^2*b^2*d^2+1/2/f*ln(1+tan(f*x+e)^2)*B*a^2*c^2+1/4/f*B*tan(f*x+e)^4*b^2*d^2-1/f
*C*b^2*c^2*tan(f*x+e)+1/f*A*b^2*c^2*tan(f*x+e)-1/f*A*b^2*d^2*tan(f*x+e)-1/f*a^2*C*d^2*tan(f*x+e)+1/5/f*C*b^2*d
^2*tan(f*x+e)^5+1/3/f*C*tan(f*x+e)^3*b^2*c^2-1/3/f*C*tan(f*x+e)^3*b^2*d^2+1/f*A*arctan(tan(f*x+e))*a^2*c^2-1/f
*A*arctan(tan(f*x+e))*a^2*d^2-1/f*A*arctan(tan(f*x+e))*b^2*c^2+1/f*A*arctan(tan(f*x+e))*b^2*d^2-1/f*C*arctan(t
an(f*x+e))*a^2*c^2-1/f*C*arctan(tan(f*x+e))*b^2*d^2+1/f*C*b^2*d^2*tan(f*x+e)+1/f*A*a^2*d^2*tan(f*x+e)+1/f*C*ar
ctan(tan(f*x+e))*a^2*d^2+1/f*C*arctan(tan(f*x+e))*b^2*c^2-1/2/f*ln(1+tan(f*x+e)^2)*B*a^2*d^2-1/2/f*ln(1+tan(f*
x+e)^2)*B*b^2*c^2+1/3/f*C*tan(f*x+e)^3*a^2*d^2+1/f*C*a^2*c^2*tan(f*x+e)+1/2/f*B*tan(f*x+e)^2*a^2*d^2+1/2/f*B*t
an(f*x+e)^2*b^2*c^2+1/2/f*ln(1+tan(f*x+e)^2)*B*b^2*d^2+1/f*A*tan(f*x+e)^2*b^2*c*d+1/2/f*C*tan(f*x+e)^4*a*b*d^2
+1/f*A*tan(f*x+e)^2*a*b*d^2+1/f*C*tan(f*x+e)^2*a*b*c^2-1/f*C*tan(f*x+e)^2*a*b*d^2-1/f*ln(1+tan(f*x+e)^2)*A*a*b
*d^2-1/f*ln(1+tan(f*x+e)^2)*A*b^2*c*d-1/f*ln(1+tan(f*x+e)^2)*C*a^2*c*d-2/f*B*a*b*d^2*tan(f*x+e)-2/f*B*b^2*c*d*
tan(f*x+e)+1/2/f*C*tan(f*x+e)^4*b^2*c*d-1/f*ln(1+tan(f*x+e)^2)*C*a*b*c^2+1/f*ln(1+tan(f*x+e)^2)*C*a*b*d^2+1/f*
ln(1+tan(f*x+e)^2)*C*b^2*c*d-2/f*B*arctan(tan(f*x+e))*a^2*c*d+2/3/f*B*tan(f*x+e)^3*a*b*d^2+2/3/f*B*tan(f*x+e)^
3*b^2*c*d-1/f*C*tan(f*x+e)^2*b^2*c*d+1/f*C*tan(f*x+e)^2*a^2*c*d+1/f*ln(1+tan(f*x+e)^2)*A*a^2*c*d+1/f*ln(1+tan(
f*x+e)^2)*A*a*b*c^2+2/f*B*a^2*c*d*tan(f*x+e)-2/f*B*arctan(tan(f*x+e))*a*b*c^2+2/f*B*arctan(tan(f*x+e))*a*b*d^2
+2/f*B*arctan(tan(f*x+e))*b^2*c*d+2/f*B*a*b*c^2*tan(f*x+e)

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Maxima [A]  time = 1.48102, size = 625, normalized size = 1.41 \begin{align*} \frac{12 \, C b^{2} d^{2} \tan \left (f x + e\right )^{5} + 15 \,{\left (2 \, C b^{2} c d +{\left (2 \, C a b + B b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )^{4} + 20 \,{\left (C b^{2} c^{2} + 2 \,{\left (2 \, C a b + B b^{2}\right )} c d +{\left (C a^{2} + 2 \, B a b +{\left (A - C\right )} b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )^{3} + 30 \,{\left ({\left (2 \, C a b + B b^{2}\right )} c^{2} + 2 \,{\left (C a^{2} + 2 \, B a b +{\left (A - C\right )} b^{2}\right )} c d +{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )^{2} + 60 \,{\left ({\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} c^{2} - 2 \,{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} c d -{\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} d^{2}\right )}{\left (f x + e\right )} + 30 \,{\left ({\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} c^{2} + 2 \,{\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} c d -{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 60 \,{\left ({\left (C a^{2} + 2 \, B a b +{\left (A - C\right )} b^{2}\right )} c^{2} + 2 \,{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} c d +{\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*C*b^2*d^2*tan(f*x + e)^5 + 15*(2*C*b^2*c*d + (2*C*a*b + B*b^2)*d^2)*tan(f*x + e)^4 + 20*(C*b^2*c^2 +
2*(2*C*a*b + B*b^2)*c*d + (C*a^2 + 2*B*a*b + (A - C)*b^2)*d^2)*tan(f*x + e)^3 + 30*((2*C*a*b + B*b^2)*c^2 + 2*
(C*a^2 + 2*B*a*b + (A - C)*b^2)*c*d + (B*a^2 + 2*(A - C)*a*b - B*b^2)*d^2)*tan(f*x + e)^2 + 60*(((A - C)*a^2 -
2*B*a*b - (A - C)*b^2)*c^2 - 2*(B*a^2 + 2*(A - C)*a*b - B*b^2)*c*d - ((A - C)*a^2 - 2*B*a*b - (A - C)*b^2)*d^
2)*(f*x + e) + 30*((B*a^2 + 2*(A - C)*a*b - B*b^2)*c^2 + 2*((A - C)*a^2 - 2*B*a*b - (A - C)*b^2)*c*d - (B*a^2
+ 2*(A - C)*a*b - B*b^2)*d^2)*log(tan(f*x + e)^2 + 1) + 60*((C*a^2 + 2*B*a*b + (A - C)*b^2)*c^2 + 2*(B*a^2 + 2
*(A - C)*a*b - B*b^2)*c*d + ((A - C)*a^2 - 2*B*a*b - (A - C)*b^2)*d^2)*tan(f*x + e))/f

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Fricas [A]  time = 1.22046, size = 1007, normalized size = 2.27 \begin{align*} \frac{12 \, C b^{2} d^{2} \tan \left (f x + e\right )^{5} + 15 \,{\left (2 \, C b^{2} c d +{\left (2 \, C a b + B b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )^{4} + 20 \,{\left (C b^{2} c^{2} + 2 \,{\left (2 \, C a b + B b^{2}\right )} c d +{\left (C a^{2} + 2 \, B a b +{\left (A - C\right )} b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )^{3} + 60 \,{\left ({\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} c^{2} - 2 \,{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} c d -{\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} d^{2}\right )} f x + 30 \,{\left ({\left (2 \, C a b + B b^{2}\right )} c^{2} + 2 \,{\left (C a^{2} + 2 \, B a b +{\left (A - C\right )} b^{2}\right )} c d +{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )^{2} - 30 \,{\left ({\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} c^{2} + 2 \,{\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} c d -{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} d^{2}\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 60 \,{\left ({\left (C a^{2} + 2 \, B a b +{\left (A - C\right )} b^{2}\right )} c^{2} + 2 \,{\left (B a^{2} + 2 \,{\left (A - C\right )} a b - B b^{2}\right )} c d +{\left ({\left (A - C\right )} a^{2} - 2 \, B a b -{\left (A - C\right )} b^{2}\right )} d^{2}\right )} \tan \left (f x + e\right )}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*C*b^2*d^2*tan(f*x + e)^5 + 15*(2*C*b^2*c*d + (2*C*a*b + B*b^2)*d^2)*tan(f*x + e)^4 + 20*(C*b^2*c^2 +
2*(2*C*a*b + B*b^2)*c*d + (C*a^2 + 2*B*a*b + (A - C)*b^2)*d^2)*tan(f*x + e)^3 + 60*(((A - C)*a^2 - 2*B*a*b - (
A - C)*b^2)*c^2 - 2*(B*a^2 + 2*(A - C)*a*b - B*b^2)*c*d - ((A - C)*a^2 - 2*B*a*b - (A - C)*b^2)*d^2)*f*x + 30*
((2*C*a*b + B*b^2)*c^2 + 2*(C*a^2 + 2*B*a*b + (A - C)*b^2)*c*d + (B*a^2 + 2*(A - C)*a*b - B*b^2)*d^2)*tan(f*x
+ e)^2 - 30*((B*a^2 + 2*(A - C)*a*b - B*b^2)*c^2 + 2*((A - C)*a^2 - 2*B*a*b - (A - C)*b^2)*c*d - (B*a^2 + 2*(A
- C)*a*b - B*b^2)*d^2)*log(1/(tan(f*x + e)^2 + 1)) + 60*((C*a^2 + 2*B*a*b + (A - C)*b^2)*c^2 + 2*(B*a^2 + 2*(
A - C)*a*b - B*b^2)*c*d + ((A - C)*a^2 - 2*B*a*b - (A - C)*b^2)*d^2)*tan(f*x + e))/f

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Sympy [A]  time = 4.66021, size = 1134, normalized size = 2.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*a**2*c**2*x + A*a**2*c*d*log(tan(e + f*x)**2 + 1)/f - A*a**2*d**2*x + A*a**2*d**2*tan(e + f*x)/f
+ A*a*b*c**2*log(tan(e + f*x)**2 + 1)/f - 4*A*a*b*c*d*x + 4*A*a*b*c*d*tan(e + f*x)/f - A*a*b*d**2*log(tan(e +
f*x)**2 + 1)/f + A*a*b*d**2*tan(e + f*x)**2/f - A*b**2*c**2*x + A*b**2*c**2*tan(e + f*x)/f - A*b**2*c*d*log(ta
n(e + f*x)**2 + 1)/f + A*b**2*c*d*tan(e + f*x)**2/f + A*b**2*d**2*x + A*b**2*d**2*tan(e + f*x)**3/(3*f) - A*b*
*2*d**2*tan(e + f*x)/f + B*a**2*c**2*log(tan(e + f*x)**2 + 1)/(2*f) - 2*B*a**2*c*d*x + 2*B*a**2*c*d*tan(e + f*
x)/f - B*a**2*d**2*log(tan(e + f*x)**2 + 1)/(2*f) + B*a**2*d**2*tan(e + f*x)**2/(2*f) - 2*B*a*b*c**2*x + 2*B*a
*b*c**2*tan(e + f*x)/f - 2*B*a*b*c*d*log(tan(e + f*x)**2 + 1)/f + 2*B*a*b*c*d*tan(e + f*x)**2/f + 2*B*a*b*d**2
*x + 2*B*a*b*d**2*tan(e + f*x)**3/(3*f) - 2*B*a*b*d**2*tan(e + f*x)/f - B*b**2*c**2*log(tan(e + f*x)**2 + 1)/(
2*f) + B*b**2*c**2*tan(e + f*x)**2/(2*f) + 2*B*b**2*c*d*x + 2*B*b**2*c*d*tan(e + f*x)**3/(3*f) - 2*B*b**2*c*d*
tan(e + f*x)/f + B*b**2*d**2*log(tan(e + f*x)**2 + 1)/(2*f) + B*b**2*d**2*tan(e + f*x)**4/(4*f) - B*b**2*d**2*
tan(e + f*x)**2/(2*f) - C*a**2*c**2*x + C*a**2*c**2*tan(e + f*x)/f - C*a**2*c*d*log(tan(e + f*x)**2 + 1)/f + C
*a**2*c*d*tan(e + f*x)**2/f + C*a**2*d**2*x + C*a**2*d**2*tan(e + f*x)**3/(3*f) - C*a**2*d**2*tan(e + f*x)/f -
C*a*b*c**2*log(tan(e + f*x)**2 + 1)/f + C*a*b*c**2*tan(e + f*x)**2/f + 4*C*a*b*c*d*x + 4*C*a*b*c*d*tan(e + f*
x)**3/(3*f) - 4*C*a*b*c*d*tan(e + f*x)/f + C*a*b*d**2*log(tan(e + f*x)**2 + 1)/f + C*a*b*d**2*tan(e + f*x)**4/
(2*f) - C*a*b*d**2*tan(e + f*x)**2/f + C*b**2*c**2*x + C*b**2*c**2*tan(e + f*x)**3/(3*f) - C*b**2*c**2*tan(e +
f*x)/f + C*b**2*c*d*log(tan(e + f*x)**2 + 1)/f + C*b**2*c*d*tan(e + f*x)**4/(2*f) - C*b**2*c*d*tan(e + f*x)**
2/f - C*b**2*d**2*x + C*b**2*d**2*tan(e + f*x)**5/(5*f) - C*b**2*d**2*tan(e + f*x)**3/(3*f) + C*b**2*d**2*tan(
e + f*x)/f, Ne(f, 0)), (x*(a + b*tan(e))**2*(c + d*tan(e))**2*(A + B*tan(e) + C*tan(e)**2), True))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out